Reinforced Concrete 6r6y6i

Reinforced Concrete Design

Concrete Properties Modulus of Elasticity For concrete weighing from 1,500 to 2500 kg/m3

Ec  w 0.043 fc' 1.5 c

fc’ = 28 day compressive strength in MPa Wc = unit weight of concrete in kg/m3 For Normal weight concrete

Ec  4700 fc' fc’ = 28 day compressive strength in MPa Wc = unit weight of concrete in kg/m3

Reinforcing Bars GRADES AND STRENGTH OF REINFORCING BARS (Non Prestressed) GRADE

Min.Yeild Strength (ksi)

Min.Yeild Min Strength Tensile (MPa) Strength (ksi)

Min Tensile Strength (MPa)

A615

40 60

40 60

276 414

70 90

483 620

A616

50 60

50 60

345 414

80 90

552 620

A617

40 60

40 60

276 414

70 90

483 620

A615

60

60

414

80

552

ASTM SPECS

Modulus of Elasticity of steel shall be taken as 200,000 MPa

CONCRETE PROTECTION FOR REINFORCEMENT Minimum ( NON PRESTRESSED) CAST IN PLACE CONCRETE Cover,mm a) Concrete cast against and permanently exposed to earth

75

b) Concrete exposed to earth or weather: 20 mm bars through 36 mm bars 16 mm bar, W31 or D31 wire, and smaller

50 40

c) Concrete not exposed to earth or weather or in with ground: Slabs, walls, joist 32 mm bar and smaller Beams,columns Primary reinforcement ,ties,stirrups,spirals Shells,folded plates : 20 mm bars and larger 16 mm bars,W31or D31 wire,or smaller

20 40 20 15

PRECAST IN PLACE CONCRETE ( manufactured under plant Conditions)

Minimum Cover,mm

a) Concrete exposed to earth or weather:

Wall s 32 mm bar and smaller Other 20 mm bars through 32 mm bars 16 mm bars,W31or D31 wire,or smaller

20 40 30

b) Concrete not exposed to earth or weather or in with ground: Slabs, walls, joist 32 mm bar and smaller Beams,columns Primary reinforcement ties,stirrups,spirals Shells,folded plates : 20 mm bars and larger 16 mm bars,W31or D31 wire,or smaller

15 10 15 10

ULTIMATE STRENGTH DESIGN(USD) Required

Strength ( Load Factors)

1.Required strength U to resist dead load D and live load L shall at least be equal to : U =1.4D +1.7L If resistance to structural effects of specified wind load W, are included in the design,the following combinations of D,L and W shall be investigated to determine the greatest required strength U a) U =0.75(1.4D + 1.7L + 1.7W) Where load combination shall include full value and zero value of L to determine the most severe condition, and b) U = 0.9D +1.3L For any combination of D, L and W required strength U shall not be less than c) U =1.4D +1.7L

2.

If resistance to structural effects of specified earthquake load E, are included in the design,the following combinations of D,L and E shall be investigated to determine the greatest required strength U a) U =1.32D + 1.1(f1)L + 1.1E Where load combination shall include full value and zero value of L to determine the most severe condition, and b) U = 0.99D +1.1E For any combination of D, L and E required strength U shall not be less than c) U =1.4D +1.7L 4. If resistance to earth pressure H, are included in the design,the following combinations of D,L and H shall be investigated to determine the greatest required strength U a) U =1.4D + 1.7L + 1.7H Where D or L reduces the effect of H b) U = 0.9D +1.7H For any combination of D, L and E required strength U shall not be less than c) U =1.4D +1.7L 3.

5. If resistance to loadings due to weight and pressure offluids with well defined densities and controllable height F are included in the design, such loading shall have a factor of 1.4 and be added to all loading combinations that include live load. 6. If resistance to impact effects are taken into in design,such effect shall be included with live load L. 7. Where structural effects T of differential settlement, creep, shrinkage,expansion of creep compensating concrete or temperature change maybe significant in design, required strength U shall be a least equal to a) U =0.75(1.4D + 1.7L + 1.4T) but required strength U shall not be less than b) U =1.4D +1.7L

STRENGTH REDUCTION FACTOR Ф

STRENGTH REDUCTION FACTOR Ф shall be as follows: 1. Flexure, without axial load 0.90 2. Shear and Torsion 0.85 3. Bearing on concrete except on Post tension anchorage zone 0.70 4. Post tension anchorage zone 0.85 5. Axial tension and axial tension with flexure 0.90 6. Axial load and axial load with flexure Both axial load and moment shall be multiplied by Ф 7. Axial compression and axial compression with flexure a) with spiral reinforcement 0.75 b) Other reinforced 0.70 except that for low values of axial compression Ф shall be permitted to increase in accordance with the following: For in which fy does not exceed 415 MPa with symmetric reinforcement,and with ( h –d’-d)/h not less than 0.7, Ф shall be permitted to increase linearly to 0.9, as ФPn decreases from 0.10fc’Ag to zero. For other reinforced , Ф shall be permitted to increase linearly to 0.9, as ФPn decreases from 0.10fc’Ag or ФPb whichever is smaller to zero.

DESIGN AND ANALYSIS FOR FLEXURE (BEAMS)  

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Basic Assumptions 1. Strain in concrete and the reinforcement shall be assumed directly proportional to the distance from the neutral axis ,except, for deep flexural with overall depth-to-clear span ratio greater than 2/5 for continuous spans and 4/5 for simple span a non linear distribution of strain shall be considered. 2. Maximum usable strain at extreme concrete compression fiber shall be 0.003 3. Stress in reinforcement below specified yield strength fy for grade of reinforcement used shall be taken Es times steel strain. For strain greater than corresponding to fy, stress in the reinforcement shall be considered independent of strain and equal to fy. 4. Tensile strength of concrete shall be neglected in axial and flexural calculations.

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5. Relationship between concrete compressive stress distribution and concrete strain shall be assumed to be rectangular.trapezoidal,parabolic or any other assumed shape that result in prediction of strength in substantial agreement with results of comprehensive tests. 6. Requirements of 5 may be considered satisfied by an equivalent rectangular stress distribution defined by the following: Concrete stress distribution of 0.85fc’ shall be assumed uniformly distributed over an equivalent compression zone bounded by the edges of the cross section and a straight line located parallel to the neutral axis at a distance “a” from the fiber of maximum compressive strain. Distance c from fiber of maximum strain to the neutral axis shall be measured in a direction perpendicular to the neutral axis.

Compression Zone

(stress

in concrete)

(maximum usable strain of concrete)

0.85fc’

0.003

a

c

NA

a  1c

εs (strain of steel)

7. Factor β1 shall be taken as follows: β1 = 0.85 if fc’≤ 30 MPa β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not be less than 0.65

SYMBOLS AND NOTATIONS            

a = depth of equivalent rectangular stress block, mm c = distance from extreme compression face, mm As = area of non prestressed tension reinforcement, mm2 As’ = area of non prestressed compression reinforcement, mm2 b = width of compression face of the member, mm bw = width of the web, mm d = distance from extreme compression face to center of tension reinforcement, mm d’ = distance from extreme compression face to center of compression reinforcement, mm fc’ = specified compressive strength of concrete, MPa fy = specified yield strength of non prestressed reinforcement,MPa fs = calculated tensile stress in reinforcement at service loads,MPa fs’ = calculated compressive stress in reinforcement at service loads,MPa

           

Mu = factored moment at section; ultimate moment capacity,design strength Mn = nominal moment capacity Ф = strength reduction factor pb = reinforcement ratio producing balance strain condition p = ratio of non prestressed tension reinforcement = As/bd p’ = ratio of non prestressed compression reinforcement = As’/bd pmin = minimum required ratio of non prestressed tension reinforcement pmin = 1.4/fy Ec = modulus of elasticity of concrete,MPa Es = modulus of elasticity of reinforcement,MPa bf = flange width of T –beams t = flange thickness of T- beams



Balanced strain condition This exist at a cross section when tension reinforcement reaches the strain corresponding to its yield strength fy just as concrete compression reaches its assumed ultimate strain of 0.003. Stress Diagram 0.85fc’

b Compression Zone

ab

C = 0.85fc’abb

d

Strain Diagram 0.003 cb

T = Asbfy Asb = balance steel area

s 

fy Es

From the Stress Diagram ∑ Fx = 0 C= T= A f 0.85fc’a b b sb y

EQ.1

divide both sides of EQ.1 by bdfy

0.85 fc' ab Asb  df y bd Let

Asb pb  bd

then

0.85 fc' ab pb  df y

From the strain diagram

cb 0.003  d 0.003   s cb 

ab

1

Substituting EQ 3 in EQ 2

s 

EQ. 2

fy 200000

ab 600 1  d 600  fy

0.85 fc ' 1 600 pb  (600  f y ) f y

EQ. 3

Maximum permissible tensile steel ratio pmax = 0.75pb This limitation is to ensure that the steel reinforcement will yield first to ensure ductile failure.

Minimum permissible tensile steel ratio pmin = 1.4/fy The provision for minimum amount of reinforcement applies to beams which for architectural and other reasons are much larger in cross section as required by strength consideration. With very small amount of tensile reinforcement, the computed moment strength as a reinforced concrete member is smaller than that of the corresponding plain concrete section computed from its modulus of rupture. Failure in this case is quite sudden.

Overreinforced beam A design in which the steel reinforcement is more than that required for balanced strain condition. If the beam is overeinforced, the steel will not yield before failure. As the load is increased, deflections are not noticeable although the compression concrete is highly stressed, and failure occurs suddenly without warning to the of the structure. Underreinforced beam A design in which the steel reinforcement is lesser than that required for balanced strain condition. If the ultimate load is approached , the steel will begin to yield although the compression concrete is understressed. As the load is increased, the steel will continue to elongate, resulting into appreciable deflections and large visible cracks in the tensile concrete. Failure under this condition is ductile and will give warning to the of the structure to decrease the load or apply remedial measure.

SPACING LIMITS OF REINFORCEMENT

 





Beams The minimum clear spacing between parallel bars in a layer should be db( bar diameter) but not less 25 mm. Where parallel reinforcement is placed in two or more layers,bars in the upper layer should be directly placed above bars in the bottom layer with clear distance between layers not less than 25 mm. Columns In spirally reinforced or tied reinforced compression , clear distance between longitudinal reinforcement shall not be less than 1.5db nor 40 mm. Walls and Slabs In walls and slabs other than concrete joist construction, primary reinforcement shall be spaced not farther than three times the slab or wall thickness nor 450 mm.

MINIMUM THICKNESS OF NON-PRESTRESSED BEAMS AND ONE WAY SLABS UNLESS DEFLECTIONS ARE COMPUTED Member

Solid One-Way Slab Beams or ribbed one way slab

Simply ed

L/20

One end continuous

Both ends continuous

Cantilever

L/24

L/28

L/10

L/18.5

L/21

L/8

L/16

Span Length L in millimeters Values given shall be used directly for with normal density concrete (Wc = 2300 kg/m3) and Grade 60 (415 MPa) reinforcement. For other conditions, the values shall be modified as follows: For structural lightweight concrete having unit weights of 1500 -2000 kg/m3 the values shall be multiplied by ( 1.65 – 0.0005 Wc) but not less than 1.09, where Wc is the unit mass in kg/m3. For fy other 415 MPa, the values shall be multiplied by ( 0.4 +fy/700)

FLEXURAL ANALYSIS : BEAMS REINFORCED FOR TENSION

 Case I : Steel yields at failure (pmax ≥ p , fs ≥ fy) Stress Diagram

b

0.85fc’

Compression zone

a d

C = 0.85fc’ab (d – a/2)

Mu

As

T = Asfy Depth of concrete stress block

a

As f y

0.85 fc' b

Ultimate moment capacity

Mu = Ф 0.85fc’ab(d – a/2) Mu = Ф Asfy (d – a/2)

General Procedure for Analysis : Case I Given: b,d,As,fc’,fy Required : Mrequirements 1. Check for ductility u

0.85 fc '  1 600 p = As/bd pb  (600  f y ) f y β1 = 0.85 if fc’≤ 30 MPa β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not be less than 0.65 pmax = 0.75pb pmin =1.4/fy As f y pmin ≤ p≤ pmax a 2. Solve for the depth of the concrete stress block 0.85 fc' b 3. Check for minimum depth if necessary 4. Determine MU Mu = Ф 0.85fc’ab(d – a/2) or Mu = Ф Asfy (d – a/2) Units: If As is in mm2, fc’ and fy in MPa, a,b and d in mm then Mu is in N.mm. Dividing this by 106 changes N.mm to kN.m 5. Solve for any other requirement if there are any.

Problems  



CE Board May 1981 A rectangular beam with b =250 mm and d =460 mm is reinforced for tension only with 3 – 25 mm bars. The beam is simply ed on a span of 6 m and carries a uniform dead load of 12 kN/m. Calculate the uniform live load the beam can carry. Concrete weighs 23 kN/m3 and steel covering is 60 mm. fc’ = 20.7 MPa, fy =276 MPa. Also check for minimum depth requirement. Solution

3 ( 25) 2 As   1472.62mm 2 4 A 1472.62 p s   0.013 bd 250( 460) 1.4 1.4 pmin    0.005 fy 276

pb 

0.85 fc' 1 600 0.85(20.7)0.85(600)   0.03711  600  f y  f y (600  276)276

pmax  0.75 pb  0.75(0.03711)  0.0278  p

Steel yields at failure

As f y

1472.62(276) a   92.4mm 0.85 fc' b 0.85(20.7)250

92.4 0.9(1472.62)276(460  ) a 2  151.37kN .m M u  As f y (d  )  2 (10) 6 Wu L2 Mu  8 2 Wu (6) 151.37  8 Wu  33.64kN / m

Wu  1.4WD  1.7WL

460

60

Weight of the beam WB=bDWc

WB  0.25(0.52)23  3kN / m Total dead load

WD  12  3  15kN / m

250

33.64  1.4(15)  1.7WL WL  7.42kN / m Minimum required depth

d min

fy L 6000 276  (0.4  ) (0.4  )  297.85mm  520mm 16 700 16 700

CE Board May 1985 A 350 mm x 500 mm rectangular beam is reinforced for tension only with 5 of 28 mm diameter bars. The beam has an effective depth of 446 mm. The beam carries a uniform dead load of 4.5 kN/m ( including its own weight), a uniform live load of 3 kN/m, and a concentrated dead load of P and 2P as shown in the figure. fc’ =34.5MPa, fy = 414 MPa. Determine the following : a) Ultimate moment capacity in kN.m b) the maximum value of P in kN

Figure 2P

2m

P

2m

2m

350

As

5 (28) 2 As   3078.76mm 2 4 As 3078.76 p   0.0197 446 bd 350(446) 1.4 p min   0.00338 414 1  0.85  0.008( fc '30) 1  0.85  0.008(34.5  30)  0.814

0.85 fc ' 1 600 0.85(34.5)0.814(600) pb    0.03412  600  f y  f y (600  414)414 pmax  0.75 pb  0.75(0.03412)  0.0256  p

Steel yields at failure

a

As f y 0.85 fc ' b



3078.76(414)  124.18mm 0.85(34.5)350

124.18 0.9(3078.76)414(446  ) a 2 M u  As f y (d  )   440.4kN .m 6 2 (10) 2P

2m

P

2m

2m

w  1.4wD  1.7 wL  1.4(4.5)  1.7(3)  11.4kN / m 1.4(2 P)  2.8P 1.4 P

2.8P

1.4P

M

11.4 kN/m

A

2m

0

R(4)  2.8P (2)  1.4 P(2)  11.4(6)1  0 R  0.7 P  17.1

B

2m

B

2m

R Maximum Positive Moment

M u  R (2)  11.4(2)1 M u  (0.7 P  17.1)2  11.4(2) 440.4  1.4 P  17.1(2)  11.4(2) 440.4  11.4(2)  17.1(2) P  306.43kN 1 .4

Maximum Negative Moment

M u  1.4 P(2)  11.4(2)1 M u  2.8 P  22.8 440.4  2.8 P  22.8 440.4  22.8 P  149.14kN 2 .8

Use P =149.14 kN

Case II : Steel does not yield at failure ( pmax < p , fs < fy ) General Procedure for Analysis : Case II Given: b,d,As,fc’,fy Required : Mu 1. Check for ductility requirements p = As/bd β1 = 0.85 if fc’≤ 30 MPa β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not

pb 

be less 0.65 0.85 fcthan ' 1 600

(600  f y ) f y

pmax = 0.75pb

p > pmax 2. Using the stress and strain diagram solve for fs and a. Stress Diagram Strain Diagram b

0.003

0.85fc’ C=0.85fc’ab d

d-a/2

Mu

c NA d-c

As T =Asfs From the stress Diagram : ∑F =0 C=T

εs

0.85fc’ab = Asfs

EQ 1

From the strain diagram :

s d c  0.003 c

substitute

EQ 2

fs s  200000 in EQ 2, combine with EQ1 to solve for fs and a a c 1

3. Determine Mu Mu = Ф 0.85fc’ab(d – a/2) Mu = Ф Asfs (d – a/2) Problem: A rectangular beam has b =300 mm, d = 500 mm, As = 6 of 32 mm, fc’ =27.6 MPa,fy =414 MPa. Calculate the ultimate moment capacity.

6 (32) 2 As   4825.49mm 2 4

300

500 As

As 4825.49 p   0.032 bd 300(500) 1.4 p min   0.00338 414

0.85 fc ' 1 600 0.85(27.6)0.85(600) pb    0.02850  600  f y  f y (600  414)414 pmax  0.75 pb  0.75(0.02850)  0.0214  p ∑F =0

C=T

0.85fc’ab = Asfs

0.85(27.6)300a  4825.49 f s f s  1.46a  EQ.1

Tension Steel does not yield at failure

s d c  0.003 c fs  200000(0.003)

fs  600

500 

a 0.85

fs s  200000 a c 1 500  a

a

1

1

a 0.85 500(0.85)  a fs 0.85  a 600 0.85 425  a f s  600( )  EQ.2 a

EQ.1  EQ.2 600 1.46a  (425  a ) a 1.46a 2  255000  600a a 2  410.96a  174657.53  0  410.96  (410.76)  4(174657.53) a  260.22mm 2 2

600( 425  260.22) fs   379.94MPa  414 MPa 260.22

260.22 0.9(4825.49)379.94(500  ) a 2 M u  As f s (d  )  2 (10) 6 M u  610.34kN .m

Plate #1 : Beams Reinforced for tension 1.

A simply ed beam 6 m long is 350 mm wide has an effective depth of 500 mm. It s a uniform dead load of 12 kN/m and a concentrated live load applied at the midspan. If it is reinforced with 6 of 22 mm diameter bars, fc’ = 20.7 MPa, fy = 414 MPa, concrete weighs 23 kN/m3, determine the maximum value of this concentrated load . Use concrete cover of 70 mm.

2.

A rectangular beam reinforced for tension has b = 300 mm, d = 480 mm The beam is reinforced with 7 of 25 mm bars with fc’ = 21 MPa, fy =415 MPa. If the beam is a cantilever beam 3 m long and s a uniform dead load of 15 kN/m( including its own weight) applied along its entire length, calculate the maximum value of the concentrated live load that can be applied at the free end.

3.

A reinforced concrete beam rectangular beam 300 mm wide has an effective depth of 450 mm and is reinforced for tension only. Assuming fc’ = 27 MPa. fy = 350 MPa, determine the required steel area that would produce balance strain condition.

Solution to #1

 (22) 2 As  6  2280.8mm 2 4 As 2280.8 p   0.013 bd 350(500) pmin

1.4 1.4    0.00338  p f y 414

a M u  As f y (d  )  2

0.85(20.7)0.85(600) pb   0.02137 (600  414)414 pmax  0.75 pb  0.016  p

Tension steel Yields at failure

As f y

2280.8(414) a   153.28 0.85 fc' b 0.85( 20.7)350

0.9(2280.8) 414(500  (10) 6

153.28 ) 2  359.78kN .m

Weight of the beam

Total Dead

WB  23(0.35)(0.57)  4.59kN / m

WD  12  4.59  16.59kN / m

WD L2 PL L M u  1.4M D  1.7 M L  1.4  1.7 8 4 (16.59)(6) 2 PL (6) 359.78  1.4  1.7 8 4

PL  100.1kN

Solution to #2

 (25) 2 As  7  3436.1mm 2 4 A 3436.1 p s   0.024 bd 300(480) 0.85fc’ab = Asfs

pb 

0.85(21)0.85(600)  0.02161 (600  415)415

pmax  0.75 pb  0.0162  p

0.85(21)300a=3436.1fs

Tension steel does Not Yield at failure

fs =1.56a

EQ.1

s d c  0.003 c a d fs 1  a (200000)0.003 1 [(0.85)(480)  a] fs  600 a

fs s  200000 a c 1 ( 1d  a ) fs  600 a 244800  600a fs  a

EQ. 1 = EQ.2

244800  600a 1.56a  a

a  156,923  384.6a 2

1.56a 2  244800  600a a 2  384.6a  156,923  0

EQ 2

a 2  384.6a  156,923  0  384.6  (384.6) 2  4(156923) a  248.04mm 2 244800  600(248.04) fs   386.94 MPa  f y 248.04 a M u  As f s (d  ) 2 a M u  As f s (d  )  2

0.9(3436.1)386.94( 480  (10) 6

248.04 ) 2  425.97 kN .m

PL 3m 15 kN/m

WD L2 M u  1.4 M D  1.7 M L  1.4  1.7 PL L 2 15(3) 2 425.97  1.4  1.7 PL 3 2

PL  65kN

FLEXURAL ANALYSIS: BEAMS REINFORCED FOR TENSION & COMPRESSION ( DOUBLY REINFORCED BEAMS ) Criteria for adding compression reinforcement : p > 0.75pb b

STRESS DIAGRAMS

STRAIN DIAGRAM 0.003

0.85fc’ As’

d

d’

Mu As

a

C1=0.85fc’ab d-a/2 As1

As’

d-d’

Mu1 T1=As1fy

Compression is resisted by concrete

C2=As’fs’ c

εs’

Mu2

Compression is resisted by As’

c-d’ d-c

As2

T2=As2fy

d’

εs

Compression reinforcement is provided to ensure ductile failure ( tension steel must yield) thus the stress in tension steel must always be equal to fy.

On the other hand the stress in compression

steel may be equal to or less than fy. This stress must always be checked.

Maximum permissible tensile steel area – NS states that for with compression reinforcement, the portion of pb equalized by compression reinforcement need not be multiplied by the 0.75 factor thus

As max

, f  0.75 pb bd  As' s fy

Stress in compression steel

From the strain diagram

also

 s' c  d'  0.003 c

' f s  s'  200000

then

and

a c 1

' 600 ( a   d 1 ) f s'  a

Other double reinforced beam formulas ( derived from stress diagrams) Mu = Mu1 +Mu2 As = As1 + As2 C 1 = T1 0.85fc’ab =As1fy C 2 = T2 As2 fy = As’fs’ if fs’ = fy Mu1 =ФO.85fc’ab(d-a/2) Mu2 =ФAs2fy(d-d’)

As2 =As’ Mu1 = ФAs1fy(d-a/2) Mu2 =ФAs’fs’(d-d’) If fs’=fy Mu2 =ФAs’fy(d-d’)

FLEXURAL ANALYSIS :DOUBLY REINFORCED BEAMS Case 1 :Compression and Tension steel yields at failure Given :b,d,d’,As,As’,fc’,fy Req’d :Mu 1. Assume that compression steel yields at failure fs’ = fy As’ = As2 As1 = As - As’ 2. Solve for a

a

As1 f y

0.85 fc' b

' 600 ( a   d ' 1 ) f  3. Solve for fs’ s a

4.

If

fs’ ≥ fy

Mu1 = ФAs1fy(d-a/2) Mu2 =ФAs’fy (d-d’) Mu = Mu1 +Mu2 5. Check for yielding of tension steel

As max  0.75 pbbd  As'

≥ As

Problem#1: Determine the permissible ultimate moment capacity of the beam shown in figure. fc’= 20.7 MPa, fy = 345 MPa. 350 mm 2 of 28 mm

60 mm 600 mm

4 of 36 mm

Case 2 : Compression steel does not yield at failure tension steel yields at failure Given :b,d,d’,As,As’,fc’,fy Req’d :Mu

1. Assume that compression steel yields at failure fs’ = fy As’ = As2 As1 = As - As2 2. Solve for a a  As1 f y

0.85 fc' b

' 600 ( a   d ) 3. Solve for fs’ ' 1 fs  a

4. If fs’< fy From the stress diagrams ∑F = 0 C1 + C2 = T1 + T2 0.85fc’ab + As’fs’ = As1fy + As2fy

0.85fc’ab + As’fs’ = Asfy 600(a   1 d ' ) ' fs 

a

5.Using EQ. A and EQ. B solve for a and fs’

As1fy + As2fy = Asfy

EQ.A EQ.B

6. Solve for Mu Mu1 = Ф0.85fc’ab(d-a/2) Mu2 =ФAs’fs’(d-d’) Mu = Mu1 +Mu2 7. Check for yielding of tension steel '

As max  0.75 pb bd  As

≥

As

Problem#2: Determine the permissible ultimate moment capacity of the beam shown in figure. fc’= 27.5 MPa, fy = 345 MPa. 350 mm As’ = 775mm2

63 mm 600 mm

As =3625 mm2

Problem#1: Determine the permissible ultimate moment capacity of the beam shown in figure. fc’= 20.7 MPa, fy = 345 MPa. 350 mm 2 of 28 mm

Solution to Problem #1

60 mm 600 mm

4 of 36 mm

 (36) 2 As  4  4071.5mm 2 4  (28) 2 As'  2  1231.5mm 2 4

Assume that compression steel yields at failure

As1  As  As '  4071.5  1231.5  2840mm 2

As1 f y

2840(345) a   159.1mm 0.85 fc' b 0.85(20.7)350 600(a  1d ' ) 600(159.1  [0.85(60)]) f    407.67 MPa  f y a 159.1 ' s

Compression steel yields at failure

159.1 2840(345)(600  ) a 2 Mu1  As1 f y (d  )  0.9  458.94kN .m 6 2 10 Mu2  As ' f y (d  d ' )  0.9

1231.5(345)(600  60 )  206.48kN .m 6 10

Mu  Mu1  Mu2  458.94  206.48  665.42kN .m

Solution to #2

As1  As  As '  3625  775  2850mm 2

As1 f y

2850(345) a   120.18mm 0.85 fc' b 0.85(27.5)350 600(a  1d ' ) 600(120.18  [0.85(63)]) f    332.65MPa  f y a 120.18 ' s

Compression steel does not yields at failure ∑Fx =0

0.85fc’ab + As’fs’ = Asfy 0.85(27.5)350a + 775fs’ = 3625(345) 10.56a +fs’ = 1613.7 fs’ =1613.7 – 10.56 a EQ.1

' 600 ( a   d 600(a  [0.85(63)]) ' ' 1 ) f fs   s  334.7 MPa  f y a a a Mu1  As1 f y (d  ) 600(a  53.55) ' 2 fs   EQ.2 121.12 a 0.9( 2850)345)(600  )

EQ.1  EQ.2 600(a  53.55) 1613.7  10.56a  a 1613.7 a  10.56a 2  600a  32130

Mu1 

10 6 Mu1  477.36kN .m

2

Mu2  As ' fs ' (d  d ' ) 0.9(775)334.7(600  63) 10 6 Mu2  125.36kN .m Mu2 

10.56a 2  1013.7 a  32130  0 a 2  96a  3042.6  0 M u  Mu1  Mu2  602.72kN .m 0.85( 27.5)334.7(600) 96  (96) 2  4(3042.6) p   0.0365 a  121.12mm b (600  345)345 2 ' fs ' 2 A  0 . 75 p bd  A  6500 . 6 mm s max b s 600(121.12  53.55) ' fy fs  121.12

T – BEAMS Reinforced concrete floor systems, roof ,decks etc. are almost always monolithic. Forms are built for beam soffits and sides and for the underside of the slabs, and the entire construction is poured at once. Beam reinforcement and stirrups extend up into the slab thus part of the slab will act with the upper part of the beam to resist longitudinal compression. The resulting shape of the beam is in the form of a T rather than rectangular. Figure below shows the stress and strain diagrams of a T- beam.

Stress Diagrams (bf –bw)

bf

0.85fc’

t

As bw

Mu

Cf =0.85fc’(bf-bw)

d

d-t/2 Asf

bw t

Muf Tf =Asffy

Strain Diagram a

d-a/2 Asw

0.003

0.85fc’

Cw=0.85fc’abw

Muw T w =Aswfy

εs

Compression is resisted by the Compression is resisted by the overhanging flange Web Where bf = flange width bw = width of the web t = thickness of the slab

c

For yielding of tension steel

p w max  0.75( p b  p f )  p w Where

0.85 fc'  1 600 pb  (600  f y ) f y

As pw  bw d

pf 

Asf bw d

Code requirements of T- beams 1. In T-beam construction, the flange and the web shall be built integrally or effectively bonded together 2. The width of the flange effective as a T- beam shall not exceed ¼ of the span , and the effective overhanging flange on each side of the web shall not exceed : a) eight times the slab thickness b) ½ the clear spacing to the next web 3. For beams with slab on one side only, the effective overhanging flange shall not exceed : a) 1/12 the span length of the beam b) 6 times the slab thickness c) ½ the clear distance to the next web

bf’

bf t

bw’

S3

bw

S2

bw

S1

For interior beam: bf is the smallest of 1. bf = L/4 2. bf = bw + 16t 3. bf = S1 /2 + S2 /2 + bw For End beams: bf’ is the smallest of 4. bf’ = L/12 + bw’ 5. bf’ = bw’ + 6t 6. bf’ = S3/2 + bw’ For Symmetrical interior beam (S1 =S2 = S3) bf is the smallest of 7. bf = L/4 8. bf = bw + 16t 9. bf = center to center spacing of beams

T- BEAMS FORMULAS 

M u  M uf  M uw

From the stress diagrams

As  Asf  Asw

C f T f

C w  Tw

Asf f y  0.85 fc' (b f  bw )t

0.85 fc ' abw  Asw fy

Asf 

0.85 fc' (b f  bw )t fy

t M uf   0.85 fc' (b f  bw )t (d  ) 2 t M uf  Asf fy (d  ) 2

Asw fy a 0.85 fc ' bw M uw M uw

a   0.85 fc' abw (d  ) 2 a  Asw f y (d  ) 2

The compression block of a T- beam can fall either within the flange only or partly in the web. If it falls within the flange, the rectangular beam formulas apply, if it falls partly on the web the beam should be considered as a T- beam . z

NA

t

bf

z NA

As

As

Criterion for selection of analysis: 0.85fc’bfz = Asfy

z

As f y 0.85 fc ' b f

if z < t

wide rectangular beam

if z > t

T-beam

Flexural analysis of T- beams ( z > t)



Given: bw,t,As,d,center to center spacing of beams (assuming symmetrical interior beam), L, fc’,fy Required: MU 1. Determine the effective flange width bf bf = L/4 bf = bw + 16t bf = center to center spacing of beams Use the smallest value of bf Note: if bf is given omit step 1

2. Determine if it is to be analyzed as T- beam

z

As f y 0.85 fc' b f

z>t 3. Solve for Asf

Asf 

0.85 fc' (b f  bw )t fy

4. Check for yielding of tension steel

As pw  bw d pf 

Asf bwd

5. Solve for Muf

0.85 fc '  1 600 pb  (600  f y ) f y

pw max  0.75( pb  p f )  pw t M uf   0.85 fc' (b f  bw )t (d  ) 2 t M uf  Asf fy (d  ) 2

6. Solve for Asw 7. Solve for a 8. Solve for Muw

Asw  As  Asf a

As f y

0.85 fc' bw a M uw   0.85 fc' abw (d  ) 2

M uw 9. Solve for Mu

a  Asw f y (d  ) 2

M u  M uf  M uw

Problem : A reinforced concrete T- beam spaced at 2.0 m on centers has a span of 3.0 m with a slab thickness of 100 mm. The effective depth is 750 mm and the width of the web is 350 mm. The beam is reinforced with steel of area 5200 mm2. If fc’ = 20.7 MPa and fy= 345 MPa, calculate the ultimate moment capacity. Solution Determine the effective flange width bf bf = L/4 = 3000/4 = 750 mm bf = bw + 16t = 350+16(100) = 1950 mm bf = center to center spacing of beams = 2000 mm Use bf = 750 mm

Determine if it is to be analyzed as T- beam

As f y

5200(345) z   135.94mm  100mm 0.85 fc' b f 0.85(20.7)(750) Analyze as T - beam

Asf 

0.85 fc ' (b f  bw )t fy



0.85(20.7)(750  350)100  2040mm 2 345

As 5200 pw    0.0198 bw d 350(750) Asf

2040 pf    0.0078 b w d 350(750) 0.85 fc ' 1 600 0.85( 20.7)0.85(600) pb    0.0275 (600  f y ) f y (600  345)345

pw max  0.75( pb  p f )

pw max  0.75(0.0275  0.0078)  0.02649  pw tension steel yields at failure

t M uf  Asf fy (d  )  2

0.9(2040)345(750  10 6

100 ) 2

M uf  443.39kN .m

Asw  As  Asf  5200  2040  3160mm 2 Asw f y

3160(345) a   177.03mm  100mm 0.85 fc ' bw 0.85( 20.7)350 177.03 0.9(3160)345(750  ) a 2 M uw  Asw f y ( d  )   649.03kN .m 6 2 10 M u  M uf  M uw  443.39  649.03  1092.42kN .m

Flexural analysis of T- beams ( z < t) Given: bw,t,As,d,center to center spacing of beams(assuming symmetrical interior beam), L, fc’,fy Required: MU 1. Determine the effective flange width bf bf = L/4 bf = bw + 16t bf = center to center spacing of beams Use the smallest value of bf Note: if bf is given omit step 1 2. Determine if it is to be analyzed as T- beam

z

As f y 0.85 fc' b f z
3.

Solve for MU

z M u  As f y (d  ) 2 4. Check for yielding of tension steel

As p bd

0.85 fc '  1 600 pb  (600  f y ) f y

p max  0.75 pb  p Problem : A reinforced concrete T- beam has an effective flange width of 1500 mm span, slab thickness of 100 mm, effective depth of 600 mm and the width of the web is 250 mm. The beam is reinforced with steel of area 4500 mm2. If fc’ = 20.7 MPa and fy= 345 MPa, calculate the ultimate moment capacity.

solution

As f y

4500(345) z   58.8mm  100mm 0.85 fc' b f 0.85(20.7)(1500) Analyze as wide rectangular beam

58.8 0.9(4500)345(600  ) z 2  797.3kN .m M u  As f y (d  )  6 2 10 As 4500 p   0.005 bd 1500(600) 0.85 fc' 1 600 0.85( 20.7)0.85(600) pb    0.0275 (600  f y ) f y (600  345)345

p max  0.75 pb  0.0206  p tension steel yields at failure

Plate #2

Doubly Reinforced beams

1.

A rectangular concrete beam has a width of 300 mm and an effective depth to bottom bars of 450 mm. The beam is reinforced with six 32 mm bars and two 28 mm top bars located 65 mm from the top of the beam. If fc’=34.5 MPa, fy =345 MPa, calculate the ultimate moment capacity of the beam.

2.

Calculate the ultimate moment capacity of the beam shown in figure. fc’=34.5 MPa, fy = 415 MPa. 400 mm As’ = 1850mm2

80 mm 700 mm

As =4820 mm2

Non- Rectangular Beams To deal with these beams, the code requirements and principles of rectangular beams are applied. Stress and strain diagrams plays an important role in establishing the formulas that will analyze these beams.

Problem 375 mm

3 of 16 mm 75 mm

375 mm

Compute the ultimate moment capacity Of the beam shown in figure. Assume fc’=21 MPa, fy = 345 MPa

Stress diagram

a

Ac

375 mm

x

Strain Diagram 0.003 c

C=0.85fc’Ac 375 - 2a/3

3 of 16 mm 75 mm

T=Asfy

375 - c

s

375 mm

Solution

3 (16) 2 As   603.2mm 2 4

Assume that steel yields at failure( subject to checking) C=T 0.85fc’Ac=Asfy

0.85(21)Ac =603.2(345)

Ac =11,658.48 mm2

By similar triangles

x 375 x  0.833a  EQ.1  a 450 1 1 a  167.3mm 11658 . 48  a (0.833a) Ac  ax 2 2 a 167.3 c   196.82mm 1 0.85 From the strain diagram

s

fs 375  c s   200000 0.003 c fs 375  196.82  (200000)0.003 196.82

f s  543.18MPa  f y

2a M u  Asf fy (375  )  3

tension steel yields at failure

0.9(603.2)345(375  106

[2]167.3 ) 3  49.34kN .m

Problem Compute the ultimate moment capacity of the beam shown in figure. Assume fc’=20.7 MPa, fy = 345 MPa

400 mm

3 of 16 mm 75 mm

400 mm

Stress diagram

a

Ac

400 mm

x

Strain Diagram 0.003

C=0.85fc’Ac 400 - 2a/3

3 of 16 mm 75 mm

T=Asfy

c

400 - c

s

400 mm

Solution

3 (16) 2 As   603.2mm 2 4

Assume that steel yields at failure( subject to checking) C=T 0.85fc’Ac=Asfy

0.85(20.7)Ac =603.2(345)

Ac =11,829.8 mm2

By similar triangles

x 400 x  0.842a  EQ.1  a 475 1 1 a  167.63mm 11829 . 8  a (0.842a) Ac  ax 2 2 a 167.63 c   197.21mm 1 0.85 From the strain diagram

s

fs 375  c s   200000 0.003 c fs 400  197.21  (200000)0.003 197.21

f s  616.98MPa  f y

2a M u  Asf fy (400  )  3

tension steel yields at failure

0.9(603.2)345(400  106

[2]167.63 ) 3  53.98kN .m

Beam Deflections Unless stiffness values are obtained by a more comprehensive analysis,immediate deflection shall be computed with the modulus of elasticity of concrete and with an effective moment of inertia as follows, but not greater than Ig .

  M   M cr  cr   Ie    Ig   1    M a   Ma  fr I g Where M cr  yt 3



3

 I cr 

f r  0.7

fc '

fc’ in MPa fr = modulus of rupture of concrete

Ma = maximum moment in member at stage deflection is computed. Ig = moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement. Icr = moment of inertia of cracked section transformed to concrete yt = distance from centroidal axis of gross cross section , neglecting reinforcement, to extreme fiber in tension TRANSFORMED SECTION b b x d d-x As

n = modular ratio

Es n Ec

nAs To Locate nuetral axis: Moment of area of Moment of area of concrete about NA = steel about NA

x2 b  nAs (d  x) 2

To determine Icr

3

bx I cr   nAs (d  x) 2 3

yt = distance from centroidal axis of gross section neglecting reinforcement, to extreme fiber in tension. Unless stiffness values are obtained by a more comprehensive analysis, additional long term deflection resulting from creep and shrinkage of flexural shall be determined by multiplying the immediate deflection caused by the sustained load considered, by the factor

  1  50 p ' where p’ shall be the value of reinforcement ratio for non prestress compression reinforcement at midspan for simple and continuous spans, and at for cantilever. It is permitted to assume the time-dependent factor ξ for sustained load to be equal to

5 years or more 12 months 6 months 3 months

………………………….. ………………………….. ………………………….. …………………………..

2 1.4 1.2 1.0

Problem: A concrete beam 6 m long is 300 mm wide and 600 mm deep and carries a dead load of 9 kN/m and live load of 12 kN/m. The beam is reinforced for tension only with four 25 mm bars with an effective depth to tension bars of 530 mm. fc’ = 20.7 MPa , fy = 345 MPa , fr = 2.832 MPa, Ec = 20,000 MPa, Es = 200,000 MPa. Covering of bars is 70 mm. a) Calculate the maximum instantaneous deflection due to dead load and live load. b) Calculate the deflection due to the same loads after five years assuming that 30% of the live load is sustained.

Gross moment of inertia

bh 3 300(600)3 Ig    5400 x106 mm 4 12 12

M cr 

fr I g yt

2.832(5400)10 6   50.97 x10 6 N .mm 300 M cr  50.97 kN .m

Transformed section

b

b=300

x 530-x

4 (25) 2 As   1963.2mm 2 4

d=530 As

nAs

Es 200000   10 Ec 20000

n

x2 b  nAs (d  x) 2

x2 b  nAs (d  x) 2 x 2  69377  130.9 x

x2 300  10(1963.5)(530  x) 2

x 2  130.9 x  69377  0

 130.9  (130.9) 2  4(69377) x  205.96mm 2

bx 3 300(205.96) 3 2 I cr   nAs (d  x)   10(1963.5)(530  205.96) 2 3 3

I cr  2,935.38 x106 mm 4 a)

WT L2 (12  9)(6) 2 Ma    94.5kN .m 8 8

Effective moment of inertia

  M    M cr  cr    I cr  Ie    Ig  1    M a    Ma  3

 50.97   50.97   6 6 Ie   5400 x 10  1  2935 . 38 x 10       94.5    94.5   6 4 3



3

3

I e  3,322.1x10 mm

a) Instantaneous deflection

5WT L4 5(21)(6000) 4    3.44mm 6 384 Ec I e 384(20000)3322.1(10)

a)

Long term deflection Since only 30% l of the live load is sustained

WT  12  0.3(9)  14.7kN / m 5WT L4 5(14.7)(6000) 4  '   2.41mm 6 384 Ec I e 384(20000)3322.1(10) 

 2  2 ' 1  50 p 1  50(0)

Long term deflection

 L     '  3.44  2( 2.41)  8.26mm

Design for Flexure : Beams Reinforced for tension Derivation of deg formulas

Stress Diagram 0.85fc’

b Compression Zone

a d

Strain Diagram 0.003

C = 0.85fc’ab c d-a/2 Mu

T = Asfy

a M u  As f y (d  )  EQ.1 2

s 

a

fy Es

As f y 0.85 fc' b

 EQ.2

EQ.2 in EQ.1

M u  As f y (d 

Mu  

As f y bd

As f y 2[0.85 fc ' b]

bd ( d 

)

As f y d 2[0.85 fc' b]d

)

fy As As Mu   bdf y ( d  d) bd bd 2[(0.85 fc ' ] Let

As p bd

and

m

fy 0.85 fc '

pm M u  pbdf y ( d  d) 2 pm 2 M u  bd pf y (1  ) 2 pm ) Let Ru  pf y (1  2

M u  bd Ru 2

pm Ru  pf y (1  ) 2 p 2m Ru  pf y 1  fy 2

Coeffecient of resistance

For proportioning of section

p 2m 2 {Ru  pf y 1  f y} 2 mfy 2 2p Ru   p2 mf y m 2p 2 p   Ru  0 m mf y 2

p

2 Ru 2 2 2  ( ) 4 m m mf y 2

p

p

2mRu 2 2 2  ( ) 4 2 m m m fy 2 2mRu 2 2  1 m m fy 2

2mRu 1 p  (1  1  ) m fy

p

p

2mRu 2 2 2  ( ) [1  ] m m fy 2

2mRu 2 [1  1  ] m fy 2

Actual tensile steel ratio

NS COEFFECIENTS FOR CONTINUOUS BEAMS AND SLABS NS states that in lieu of frame analysis, the following approximate moments and shear are permitted for design of continuous beams and one way slabs provided that : There are two or more spans Spans are approximately equal,with the large of two adjacent spans not greater than the shorter by more than 20 % Loads are uniformly distributed Unit live load does not exceed three times the unit dead load The are prismatic Positive moment End spans Discontinuous end unrestrained Discontinuous end integral with Interior spans

W ULn2/11 WULn2/14 W ULn2/16

Negative moment at exterior face of first interior Two spans W ULn2/9 More than two spans W ULn2/10 Negative moment at other faces of interior s W ULn2/11 Negative moment at face of all s for Slabs with span not exceeding 3 m; and beams where ratio of column stiffness to beam stiffness exceeds eight at each end of the span W ULn2/12 Negative moment at interior face of exterior s for built integrally with s Where is a spandrel beam Where is a column

W ULn2/24 W ULn2/16

Shear in end at face of first interior 1.15W ULn/2 Shear at face of all other s

W ULn/2

Where Ln = the clear span for positive moment or shear and average of adjacent clear spans for negative moments.

column

column L1

L2

Shear

1.15wL2 2

wL1 2 2 1

wL 14

1.15wLn 2

wL22 11 Moment

 wL12 16

 wL2n 9

Ln 

L1  L2 2

Shear and moment for continuous beams or slab with two spans discontinuous edge integral with ,discontinous end unrestrained

column

column L1

column L3 Spandrel beam

L2

Shear

wL1 2 wL12 14

1.15wLn 2

wL22 16

1.15wLn 2

wL3 2 2 wL3 14

Moment

 wL12 16

 wL 10 L  L2 Ln  1 2 2 n

 wL2n 10 L2  L3 Ln  2

 wL23 24

Shear and moment for continuous beams or slab with more than two spans and discontinuous end integral with

column L1

column L3

L2

Shear

wL1 2 wL12 11

1.15wLn 2

wL22 16

1.15wLn 2

wL3 2 2 wL3 11

Moment

 wL 10 L  L2 Ln  1 2 2 n

 wL2n 10 L2  L3 Ln  2

Shear and moment for continuous beams or slab with more than two spans and discontinuous end unrestrained

Case 1 : Design for balanced strain condition with given dimensions Given : b,d, fc’ and fy Required : Steel area that would produce balance strain condition General Procedure: 1. Solve for pb

pb 

0.85 fc '  1 600 (600  f y ) f y

2. Solve for Asb Asb = pbbd Problem: A reinforced concrete rectangular beam 300 mm wide has an effective depth of 460 mm and is reinforced for tension only. If fc’=20 MPa, fy = 300 MPa, determine the balance steel area in mm2. Solution

pb 

0.85 fc ' 1 600 0.85(20)0.85(600)   0.032 (600  f y ) f y (600  300)300

Asb = pbbd=0.032(300)460 = 4416 mm2

Case 2 : Design of cross section and reinforcement from given loads and type of beam dimensions Given : fc’ , fy, Loads, type of beam, Wc Required : cross section dimension ,size and number of steel bars General Procedure: •Assume the weight of the beam ( DL) as 20 to 25% of (DL + LL). add this to the given dead load. •By any available method, determine deg moment M u. •Assume a value of steel ratio p ( 0.3pb to 0.6pb but not less than pmin). This will provide enough allowance for the rounding off of values of number of bars to be used. •Solve for the following design constants :

m 5.

Solve for bd

2

fy 0.85 fc'

Mu bd  Ru 2

Ru  pf y (1 

pm ) 2

6. Try a ratio of d/b ( from d= b to d = 2b ) , then solve for b and d. Alternatively b may be assumed until a reasonable value of d is attained. Provide concrete cover to get total thickness. Check for minimum depth if required. 7. Compute actual weight ( W = bDWc) and compare with assume Weight. 1 2mR 8. Solve for actual p p  m (1  1  f u ) y 9.Solve for As: As = pbd 10.Solve for the number of bars by dividing As by the area of one bar to be used. Round off this number to the next integer. Problems : Design a rectangular beam for a 6 m simple span to a uniform dead load of 15 kN/m and uniform live load of 24 kN/m applied along it’s entire length. fc’=20 MPa, fy = 350 MPa,Wc=23.5 kN/m3. Design a rectangular beam reinforced for tension only to a service dead load moment of 85 kN.m ( including its weight) and service live load moment of 102 kN.m. Use p =0.6pb, d/b = 1.75 , fc’=28 MPa, fy = 276 MPa.

Problems : Design a rectangular beam for a 6 m simple span to a uniform dead load of 15 kN/m and uniform live load of 24 kN/m applied along it’s entire length. fc’=20 MPa, fy = 350 MPa,Wc=23.5 kN/m3

wB  0.20(24  15)  7.8

WD  15  7.8  22.8kN / m 2

2

22.8(6) 24(6) M u  1.4  1.7  327.4kN .m 8 8 0.85(20)0.85(600) p  0.4 pb  0.4  0.0104 (600  350)350 fy 350 m   20.59 0.85 fc' 0.85(20) pm 0.0104(20.59) Ru  pf y (1  )  0.0104(350)[1  ]  3.25 2 2

Mu bd  Ru 2

Mu 327.24(10) 6 d   570mm Ru b 0.9(3.25)350 Trial section

350 mm x 570mm effective depth, total depth 650 mm

Actual weight

wB  0.35(0.65)23.5  5.35kN / m  7.8kN / m 6

Mu 327.24(10) Ru    3. 2 2 2 bd 0.9(350)(570)

2mRu 1 p  (1  1  ) m fy 1 2( 20.59)(3.2) p (1  1  )  0.102 20.59 350 2 As  pbd  0.0102(350)570  2035mm try 20mmbars 2035 N 7 2 ( 20)  4

Plate # 3 Design of beams reinforced for tension

Problems : Design a rectangular beam for a 5 m simple span to a uniform dead load of 12 kN/m and uniform live load of 20 kN/m applied along it’s entire length. fc’=20 MPa, fy = 400 MPa,Wc=23.5 kN/m3. Design a rectangular beam reinforced for tension only to a service dead load moment of 65 kN.m ( including its weight) and service live load moment of 80 kN.m. Use p =0.45pb, d/b = 1.5 , fc’=20 MPa, fy = 300 MPa. A reinforced concrete T- beam spaced at 3.0 m on centers has a span of 4.0 m with a slab thickness of 75 mm. The effective depth is 750 mm and the width of the web is 300 mm. The beam is reinforced with steel of area 4200 mm2. If fc’ = 20.7 MPa and fy= 345 MPa, calculate the ultimate moment capacity.

Case 3 : Design of reinforcement of a beam with given moment and cross sectional dimension Given : b,d, Mu ,fc’ , fy, Required : Number of steel bars General Procedure : 1. Solve for Ru and m

Mu Ru  bd 2

m

fy 0.85 fc'

2. Solve for p

p

1 (1  m

1

2mRu ) fy

3. Check for yielding of tension steel and pmin if required. 4. Solve for As : As = pbd

Problem : Determine the required tension steel area for a rectangular beam with b =250 mm, d =330mm, fc’ =20.7 MPa, fy = 414 MPa. The beam is required to a factored moment of 110 kN.m. Solution 6

Mu 110 (10) Ru    4.49 2 2 bd 0.90(250)(330) fy

414 m   23.53 0.85 fc' 0.85(20.7) p

1 (1  m

1

1 p (1  23.53

2mRu ) fy

Note : If p > 0.006 no need to check for pmin p < 0.02 no need to check for pmax

As = pbd As = 0.01276(250)330 As = 1052.7 mm2

2( 23.53) 4.49 1 )  0.01276 414

DESIGN OF CONTINUOUS BEAM In the design of continuous beam, the cross section is determined by the maximum moment obtained by any structural analysis method or by its equivalent NS coefficients. The reinforcements are designed from the moment obtained at the different sections of maximum positive and negative moments. Problem: Figure shows a continuous beam of three spans with the left and right ends discontinuous and integral with the . Design the section and reinforcements at critical sections using the given service uniform loading. Given dead loads includes the weight of the beam. fc’= 28 MPa, fy =350 MPa . Use NS coefficients to determine the moments.

DL = 12 kN/m LL =16 kN/m

4m

A

B

C

DL = 15 kN/m LL = 18 kN/m

DL = 20 kN/m LL = 24 kN/m

5m

6m

D

E

F

G

Factored loads W1 = 1.4(12)+1.7(16)= 44 W2=1.4(15)+1.7(18)= 51.6 W3 = 1.4(20)+1.7(24)=68.8 Design moments by NS coeffecients

 w1 L12  44(4) 2 MA    44kN .m 16 16 w1 L12 44(4) 2 MB    50.28kN .m 14 14

 w2 L2n1 MC   10

45 2 ) 2  104.49kN .m 10

 51.6(

w2 L22 51.6(5) 2 MD    80.63kN .m 16 16  w3 L2n 2 ME  10

65 2  68.8( ) 2   208.12kN .m 10

w3 L23 68.8(6) 2 MF    176.91kN .m 14 14  w3 L23  68.8(6) 2 MG    157.9kN .m 16 16

Proportioning of uniform beam size Note: use the biggest computed design Moment

Mu =208.12 kN.m

0.85 fc' 1 600 0.85(28)0.85(600) pb    0.0365 (600  f y ) f y (600  350)350 p  0.5 pb  (0.5)0.0365  0.01825 fy

350 m   14.71 0.85 fc' 0.85(28)

pm Ru  pf y (1  ) 2

0.01825(14.71) Ru  0.01825(350)(1  )  5.53MPa 2

Try b = 250 mm

Mu d  Ru b pmin

208.12(10) 6  410mm 0.9(5.53)(250)

1.4 1.4    0.004 fy 350

Section A

M u  44kN .m

Mu 44(10) 6 Ru    1.16 2 2 bd 0.9( 250)(410) 1 2mRu 1 2(1.16)14.71 p  (1  1  ) (1  1  )  0.0034 m fy 14.71 350

Use p = 0.004

As  pbd  0.004(250)410  410mm 2 Try 20 mm bars

N 

410( 4)  2 pcs 2  ( 20)

Section B

Top bars

M u  50.28kN .m

Mu 50.28(10) 6 Ru    1.32 2 2 bd 0.9(250)(410) p

1 2mRu 1 2(1.32)14.71 (1  1  ) (1  1  )  0.0034 m fy 14.71 350

Use p = 0.004

As  pbd  0.004(250)410  410mm 2 Try 20 mm bars

N

410( 4)  2 pcs 2  ( 20)

Section C

bottom bars

M u  104.49kN .m

Mu 104.49(10) 6 Ru    2.74 2 2 bd 0.9(250)(410) p

1 2mRu 1 2(2.74)14.71 (1  1  ) (1  1  )  0.00834 m fy 14.71 350

As  pbd  0.00834( 250)410  854.86mm 2 Try 20 mm bars

N

854.86(4)  3 pcs 2  (20)

Section D

top bars

M u  80.63kN .m

Mu 80.63(10) 6 Ru    2.11 2 2 bd 0.9(250)(410) 2mRu 1 1 2(2.11)14.71 p  (1  1  ) (1  1  )  0.00632 m fy 14.71 350

As  pbd  0.00632(250)410  648.06mm 2 Try 20 mm bars

N

648.06(4)  3 pcs 2  (20)

Section E

bottom bars

M u  208.12kN .m

Mu 208.12(10) 6 Ru    5.44 2 2 bd 0.9(250)(410) 2mRu 1 1 2(5.44)14.71 p  (1  1  ) (1  1  )  0.018 m fy 14.71 350

As  pbd  0.018(250) 410  1845mm 2 Try 20 mm bars

N

1845(4)  6 pcs 2  (20)

Section F

Top bars

M u  176.91kN .m

Mu 176.91(10) 6 Ru    4.62 2 2 bd 0.9(250)(410) 2mRu 1 1 2(4.62)14.71 p  (1  1  ) (1  1  )  0.0148 m fy 14.71 350

As  pbd  0.0148(250) 410  1517 mm 2 Try 20 mm bars

N

1517(4)  5 pcs 2  (20)

Section G

bottom bars

M u  157.9kN .m

Mu 157.9(10) 6 Ru    4.12 2 2 bd 0.9(250)(410) 2mRu 1 1 2(4.12)14.71 p  (1  1  ) (1  1  )  0.013 m fy 14.71 350

As  pbd  0.013(250)410  1332.5mm 2 Try 20 mm bars

N

1332.5(4)  5 pcs 2  (20)

top bars

Gross moment of inertia using 100 mm covering

bh 3 300(600)3 Ig    5400 x106 mm 4 12 12

M cr 

fr I g yt

3.1(5400)10 6   55.79 x10 6 N .mm 300

Placement of bars Requirement

2 of 20 2 of 20

A

B

3 of 20

3 of 20

C

D

6 of 20

E

5 of 20

F

5 of 20

G

Layout 2 of 20 1 of 20 2 of 20

A

B

C

3 of 20 1 of 20 2 of 20

D

3 of 20 3 of 20

E

3 of 20 2 of 20 2 of 20

F

G

Section at A and B 250 mm

Section at C

Section at D

250 mm

250 mm

500 mm

500 mm

Section at F

Section at E 250 mm

Section at G

250 mm

500 mm

500 mm

250 mm

500 mm

500 mm

Plate # 4: Design of continuous beams

DL = 9 kN/m LL =12 kN/m

DL = 12 kN/m LL = 14 kN/m

3.6 m

4.0 m

DL = 15kN/m LL = 18 kN/m

4.5 m

Problem Design the uniform size and reinforcements at critical section of the continuous beam shown above. fc’= 20 MPa, fy = 300 MPa. Given dead loads includes the weight of the beam.

DESIGN FOR FLEXURE :DOUBLY REINFORCED BEAMS Given :b,d,d’, Mu ,fc’,fy Req’d : As, As’ General Procedure 1. Solve for Ru

Mu Ru  bd 2

2. Solve for p

2mRu 1 p  (1  1  ) m fy

3. Check if the beam needs compression reinforcement

0.85 fc '  1 600 pb  (600  f y ) f y

p max  0.75 pb

if p > pmax then compression reinforcement is necessary

4. Solve for As1

9. Solve for fs’

As1  p max bd

' 600 ( a   d ' 1 ) fs  a

5. Solve for a

a 6. Solve for Mu1

As1 f y 0.85 fc' b

a M u1  As1 f y (d  ) 2

7. Solve for Mu2

M u 2  M u  M u1 8. Solve for As2

M u2 As 2  f y (d  d ' )

Case 1 If fs’ ≥ fy then fs’ = fy ( compression steel yields at failure)

As'  As 2

Case 2 If fs’ < fy then use fs’ ( compression steel does not yield at failure)

A  ' s

As 2 f y f s'

10. Solve for As

As  As1  As 2

Problem: Design the reinforcement of a rectangular beam to carry a factored moment of 272 kN.m. The beam width is 250 mm,effective depth 400mm. Use fc’ = 20.7 MPa, fy = 345 MPa, d’ =60 mm. Solution

Mu 272(10) 6 Ru    7.55 2 2 bd 0.9(250)(400) fy 345 m   19.61 0.85 fc ' 0.85( 20.7)

1 2(19.61)7.55 p (1  1  )  0.0317 19.61 345 0.85 fc ' 1 600 0.85(20.7)0.85(600) pb    0.0275 (600  f y ) f y (600  345)345 pmax  0.75(0.0275)  0.0206 p > pmax compression reinforcement is necessary

As1  pmax bd  0.0206(250)400  2060mm 2 As1 f y 2060(345) a   161.57mm 0.85 fc' b 0.85(20.7)250 161.57 0.9(2060)345(400  ) a 2 M u1  As1 f y (d  )   204.18kN .m 6 2 10

M u 2  M u  M u1  272  204.18  67.82kN .m M u2 67.82(10) 6 As 2    642.42mm 2 f y (d  d ' ) 0.9(345)(400  60) 600(a  1d ' ) 600(161.57  [0.85(60)]) f    410.6MPa a 161.57 ' s

As'  As 2  642.42mm 2

Compression Steel yields at failure

As  As1  As 2  2060  642.42  2702.42mm 2

Plate # 5 : DESIGN FOR FLEXURE :DOUBLY REINFORCED BEAMS Design the reinforcement of a rectangular beam to resist a dead load moment of 200 kN.m(including its own weight) and a live load moment of 300 kN.m . The beam is limited in size to 350 mm by 600mm overall depth. Steel covering ( from centroid of bars to outermost fiber is 100 mm for both tension and compression reinforcement. Use fc’ = 27.5 MPa, fy = 414 MPa

DESIGN FOR BENDING : T-BEAMS (z > t) Given: bw,t,Mu,d,center to center spacing of beams(assuming symmetrical interior beam), L, fc’,fy Required: As 1. Determine the effective flange width bf bf = L/4 bf = bw + 16t bf = center to center spacing of beams Use the smallest value of bf Note: if bf is given omit step 1 2. Solve for Asf

Asf  3. Solve for Muf

M uf

0.85 fc' (b f  bw )t fy

t  Asf fy (d  ) 2

If Mu > Muf then design as T- beam

4. Solve for Muw

M uw  M u  M uf

5. Solve for a

a M uw   0.85 fc' abw (d  ) 2 6. Solve for Asw 0.85 fc' abw Asw  fy

7. Solve for As

As  Asf  Asw

8. Check for yielding of tension steel

As pw  bw d pf 

Asf bwd

0.85 fc' 1 600 pb  (600  f y ) f y

pw max  0.75( pb  p f )  pw

Design the reinforcement of a T- beam to resist a factored moment of 750 kN.m. Properties of the T- beam are as follows : bf = 550 mm, bw = 300 mm, d = 600 mm, t = 110 mm, fc’ = 20.7 MPa, fy = 345 MPa. solution

Asf 

M uf

0.85 fc ' (b f  bw )t fy



0.85(20.7)(550  300)110  1402.5mm 2 345

110 0.9(1402.5)345(600  ) t 2  237.33kN .m  Asf fy (d  )  2 106

Mu > Muf design as T- beam

M uw  M u  M uf  750  237.33  512.67kN .m

a M uw   0.85 fc' abw (d  ) 2 a 512.67(10) 6  (0.9)0.85( 20.7)300a (600  ) 2 a2 107915.8  600a  2 a 2  1200a  215831.6  0 1200  (1200) 2  4( 215831.6) a  220.3mm  110 mm 2 0.85 fc ' abw 0.85(20.7)220.3(300) Asw    3370.6mm 2 fy 345

As  Asf  Asw  1402.5  3370.6  4773.1mm 2

Check for yielding of tension steel

As 4773.1 pw    0.0265 bw d 300(600) Asf

1402.5 pf    0.008 b w d 300(600) 0.85 fc' 1 600 0.85(20.7)0.85(600) pb    0.0275 (600  f y ) f y (600  345)345

pw max  0.75( pb  p f )  0.75(0.0275  0.008)  0.0266  pw Tension steel yields at failure

DESIGN FOR BENDING : T-BEAMS (z < t) Given: bw,t,Mu,d,center to center spacing of beams(assuming symmetrical interior beam), L, fc’,fy Required: As 1. Determine the effective flange width bf bf = L/4 bf = bw + 16t bf = center to center spacing of beams Use the smallest value of bf Note: if bf is given omit step 1 2. Solve for Asf

Asf 

0.85 fc' (b f  bw )t fy

3. Solve for Muf

M uf

t  Asf fy (d  ) 2

If Mu < Muf then z < t ,design as wide rectangular beam 4. Solve for a

a M u   0.85 fc' ab f (d  ) 2

5. Solve for As

As 

0.85 fc ' b f a fy

6. Check for yielding of tension steel

As p bf d

0.85 fc' 1 600 pb  (600  f y ) f y

p max  0.75 pb  p

Problem : A reinforced concrete T-beam with bf = 810 mm, d = 300 mm, bw = 200 mm , t = 100 mm, fc’=20.7 MPa, fy = 414 MPa is to be designed to carry an ultimate moment of 221 kN.m Determine the required steel area. solution

Asf 

M uf

0.85 fc ' (b f  bw )t fy



0.85(20.7)(810  200)100  2683.2mm 2 414

100 0.9(2683.2)414(300  ) t 2  249.9kN .m  Asf fy (d  )  2 10 6

Design as wide rectangular beam

a M u   0.85 fc' ab f (d  ) 2

a 221(10)   0.9 0.85(20.7)810a(300  ) 2 2 a 17229.6  300a  2 2 a  600a  34459.2  0 6

600  (600) 2  4(34459.2) a  64.33mm 2 0.85 fc ' b f a 0.85( 20.7)(810)64.33 As    2214.5mm 2 fy 414

As 2214.5 p   0.0091 b f d 810(300)

0.85(20.7)0.85 600  pb   0.0213 (600  414)414

pmax  0.75(0.0213)  0.016  0.0091

Tension steel yields at failure

Plate # 6: Flexural design of T-beams A reinforced concrete T-beam with d = 550 mm, bw = 300 mm , slab thickness =100 mm is 4.8 m long and spaced 3 m on centers. The beam a service dead load moment of 400 kN.m (including its weight) and service live load moment of 500 kN.m. If fc’=27.5 MPa, fy = 414 MPa , determine the required steel area. Design the reinforcement of a T- beam to a uniform service dead load of 25 kN/m and service live load of 30 kN/m on a simple span of 8 m . Properties of the T- beam are as follows : bf = 1500 mm, bw = 250 mm, d = 600 mm, t = 100 mm, fc’ = 20.7 MPa, fy = 345 MPa.

ONE WAY SLAB

Reinforced concrete slab are large flat plates that are ed at its sides by reinforced concrete beams, walls, column, steel beams or by the ground. If the slabs are ed on opposite sides only they are called one way slab since bending will occur on one direction only. A one way slab is considered as a wide & shallow rectangular beam. Reinforcing steel is usually spaced uniformly over its width. One way slabs are analyzed by considering a one meter strip, which is assumed independent of the adjacent strips. Maximum spacing of reinforcement Flexural reinforcement shall not be spaced farther apart than 3 times the slab thickness nor 450 mm. Minimum size of flexural reinforcement = 12 mm

Shrinkage and temperature reinforcement The area of shrinkage reinforcement shall be where Grade 275 deformed bars are used… 0.002bt where Grade 415 deformed bars are used… 0.0018bt where reinforcement with fy > 415 MPa measured at yield strain of 0.35% are used …. 0.0018(400)bt/fy Shrinkage reinforcement shall not be spaced farther apart than 5 times the slab thickness nor 450 mm. Minimum size of shrinkage and temperature bars = 10 mm

Design of one way slab Given: Loads, type of slab, fc’,fy,Wc Req’d: t, size and spacing of main bars and temperature bars General Procedure Determine the minimum slab thickness t using table for minimum thickness of non prestressed beams and one way slab. This thickness should be at least 75 mm 2. Compute the weight of the slab ( this is to be added to the given dead load) 3. Calculate the design moment Mu 4. Compute the effective depth d d = t – covering – ½ bar diameter ( minimum of 12 mm) 1.

5. Compute design constants

Mu Ru  bd 2

m

fy

0.85 fc '

1 2mRu p  (1  1  ) m fy

p > pmin 5. Solve for As As = pbd 6. Solve for the spacing of bars

a)S

A1 S  1000 As

where : A1 = area of 1 bar Use the smallest of the following b)3t c) 450 mm

7. Solve for area of temperature bars At = 0.002bt , At = 0.0018bt, At =0.0018(400)bt/fy

A1 1000 8. Solve for the spacing of temperature bars St  At where : A1 = area of 1 temperature bar ( minimum of 10mm dia.) Use the smallest of the following a) St b) 5t c)450 mm

Problem: Design a one way slab having a simple span of 3.0 m. The slab is to carry a uniform dead load of 2.5 KPa and uniform live load of 4.2 kPa. fc’ = 27.6 MPa , fy = 276 MPa for main bars and temperature bars. Concrete weighs 23.5 kN/m3 Solution

Slab thickness

fy L t (0.4  ) 20 700 3000 276 t (0.4  )  120mm  75mm 20 700

Weight of Slab (assuming 1 m wide strip)

Ws  23.5(1)0.12  2.82kN / m Total dead load

WD  2.5  2.82  5.32kN / m

Factored uniform load

Wu  1.4WD  1.7WL  1.4(5.32)  1.7(4.2)  14.59kN / m Design Moment

L2 14.59(3) 2 M u  Wu   16.41kN .m 8 8 Effective depth assuming 12 mm bar

1 12 d  t  20    120  20   94mm 2 2 Mu 14.59(10) 6 Ru    1.8346 MPa 2 2 bd 0.9(1000)(94) m

fy 0.85 fc'



276  11.76 0.85(27.6)

p

1 2(11.76)(1.8346) (1  1  )  0.00693 11.76 276

pmin 

1.4 1.4   0.00507 f y 276

As  pbd  0.00693(1000)94  651.42mm 2 Using 12 mm bars

 (12) 2 A 4 (1000)  173.6 say170mm  3t  360mm S  1 1000  As 651.42 Temperature bars

At  0.002bt  0.002(1000)120  240mm 2 Using 10 mm bars

 (10) 2 A1 4 (1000)  327 say 320mm  5t  600mm S t  1000  At 240

10 mm temp bars @ 320 mm o.c 12 mm main bars @ 170 mm o.c

120 mm

Problem: Design a 4 m long one way slab with one end discontinuous as shown in the figure . The slab is to carry a uniform dead load of 3.6 KPa and uniform live load of 4.0 kPa. fc’ = 20.7 MPa , fy = 415 MPa for main bars and fy = 276 MPa for temperature bars. Concrete weighs 22.56 kN/m3. column column column 4m

4m

C

t

B

A

L 4000   166.67 say170mm 24 24

Weight of Slab (assuming 1 m wide strip)

Ws  22.56(1)0.17  3.83kN / m Total dead load

WD  3.6  3.83  7.43kN / m Factored uniform load

Wu  1.4WD  1.7WL  1.4(7.43)  1.7( 4.0)  17.2kN / m Effective depth assuming 12 mm bar

1 12 d  t  20    170  20   144mm 2 2

Design Moments

L2 17.2(4) 2 M A  Wu   17.2kN .m 16 16 L2 17.2(4) 2 M B  Wu   19.66kN .m 14 14

17.2(10) 6 Ru   0.92 MPa 2 0.9(1000)(144) 19.66(10) 6 Ru   1.05MPa 2 0.9(1000)(144)

30.58(10) 6 L2 17.2(4) 2  1.635MPa M C  Wu   30.58kN .m Ru  2 0.9(1000)(144) 9 9

fy

415 m   23.58 0.85 fc' 0.85(20.7)

pmin 

1.4 1.4   0.00337 f y 415

1 2(23.58)(0.92) pA  (1  1  )  0.00228use0.00337 23.58 415 1 2(23.58)(1.05) pB  (1  1  )  0.00261use0.00337 23.58 415 pC 

1 2(23.58)(1.635) (1  1  )  0.00415 23.58 415

AsA  AsB  pmin bd  0.00337(1000)144  485.28mm 2 Using 12 mm bars

 (12) 2 A1 4 (1000)  233say 230mm  450mm S 1000  As 485.28

AsC  pc bd  0.00415(1000)144  597.6mm 2 Using 12 mm bars

 (12) 2 A 4 (1000)  189 say180mm  450mm S  1 1000  As 597.6 Temperature bars

At  0.002bt  0.002(1000)170  340mm 2 Using 10 mm bars

 (10) 2 A1 4 (1000)  230.9 say 230mm  450mm S t  1000  At 340

170mm

C

B

A

12 mm extra bars at 230 mm 0.C 10 mm temperature bars at 230 mm 0.C

12 mm continuous bent bars at 230 mm 0.C

Plate # 7: Design of one way slab Design a one way cantilever slab of 2.0 m span . The slab is to carry a uniform dead load of 2.4 KPa and uniform live load of 3.6 kPa. fc’ = 27.6 MPa , fy = 415 MPa for main bars and temperature bars. Concrete weighs 22.56 kN/m3. Draw layout of bars.

Design a 4.5 m long one way slab with one end discontinuous as shown in the figure . The slab is to carry a uniform dead load of 4.2 KPa and uniform live load of 4.5 kPa. fc’ = 20.7 MPa , fy = 345 MPa for main bars and fy = 276 MPa for temperature bars. Concrete weighs 23.5 kN/m3. Draw layout of bars. column

4.5m

column

4.5m

column

Prelim Exam A rectangular beam has b =300 mm, d = 500 mm, As = 6 of 32 mm, fc’ =27.6 MPa,fy =414 MPa. If the beam is simply ed on a span of 6 m, determine the concentrated live load that could be applied at the third points on the beam if steel covering is 80 mm and concrete weighs 23.5 kN/m3. A doubly reinforced rectangular concrete beam has b =350 mm,d =600mm, fc’=27.5 MPa, fy = 345 MPa, As =3625 mm2, As’ = 775mm2 ,covering for tension and compression bars 80 mm and 63 mm respectively. If the beam is an interior span of a three span continuous beam ing a service dead load of 20 kN/m (weight included) determine the maximum uniformly distributed live load it can on an average clear span of 5.0 m. Use NS moment coeffecients.

400 mm

3 of 16 mm 75 mm 400 mm

Determine the ultimate moment capacity of the triangular beam shown in figure. fc’ =20.7 MPa, fy = 345 MPa.

Solution to #2

As1  As  As '  3625  775  2850mm 2

As1 f y

2850(345) a   120.18mm 0.85 fc' b 0.85(27.5)350 600(a  1d ' ) 600(120.18  [0.85(63)]) f    332.65MPa  f y a 120.18 ' s

Compression steel does not yields at failure ∑Fx =0

0.85fc’ab + As’fs’ = Asfy 0.85(27.5)350a + 775fs’ = 3625(345) 10.56a +fs’ = 1613.7 fs’ =1613.7 – 10.56 a EQ.1

' 600 ( a   d 600(a  [0.85(63)]) ' ' 1 ) f fs   s  334.7 MPa  f y a a a Mu1  As1 f y (d  ) 600(a  53.55) ' 2 fs   EQ.2 121.12 a 0.9( 2850)345)(600  )

EQ.1  EQ.2 600(a  53.55) 1613.7  10.56a  a 1613.7 a  10.56a 2  600a  32130

Mu1 

10 6 Mu1  477.36kN .m

2

Mu2  As ' fs ' (d  d ' ) 0.9(775)334.7(600  63) 10 6 Mu2  125.36kN .m Mu2 

10.56a 2  1013.7 a  32130  0 a 2  96a  3042.6  0 M u  Mu1  Mu2  602.72kN .m 0.85( 27.5)334.7(600) 96  (96) 2  4(3042.6) p   0.0365 a  121.12mm b (600  345)345 2 ' fs ' 2 A  0 . 75 p bd  A  6500 . 6 mm s max b s 600(121.12  53.55) ' fy fs  121.12

#1 p=0.03217 pmax = 0.021 tension steel does not yield at failure a = 260.22 mm fs =379.92 MPa Mu = 610.31 kN.m WD = 4.1 kN/m P = 171.9 kN #2 a = 121.12 mm compression steel does not yield at failure fs’=334.7 MPa Mu = 602.72 kN.m WL = 125.34 kN.m

0.003 a

Ac

400mm

x

C=0.85fc’Ac 375 - 2a/3

3 of 16 mm 75 mm

T=Asfy

c

375 - c

s

400 mm

Solution

3 (16) 2 As   603.2mm 2 4

Assume that steel yields at failure( subject to checking) C=T 0.85fc’Ac=Asfy

0.85(20.7)Ac =603.2(345)

Ac =11,827.45 mm2

By similar triangles

x 400 x  0.842a  EQ.1  a 475 1 1 a  167.6mm 11827 . 45  a (0.842a) Ac  ax 2 2 a 167.6 c   197.17 mm 1 0.85 From the strain diagram

s

fs 400  c s   2000000 0.003 c fs 400  197.17  (200000)0.003 197.17

f s  617.22MPa  f y

2a M u  As f y (375  )  3

tension steel yields at failure

0.9(603.2)345(400  10 6

[2]167.6 ) 3  53.91kN .m

Shear and Diagonal Tension Another type of beam failure other than bending is shear failure. Shear failure are very dangerous if it happens before flexure failure because they can occur without warning BASIC CODE REQUIREMENT Factored shear strength Vu shall be equal or less than design shear ФVn

Vu  Vn where:

V n  Vc  V s Vc = shear carried by concrete Vs = shear carried by the stirrups Vu = factored shear strength

Shear Strength provided by concrete

Shear strength provided by concrete subject to shear and flexure only;

1 Vc  6

fc'bw d

or in more detailed calculation

 

Vc  

Vu d  fc'  120 p w  Mu 

where :

7

bw d  0.3 fc'bw d

Vu d  1. 0 Mu

fc' is in MPa and shall not exceed 0.7 MPa

bw = width of the beam web for T-beams, width of the beam for rectangular beams d =effective depth of the beam pw = As/bwd

Spacing limits of shear reinforcement Spacing S of shear reinforcement placed perpendicular to the axis of the member shall not exceed d/2 for nonprestressed and 3/4 h for prestressed , nor 600mm. When V s  0.33 fc 'bw d maximum spacing given by the above limits shall be reduced by one half. Minimum shear reinforcement Vc , minimum area of shear reinforcement When

Vu 

2

shall be provided in all reinforced concrete flexural except in the following conditions: a)slabs and footings b)concrete joist construction c)beams with total depth not greater than 250 mm,2.5 times flange thickness or half the width of the web whichever is greatest.

Where shear reinforcement is required, the minimum area of shear reinforcement shall be computed by:

Where:

bw S Av  3fy Av =cross sectional area of the stirrups taken twice for u-shaped stirrups

Shear strength provided by reinforcement a) When shear reinforcement perpendicular to the axis of

the member is used

Av f y d 2 Vs   S 3

fc'bw d

b) When inclined stirrups are used as shear reinforcement

Vs 

Av f y d (sin   cos  ) S

2  3

fc 'bw d

CRITICAL SECTION FOR BEAM SHEAR Maximum factored shear force Vu maybe computed in accordance with the following provided that: a) the reaction ,in the direction of the applied shear introduces compression on the end regions of member no concentrated load occur between the face of the and location of the critical section. b) For non prestressed ,sections located less than a distance of d from the face of the maybe designed for the same shear Vu as that computed at a distance of d Size of stirrups Main bars smaller than or equal to 32 mm diameter: 10 mm Main bars greater than 32 mm diameter : 12 mm Shear carried by stirrups

Vs 

Vu



 Vc

Problems: Determine the minimum cross section required for a rectangular beam to satisfy the condition that web reinforcement be neglected Vu = 72 kN,fc’ =27.6MPa. Assume d = 1.6b.

Vu  Vu  Vu 



Vc 2



fc 'bd ( 6) 2

b

72000(12) 0.86 27.6 (1.6)

b  350mm d  (1.6)350  560mm

fc 'b(1.6b) (6) 2

0.85 27.6 (1.6b 2 ) 72(1000)  (6)2

A rectangular beam with b = 270 mm, d = 500 mm is provided with 10 mm vertical stirrups with fy = 276 MPa. Assuming fc’ = 21 MPa; a) Determine the required spacing if Vu = 40 kN b) Determine the required spacing if Vu = 92 kN c) Determine the required spacing if Vu = 236 kN d) Determine the required spacing if Vu = 473 kN Solution Shear carried by concrete

1 Vc  6

fc'bw d

1 Vc  21(270)500  103,108 N 6 Vc 0.85(103108)   43821N 2 2

a) b)

Vc Stirrups not necessary Vu  42000 N  2 Vu  92000 N Vu Vs   Vc  92000 Vs   103108  5127.3N 0.85 2 (10) 2 Av   157.1mm 2 4 S

Av f y d Vs

157.1(276)500   4228.3mm 5127.3

1 1 fc 'bw d  21(270)(500)  206216 N 3 3 1 Vs  fc 'bw d 3 d 500 S   250mm Use S = 250 mm 2 2 c)

Vu  236000 N Vu Vs   Vc  236000 Vs   103108  174539 N 0.85

S

Av f y d Vs

157.1( 276)500   124mmsay120mm 174539

1 fc 'bw d  206216 N 3 1 Vs  fc 'bw d 3 d 500 S   250mm 2 2

2 3

fc 'bw d  412432 N

2 Vs  3

Beam size is inadequate for shear

Use S = 120 mm d)

fc 'bw d

Vu  473000 N Vu Vs   Vc  473000 Vs   103108  453362.5 N 0.85 1 fc 'bw d  206216 N 3

Design of vertical stirrups Given :bw or b,d,fc’,fy, beam loading & span, Required: size and spacing of stirrups General Procedure 1.Calculate factored shear force VU at the critical section. 2.Calculate shear strength of concrete:

1 Vc  6

fc'bw d

Vc Vu   2 Vc Vu   2

provide stirrups

stirrups not necessary

Vc Vu   2

Assuming that

3. Calculate the shear strength provided by the stirrups

Vu V s   Vc  Note: if

2 Vs  3

fc'bw d

adjust the beam size

4. Calculate the required spacing of stirrups Spacing is the smallest of: a)

Av f y d S Vs

Calculate

1 3

fc 'bw d

b)

S = d/2

when

c)

S = d/4

when

1 fc'bwd 3 1 Vs  fc'bw d 3 Vs 

5. Check for minimum required area of stirrups

Av 

bw S 3fy

Note: Av must be less than or equal to the actual area of Stirrups

Problem: A simply ed reinforced concrete beam 230 mm wide with an effective depth of 500 mm has a span of 6m. The beam carries a dead load of 9 kN/m ( including its own weight) and live load of 18 kN/m applied throughtout its entire span. Determine the required spacing of 10 mm stirrups. fc’ = 28 MPa ,fy= 345 MPa.

1 Vc  6

1 fc'bw d  28 (230)500  101420 N 6

Wu =1.4(9)+1.7(18)=43.2 kN/m wu

6m R

R =Wu(3)=43.2(3)=129.6kN

43.2 kN/m Shear force at crtical section VU

Vu = 129.6-0.5(43.2)=108kN =108000N

0.5 R=129.6kN

Vc Vc 0.85(101420) Vu    43103.5 N 2 2 2 Vu 108000 Vs   Vc   101420  0.85 2 (10) 2 2 A   157 . 1 mm Vs  25639 N v 4 S

Av f y d Vs

157.1(345)500   1056mm 25639

1 fc 'bw d  202840 N 3 1 Vs  fc 'bw d 3 d 500 S   250mm 2 2 Use

S  250mm

Minimum required area

bw S Av  3fy

230(250) Av   55.56mm 2  157.1mm 2 3(345) Use 10 mm u shaped stirrups spaced at 250 mm on centers

PLATE # 7 : SHEAR

A simply ed reinforced concrete beam 250 mm wide with an effective depth of 600 mm has a span of 7.5m. The beam carries a dead load of 12 kN/m ( including its own weight) and live load of 24 kN/m applied throughtout its entire span. Determine the required spacing of 10 mm stirrups. fc’ = 28 MPa ,fy= 345 MPa.

A rectangular beam with b = 300 mm, d = 550 mm is provided with 10 mm vertical stirrups with fy = 276 MPa. Assuming fc’ = 21 MPa; a) Determine the required spacing if Vu = 50 kN b) Determine the required spacing if Vu = 220 kN c) Determine the required spacing if Vu = 360 kN d) Determine the required spacing if Vu = 500kN

Design of beams for bending shear and deflection General procedure 1. 2. 3.

Design section and reinforcement by bending Design stirrups by shear Check adequacy of design by deflection

Problem : Design a rectangular beam for a 6 m simple span to a uniform dead load of 18 kN/m(weight included) and uniform live load of 12 kN/m applied along it’s entire length. fc’=20 MPa, fy = 345 MPa for main bars and stirrups,Wc=23.5 kN/m3 ,p=0.6pmax Consider immediate deflection due to live load only with an allowable of 1/360 of span length. fr=3.1 MPa n=10,Ec=20000 MPa. Use 100 mm covering.

Solution

Wu  1.4(18)  1.7(12)  45.6kN / m Wu L2 45.6(6) 2 Mu    205.2kN .m 8 8

0.85 fc ' 1 600 0.85(20)0.85(600) pb    0.02659 (600  f y ) f y (600  345)345 p  (0.6)(0.75 pb )  0.6(0.75)0.02659  0.012 fy

345 m   20.29 0.85 fc' 0.85(20) Ru  pf y (1 

pm 0.012[20.29] )  0.012(345)(1  )  3.635MPa 2 2

Try b = 250 mm

Mu 202.5(10) 6 d   500mm Ru b 0.9(3.635)250 Mu 202.5(10) 6 Ru    3 .6 2 2 bd 0.9( 250)(500) 1 2mRu 1 2(20.29)3.6 p  (1  1  ) (1  1  )  0.01186 m fy 20.29 345

As  pbd  0.01186 (250)500  1,482.5mm 2 1482.5 N  5 pcs 2  (25) 4

1 Vc  6

1 fc'bw d  20 (250)500  93169 N 6

R =Wu(3)=45.6(3)=136.8kN wu

6m R 45.6 kN/m Shear force at critical section VU 0.5 R=136.8

Vu = 136.8- 0.5(45.6)=114kN =114000N

Vc Vc 0.85(93169) Vu    39596.8 N 2 2 2 Vu 114000 Vs   Vc   93169  0.85

Vs  40949 N Using 10 mm u shape stirrups

2 (10) 2 Av   157.1mm 2 4

S

1 3

Av f y d Vs

157.1(345)500   661mm 40949

fc 'bw d  186338 N

d 500 S   250mm 2 2 Use

S  250mm

Minimum required area

bw S Av  3fy

230(250) Av   55.56mm 2  157.1mm 2 3(345)

Gross moment of inertia

bh 3 250(600) 3 Ig    4500 x106 mm 4 12 12

M cr 

fr I g yt

3.1(4500)10 6   46.5 x10 6 N .mm  46.5kN .m 300

Transformed section

b

b=250

x d=500 500-x As

nAs 100

5 (20) 2 As   1570mm 2 4

x2 b  nAs (d  x) 2 x 2  62800  125.6 x

x2 250  10(1570)(500  x) 2

x 2  125.6 x  62800  0

 125.6  (125.6) 2  4(62800) x  195.6mm 2 3 bx 3 250 ( 195 . 6 ) I cr   nAs (d  x) 2   10(1570)(500  195.6) 2 3 3

I cr  2,078.3x106 mm 4 a)

WL L2 (12)(6) 2 Ma    54kN .m 8 8

Effective moment of inertia

  M    M cr  cr    I cr  Ie    Ig  1    M a    Ma  3

3



 46.5   46.5   6 6 Ie   4500 x 10  1  2078 . 3 x 10       54    54   3

3

I e  3,624.6 x106 mm 4

Instantaneous deflection due to live load

5WL L4 5(12)(6000) 4    2.79mm 6 384 Ec I e 384(20000)3624.6(10) allowable deflection due to live load

 allowable

L 6000    16.67mm  2.79mm 3 360 360 Section is adequate

Plate #8:Design of beams for bending shear and deflection

Make a complete design of a rectangular beam reinforced for tension only for a 7.5 m simple span to a uniform dead load of 24 kN/m(weight included) and uniform live load of 18 kN/m applied along it’s entire length. fc’=20 MPa, fy = 300 MPa for main bars and stirrups,Wc=23.5 kN/m3 ,p=0.18(fc’/fy). Consider immediate deflection 0.7 fcof' span length. due to total load with an allowablef rof1/360 Es = 200000MPa,Ec=18500 MPa, . Use 100 mm covering.

BOND and DEVELOPMENT LENGTH Bond In reinforced concrete, concrete and steel act as a unit. For this to happen, there must be absolutely no slippage of the bars in relation to the surrounding concrete. The steel and concrete must stick or bond together so that there will be transfer of stress from steel to concrete and vice-versa. Failure of transfer of stress makes the concrete an unreinforced member thus it will be subject to collapse. Development Length Bar development length is the embedment necessary to assure that the bar can be stressed to its yield point with some reserved to ensure member toughness.

Basic concept of development length

 2 d b Ld  d b f y 4 fy Ld  db 4 : μ = average bond stress db

F Ld

ΣF = 0 F=T

T = Abfy

Where Ld = minimum development length

The code however provides the basic development length Ldb for various conditions. The values provided are modified for different conditions. Thus the minimum development length provided by the code is; Ld = Ldb(applicable modification factors) ≥ 300 mm

Basic Development Length of bars in tension For 32 mm bar & smaller and deformed wire

Ldb 

0.02 Ab f y fc '

For 36 mm bar

Ldb 

25 f y fc '

For deformed wire

Ldb 

3d b f y

8

but not to be taken less than 0.6dbfy

Where Ldb = basic development length ( mm) Ab = area of one bar (mm2) db = diameter of one bar (mm) fc’,fy in MPa

fc '

Basic Development Length of bars in compression

Ldb 

0.24d b f y fc'

but not to be taken less than 0.04dbfy

Modification Factors for bars in tension Condition

Modification Factor,m

a) For bars in beams or column with a minimum cover not less than specified by the code

1.0

b)

1.0

For bars in beams or column with transverse reinforcement satisfying the requirement of the code

c) Bars in beams or column with a clear spacing not less than 3db

1.0

d) Bar in the inner layer of slab or wall reinforcement and with a clear spacing not less than 3db

1.0

e) Any bars with cover of not less than 2db and with a clear

1.0

spacing not less than 3db

f) For bars with cover of db or less with a clear spacing of 2db or less

2.0

g) For bars not included in items a to f

1.4

h) For 32 mm bar and smaller with clear spacing not less than 5db and with cover from face of the member to edge bar, measured in the plane of the bar, not less than 2.5db, the factors from items a to g may be multiplied by 0.8

0.8

i) Top reinforcement

1.3

j) Lightweight aggregate concrete K)Lightweight aggregate concrete when fct is specified l) For reinforcement enclosed within special reinforcement not less than 6 mm diameter and not more than 100 mm pitch, within 12 mm or larger circular ties spaced at not more than 100 mm on center or larger ties or stirrups spaced not more than 100 mm on center and arranged such that alternate bars shall have ed by the corner of a tie hoop with an included angle not more than 1350,the factors in items a

through g maybe multiplied by 1.8

m) Excess Reinforcement. Development length maybe reduced where reinforcement in a flexural member is more than required by analysis

fc1.3 ' 1.8 f ct

1.8 As required As provided

Modification Factors for bars in Compression Condition

Modification Factor,m As required

a)

Excess reinforcement. Reinforcement more than that required by analysis

b) Spiral and Ties. Reinforcement enclosed within spiral reinforcement not less than 6 mm diameter and not more than 100 mm pitch or within 10 mm ties and spaced not more than 100 mm on center.

As provided 0.75

Problems A rectangular beam 200 mm wide and 400 mm deep is reinforced with 3 of 22 mm tension top bars with fc’ = 20.7 MPa and fy = 275 MPa. Calculate the required development length.

Solution

Ldb 

Ldb 

0.02 Ab f y fc ' 0.02

 ( 22) 2 4 20.7

275

 460mm

Ldb  0.06d b f y Ldb  0.06( 22)(275)  363mm

Use

Modification Factor Top bar = 1.3 Required development length

Ld  1.3Ldb  1.3(460)  598mm

Ldb  460mm

A rectangular beam 250 mm wide and 500 mm deep is reinforced with 4 of 25 mm with fc’ = 27 MPa and fy = 345 MPa. If the member is made up of lightweight aggregate with fct = 2.88 MPa ,Calculate the required development length. Solution

Ldb 

Ldb 

Modification Factor Lightweight concrete With specified fct

0.02 Ab f y fc ' 0.02

 ( 25) 2 4 27

345

 652mm

Ldb  0.06d b f y Ldb  0.06( 25)(345)  518mm Use

Ldb  652mm

fc' 27   1.00 1.8 f ct 1.8(2.88) Ld  1.0 Ldb Ld  1.0(652) Ld  652mm

A cantilever beam 320 mm wide and 500 mm deep is reinforced with 3 of 36 mm straight top bars with fc’ = 27 MPa and fy = 345 MPa. Calculate required development length. Solution

Ldb 

Ldb 

25 f y

fc ' 25(345)

27

 1660mm

Modification Factor Top bar = 1.3 Required development length

Ld  1.3Ldb  1.3(1660)  2158mm

A rectangular beam 250 mm wide and 410 mm deep is reinforced with 3 of 20 mm compression bars with fc’ = 20.7 MPa and fy = 275 MPa. Calculate the required development length.

Ldb 

0.24d b f y fc'

0.24(20)275 Ldb   254mm 20.7 Ldb  0.04d b f y Ldb  0.04( 20)(275)  220mm Use

Ldb  254mm

No applicable modification factor

Ld  254mm

Plate # 9 Development Length A rectangular beam 300 mm wide and 450 mm deep is reinforced with 4 of 25 mm tension top bars with fc’ = 20.7 MPa and fy = 345 MPa. Calculate the required development length. A rectangular beam 200 mm wide and 350 mm deep is reinforced with 3 of 20 mm with fc’ = 30 MPa and fy = 415 MPa. If the member is made up of lightweight aggregate with fct = 2.4 MPa ,Calculate the required development length. A cantilever beam 300 mm wide and 450 mm deep is reinforced with 6 of 22 mm straight top bars with fc’ = 20.7 MPa and fy = 415 MPa. Calculate required development length. A rectangular beam 350 mm wide and 600 mm deep is reinforced with 4 of 32 mm compression bars with fc’ = 20.7 MPa and fy = 275 MPa. Calculate the required development length.

Development Length of Flexural Reinforcement Tension reinforcement in flexural maybe developed by : a) bending across the web to be anchored b) made continues with reinforcement on opposite face ofthe member Critical points for development length in flexural are at points of maximum stress and at points where the adjacent reinforcements terminates or is bent. Reinforcement shall extend beyond the point at which it isno longer required to resist flexure for a distance equal to the effective depth of member or 12db whichever is greater, except at s of simple beam and free end of cantilevers. Continuing reinforcement shall have an embedment length not less than the development length beyond the point where bent or terminated tension reinforcement is no longer required to resist flexure

Development of positive moment bars The code stipulates that at least one third the positive reinforcement in simple and one fourth the positive reinforcement in continuous shall extend along the same face in the member into the s. In beams such reinforcement shall extend into the at least 150mm. At simple s and at points of inflection,positive moment tension reinforcement shall be limited to a diameter such that Ld computed by

Mn Ld   la Vu Where: Mn = nominal moment strength assuming that all reinforcement at section are stressed to specified yield strength f y Vu = factored shear force at s for simple beams and at points of inflection for continuous beams

la = end anchorage ;at shall be the embedment length beyond the center of ; at point of inflection limited to the effective depth or 12db whichever is greater. Value of Mn/Vu maybe increased by 30% when the ends of the reinforcement are confined by a compressive reaction such as a column below but not when the beam frame into the girder.

Mn Ld  1.3 l a Vu When

Where:

Mn Ld  1.3  la Vu

or

a M n  As f y (d  ) 2

Mn Ld   la Vu a

As f y 0.85 fc' b

Use smaller bar Size or increase End anchorage la

Max Ld

la

1.3Mn/Vu

at least 1/3 of positive reinforcement

VU

Max Ld

tension bars

Mn/Vu

la

Max Ld la is the larger Value of d or 12db

la

Mn/Vu

tension bars

150 mm min

at least ¼ of positive moment reinforcement .

CL

Point of inflection

Development length for Negative moment Reinforcement Negative moment reinforcement should have an embedment length into the span to develop the calculated tension in the bar,or a length equal to the effective depth of the member or 12db whichever is the greatest. At least one third of the total negative reinforcement should have an embedment length beyond the point of inflection not less than the effective depth of the member or 12 db or 1/16 of the clear span whichever is greatest. Ld `

Development length for Negative moment the larger value of 12db,d or Ln/16

at least 1/3 of the total Negative moment reinforcement Clear span =Ln

Point of inflection

Recommended bar details for continuous beams

Ln1/4

0.3 Ln1

Ln1/8

150 mm min

0.3 Ln2

Ln1/8

Ln1

150 mm min

Ln2/8

Ln2

Hooks If sufficient space is not available to anchor tension bars by running them straight for the required development length as required by the code, hooks maybe used. 12db

r db

r 90 hook

D

0

bend diameter,D =2r

D =6db for 10 mm through 25 mm bars D =8db for 28 mm through 32 mm bars D =10db for 36 mm bars

4db

1800 hook

STANDARD HOOKS

Ldb 65 mm min

12db

4db

4db for 10 mm through 25 mm bars 5db for 28 mm through 32 mm bars 6db for 36 mm bar

Development of standard hooks Basic Development Length of standard hooks

l hb 

100d b fc '

Actual Development Length of standard hooks ldh = lhb(applicable modification factors) ≥8db nor 150 mm

Modification Factors 1. If the reinforcing bar has an fy other 415 MPa, lhb is to be multiplied by fy/415 2. When 900 hooks and 32 mm or smaller bar are used and when 60 mm or more of side cover normal to the hook is present, together with at least 50 mm cover for the bar extension, lhb is to be multiplied by 0.7. 3. When hooks made of 32 mm and smaller are enclosed vertically and horizontally within ties or stirrups ties spaced no farther apart than 3db, lhb is to be multiplied by 0.8 4. Where the amount of flexural reinforcement exceeds the theoretical amount required and where the specifications being used do not specifically require that development lengths be based on fy the value of lhb is to be multiplied by Asrequired/As provided. 5. When light weight concrete are used, apply a modification factor of 1.3 6. For bars being developed by standard hook at discontinuous end with side cover and top or bottom cover over hook less than 60 mm,hooked bar shall be enclosed within ties or stirrups spaced along the full development length ldh not greater than 3db where db is the diameter of hooked bar. For this case, the factor mentioned in item 3 shall not apply.

Splices of Reinforcement Splicing maybe done by welding, by mechanical connections or most frequently by lapping bars. Lapped bars are usually tied in . Lap splice must not be used for bars larger than 32 mm. Splices in tension The minimum length of lap for tension lap splice shall be as required for class A or class B, but shall not be less than 300 mm, where Class A splice 1.0Ld Class B splice 1.3Ld Lap splices of deformed bars and wires in tension shall be class B splice except that class A splice are allowed when •the area of the reinforcement provided is at least twice than that required by analysis. •One half or less of the total reinforcement is spliced within the required lap strength. Splices of deformed bars in compression Compression bars maybe spliced by lapping,end bearing, welding or mechanical devices. The minimum length of such bars should be the development length Ld but may not be less than 0.07d bfy for fy of 415 MPa or less, or (0.13fy -24)db for fy greater than 415 MPa.

Problems A simply ed beam is reinforced with three of 28 mm bars with fc’= 27.6 MPa and fy = 275 MPa. Assuming that side, bottom and top cover to be greater than 60 mm, determine the following: a) the required development length if a 900 hook is used b) the required development length if a 1800 hook is used

Solution

Using a 900 hook

l hb  lhb 

ldh

ldh

100d b fc'

100(28)  533mm 27.6

Modification factor for fy other than 415 MPa

m

fy 415



275  0.6626 415

Modification factor for 900 hook = 0.7 Required development length Ldh =0.6626(533)0.7=247.2 say 250 mm

ldh Solution Using a 1800 hook

l hb  lhb 

100d b fc'

100(28)  533mm 27.6

Modification factor for fy other than 415 MPa

m

fy 415



275  0.6626 415

Required development length Ldh =0.6626(533)=353.17 say 355 mm

Problem For the simply ed beam shown in figure below, investigate whether the bars size is satisfactory for the required development length. The beam is reinforced with 4 of 25 mm bars . fc’ = 20.7 MPa and fy = 414 MPa, Vu = 270 kN. The beam is made up of normal sand concrete and the reaction produces compression on concrete.

300mm

600 mm

Ldb  0.06d b f y Solution 175 mm Basic development length

Ldb  Ldb 

Ldb  0.06( 25)(414)  621mm Use

0.02 Ab f y

0.02

Since there is no applicable modification factor

fc '  ( 25) 2 4 20.7

Ldb  894mm

414

 894mm

Ld  894mm

a

As f y

425.14 x106 1.3  175  2221.2mm 270(1000)

0.85 fc' b  (25) 2 4 414 4 a  154mm 0.85(20.7)300

a M n  As f y (d  ) 2

 (25) 2 154 Mn  4 414(600  ) 4 2 M n  425.14 x106 N .mm

Mn 1.3  la Vu

894mm  2221.2mm Mn Ld  1.3  la Vu bars are adequate

A rectangular beam has b = 380 mm and d = 500 mm. The beam is simply ed over a length of 6 m measured from the center of the and is reinforced for tension with 6 of 25 mm diameter bars. Assuming fc’ = 21 MPa, fy = 276 MPa, draw the details of the bar showing the cut off points of each bar. The beam carries a uniformly distributed load throughout its length and is made up of normal sand- gravel concrete. The reactions at the ends produces compression on concrete. Maximum value of la is 150 mm. Vu = 210 kN at .

Ldb  0.06d b f y Solution Basic development length

Ldb  Ldb 

Ldb  0.06( 25)(276)  414mm Use

0.02 Ab f y

0.02

Since there is no applicable modification factor

fc '  ( 25) 2 4 21

Ldb  591mm

276

 591mm

Ld  591mm

BAR LAYOUT & CUT OFF POINTS L3 L1

2 of 25

L2

CL

x2

x1 3m

y 2y 3 3

y

Parabolic Moment diagram

Location of cut off points By square property of parabola

2y x12 3  32 y

x1  2.45m

y x22 3  2 3 y

x2  1.73m

Available development length

Reinforcement shall extend beyond the point at which it is no longer required to resist flexure for a distance equal to the effective depth of member or 12db whichever is greater

12(25)  300mm

d  500mm

Use extension = 500 mm

L1  x1  500  2450  500  2950mm  Ld L2  x2  500  1730  500  2230mm  Ld

No need for hook

The code stipulates that at least one third the positive reinforcement (2 bars) in simple shall extend along the same face in the member into the s

For the last two bars the code requires that the be bars be limited to a diameter so that

Ld  1.3

Mn l a Vu

2 (25) 2 AS   981.75mm 2 4 As f y a 0.85 fc ' b 981.75(276) a  40mm 0.85(21)380

a M n  As f y (d  ) 2

130.1x106 1.3  150  955mm  Ld 210(1000)

L3  3000  150  3150mm  Ld

40 M n  981.75(276)(500  ) 2

M n  130.1x106 N .mm

2230 2950

2230 2950

3150

3150 DETAILS

Problem Figure shows the first interior of a continuous beam reinforced with 6 of 25 mm bars to resist a negative moment of 496 kN.m for which the calculated steel area required is 2600 mm2. Determine and draw the details of the lengths of the bars required if normal gravel-sand concrete is used. fy =414 MPa ,fc’ = 20.7 MPa, b = 350mm,d= 600 mm. 7m

10 m

2.4 m

2.1 m

350mm 320mm Mu for 4 bars

point of inflection

496 kN.m

point of inflection

7m

10 m

Bar 3 Bar 2 Bar 1 Ld1 L1

Ld2 L2

L3

L4

2.4 m

2.1 m Mu for 2 bars

350mm 320mm

Mu for 4 bars point of inflection

496 kN.m (Mu for 6 bars)

point of inflection

Solution Basic development length

Ldb  Ldb 

0.02 Ab f y

0.02

Ldb  0.06d b f y

fc'  ( 25) 2 4 20.7

For bar 1

Ldb  0.06( 25)(414)  621mm 414  893mm Use Ldb  893mm

(As at section = 2 of 25 mm)

At least one third of the total negative reinforcement should have an embedment length beyond the point of inflection not less than the effective depth of the member or 12 db or 1/16 of the clear span whichever is greatest. 1/3 of 6 = 2 bars 12db = 12(25)=300 mm d=600 mm

1/16 of 7000=438 mm 1/16 of 10000=625mm

7m span

L3  2400  600  3000mm

10m span

L4  2100  625  2725mm

for 7 m span use 600 mm for 10 m span use 625 mm

For bar 2

(As at section = 4 of 25 mm)

Negative moment reinforcement should have an embedment length into the span to develop the calculated tension in the bar,or a length equal to the effective depth of the member or 12db whichever is the greatest 12db = 12(25)=300 mm d=600 mm use 600 mm

L1  350  600  950mm For bar 3

As provided 

L2  320  600  920mm

(As at section = 6 of 25 mm) Modification factor 6 ( 25) 2

4

 2945.24

Asrequired  2600mm 2

m

Asrequired

As provided



2600  0.88 2945.24

Ld1  0.88(893)  785mm Ld 2  0.88(893)  785mm

Plate # 10. BAR DETAILING A rectangular beam has b = 350 mm and d = 450 mm. The beam is simply ed over a length of 5 m measured from the center of the and is reinforced for tension with 7 of 20 mm diameter bars. Assuming fc’ = 27 MPa, fy = 345 MPa, draw the details of the bar showing the cut off points of each bar. The beam carries a uniformly distributed load throughout its length and is made up of normal sand- gravel concrete. The reactions at the ends produces compression on concrete. Maximum value of la is 150 mm. Vu = 150 kN at . 350 mm

450 mm

Hint : extend 3 bars beyond the

Figure shows the first interior of a continuous beam reinforced with 6 of 25 mm bars to resist a negative moment of 450 kN.m for which the calculated steel area required is 2400 mm2. Determine and draw the details of the lengths of the bars required if normal gravel-sand concrete is used. fy =414 MPa ,fc’ = 20.7 MPa, b = 350mm,d= 600 mm. 6m

9m

2.5 m

2.0 m

330mm 300mm

point of inflection

450 kN.m

point of inflection

Axially Loaded Columns 





Columns are classified as pedestal, short column and long column. Pedestal is a column whose height is less than three times its least lateral dimension. They maybe designed without reinforcement with a maximum permissible compressive strength of Φ0.85fc’ where Φ =0.70. If the column does not qualify as pedestal then it maybe classified as a short column. They fail due to initial material failure. The load on the column depends on the dimension and the strength of the material it is made. If the length of the column is increased, chances of lateral buckling increases. Column that fails due to lateral bucking are classified as long columns.

AXIALLY LOADED SHORT TIED COLUMNS

The axial load capacity of tied columns is given by: Pu =Ф0.80{0.85fc’(Ag-Ast) + Astfy} Ф = 0.70 Ag = gross area of the column Ast = area of reinforcement Ast Limits of reinforcement for tied columns pg  1. Pg ranges from 0.01 to 0.06 Ag 2. The minimum number of longitudinal bars is 4 for bars within rectangular or circular ties, 3 for bars within triangular bars. Sizes and spacing of main bars and ties 1. Clear distance between longitudinal bars shall not be less than 1.5dbnor 40 mm. 2. Use 10 mm diameter ties for 32 mm bars or smaller and at least 10 mm in size for 36 mm and bundled longitudinal bars. Vertical spacing of ties shall be the smallest of the following:   

16 times longitudinal bar diameter 48 times tie diameter least dimension of the column

Axially loaded short spiral columns The axial load capacity of spiral columns is given by: Pu =Ф0.85{0.85fc’(Ag-Ast) + Astfy} Ф = 0.75 Limits of reinforcement for spiral columns 1. Pg ranges from 0.01 to 0.06 2. The minimum number of longitudinal bars is 6 Sizes and spacing of main bars and ties 1. Clear spacing between spirals shall not exceed 75 mm, nor less than 25 mm 2. For cast in place construction, size of spirals shall not be less than 10 mm. 3. The percentage of spiral reinforcement is computed by

4a s ( Dc  d b ) ps  SDc2

Where: Dc = diameter of the concrete core as = cross sectional area of the spiral S = spacing of the spirals db = diameter of main bars 4. The minimum spiral percentage is given by

p s min

Ag 0.45 fc' (  1) Ac  fy

Where: Ac = area of the concrete core

D A  4

2 c

c

Clear cover ( min. of 40 mm)

DC D Clear cover ( min. of 40 mm)

Dc = D – 2 ( clear cover )

Problems A square tied column 350 mm by 350 mm is reinforced with 6 of 25 mm bars with fc’ = 20.7 MPa and fy = 345 MPa. Determine the following : a) Ultimate axial load capacity of the column. b) spacing of 10 mm lateral ties Solution 

 (25) 2 Pu =Ф0.80{0.85fc’(Ag-Ast) + Astfy} Ast  6  2945.25mm 2 4 0.7(0.80){0.85(20.7)[(350)(350)  2945.25]  2945.25(345)} Pu   1747 kN 1000 Spacing of 10 mm ties

S  16(25)  400mm S  48(10)  480mm S  350mm

Use S = 350 mm

Problems 

A circular spiral column 400 mm in diameter is reinforced with 8 of 25 mm bars with fc’ = 20.7 MPa and fy = 345 MPa. Determine the following : a) ultimate axial load capacity of the column. b) the spacing of 10 mm spirals assuming clear covering of 40 mm

Solution Pu =Ф0.85{0.85fc’(Ag-Ast) + Astfy}

 (25) 2 Ast  8  3927 mm 2 4

 ( 400) 2 Ag   125664mm 2 4 0.75(0.85){0.85(20.7)[(125664  3927]  3927(345)} Pu   2229.2kN 1000

Diameter ot the concrete core Dc = D – 2 ( clear cover )=400-2(40) = 320 mm

Dc2  (320) 2 Ac    80425mm 2 4 4 0.45 fc' ( ps 

fy

Ag Ac

 1)

125664 0.45(20.7)(  1) 80425   0.0152 345

Ds2  (10) 2 As    78.54mm 2 4 4 4as ( Dc  d b ) 4(78.54)(320  25) S   60mm  25mm  75mm 2 2 Ps Dc 0.0152(320) Use s = 60 mm

Short Composite column Design strength of composite





Pu   0.85 0.85 fc ' Ac  Ast f y  Ass Fy Where: Ф= 0.75 for composite with spiral reinforcement Ф= 0.70 for other reinforcement Ast = area of reinforcing bars of strength fy Ass = area of structural shape of strength Fy Ac = net concrete area For evaluation of slenderness effect, radius of gyration of composite should not exceed

Ec I g r

5 E c Ag 5

 Es I t  E s At

Structural steel encased concrete core

Steel pipe filled with concrete

t Dc D

Concrete Core

t

t min  D

fy 8Es

Steel tubing filled with concrete t2

Concrete Core

b2 t1

b1

t1min  b1

fy 3E s

and

t 2 min  b2

fy 3E s

Calculate the maximum axial load that the concrete-filled pipe shown can resist. fc’ = 25 MPa, Fy =248 MPa for steel pipe. Check if the section complies with the code. 10 mm Concrete core

300 mm

 [(300) 2  (280) 2 ] Ap   9111mm 2 4  [(280) 2 ] Ac   61575mm 2 4



Pu   0.85 0.85 fc' Ac  Ap Fy



Pu  (0.7)0.85 0.85(25)(61575)  9111(248)  2122958 N Minimum required thickness of pipe

t min

fy

248 D  300  3.74mm  10mm 8Es 8(200000)

Calculate the ultimate axial load capacity of the composite column shown below. fc’ = 21 MPa, Fy = 248 MPa 500 mm

Properties of W 14 x 210 Area = 40000 mm2 Depth = 400 mm

500 mm

Ac  500(500)  40000  210000mm 2 W14 x 210



Pu   0.85 0.85 fc' Ac  AWF Fy



Pu  (0.7)0.85 0.85(21)(210000)  40000(248)  8132757 N

Calculate the ultimate axial load capacity of the composite column shown below .fc’=21 MPa, Fy=248 MPa,fy =276MPa. Reinforcing bars consist of 8 of 20 mm diameter bars. 500 mm

500 mm

Properties of W 14 x 210 Area = 40000 mm2 Depth = 400 mm

8 [(20) 2 ] As   2513mm 2 4

W14 x 210

Ac  500(500)  40000  2513  207487 mm 2



Pu   0.85 0.85 fc ' Ac  As f y  AWF Fy



Pu  (0.7)0.85 0.85(21)(207487)  2513(276)  40000(248) Pu  8518752 N

Design of axially Loaded Tied Columns Given :axial load, fc’, fy Required : Column size, size of main bars, size and spacing of lateral ties General Procedure 1.Solve for design axial load Pu 2.Assume pg 0.01 to 0.06 3. Solve for required gross area to get column size

Pu Ag  (0.7)0.8 0.85 fc ' (1  p g )  p g f y  4. Solve for Ast

Ast = pgAg 5. Using adjusted values of column size and Ast, check column capacity (optional) Pu =Ф0.80{0.85fc’(Ag-Ast) + Astfy} 6. Assume size ot ties and solve for the spacing Use the smallest value from the ff: • 16 times longitudinal bar diameter • 48 times tie diameter c) least dimension of the column

Problem: Design a square tied column to an axial dead load of 600 kN and axial live load of 750 kN. fc’ =20.7 MPa, fy = 345 MPa.

Pu  1.4 PD  1.7 PL  1.4(600)  1.7(750)  2115 kN Assume pg = 0.02

Pu Ag  (0.7)0.8 0.85 fc ' (1  p g )  p g f y  2115(1000) Ag   156435mm 2 (0.7)0.8 0.85( 20.7)(1  0.02)  0.02(345) t

Ag  156435  395mm

Say 400 mmx 400 mm square column

As  p g Ag  0.02(156435)  3128mm 2 Using 20 mm bars

No. 

3128( 4)  10 pcs 2  ( 20)

Note: No need to check column capacity Spacing of 10 mm ties

S  16( 20)  320mm S  48(10)  480mm S  400mm

Use S = 320 mm

Design of axially Loaded Short Spiral Columns Given :axial load, fc’, fy Required : Column size, size of main bars, size and spacing of spirals General Procedure 1.Solve for design axial load Pu 2. Assume pg 0.01 to 0.06 3. Solve for required gross area to get column size

Pu Ag  (0.75)0.85 0.85 fc' (1  p g )  p g f y 

4. Solve for Ast

Ast = pgAg 5. Using adjusted values of column size and Ast, check column capacity (optional) Pu =Ф0.85{0.85fc’(Ag-Ast) + Astfy} 6. Solve for the diameter of the concrete core Dc = D – 2 ( clear cover) 7. Solve for the spiral percentage

0.45 fc ' ( Ag  Ac ) ps  fy

8. Assume size of spirals and solve for required spacing

4as ( Dc  d b ) S Ps Dc2

where: S ≥ 25 mm, S≤ 75 mm

Problem: Design a circular spiral column to an axial dead load of 600 kN and axial live load of 750 kN. fc’ =20.7 MPa, fy = 345 MPa.

Pu  1.4 PD  1.7 PL  1.4(600)  1.7(750)  2115 kN Assume pg=0.02

Pu Ag  (0.75)0.85 0.85 fc ' (1  p g )  p g f y  2115(1000) Ag   137427 mm 2 (0.75)0.85 0.85(20.7)(1  0.02)  0.02(345)

D 2  137427 mm 2 4

D  420mm

Ast  p g Ag  0.02(137427)  2749mm 2

25 mm main vertical bars

Using 25 mm bars

N

2749(4) 6 2  (25)

40

Diameter ot the concrete core Dc = D – 2 ( clear cover )=420-2(40) = 340 mm

Dc2  (340) 2 Ac    90792mm 2 4 4 0.45 fc' ( ps 

fy

Ag Ac

 1)

420

40 10 mm spirals at 65 mm pitch

137427 0.45(20.7)(  1) 90792   0.014 345

Using 10 mm spirals

Ds2  (10) 2 As    78.54mm 2 4 4

S

4as ( Dc  d b ) 4(78.54)(340  25)   65mm  25mm  75mm 2 2 Ps Dc 0.014(340) Use 10 mm spirals at 65 mm pitch

Plate # 11 : Axially loaded short columns A square tied column 400 mm by 400 mm is reinforced with 8 of 25 mm bars with fc’ = 20.7 MPa and fy = 345 MPa. Determine the following : a) Ultimate axial load capacity of the column. b) spacing of 10 mm lateral ties A circular spiral column 450 mm in diameter is reinforced with 10 of 25 mm bars with fc’ = 27 MPa and fy = 415 MPa. Determine the following : a) ultimate axial load capacity of the column. b) the spacing of 10 mm spirals assuming clear covering of 40 mm Calculate the maximum axial load that the concrete-filled pipe shown can resist. fc’ = 20 MPa, Fy =250 MPa for steel pipe. Check if the section complies with the code. 12mm Concrete core

350 mm 12mm

Calculate the ultimate axial load capacity of the composite column shown below .fc’=25 MPa, Fy=248 MPa,fy =345MPa. Reinforcing bars consist of 8 of 20 mm diameter bars. 450 mm

450 mm

Properties of W 12 x 150 Area = 30000 mm2 Depth = 300 mm

W12 x 150

Design a circular spiral column to an axial dead load of 450 kN and axial live load of 600 kN. fc’ =20.7 MPa, fy = 345 MPa. Use pg =0.025 and 10 mm spirals Design a square tied column to an axial dead load of 450 kN and axial live load of 600 kN. fc’ =20.7 MPa, fy = 345 MPa. Use pg =0.03 and 10 mm ties.

SHORT COLUMNS SUBJECT TO AXIAL LOAD AND BENDING

( Eccentrically loaded columns) 

All columns are subject to axial force and bending and they must be proportioned to resist these forces. Eccentricities of 0.10h for tied column and 0.05h for spiral columns can be permitted for axially loaded columns. Beyond this, analysis for the effect of axial load and bending must be undertaken.



Columns will tend to bend under the action of moment and produces compression on one side and tension on the other side. The following failures are possible under combined axial load and bending.

Pn 1.

Plastic centroid

e

Pn

Large axial load and negligible moment. Failure under this condition occur by sudden crushing of concrete with all bars reaching their yeild stress simultaneuosly

2. Large axial load and small moment. with entire cross section in compression . Failure under this Plastic centroid condition occur by sudden crushing of concrete with all bars are in compression

Pn e

Plastic centroid

3. Large axial load with moment bigger in 2. Bars in the far side in tension has not yielded. Failure occur by crushing of concrete

e

Pn Plastic centroid

e

4. Balance loading condition – bars on the tensile side yield at the same time that concrete crushes at 0.85fc’

Pn

Plastic centroid

5. Large moment, small axial load Failure initiated by tensile yielding of tension bars

M Plastic centroid

6. Moment, no axial load - Fails as a beam

Plastic centroid – the point in the column through which the resultant of the column load es through to produce uniform strain at failure. It is the location of the resultant force of concrete and steel. In locating this point, all concrete is assumed to be stressed in compression to 0.85fc’ and all steel bars in compression at (fy – 0.85fc’). For symmetrical cross sections the plastic centroid coincides with the centroid of the cross section. Eccentricity of a column is the distance from the load to the plastic centroid of the cross section. eccentricity

Pn Plastic centroid centroid of cross section Plastic centroid Centroid of cross section

Problem: The T shaped cross section shown below is reinforced with 4 of 32 mm bars with fc’= 20.7 MPa and fy = 345 MPa. Determine the location of the plastic centroid measured from the 450 mm side. 150 mm 200 mm 100 mm 450mm

250 mm 100 mm 75mm

75mm

150 mm 200 mm 100 mm

1 450mm

250 mm

2

100 mm 75mm 75

75mm

P1 P 2

P4

75 275

P

x

P2  0.85 fc' A2  0.85(20.7)200(250)  879.75kN  (32) 2 P3  ( fy  0.85 fc' ) As  {345  0.85(20.7)}2  526.63kN 4  (32) 2 P4  ( fy  0.85 fc' ) As  {345  0.85(20.7)}2  526.63kN 4 P  P1  P2  P3  P4  3120.7kN

P x   Px

P x  P1 x1  P2 x2  P3 x3  P4 x4

250

P3

P1  0.85 fc' A1  0.85(20.7)150(450)  1187 .63kN

3120.7 x  1187 .63(75)  879.75(250)  526.63(75)  526.63 275

x  158mm

Balanced Loading condition



Balanced loading occur when the tension steel just reaches its yield strain as concrete is strained to 0.003. Every column has always have a loading situation where an ultimate load Pnb placed at eccentricity eb will produce a moment Mnb. If the eccentricity of the column is less than eb (e < eb) , the column fails in compression( compression controlled column); if e > eb the column fails in tension( tension controlled column)

Balanced loading condition d-d’ 2

Pnb

eb

Determination of Pnb From stress Diagram ΣFy =0 Pnb - C1 - C2+ T = 0

d b

Axis of bending d – d’

T = Asfy

s 

Es

0.85fc’

C1= 0.85fc’abb

εs’

EQ. 1

From strain Diagram

C2 = As’fy ab

cb

If As = As’ C2 = T

Pnb =C1 Pnb =0.85fc’abb

d’

d-ab/2 fy

Stress Diagram

Strain Diagram 0.003

600  1 d ab  600  f y

EQ. 2

To determine eb ΣM @ center of tension steel = 0 Pnb( eb + ½{d-d’}) –C1( d-ab/2) – C2(d –d’)= 0 Pnb( eb + ½{d-d’}) –0.85fc’abb( d-ab/2) – As’fy(d –d’)= 0 ' ' ab As f y (d  d ) (d  d ' ) eb  ( d  )   2 Pnb 2

Balanced Moment

EQ. 3

Mnb = Pnbeb

Determination of eb, Pnb and Mb of column with given section and reinforcement. 1. Solve for ab

ab 

600  1 d 600  f y

2. Solve for Pnb

Pnb =0.85fc’abb

3. Solve for eb

' ' ab As f y (d  d ) (d  d ' ) eb  ( d  )   2 Pnb 2

4. Solve for the balanced Moment

Mnb = Pnbeb

Problem: The column shown is reinforced with 6 of 25 mm diameter bars. If fc’ = 21 MPa, fy = 345 MPa determine P nb, eb and Mnb. 65 mm

500 mm

Axis of bending

65 mm 400 mm

3 (25) 2 As  As '   1472.6mm 2 d  500  65  435mm 4 ab 

600 1d 600(0.85)435   234.76mm 600  f y 600  345

Pnb =0.85fc’abb Pnb =0.85(21)234.76(400)= 1,676,186 N ' ' A f ( d  d ) (d  d ' ) ab s y eb  ( d  )   2 Pnb 2 234.76 1472.6(345)(435  65) (435  65) eb  (435  )  2 1676186 2

eb  244.76mm M nb  Pnb eb  (1676186)(244.76)  410.26 x10 6 N .mm M nb  410.26kN .m

ANALYSIS OF ECCENTRICALLY LOADED COLUMNS: ANALYTICAL METHOD

e e eb

tension controlled columns Compression controlled columns

Plastic centroid

Analysis of tension controlled columns

e Plastic centroid

Large moment, small axial load Failure initiated by tensile yielding of tension bars

eb  e fs  f y fs ' f y

 c  0.003

Tension steel yields at failure Compression steel yields at failure

d-d’ 2

Pn

e

1. From stress Diagram ΣFy =0 Pn - C1 - C2+ T = 0 d b

Axis of bending d – d’

T = Asfy

Stress Diagram

Pn =0.85fc’ab

d’

C2 = As’fy a 0.85fc’

s 

fy Es

C1= 0.85fc’ab

c

εs’

If As = As’ C2 = T

Strain Diagram 0.003

EQ. 1

2. ΣM @ center of tension steel = 0 Pn( e + ½{d-d’}) –C1( d-ab/2) – C2(d –d’)= 0 Pn( e + ½{d-d’}) –0.85fc’ab( d-a/2) – As’fy(d –d’)= 0 0.85fc’ab( e + ½{d-d’}) –0.85fc’ab( d-a/2) – As’fy(d –d’)= 0

Solve for a

3. Check for yielding of compression steel

c

a 1

s' cd  0.003 c

4. Solve for Pn

Pn =0.85fc’ab

5. Solve for Pu

Pu  Pn

6. Solve for Mu

M u  Pu e

'

' ( c  d ) f s'  600  fy c

Plate #12: Eccentrically loaded columns The column shown is reinforced with 6 of 32 mm diameter bars. If fc’ = 27 MPa, fy = 345 MPa , e = 370 mm, determine Pu, and Mu . 65 mm

600 mm

65 mm 450 mm

Axis of bending

Plate #12: Eccentrically loaded columns The column shown is reinforced with 6 of 28 mm diameter bars. If fc’ = 27 MPa, fy = 345 MPa , e = 150 mm, determine Pu, and Mu . 65 mm

500 mm

65 mm 400 mm

Axis of bending

Problem: The column shown is reinforced with 6 of 25 mm diameter bars. If fc’ = 21 MPa, fy = 345 MPa , e = 200 mm, determine Pu, and Mu . 65 mm

500 mm

Axis of bending

65 mm 400 mm

3 (25) 2 As  As '   1472.6mm 2 d  500  65  435mm 4

From the preceding problem

eb  244.76mm  260mm

Tension controls

Pn =0.85fc’ab Pn=0.85(21)400a=7140a

EQ.1

ΣM @ center of tension steel = 0 0.85fc’ab( e + ½{d-d’}) –0.85fc’ab( d-a/2) – As’fy(d –d’)= 0

1 a 7140a{260  (435  65)}  7140a (435  )  1472.6(345)(435  65)  0 2 2 2 a 445a  435a   26327.4  0 2

a 2  20a  52654.8  0  20  (20) 2  4(52654.8) a  219.7 mm 2 a 219.7 c   258.47 mm 1 0.85

compression steel (258.47  65) f  600  449.1MPa  f y yields at failure 258.47 ' s

Pn  7140a  7140(219.7)  1568658 N  1568.65kN Pu  (0.7)1568.65  1098kN M u  Pu e  1098 0.26   285.48kN .m yeilding of tension steel( not required)

s d c  0.003 c

(d  c) f s  600  fy c

(435  258.47) tension steel f s  600  409.8MPa  f y yields at failure 258.47

Analysis of compression controlled columns

e Plastic centroid

2. Large axial load and small moment. with entire cross section in compression. Failure under this condition occur by sudden crushing of concrete with all bars are in compression 3. Large axial load with moment bigger in 2. Bars in the far side in tension has not yielded. Failure occur by crushing of concrete

eb  e fs  f y fs ' f y

 c  0.003 Tension steel does not yield at failure Compression steel yields at failure

d-d’ 2

Pn

e

d b

Axis of bending

d – d’

T = Asfs

d s 

fy Es

a 2

Stress Diagram

d’

Pn  0.85 fc' ab  As' f y  As f s  0

C2 = As’fy a 0.85fc’

εs’

Pn  0.85 fc' ab  As' f y  As f s  EQ.1 Note: unknowns are Pn , a &

C1= 0.85fc’ab

c

1. From stress Diagram ΣFy =0 Pn - C1 - C2+ T = 0

Strain Diagram 0.003

fs

1 a Pn {e  (d  d ' )} C 1(d  )  C2 (d  d ' )  0 2 2 1 a Pn {e  (d  d ' )}  0.85 fc' ab(d  )  As' f y (d  d ' )  0 2 2 As' f y (d  d ' ) 0.85 fc' ab(d  0.5a ) Pn    EQ.2 {e  0.5(d  d ' )} {e  0.5(d  d ' )}

 M centerofT .S  0

EQ.1  EQ.2 Pn  0.85 fc' ab  As' f y  As f s  EQ.1 As' f y (d  d ' )

0.85 fc' ab(d  0.5a) 0.85 fc' ab  A f y  As f s    EQ.3 {e  0.5(d  d ' )} {e  0.5(d  d ' )} ' s

a 600(d  ) (d  c) 1 f s  600  a c 1

From the strain diagram

( 1d  a ) f s  600  EQ.4 a

0.85 fc' ab  As' f y  As 600

As' f y (d  d ' )

( 1d  a ) 0.85 fc ' ab(d  0.5a)   a {e  0.5(d  d ' )} {e  0.5(d  d ' )}

Solve for a Check for yielding of compression steel

a c 1 Solve for fs Solve for Pn Solve for Pu Solve for Mu

s' cd  0.003 c

'

(c  d ' ) f  600  fy c ' s

( 1d  a ) f s  600 a Pn  0.85 fc' ab  As' f y  As f s  EQ.1 Pu  Pn

M u  Pu e

Problem: The column shown is reinforced with 6 of 25 mm diameter bars. If fc’ = 21 MPa, fy = 345 MPa , e = 200 mm, determine Pu, and Mu . 65 mm

500 mm

Axis of bending

65 mm 400 mm

3 (25) 2 As  As '   1472.6mm 2 4

d  500  65  435mm

From the preceding problem

eb  244.76mm  200mm

Compressions controls

0.85(21)400a  1472.6(345)  1472.6(600)

{0.85 435  a} 0.85(21) 400a (435  0.5a ) 1472.6(345)(435  65)   a {200  0.5( 435  65)} {200  0.5(435  65)}

{369.75  a} a(435  0.5a) a  71.16  123.75   68.38 a 385 385a 2  27396.6a  17616276.5  47643.75a  435a 2  0.5a 3  26326.3a

a 3  100a 2  97428.1a  35232553  0 a  256.28mm

Can be solved by Newton’s Method of approximation or by trial and error

(256.28) 3  100(256.28) 2  97428.1(256.28)  35232553  0 703.1

0k

( 1d  a ) { 0.85 435  256.28} f s  600  600  214.73MPa a 256.28

c

256.28  301.5mm 0.85

(301.5  65 ) f  600  470.65MPa  f y 301.5 ' s

Compression steel yields

Pn  0.85 fc' ab  As' f y  As f s  EQ.1 Pn  0.85(21)256.28 400   1472.6(345)  1472.6(214.73) Pn  2021682 N  2021.68kN Pu  (0.7)2021.68  1415.2kN M u  1415.2(0.2)  283.04kN .m

COLUMN INTERACTION DIAGRAM It was observed that the use of analytical method is very complicated and requires accuracy of manipulation of algebraic equations. An option is to analyze the column by using the column interaction diagram. Interaction diagram is the graph of the axial load capacity and the moment capacity of the column of the M-P axes. The diagram is very useful for studying the strength of the column with varying proportions of the loads and moments.

Pn Pn e

A

Pure axial load B

Compression Failure Balanced Loading

Compression zone C

Pn

Mn

Tension failure Tension zone Pure bending Mn D

The diagram is made by plotting the axial load capacity when the moment is zero( point A), then the balanced loading( point C) and the moment capacity when the axial load is zero( point D). In between points A and D , the column fail due to axial and bending. Point C the balanced point. In reference to Point B, the vertical and horizontal dotted line represents the particular load combination of axial load and moment.

How to use the column interaction Diagram Process Given Required calculate

Design Loads, Pu and e section dimension Amount of reinforcement

Pn Ag

and

M n Ag h

Intersection of

Locate

Pn Ag

and

M n Ag h

and determine pg obtain

Ast

Analysis Section properties eccentricity axial load and moment capacity

Ast pg  Ag



e/h

Intersection of pg & e/h project horizontally to The left to get Pn

Ag project vertically M n down to get Ag h Pn or Pu

Mn or Mu

Problem Calculate the ultimate axial load capacity of the column section shown if the eccentricity is 200 mm. fc’ = 20.7 MPa, fy = 413.7 MPa. Solve using the column interaction diagram. 400 mm 75 mm 4 of 25 mm Axis of bending

450 mm

4 of 25 mm

75 mm

Solution 1. Solve for



,pg, and e/h

 (25) 2 8 Ast 4 pg    0.016 Ag (400)(600)

h  450  600  450   0.75

e 200   0.33 h 600 2.

Locate the intersection of pg and e/h. Project horizontally to the left Pn to obtain Pn

Ag

Ag

 8.52MPa

Pu  Pn

Pu =8.52Ag = 8.52(400)600 = 2044.8 kN

Pn (MPa) Ag 50

Interaction Diagram

0.1 0

45

e/ h= 0. 20

35

30 .33 . =0 h =0 h e/ e/ .40 0 = e/h . 50 0 = e/h

30

20 15

0.016 8.2 MPa

M n (MPa) Ag h

10

  0.75

e/h =

40

25

fc’= 3 ksi ( 20.7 MPa) fy = 60 ksi ( 413.7 MPa)

pg=0.08 0.07 0.06 0.05 04 0.03 0.02 0.01

h

h Pn e

0 e/h=1.

5 0

1

2

3

4

5

6

7 8

9 10 11 12 13 14 15

Problem Calculate the nominal axial load capacity of the colum shown applied at an eccentricity of 400 mm. fc’=20.7 MPa,fy=413.7 MPa. Reinforcement consist of 16 of 28 mm bars. 500 mm 70 mm Axis of bending

360 mm

70 mm

 (28) 2 10 Ast 4 pg    0.0394 Ag (500)(500)

h  360  500  360   0.72

Since there is no available interaction diagram for   0.72 Use the interaction diagrams for   0.60 and   0.75 With fy = 413.7 MPa and fc’=20.7 MPa and interpolate the values obtained e/h = 400/500 = 0.80 From column interaction diagram for fc’=20.7 MPa

  0.60

Pn  4.6 MPa Ag

f y = 413.7 MPa and

From column interaction diagram for  f y = 413.7 MPa and fc’=20.7 MPa

 0.75

Pn  5.8MPa Ag Interpolating values

x  5.8 0.72  .075  4.6  5.8 0.60  0.75 X = 5.56 MPa

Pn  5.56 MPa Ag

 0.60 0.72 0.75

0.7 Pn  5.56 (500)(500)

Pn = 1,985,714 N = 1,985.7 kN

Pn Ag 4.6 x 5.8

50

Interaction Diagram

0.1 0

45

0.039

20

0.08 0.07 0.06 0.05 0.04

15 0.03 10

4.6 MPa

5 0

0.02 0.01

e/ h= 0. 20

35

25

  0.60

e/h =

40

30

fc’= 3 ksi ( 20.7 MPa) fy = 60 ksi ( 413.7 MPa)

0 .3 0 h= / e 40 . 0 = e/h . 50 0 = e/h

h

h Pn e

0.8 e/h =

0 e/h=1.

50

Interaction Diagram

0.1 0

45

30 25

0.039

15 10

5.8 MPa

5 0

0.0 4

e/ h= 0. 20

35 0.0 8

  0.75

e/h =

40

20

fc’= 3 ksi ( 20.7 MPa) fy = 60 ksi ( 413.7 MPa)

30 . =0 h e/ 40 . 0 = e/h 50 . 0 = e/h

h

h Pn e

0. e/h =

0 e/h=1.

8

Plate # 13: Analysis of eccentric columns using interaction diagram Calculate the ultimate axial load capacity of the column section shown if the eccentricity is 300 mm. fc’ = 27.57 MPa(4 ksi), fy= 413.MPa(60ksi). Solve using the column interaction diagram.

350 mm 75 mm 1ksi =6.8928 MPa

4 of 25 mm Axis of bending

400 mm

4 of 25 mm

75 mm

Calculate the nominal axial load capacity of the colum shown applied at an eccentricity of 400 mm. fc’=27.57 MPa,fy=413.7 MPa. Reinforcement consist of 16 of 22 mm bars.

450 mm 70 mm Axis of bending

330 mm

70 mm

Problem CE Board May 2000 For the spiral column shown, determine the maximum nominal axial load capacity if the eccentricity is 150 mm. fc’= 27.6 MPa, fy = 413.7 MPa. Use the column interaction diagram. 1 ksi = 6.8928 MPa

8 of 22 mm

375 mm 500 mm

 (22) 2 8 Ast 4 pg    0.0155 2  (500) Ag 4

h  375  500  375   0.75

e 150   0.3 h 500

Pn  1.3ksi  1.3 x6.8928  8.961MPa Ag

6

ΦPn/Ag(ksi)

0.08

  0.75

0. 20

5

INTERACTION DIAGRAM fc’ = 4ksi(27.6 MPa) fy = 60 ksi ( 413.7 MPa)

0.10

e/h=0. 05

7

0.07

h e/

0.06

4

0.05 0.04

30 . =0

Pn

40 0.

0.03

3

h h

0.02

e

0 0.5

0.0155 0.01

2 1.0

1.3 ksi 1 0

0.2

0.4

0.6

0.8

1.0

1.2

ΦMn/Agh(ksi)

1.4

1.6

1.8

2.0

Determination of steel area A rectangular column is as shown in the figure. It is acted upon by the following loads: axial load due to dead load = 450 kN axial load due to live load = 500 kN Moment due to dead load = 134 kN.m Moment due to live load = 150 kN.m The column is to be analyzed and designed by ultimate strength design method. Use fc’ = 20.7MPa, fy = 413.7 MPa. Determine a) Eccentricity of the column b) Area of steel bars required

400 mm 75 mm Axis of bending

450 mm

75 mm

Solution: Pu = 1.4PD + 1.7PL Pu =1.4 (450) + 1.7(500) = 1480 kN Pn = Pu/Φ Pn = 1480/0.7 Pn = 2114.28 kN

h  450  600  450   0.75 e 300   0.50 h 600

Pn 0.7(2114 .28)(1000)   6.17 MPa Ag (400)(600)

6  M 0 . 7 ( 632 . 28 )( 10 ) n Mu =1.4MD + 1.7ML   3.07 MPa (600)(400)(600) Mu =1.4(134)+1.7(150) = 442.6 kN.m Ag h Mn =Mu/Φ = 442.6/0.7 = 632.28 kN.m Pn M n Intersection of and Eccentricity Ag Ag h e= M /P = 632.28/ 2114.28

n

n

e = 0.3 m = 300 mm

pg =0.017 Ast =pgAg= 0.017(400)(600) Ast = 4080 mm2

50

Interaction Diagram

0.1 0

45

e/ h= 0. 20

35 30 25

15

Pn e

0 e/h=1.

01 0.

Pn  6.17 MPa Ag

10

h

30 . =0 h e/ 40 . 0 = e/h . 50 0 = e/h

0. 0. 08 0. 07 0. 06 0.0 05 0.0 4 0. 3 02

20

h

  0.60

e/h =

40

pg=0.017

fc’= 3 ksi ( 20.7 MPa) fy = 60 ksi ( 413.7 MPa)

5 0 1

2

3

4

5

6

M n  3.07 MPa Ag h

7

8

9

10

11 12

13

14

15

Design of Eccentric columns( section & reinforcement) 

The size of column can be obtained by assuming an average compressive stress of concrete from 0.5fc’ to 0.6 fc’ under the load PU. Design of reinforcement follows the procedure of the preceding topic. If value of pg is not reasonable, section maybe revised.

Design a square tied column to carry a factored axial load of 2,000 kN And a factored moment of 190 kN.m. Use 25 mm bars to be placed uniformly along the faces of the column. Assume fc’=20.7MPa and fy =413.7 MPa.

Solution Ag = Pu/0.5fc’ Ag = (2000)(1000) /0.5(20.7) Ag = 193,237 mm2 Using a square column h = 439 mm say 450 mm Proposed Layout 450 mm 62.5 mm Axis of bending

325 mm

62.5 mm

e = Mu/Pu e =190/2000 = 0.095 m = 95 mm e/h = 95/450 = 0.211

h  325  450  325   0.722

Interpolate between   0.60 and   0.75 Pn =Pu/Ф= 2000/0.7=2,857 kN ФPn/Ag = 0.7(2857)1000/(450)(450) ФPn/Ag =9.876 MPa/6.8928 = 1.43 ksi

  0.60

Intersection of e/h and ФPn/Ag

  0.75 Intersection of e/h and ФPn/Ag

 0.60 0.722 0.75

pg =0.018 pg =0.016

pg 0.018 x 0.016

x  0.016 0.722  0.75  0.018  0.016 0.6  0.75

x = pg = 0.01637

As =PgAg As = 0.01637(450)(450) As = 3315 mm2

7 0.10

ΦPn/Ag(ksi)

5

3

1.43 ksi

0.0 0. 8 0 0.0 7 0.0 6 0.0 5 0. 4 03

h e/

h

30 . =0

h

Pn

e 40 . 0 0 0.5

02 0.

2

  0.75

0. 20 e/ h= 0. 21 1

6

4

INTERACTION DIAGRAM fc’ = 3 ksi(20.7 MPa) fy = 60 ksi ( 413.7 MPa)

1.0

0.0 1

1 0

0.2

0.4

0.6

pg=0.016

0.8

1.0

1.2

1.4

ΦMn/Agh(ksi)

1.6

1.8

2.0

7 0.10

6

  0.60

0. 2 e/ h= 0 0. 21 1

5 4

e/

0 0.

3

0

h

Pn

e

0 0.5

2 0. 0 0 1 0.

2

.3 0 h=

h

40 0.

8

ΦPn/Ag(ksi)

INTERACTION DIAGRAM fc’ = 3ksi(20.7 MPa) fy = 60 ksi ( 413.7 MPa)

1.0

1.43 ksi 1 0

0.2

0.4

0.6

pg=0.018

0.8

1.0

1.2

ΦMn/Agh(ksi)

1.4

1.6

1.8

2.0

Plate # 14 : Design of eccentric columns by using interaction diagram For the spiral column shown, determine the maximum nominal axial load capacity if the eccentricity is 200 mm. fc’= 27.6 MPa, fy = 413.7 MPa. Use the column interaction diagram. 1 ksi = 6.8928 MPa

8 of 22 mm

400 mm 530 mm

Design a square tied column to carry a factored axial load of 2,500 kN And a factored moment of 300 kN.m. Use 28mm bars to be placed uniformly along the faces of the column. Assume fc’=20.7MPa and fy =413.7 MPa.

A rectangular column is as shown in the figure. It is acted upon by the following loads: axial load due to dead load = 500 kN axial load due to live load = 600 kN Moment due to dead load = 140 kN.m Moment due to live load = 180 kN.m The column is to be analyzed and designed by ultimate strength design method. Use fc’ = 20.7MPa, fy = 413.7 MPa. Determine a) Eccentricity of the column b) Area of steel bars required

450 mm 75 mm Axis of bending

500 mm

75 mm

Axial Load and Biaxial Bending In some cases, columns are subject to bending on both axes. This is the usual case of corner columns where beams or girder frames into the column from both directions . ex ey My Mx

Bresler Equation Biaxially loaded columns can be analyzed by Bresler’s equation.

1 1 1 1    Pn Pnx Pny Pno Where: Pnx = axial load capacity of the column for eccentricity ex with ey =0 Pny = axial load capacity of the column for eccentricity ey with ex =0 Pno = axial load capacity of the column when ex = 0 and ey = 0 Pno = 0.85fc’Ag + Asfy

Problem 

Calculate the permissible axial load capacity Pu of the short tied column shown shown if ex= 200 mm and ey = 100 mm. Use fc’ =20.7 MPa, fy = 413.7 MPa

75 mm

8 of 28 mm

225 mm 375 mm 75 mm

75 mm

450 mm 600 mm

75 mm

Solution

 (28) 2 8 Ast 4 pg    0.022 Ag (375)(600) For eccentricity ex ( bending about the y axis)

h  450  600  450   0.75

Pn  1.28ksi Ag

e 200   0.333 h 600

Pn  8.823MPa Ag

Pnx 

8.823(600)(375) 0.7(1000)

Pnx =2836 kN

For eccentricity ey ( bending about the x axis)

h  225  375  225   0.60 e 100   0.267 h 375

Pn  1.35ksi Ag Pn  9.305MPa Ag

9.305(600)(375) Pny  0.7(1000) Pnx =2991 kN

Pno = 0.85fc’Ag + Asfy = 5,997kN

1 1 1 1    Pn Pnx Pny Pno 1 1 1 1    Pn 2,836 2,991 5,997

Pn = 1,922 kN

7 0.10

6

ΦPn/Ag(ksi)

  0.75

0. 20

5

h e/

h

30 . =0

h

33 3 . =0 h / e

4

02 0.

0.0 0. 8 0 0.0 7 p3 0.0 6 g= 0.0 5 0.0 22 0.0 4 3 2 0.0 1

1.28ksi

INTERACTION DIAGRAM fc’ = 3 ksi(20.7 MPa) fy = 60 ksi ( 413.7 MPa)

Pn

e

40 . 0 0 0.5 1.0

1 0 ΦMn/Agh(ksi)

7 0.10

6

  0.60

0. 20

5

0 0.

4

h e/

8

e/

3

=

h

67 2 0.

h

30 . 0 h=

=0 pg

ΦPn/Ag(ksi)

INTERACTION DIAGRAM fc’ = 3ksi(20.7 MPa) fy = 60 ksi ( 413.7 MPa)

Pn

e

0 0.4 0 0.5

0

22 .0 .02

0. 03

2

1.35 ksi

1.0

1 0

0.2

0.4

0.6

0.8

1.0

1.2

ΦMn/Agh(ksi)

1.4

1.6

1.8

2.0

Using the data of the preceding problem, determine Mux and Muy . Solution

Pu  Pn Pu  0.7(1922)  1345.4kN M ux  Pu e y  1345.4(0.10)  134.54kN .m M uy  Pu ex  1345.4(0.20)  269.08kN .m

Plate # 14 : Columns subject to Biaxial Bending Calculate the permissible axial load capacity Pu of the short tied column shown shown if ex= 250 mm and ey = 150 mm. Solve by analytical method. Use fc’ =20.7 MPa, fy = 413.7 MPa

75 mm

10 of 28 mm

250 mm 400 mm 75 mm

75 mm

500 mm 650 mm

75 mm

Long Columns 

CONSIDERATION OF SLENDERNESS EFFECTS For compression braced against sideways,effects of slenderness maybe neglected when:

kLu M 1b  34  12 M 2b r For compression not braced against sideways, effects of slenderness maybe neglected when

kLu  22 r

Where: M2b = value of the larger factored end moment on a compression member due to loads that result in no appreciable sidesway, calculated by conventional elastic frame analysis M1b = value of the smaller factored end moment on a compression member due to loads that result in no appreciable sidesway, calculated by conventional elastic frame analysis positive if member is bent in single curvature, negative if member is bent in double curvature M1b

Lu

M2b

Effective Length Factor Effective length factor is equal to 1 when the member is braced against sideway ; for not braced against sideway effects of cracking and reinforcement on relative stiffness must be considered and should be greater than 1. Radius of Gyration r = 0.25 D for circular columns ,D the column diameter r = 0.30 h for rectangular columns ,h the column dimension in the direction stability is being considered Uned Length of Compression Uned length Lu a of compression member is taken as the clear distance between floor slabs, beams or other capable of providing lateral for that compression member. Where column capitals or haunches are present it shall be measured to the lower extremity of capital or haunch in the place considered.

Braced and Unbraced Frame A frame maybe considered braced if the bracing elements such as shear walls, shear trusses, or other means resisting lateral movement of a storey, have a total stiffness at least, six times the sum of the stiffness of all the columns resisting lateral movement in that storey.

PROBLEM A square column having an uned length of 5 m is used in a braced frame against sidesway. The column is bent in single curvature and subject to factored end moments of 58 kN/m at the top and 42 kN.m at the bottom. Determine its minimum dimension so that slenderness effects may not be considered in the analysis.

M 2b  58kN .m Lu  5m  5000mm

M 1b  42kN .m k  1 .0

r  0.3h

M 1b kLu  34  12 M 2b r

1(5000) (42)  34  12 0.3h 58 h  660mm

+ since the column is bent in single curvature

column is braced against sidesway

MOMENT MAGNIFIER METHOD According to NS 1992 edition, compression shall be designed using the factored axial load Pu from a conventional frame analysis and a magnified factored moment Mc defined by :

M c   s M 2 s   b M 2b

EQ. 1

Where: M2b = value of the larger factored end moment on a compression member due to loads that result in no appreciable sidesway, calculated by conventional elastic frame analysis. For use in EQ.1 M2b ≥ Pu(15 + 0.03 h) 15 + 0.03h in mm M1b = value of the smaller factored end moment on a compression member due to loads that result in no appreciable sidesway, calculated by conventional elastic frame analysis positive if member is bent in single curvature, negative if member is bent in double curvature

M2s = value of the larger factored end moment on a compression member due to loads that result in appreciable sidesway ( wind, earthquake and other lateral loads) , calculated by conventional elastic frame analysis . For use in EQ.1 M2b ≥ Pu(15 + 0.03 h) 15 + 0.03h in mm

b =

moment magnification factor that takes into the effect of member curvature in a frame braced against sideway.

Cm b   1 .0 Pu 1 Pc

M 1b Cm  0.6  0.4 M 2b

but not less than 0.4, For all other cases Cm = 1.

 s = moment magnification factor that takes into

the lateral drift of a column caused by lateral or gravity loading the frame not braced against sidesway

1 s   1.0  Pu 1   Pc

 s  1.0  Pu  Pc

 2 EI Pc  ( kLu ) 2

for columns braced against sidesway summation for all columns in a storey.

EQ 2

For frames not braced against sidesway, both  band  s should be computed. For frames braced against sidesway,  s should be taken as 1.0. In calculating Pc ,EI should be determined either by

Ec I g

Ec I g

 Es I se

EI  5 1 d

or

EI  2.5 1 d

Where βd = the ratio of factored dead load moment to maximum total factored moment ( always positive), where the load is due to gravity only in the calculation of Pc in computing  bor the ratio of the maximum sustained lateral load to the maximum total factored lateral load in that story in the calculation of Pc in computing  s

Ig = moment of inertia of gross section about centroidal axis neglecting reinforcement Ise = moment of inertia of reinforcement about centroidal axis of member cross section Ec =modulus of elasticity of concrete E c  4700 fc ' Es =modulus of elasticity of reinforcement Moment Magnifier for biaxial bending For compression subject to moment on both axes, moment about each principal axis shall be magnified by  computed from the corresponding restraint about that axis.

Alignment Charts

Effective Length Factor k are obtained using the Jackson and Moreland alignment chart. To use this chart a parameter ψ A for end A of column AB and ψB for end B are computed. A straightedge is placed between A and B. The point where the straightedge cross the middle monograph is k. The parameter ψ at one end of the column equals the sum of the stiffnesses of the columns (including the column under consideration) meeting at the t divided by the sum of all the stiffnesses of the beams meeting at that t. ∑EI/L of Columns Ψ= ∑EI/L of Beams

ALIGNMENT CHARTS ψA ∞ 50

K

1.0

10

5 3.0 2.0 1.0 0.8 0.6 0.5

10

0.9 0.8

0.7

0.4 0.3 0.2

5 3.0 2.0 1.0 0.8 0.6 0.5 0.4

0.6

0.1 0

ψB 5∞ 0

0.3 0.2 0.1

0.5

BRACED FRAME

0

ψA ∞ 100.0

K

50.0 30.0 20.0 10.0 8.0

∞ 20.0 10.0 5.0 4.0 3.0

ψB ∞ 100.0 50.0 30.0 20.0 10.0

8.0

6.0 5.0 4.0

2.0

3.0

6.0 5.0 4.0 3.0

2.0

1.5

1.0

2.0 1.0

0

1.0

0

UNBRACED FRAME

Note: Ψ ═ ∞ for pinned ends and 1.0 for fixed end

PROBLEMS 1. A rectangular concrete column has a dimension of 300 mm by 380 mm. The column length is 4.5 m and is to be used in a frame braced against sidesway ( k =1.0, EI = 34 MN.m2). The column is bent in single curvature by ultimate factored moment of 110 kN.m at the top and 120 kN.m at the bottom. The column carries factored axial dead load of 150 kN and factored axial live load of 350 kN. Considering the longer dimension of the column, is slenderness effects be considered in the analysis ? If so, what is the magnified moment capacity of this column ? 2. Solve the previous problem considering the effect of wind loads producing an unfactored axial load of 130 kN an unfactored moment of 100 kN.m

Solution to #1

M 2b  120kN .m M 1b  110 kN .m M 2s  0 r  0.3h  0.3(380)  114 mm Lu  4.5m  4500mm kLu 1(4500)   39.47 r 114 M 1B 110 34  12  34  12  23 M 2B 120 kLu M 1B  34  12 r M 2B

Column is Long

+ since the column is bent in single curvature

M 1b 110 Cm  0.6  0.4  0.6  0.4  0.9668 M 2b 120 Pu  150  350  500kN

 2 EI  2 34(10) 6 Pc    16,571,188 N  16,571.18kN 2 2 (kLu ) {(1)(4.5)} Cm 0.9668 b    1.0103 Pu 500 1 1 0.7(16571.18) Pc M c   b M 2b   c M 2 s  1.0103(120)   s (0)  121.24kN .m

Solution to #2

M 2 s  0.75[1.4M D  1.7 M L  1.7 M W ] M 2 s  0.75[1.4(0)  1.7(0)  1.7(100)]  127.5kN .m [1.4 PD  1.7 PL ]  150  350  500kN

Specified in the previous problem

Pu  0.75[1.4 PD  1.7 PL  1.7 PW ]  0.75[500  1.7(130)]  540.7 kN Cm 0.9668 b    1.014 Pu 540.75 1 1 0.7(16571.18) Pc

 s  1.0  Frame is braced against sidesway M c   b M 2b   s M 2 s  1.014(120)  1.0(127.5)  249.18kN .m

Plate No.15

Long Columns ( Part I)

1.A square column having an uned length of 4.5 m is used in a frame braced against sidesway. The column is bent in single curvature and subject to factored end moments of 52 kN/m at the top and 36 kN.m at the bottom. Determine its minimum dimension so that slenderness effects may not be considered in the analysis. 2. A square concrete column has a dimension of 450 mm by 450 mm. The column length is 5 m and is to be used in a frame braced against sidesway ( k =1.0, EI = 44 MN.m2). The column is bent in single curvature by ultimate factored moment of 130 kN.m at the top and 150 kN.m at the bottom. The column carries factored axial dead load of 180 kN and factored axial live load of 390 kN. Is slenderness effects be considered in the analysis ? If so, what is the magnified moment capacity of this column ? 3. Solve the previous problem considering the effect of wind loads producing an unfactored axial load of 150 kN an unfactored moment of 120 kN.m

For the column shown in figure, PD = 266.88 kN, PL = 444.8 kN, MD = 162.72kN.m, and ML = 189.84 kN.m. The column belongs to a braced frame and is bent in single curvature about the x axis by equal moments at each end of the member . The uned length is 6 m. fc’ = 20.7 MPa, fy = 413.7 MPa . Determine the following : a) Critical load Pc b) Magnified moment 375 mm c) the steel ratio required to 8 of 32 mm

375 mm 500 mm

sustain the magnified moment

Solution:

8 (32) 2 As   6433.98mm 2 4 Pu=1.4PD + 1.7PL = 1129.79 kN M1b= M2b =1.4MD + 1.7ML = 550.54 kN.m

Ec  4700 fc '  4700 20.7  21383.7 MPa

375(500)3 Ig   3906.25 x106 mm 4 12 I se  As z 2  6433.98(187.5) 2  226.194 x106 mm 4

Es  200000 MPa

1.4M D 1.4(162.72) d    0.41 Mu 550.54 Ec I g EI 

 E s I se

5 1 d

21383.71(3906.35)10 6  200000(226.194)10 6 5 EI  1  0.41

EI  4.4 x1013 N .mm 2

 2 EI Pc  (kLu ) 2  2 4.4 x1013 Pc   12062.85kN 2 (6000 ) M 1b 550.54 Cm  0.6  0.4  0.6  0.4  1.0 M 2b 550.54 Cm 1 b    1.13 Pu 1129 .79 1 1 Pc 0.7(12062.85)

M c   b M 2b  1.13(550.54)  622.11kN .m

Pn Pu 1129 .79    6.03MPa Ag Ag 375(500) M n M c (622.11)(10) 6    6.64 MPa Ag h Ag h (500)(375)(500) Using the column interaction diagram Project 6.03 MPa(0.875ksi) from

P

n

axis horizontally to the right

Ag M n Project 6.64 MPa(0.963ksi) from axis vertically up Ag h Locate intersection of projected lines to get pg = 0.05

0.10

INTERACTION DIAGRAM fc’ = 3 ksi(20.7 MPa) fy = 60 ksi ( 413.7 MPa)

6

0. 20

5

  0.75

e/

4

h

Pn

e

40 . 0 0 0.5

0 0.

2

30 . 0 h=

3 3 3 0. = e/h

0.0 0. 8 0 0.0 7 0.0 6 0.0 5 0. 4 03

3

h

1.0

2

0.0 1

1 0.875 ksi 0 0.2 2.0

0.4

0.6

0.8

1.0

1.2 1.4 1.6 ΦMn/Agh(ksi)

0.963 ksi

1.8

Problem: Design the steel reinforcement for a 7.5 m 600 mm by 600 mm square tied Column bent in single curvature in a braced frame with ψ equal 1 at one end and 2.0 at the other end. The column is used to carry the following loads axial dead load = 800 kN axial live load = 880 kN dead load moment = 135 kN. m live load moment = 175 kN. m Use fc’ = 20.7 MPa, fy = 413.7 MPa, and place the bars at two end faces With d’ = 75 mm.

600 mm

450 mm

Axis of bending

75mm

75 mm

Solution Pu =1.4 PD + 1.7PL Pu= 1.4 (800) + 1.7(880) = 2,616 kN MU = M2b = M1b = 1.4 MD + 1.7 ML MU = M2b = M1b = 1.4(135) + 1.7 (175) MU = M2b = M1b = 486.5 kN.m M1b / M2b = 1.0

r = 03 h = 0.3(600) = 180

from Alignment chart k = 0.82 KLu / r = 0.82( 7500)/180 = 34.17 34 – 12M1b / M2b =34 -12(1) = 24 Check:

ψA ∞ 50

M2b > 2616 (15 + 0.03(600))/1000 = 86.33 kN.m

1.0

10

5 3.0 2.0 1.0 0.8 0.6 0.5 0.3

0.9

0.82

0.8

0.7

5 3.0 2.0 1.0 0.8 0.6 0.5 0.4

0.6

0.1 0

ψB 5∞ 0 10

0.4 0.2

M2b > Pu (15 + 0.03h)

K

0.3 0.2 0.1

0.5

BRACED FRAME

0

M 1b Cm  0.6  0.4  0.6  0.4(1)  1.0 M 2b E c  4700 fc'  4700 20.7  21383.7 MPa

600( 600 )3 Ig   10.8 x10 9 mm 4 12 1.4( 135 ) d   0.3885 486.5 Ec I g

21,373.7( 10.8 )( 10 ) 9 2.5 EI  2.5   66.532 x10 12 N .mm 2 1  d 1  0.3885

 EI  ( 66.532 )( 10 ) Pc    17 ,361kN 2 2 ( kLu ) 0.82( 7500 ) ( 1000 ) 2

2

9

Cm 1 b    1.274 M 2s  0 Pu 2 ,616 1 1 Pc 0.7( 17 ,361 ) M c   s M 2 s   b M 2 b  0  ( 1.274 )( 486.5 )  619.8 kN .m

M u 619.8 e   0.237 m  237 mm Pu 2 ,616

h  450  600  450   0.75

e 237   0.395 h 600

Pn PU 2 ,616( 1000 )    7.27 MPa  1.0547 ksi Ag Ag 600( 600 ) Pg=0.016 As =0.016(600)600 = 5,760 mm2 Using 25 mm bars

5760 No   12 pcs 2  ( 25 ) 4

0.10

INTERACTION DIAGRAM fc’ = 3 ksi(20.7 MPa) fy = 60 ksi ( 413.7 MPa)

6

0. 20

5

  0.75

e/

4

0.0 0. 8 0 0.0 7 0.0 6 0.0 5 0. 4 03

3

0 0.

0.016 2

40 . 0 0 0.5

h

30 . 0 h=

h

Pn

5 9 3 0. = e/h

e

1.0

2

0.0 1

1.0547 ksi 1 0

0.2 2.0

0.4

0.6

0.8 1.0 1.2 ΦMn/Agh(ksi)

1.4

1.6

1.8

7 0.10

6

  0.75

0. 20

5 ΦPn/Ag(ksi)

INTERACTION DIAGRAM fc’ = 4ksi(27.6 MPa) fy = 60 ksi ( 413.7 MPa)

e/

4

h

30 . 0 h=

h

40 0.

3

Pn

e

0 0.5

2 1.0

1 0

0.2

0.4

0.6

0.8 1.0 1.2 ΦMn/Agh(ksi)

1.4

1.6

1.8

2.0

Plate # 17: Design of Long columns Design the steel reinforcement for a 6.0 m 500 mm by 500 mm square tied Column bent in single curvature in a braced frame with ψ equal 2.0 at one end and 3.0 at the other end. The column is used to carry the following loads axial dead load = 600 kN axial live load = 650 kN dead load moment = 105 kN. m live load moment = 125 kN. m Use fc’ = 20.7 MPa, fy = 413.7 MPa, and place the bars at two end faces With d’ = 75 mm.

500 mm

350 mm

Axis of bending

75mm

75 mm

For the column shown in figure, PD = 200 kN, PL = 400 kN, MD = 122kN.m, and ML = 148 kN.m. The column belongs to a braced frame and is bent in single curvature about the x axis by equal moments at each end of the member . The uned length is 6 m. fc’ = 20.7 MPa, fy = 413.7 MPa . Determine the following : a) Critical load Pc b) Magnified moment 350 mm c) the steel ratio required to 8 of 28 mm

300 mm 450 mm

sustain the magnified moment

Two Way Slabs When a rectangular reinforced concrete slab is ed on four sides,reinforcement placed perpendicular to the side may be assumed to be effective in the two directions. These slabs are known as two way slabs. Bending occurs on both directions. However if the slab is ed on all sides but the ratio of the long side to short is two or more times, the slab act as a one way slab with bending occuring in the short direction. Two methods of deg will be discussed. The direct design method and the ACI moment coeffecient method. Other methods available are the strip method and the equivalent frame method.

h =slab thickness

Beam or wall



Beam

Beam

Beam

Beam Drop Column capital

Two way edge ed slab

Two way column ed slab ( Flat Slab)

Column and Middle Strip When the design moments are computed, the moments are distributed to the column and middle strip. Column strip is a design strip with a width on each side of a column centerline equal to 0.25L1 or 0.25L2 whichever is less. Column strip includes beam if any. The middle strip is a design strip bounded by two column strips. L2

Column Strip

Middle Strip

Smaller of 0.25L1 or 0.25L2

CL of column

Column Strip

L1

Minimum thickness Minimum thickness of slabs without interior beams spanning between the s shall be in accordance with Table 1 and shall not be less than the following values: a) slabs without drop s 125 mm b) slabs with drop s 100 mm

Table 1

Yield Stress Fy,MPa(1)

275 415

With drop s(2)

Without drop s(2) Exterior s Without edge beam

With edge beam(3)

Ln /33 Ln /30

Ln /36 Ln /33

Interior s

Ln /36 Ln /33

Exterior s Without edge beam

With edge beam(3)

Ln /36 Ln /33

Ln /40 Ln /36

Interior s

Ln /40 Ln /36

(1) For values of reinforcement yield stress between 275 MPa and 415 MPa minimum thickness shall be obtained by linear interpolation (2) See definition of drop in Sections 5.13.4.7.1, 5.13.4.7.2,NS 5th Edition (3) Slabs with beam between columns along exterior edges. The value of α for the edge beam shall not be less than 0.8.

DROP Drop s and column capitals are placed around the vicinity of the column to reduce the effect of negative moment and shear Not less than 1/6L

h Not less than 1/4h

Column capital

L Sections 5.13.4.7.1 Drop shall extend in each direction from centerline of a distance not less than one sixth the span length measured center to center of s in that direction. 5.13.4.7.2 Projection of drop below the slab shall be at least one fourth of the slab thickness beyond the drop.

The minimum thickness of slabs with or without beams spanning between the s on all sides and having a ratio of long span to short span not exceeding 2 shall be h



1 36000  5000    m  0.12( 1   

but not less than

h

EQ 10.1

Ln ( 800  0.73 f y )

Ln ( 800  0.73 f y )

 ) 

EQ 10.2

36000  9000 

and need not be more than

h

Ln ( 800  0.73 f y ) 36000

EQ 10.3

The values of h obtained from EQS.10.1,10.2 and 10.3 maybe shall be modified as required by Sections 5.9.5.3.4 and 5.9.5.3.5 but in no case shall the thickness be less than

 m  2.0  m  2.0

------------------------- 125 mm ------------------------- 90 mm

Section 5.9.5.3.4 For slabs without beams, but with drop s extending In each direction from centerline of a distance not less than one sixth the span length in that direction measured center to center of s and projection of slab at least one fourth of slab thickness beyond the thickness required by EQS.10.1,10.2 and 10.3 shall be reduced by 10%. Section 5.9.5.3.5 At discontinuous edge, an edge beam shall be provided With a stiffness ratio σ not less than 0.8; or the minimum thickness required by EQS.10.1,10.2 and 10.3 shall be increased by 10% in the with a discontinuous edge.

Where Ln = length of clear span in the long direction of a two- way construction measured face to face of s in slabs without beams and face to face of beams or other s in other cases. αm = average value of α for all beams on edges of a . α = ratio of flexural stiffness of beam section to flexural stiffness a width of the slab bounded laterally by centerline of adjacent ( if any) in each side of the beam.

E cb I b  E cs I s

β = ratio of clear spans in long to short direction of two way slabs. Ecb = modulus of elasticity of beam concrete Ecs = modulus of elasticity of slab concrete Ib =moment of inertia about the centroidal axis of gross section of section defined in section 5.13.2.4 Section 5.13.2.4 For monolithic or fully composite construction, a beam includes that portion of the slab on each side of the beam extending a distance equal to the projection of the beam above or below the slab, whichever is greater, but not greater than four times the slab thickness

Is =moment of inertia about the centroidal axis of gross section of slab Is =H3 /12 times width of slab defined in notation α and βt .

Direct Design Method       

Limitations of Direct Design Method There shall be a minimum of three continuous span in each direction shall be rectangular with a ratio of longer to shorter span center to center of within a not greater than 2 Successive span lengths center to center of s in each direction shall not differ by more than 1/3 the longer Columns be offset a maximum of 10% of span ( in the direction of the offset) from either axis between centerlines of successive columns. All loads shall be due to gravity only and uniformly distributed over an entire . Live load shall not exceed three times the dead load. For a between s on all sides , the relative stiffness of beams in the two perpendicular direction, must be between 0.2 and 0.5. Where :L1 = length of the span in the direction that moments are being 2 determined, measured center to  center of s L 1 1 L2 = length of span transverse to L I, measured center to center of 2  L 2 2 s.

Moments in slabs The total moment resisted by the slab equals absolute sum of the positive and negative factored moments in each direction and shall not be less than

( wu L2 )Ln Mo  8

2

EQ. A

Where: wu is the factored load

Clear span Ln shall extend face to face of the columns,capitals,bracket or walls Ln in EQ. A shall not be less than 0.65L1. Circular or polygon s shall be treated as square s with the same area If the transverse span of s on either side of the centerline of s varies L2 in EQ. A shall be taken as the average of adjacent transverse spans. When the span adjacent and parallel to an edge is considered, the distance from edge to centerline shall be substituted for L2 in EQ. A.

Negative and Positive Factored Moments Negative factored moments shall be located at face of rectangular s. Circular or regular polygon s shall be treated as square s with the same area A. In an interior span, total static moment Mo shall be distributed as follows: Negative factored moments 0.65 Positive factored moments 0.35 B. In an end span, total factored static moment Mo shall be distributed according to the table below 1

2

3

4

Slabs without beams Exterior edge Slabs with b/w interior s unrestrained beams b/w With Without all s edge beam edge beam Interior Negative Factored Moment

5 Exterior edge Fully restrained

0.75

0.70

0.70

0.70

0.65

Positive Factored Moment

0.63

0.57

0.52

0.50

0.35

Exterior Negative Factored Moment

0

0.16

0.26

0.30

0.65

Factored Moment in Column strips Column strips shall be proportioned to resist the following portion in percent Of interior factored negative moments. TABLE A

l2 l1



l2 0 l1

l2   1.0 l1

0.5

1.0

2.0

75

75

75

90

75

45

Linear interpolation be made for the values shown

Column strips shall be proportioned to resist the following portion in percent Of exterior factored negative moments. TABLE B

l2 l1 l  2 0 l1

l  2  1.0 l1

t  0  t  2.5 t  0  t  2.5

0.5

1.0

100

100

100

75

75

75

100

100

100

90

75

45

2.0

Linear interpolation be made for the values shown Where s consists of columns or walls extending for a distance equal to or greater than three – quarter other span length L2 used to compute Mo, negative moments shall be considered to be uniformly distributed across Lo.

E cb C t  2 E cs I s Where : C = torsional rigidity of the effective transverse beam Column strips shall be proportioned to resist the following in percent of positive factored moments. TABLE C

l2 l1

0.5

1.0

2.0

l2 0 l1

60

60

60

l2  1.0 l1

90

75

45





Linear interpolation be made for the values shown

For slabs with beams between s, the slab portion of the column strip shall be proportioned to resist that portion of the column strip moments not resisted by the beams. Factored moment in beams Beams between s shall be proportioned to resist 85% of column strip l moments If  l 2  1.0 . If is between 1.0 and zero  2 ,proportion of l1 l1 of column strip moments resisted by beams shall be obtained by linear interpolation between 85 and zero percent. Factored moment in middle strip That portion of negative and positive factored moments not resisted by column strips shall be proportionately assigned to corresponding half middle strip. Each middles strip shall be proportioned to resist the sum of moment assigned to its two half middle strip. A middle strip assigned to and parallel with one edge ed by a wall shall be proportioned to resist twice the moment assigned to the half middle strip corresponding to the first row of Interior s.

Modification of Factored Moments Negative and positive factored moments maybe modified by 10 % provided the total static moment for a in the direction considered is not less than that required by EQ. A. Factored shear in slab system with beams l  2  1.0 Beams with l1 shall be proportioned to resist shear caused by factored loads on tributary areas bounded by 450 lines drawn from the corners of the and the centerlines of adjacent s parallel to the long side. Beams with  l 2  1.0 maybe proportioned to resist shear obtained by linear l1 Interpolation, assuming that the beam carries no load at α = 0

C.S for wide beam shear bw d

450 x

L2/2 L2 1

450

L1 Where d =effective depth of the slab

Problem Using NS specifications, determine the minimum slab thickness of the shown. The slab has beam between all s which are poured monolothically with the slab. fc’ =21MPa, fy = 414 MPa. Assume Ec be the same for beam, slab and column. B 5m 5m 6m

A

for design

A

6m

6m B

h

500 mm 500 mm

Section B-B

350 mm

350 mm

Solution 1. Assume

h

h Section A-A

350 mm

350 mm

Ln ( 800  0.73 f y ) 36000  9000 

LongClearSpan 6000  350    1.22 ShortClearSpan 5000  350 Ln =6000 -350 =5650 mm

5650( 800  0.73 414  ) h 36000  9000( 1.22 ) h = 133 mm say 150 mm

2. Effective flange projection of beams Section 5.13.2.4 For monolithic or fully composite construction, a beam includes that portion of the slab on each side of the beam extending a distance equal to the projection of the beam above or below the slab, whichever is greater, but not greater than four times the slab thickness a) b) 3.

500 – 150 = 350 mm 4(150) = 600 mm

use 350 mm

Moment of Inertias of beams Approximate moment of inertias Edge beams 1.4 to 1.6 of bh3/12 Interior beams 1.5 to 2 of bh3/12 a. Edge beam

b. Interior beam

350( 500 )3 I  1.4  5 ,104 x10 6 mm 4 12 350( 500 )3 I  1 .6  5 ,833 x10 6 mm 4 12

E cb I b  E cs I s

4. Values of α

Note : Ecb =Ecs

a) For edge beam with 3 m wide slab

3000( 150 ) 3 Is   843.75 x10 6 mm 4 12 5104 1   6.05 843.75 b) For interior beam with 5 m wide slab ( 2 beams)

5000( 150 )3 Is   1406.25 x10 6 mm 4 12   2

5833  4.15 1406.25

c) For interior beam with 6 m wide slab

6000( 150 )3 Is   1687.5 x10 6 mm 4 12   3

5833  3.46 1687.5

d) αm = average value of α for all beams on edges of a .

 1  2 2  3 6.05  2( 4.15 )  3.46 m    4.45 4 4 Check thickness a) Minimum thickness 5.

h

Ln ( 800  0.73 f y ) 

1 36000  5000    m  0.12( 1   

 ) 

5650( 800  0.73 414  ) h 1   36000  5000( 1.22 ) 4.45  0.12( 1  ) 1.22   h = 101 mm

< 133 mm o.k

The values of h obtained from EQS.10.1,10.2 and 10.3 maybe shall be modified as required by Sections 5.9.5.3.4 and 5.9.5.3.5 but in no case shall the thickness be less than

 m  2.0  m  2.0 Since

 m  4.45

------------------------- 125 mm ------------------------- 90 mm

hmin= 90 mm < 133 mm ok

b) thickness need not be more than

5650( 800  0.73 414  ) h  173mm133mm 36000

o.k

Design the slab given in the preceding problem using auniform dead load of 1.3 kPa excluding the weight of the slab and a uniform live load of 5.7 kPa. Use 12 mm bars , h = 150 mm and C = 4282 x 106 mm4. Solution 1. Slab weight : Ws = Wct(1) Ws =23.5(0.15) = 3.53 kPa Total uniform dead load WD = 1.3 + 3.53 = 4.83 kPa Total Factored uniform Load Wu = 1.4WD+1.7WL = 1.4(4.83) +1.7(5.7) = 16.452 kPa 2. Effective depth d = h - 1/2 bar diameter – cover d = 150 - 1/2(12) – 20 = 124 mm 3. check depth for shear l2 Beams with   1.0 shall be proportioned to resist shear caused by factored loads on l tributary areas 1 bounded by 450 lines drawn from the corners of the and the centerlines of adjacent s parallel to the long side.

5m

l2 5   ( 4.15 )  3.451.0 l1 6

1

450

x = 2.5-0.35/2 -0.124 = 2.201 m

x 0.124 m

450

C.S for wide beam shear

2.5 m

0.35 m

6m

taking b = 1 m Shear force at critical section Vu = Wu (shaded area) Vu =Wu(x)(1) Vu =16.452(2.201) Vu = 36.21 kN Shear carried by concrete

1 Vc  fc' bd 6 1 Vc  21( 1000 )( 124 ) 6

Vc  94 ,706 N  94.7 kN

Vc  0.85( 94.7 )  80.5 kN  36.21kN 4. Moment along the short span ( 5 m interior span) The total moment resisted by the slab equals absolute sum of the positive and negative factored moments in each direction and shall not be less than

( wu L2 )Ln Mo  8

2

EQ. A

Where: wu is the factored load

Clear span Ln shall extend face to face of the columns,capitals,bracket or walls Ln in EQ. A shall not be less than 0.65L 1

L1 =5 m

0.35 m

0.35 m

Ln = 5 – 0.35 = 4.65 m

( 16.452( 6 )( 4.65 ) 2 Mo  8

Ln

L2 = 6 m

M o  266.8 kN .m

Ok for shear

In an interior span, total static moment Mo shall be distributed as follows: Negative factored moments 0.65 Positive factored moments 0.35

Negative factored moment=0.65(266.8) = -173.42 kN.m Positive factored moment =0.35(266.8) = 93.38 kN.m Distribute the above negative and positive moment to the column strip, beam and middle strip l2 l2 6    3.46 (in the direction of L1) l  3.46( 1.2 )  4.15  1.0   1.2 1 l 5 1

Factored Moment in Column strips Column strips shall be proportioned to resist the following portion in percent of interior factored negative moments. Table A



l2 l1

0.5

1.0

2.0

l2  1.0 l1

90

75

45

Linear interpolation be made for the values shown

l2 l1 1.0 - 2.0

1.0 – 1.2

1.0 1.2 2.0

Percentage 75 x 45

75 - x

- 45 1.0 - 2.0 75

75 - 45

75 - x 1.0 - 1.2

x= 69

Interior Negative moment resisted by column strip =0.69(-173.42)= -119.66kN.m Factored moment in beams Beams between s shall be proportioned to resist 85% of column strip moments If

l2   1.0 l1

0.85(-119.66) = -101.71kN.m will be resisted by the beam 0.15(-119.66) = -17.95 kN.m will be resisted by the slab The remaining –(173.42 – 119.66) = - 53.76 kN.m is allotted to the middle strip Column strips shall be proportioned to resist the following in percent of positive factored moments. TABLE C



l2 l1

0.5

1.0

2.0

l2  1.0 l1

90

75

45

Linear interpolation be made for the values shown

l2 l1 1.0 - 2.0

1.0 – 1.2

1.0 1.2 2.0

Percentage 75 x 45

75 - x

- 45 1.0 - 2.0

75 - x 1.0 - 1.2

75

75 - 45

x= 69

Positive moment resisted by column strip =0.69(93.38)= 64.43kN.m 0.85(64.43) = 54.77kN.m will be resisted by the beam 0.15(64.43) = 9.66 kN.m will be resisted by the slab The remaining (93.38 – 64.43) =28.95 kN.m is allotted to the middle strip 5. Moment along the short span (along edge beam)

L1 =5 m

Ln = 5 – 0.35 = 4.65 m If the transverse span of s on either side of the centerline of s varies L2 in EQ. A shall be taken as the average of adjacent transverse spans. When the span adjacent and parallel to an edge is considered, the distance from edge to centerline shall be substituted for L 2 in EQ. A.

L2 =6/2 + 0.35/2 = 3.175 m

L2

Ln 0.35 m

0.35 m

( 16.452( 3.175 )( 4.65 ) 2 Mo  8 M o  141.18 kN .m Negative factored moment=0.65(141.18) = -91.78 kN.m Positive factored moment =0.35(141.18) = 49.9 kN.m

l2 6   1.2 l1 5

  6.05 (for edge beam)



l2  6.05( 1.2 )  7.26  1.0 l1

3000( 150 ) 3 Is   843.75 x10 6 mm 4 12

E cb C 4282 x10 6 t    2.54 6 2 E cs I s 2( 843.75 )( 10 ) Note: Ecb =Ecs

Column strips shall be proportioned to resist the following portion in percent Of exterior factored negative moments. TABLE B

l2 l1

t  0  t  2.5

l2   1.0 l1

0.5

1.0

100 90

100 75

2.0 100 45

Linear interpolation be made for the values shown

l2 l1 1.0 - 2.0

1.0 – 1.2

1.0 1.2 2.0

Percentage 75 x 45

75 - x

- 45 1.0 - 2.0 75

75 - 45

Exterior Negative Moment Column strip =0.69(-91.78) = 63.33 kN.m Beam =0.85(-63.33) = - 53.83 kN.m slab = 0.15(-63.33) = - 9.5 kN.m Middle strip =-(91.78 -63.33) = - 28.45 kN.m

x= 69

75 - x 1.0 - 1.2

Positive Moment From Table C

l2 l1

Percentage

1.0 75 75 - x 1.0 - 2.0 1.2 x 75 - 45 2.0 45 Column strip =0.69(49.4) = 34.1 kN.m Beam =0.85(34.1) = 28.99 kN.m slab = 0.15(34.1) = 5.12 kN.m Middle strip = (49.4 -34.1) = 15.30 kN.m 1.0 – 1.2

75 - x 1.0 - 1.2

x= 69

-28.45 -9.5 -26.88

1.25 m

-8.975 -8.975

+15.3

+14.475

+5.12

2.5 m

+4.83 +4.83

-28.45 -9.5 -26.88 -8.975 -8.975

1.25

3.5

1.25

1.25 m

- 45 1.0 - 2.0 75

Exterior Column strip

-17.95/2 =-8.975 +9.66/2= +4.83 -53.76/2=-26.88 +28.95/2 = 14.475 -(26.88+28.45)= -55.33 +(14.475+15.3)=+29.78

interior Column strip

6. Moment along the long span (6m end span) L2= 5 m

Ln = 6 -0.35 = 5.65 m

( 16.452( 5 )( 5.65 ) 2 Mo  8 M o  328.24 kN .m

Ln

L1 = 6 m

In an end span, total factored static moment Mo shall be distributed according to the table below

1

2

3

4

Slabs without beams Exterior edge Slabs with b/w interior s unrestrained beams b/w With Without all s edge beam edge beam Interior Negative Factored Moment

5 Exterior edge Fully restrained

0.75

0.70

0.70

0.70

0.65

Positive Factored Moment

0.63

0.57

0.52

0.50

0.35

Exterior Negative Factored Moment

0

0.16

0.26

0.30

0.65

Factored Interior Negative Moment = -0.7(328.24)= -229.77 kN.m Factored Positive Moment = -0.57(328.24)= -187.1 kN.m Factored Exterior Negative Moment = -0.16(328.24)= -52.52 kN.m

l2 5   0.83 l1 6

  4.15



l2  4.15( 00.83 )  3.44  1.0 l1

Interior Negative Moment ( Use Table A to get percentage)

l2 l1 0.5 - 1.0 0.5 – 0.83

0.5 0.83 1.0

Percentage 90 x 75

90 - x

90 - 75

Column strip =0.801(-229.77) = -184 kN.m Beam =0.85(-184) = - 156.4 kN.m slab = 0.15(-184) = - 27.6 kN.m Middle strip =-(229.77- 184) = - 45.77 kN.m

90  x 0.5  0.83  90  75 0.5  1.0 x  80.1

Exterior Negative Moment ( Use Table B to get percentage) Percentage l2 l1 0.5 90 0.5 – 0.83 90 - x 90 - 75 0.5 - 1.0 0.83 x 1.0 75 Column strip =0.801(-52.52) = -42 kN.m Beam =0.85(-42) = - 37.5 kN.m slab = 0.15(-42) = - 6.3 kN.m Middle strip =-(52.52- 42) = - 10.52 kN.m

90  x 0.5  0.83  90  75 0.5  1.0 x  80.1

Positive Moment ( Use Table C to get percentage)

l2 l1

Percentage

0.5 90 90 - x 0.83 x 90 - 75 1.0 75 Column strip =0.801(187.1) = 150 kN.m Beam =0.85(150) = 127.5 kN.m slab = 0.15(150) = 22.5 kN.m Middle strip =(187.1- 150) = 37.1 kN.m

0.5 - 1.0 0.5 – 0.83

90  x 0.5  0.83  90  75 0.5  1.0 x  80.1

Edge Beam -10.52

-3.15

-3.15

22.5/2 11.25

11.25

37.1

-27.6/2

-13.8

1.25 m

11.25

-13.8

-45.77

-13.8

1.25 m

2.5 m

1.25 m

11.25

Column strip

Column strip

-3.15

-6.3/2 -3.15

-13.8

1.25 m

Note: Since the column strip exist on both sides of the beam, the value Of the moment alloted to the slab must be divided by 2. In the middle Strip it is not divided by 2 since the same moment is also alloted by the Opposite middle strip.

+37.1

-13.8

-45.77

1.25 m

B 2.5 m

-13.8

C 1.25 m

D

-8.975

+11.25

+4.83

A

F

+11.25

E

-3.15 -9.5

-10.5

Edge Beam

-53.33

-3.15

+5.12

C

+28.97

-9.5 -53.33

B

-8.975

F E D

1.25 m 3.5 m 1.25 m

A

Row A & C Row B Across F Across E Across D Across F Across E Across D Mu b

11.25

-13.8

-10.5

37.1

-45.77

1075

1075

1075

2500

2500

2500

112

124

112

124

124

124

-3.15

d Ru

0.25955

0.75624

1.13708 0.30350

1.07238

1.32299

ρ

0.00063

0.00187

0.00284 0.00074

0.00267

0.00332

Use ρ

0.00338

0.00338

0.00338

0.00338

0.00338

0.00338

As

407.15

450.77

407.15

1048.31

1048.31

1048.31

3.6

3.99

3.6

9.27

9.27

9.27

s Use s

298.61

269.71

298.61

269.71

Position

Top

N

pmin

290

260 Bottom

1.4   0.00338 414

290 Top

269.71

269.71

260

260

260

Top

Bottom

Top

Row D & F Row E Across A Across B Across C Across A Across B Across C 5.12

-9.5

-55.33

29.78

-55.33

1075

1075

1075

3500

3500

3500

124

124

124

124

124

124

0.34417

0.6386

1.14237

0.61485

1.14237

0.00157

0.00084

0.00157 0.00285

0.00151

0.00285

Use ρ

0.00338

0.00338

0.00338

0.00338

0.00338

0.00338

As

450.77

450.77

450.77

1467.63

1467.33

1467.33

3.99

3.99

3.99

12.98

12.98

12.98

s Use s

267.71

269.71

269.71

269.71

269.71

269.71

Position

Top

Mu b

-9.5

d Ru

0.6386

ρ

N

260

260 Bottom

260 Top

260

260

260

Top

Bottom

Top

Temperature bars: At =0.0018(1000)(150) = 270 mm2 Using 10 mm S = 418 mm Maximum spacing = 2t = 300 mm say 260 mm

ACI Moment Coeffecient Method 

The complexity of the Direct Design Method particularly in meeting its requirements paved way to the ACI Moment Coeffecient Method. The method makes use of tables of moment coeffecients for a variety of conditions. These coeffecients are based on elastic analysis but also takes into inelastic distribution.



Moments at column strip and middle strip are computed by

M a  Ca wl

2 a

M b  Cb wl

2 b

Ca , Cb=Tabulated moment coeffecients

la , lb

= Clear span length in short and long directions respectively

w=uniform load in psf,Pa or KPa

la

la/4 lb/4

Column strip

lb

lb/4

Middle strip

Column strip

Width of middle strip – one-half of the Width of edge or column strip – one-fourth of the

Column strip

Column strip



la/4

lb

Table 1: Coeffecients for Negative Moments in Slabs

M a ,neg C aNeg wl a 2 Ratio

m 1.0

la

Case 1 Case 2

Case 3

Case 4

w= total uniform dead plus live load Case 5 Case 6 Case 7 Case 8 Case 9

lb

Ca,Neg Cb,Neg

0.95 Ca,Neg Cb,Neg

0.90 Ca,Neg Cb,Neg

0.85 Ca,Neg Cb,Neg 0.80 Ca,Neg Cb,Neg

0.75

Ca,Neg Cb,Neg

0.70

Ca,Neg Cb,Neg

0.65

Ca,Neg Cb,Neg

0.60

Ca,Neg Cb,Neg

0.55 Ca,Neg Cb,Neg

0.50

M b.neg  CbNeg wlb

2

Ca,Neg Cb,Neg

0.045 0.045 0.050 0.041 0.055 0.037 0.060 0.031 0.065 0.027 0.069 0.022 0.074 0.017 0.077 0.014 0.081 0.010 0.084 0.007 0.086. 0.006

0.076 0.072 0.070 0.065 0.061 0.056 0.050 0.043 0.035 0.028 0.022

0.050 0.050 0.055 0.045 0.060 0.040 0.066 0.034 0.071 0.029 0.076 0.024 0.081 0.019 0.085 0.015 0.089 0.011 0.092 0.008 0.094. 0.006

0.075 0.079

0.071 0.075

0.080

0.079

0.082

0.083

0.083

0.086

0.085

0.088

0.071 0.067 0.062 0.057 0.051 0.044

0.086

0.091

0.087

0.093

0.088

0.095

0.089

0.096

0.090

0.097

0.038 0.031 0.024 0.019 0.014

0.033 0.061 0.038 0.056 0.043 0.052 0.049 0.046 0.055 0.041 0.061 0.036 0.068 0.029 0.074 0.024 0.080 0.018 0.085 0.015 0.089. 0.010

0.061 0.033 0.065 0.029 0.068 0.025 0.072 0.021 0.075 0.017 0.078 0.014 0.081 0.011 0.083 0.008 0.085 0.006 0.086 0.005. 0.088 0.033

Table 2: Coeffecients for Dead Load Positive Moments in Slabs 2 w= total uniform dead load M apos ,dl  C a ,dl wl a2 bpos , dl b.dl b

M

Ratio

m 1.0

la

Case 1 Case 2

Case 3

 C wl

Case 4

Case 5 Case 6 Case 7 Case 8 Case 9

lb

Ca,dl Cb,dl

0.95 Ca,dl Cb,dl

0.90 Ca,dl Cb,dl

0.85 Ca,dl Cb,dl

0.80 Ca,dl Cb,dl

0.75 Ca,dl Cb,dl

0.70

Ca,dl Cb,dl

0.65

Ca,dl Cb,dl

0.60

Ca,dl Cb,dl

0.55 Ca,dl Cb,dl

0.50 Ca, dl Cb, dl

0.036 0.036 0.040 0.033 0.045 0.029 0.050 0.026 0.056 0.023 0.061 0.019 0.068 0.016 0.074 0.013 0.081 0.010 0.088 0.008 0.095 0.006

0.018 0.018 0.020 0.016 0.022 0.014 0.024 0.012 0.026 0.011 0.028 0.009 0.030 0.007 0.032 0.006 0.034 0.004 0.035 0.003 0.037 0.002

0.018 0.027 0.021 0.025 0.025 0.024 0.029 0.022 0.034 0.020 0.040 0.018 0.046 0.016 0.054 0.014 0.062 0.011 0.071 0.009 0.080 0.006

0.027 0.027 0.030 0.024 0.033 0.022 0.036 0.019 0.039 0.016 0.043 0.013 0.046 0.011 0.050 0.009 0.053 0.007 0.056 0.005 0.059 0.004

0.027 0.018 0.028 0.015 0.029 0.013 0.031 0.011 0.032 0.009 0.033 0.007 0.035 0.005 0.036 0.004 0.037 0.003 0.038 0.002 0.039 0.001

0.033 0.027 0.036 0.024 0.039 0.021 0.042 0.017 0.045 0.015 0.048 0.012 0.051 0.009 0.054 0.007 0.056 0.006 0.058 0.004 0.061 0.003

0.027 0.033 0.031 0.031 0.035 0.028 0.040 0.025 0.045 0.022 0.051 0.020 0.058 0.017 0.065 0.014 0.073 0.012 0.081 0.009 0.089 0.007

0.020 0.023 0.022 0.021 0.025 0.019 0.029 0.017 0.032 0.015 0.036 0.013 0.040 0.011 0.044 0.009 0.048 0.007 0.052 0.005 0.056 0.004

0.023 0.020 0.024 0.017 0.026 0.015 0.028 0.013 0.029 0.010 0.031 0.007 0.033 0.006 0.034 0.005 0.036 0.004 0.037 0.003 0.038 0.002

Table 3: Coeffecients for Live Load Positive Moments in Slabs 2 w= total uniform live load M apos ,ll  C a ,ll wl a2 bPos ,ll b ,ll b

M

Ratio

m 1.0

la

Case 1 Case 2

Case 3

 C wl

Case 4

Case 5 Case 6 Case 7 Case 8 Case 9

lb

Ca,ll Cb,ll

0.95 Ca,ll Cb,ll

0.90 Ca,ll Cb,ll

0.85 Ca,ll Cb,ll

0.80 Ca,ll Cb,ll

0.75 Ca,ll Cb,ll

0.70

Ca,ll Cb,ll

0.65

Ca,ll Cb,ll

0.60

Ca,ll Cb,ll

0.55 Ca,ll Cb,ll

C ,ll 0.50 Ca,ll b

0.036 0.036 0.040 0.033 0.045 0.029 0.050 0.026 0.056 0.023 0.061 0.019 0.068 0.016 0.074 0.013 0.081 0.010 0.088 0.008 0.095 0.006

0.027 0.027 0.030 0.025 0.034 0.022 0.037 0.019 0.041 0.017 0.045 0.014 0.049 0.012 0.053 0.010 0.058 0.007 0.062 0.006 0.066 0.004

0.032 0.032 0.031 0.029 0.035 0.027 0.040 0.024 0.045 0.022 0.051 0.019 0.057 0.016 0.064 0.014 0.071 0.011 0.080 0.009 0.088 0.007

0.032 0.032 0.035 0.029 0.039 0.026 0.040 0.023 0.048 0.020 0.052 0.016 0.057 0.014 0.062 0.011 0.067 0.009 0.072 0.007 0.077 0.005

0.032 0.027 0.034 0.024 0.037 0.021 0.041 0.019 0.044 0.016 0.047 0.013 0.051 0.011 0.055 0.009 0.059 0.007 0.063 0.005 0.067 0.004

0.035 0.032 0.038 0.029 0.042 0.025 0.046 0.022 0.051 0.019 0.055 0.016 0.060 0.013 0.064 0.010 0.068 0.008 0.073 0.006 0.078 0.005

0.032 0.035 0.036 0.032 0.040 0.029 0.045 0.026 0.051 0.023 0.056 0.020 0.063 0.017 0.070 0.014 0.077 0.011 0.085 0.009 0.092 0.007

0.028 0.030 0.031 0.027 0.035 0.024 0.040 0.022 0.044 0.019 0.049 0.016 0.054 0.014 0.059 0.011 0.065 0.009 0.070 0.007 0.076 0.005

0.030 0.028 0.032 0.025 0.036 0.022 0.039 0.020 0.042 0.017 0.046 0.013 0.050 0.011 0.054 0.009 0.059 0.007 0.063 0.006 0.067 0.004

Table 4: Ratio of load w in la and lb directions for shear and slabs and load 0n s Ratio

m

la

Case 3

Case 4

Case 5 Case 6 Case 7 Case 8 Case 9

lb

Wa Wb 0.95 Wa Wb 0.90 Wa Wb 0.85 Wa Wb 0.80 Wa Wb 0.75 Wa Wb lW a 0.70 Wb 0.65 Wa Wb 0.60 Wa Wb 0.55 Wa Wb 0.50 Wa Wb

1.0

Case 1 Case 2

0.50 0.50 0.55 0.45 0.60 0.40 0.66 0.34 0.71 0.29 0.76 0.24 0.81 0.19 0.85 0.15 0.89 0.11 0.92 0.08 0.94 0.06

0.50 0.50 0.55 0.45 0.60 0.40 0.66 0.34 0.71 0.29 0.76 0.24 0.81 0.19 0.85 0.15 0.89 0.11 0.92 0.08 0.94 0.06

0.17 0.83 0.20 0.80 0.23 0.77 0.28 0.72 0.33 0.77 0.39 0.61 0.45 0.55 0.53 0.47 0.61 0.39 0.69 0.31 0.76 0.24

0.50 0.50 0.55 0.45 0.60 0.40 0.66 0.34 0.71 0.29 0.76 0.24 0.81 0.19 0.85 0.15 0.89 0.11 0.92 0.08 0.94 0.06

0.83 0.17 0.86 0.14 088 0.12 0.90 0.10 0.92 0.08 0.94 0.06 0.95 0.05 0.96 0.04 0.97 0.08 0.98 0.02 0.99 0.01

0.71 0.29 0.75 0.25 0.79 0.21 0.83 0.17 0.86 0.14 0.88 0.12 0.91 0.09 0.93 0.07 0.95 0.05 0.96 0.04 0.97 0.03

0.29 0.71 0.33 0.67 0.38 0.62 0.43 0.57 0.49 0.51 0.56 0.44 0.62 0.38 0.69 0.31 0.76 0.24 0.81 0.19 0.86 0.14

0.028 0.030 0.38 0.62 0.43 0.57 0.49 0.51 0.55 0.45 0.61 0.39 0.68 0.32 0.74 0.26 0.80 0.20 0.85 0.25 0.89 0.11

0.030 0.028 0.71 0.29 0.75 0.25 0.79 0.21 0.83 0.17 0.86 0.14 0.89 0.11 0.92 0.08 0.94 0.06 0.95 0.05 0.97 0.03

Discontinuous Edge Continuous edge Cases Description Case 1 : Case 2 : Case 3 : Case 4 : Case 5 : Case 6 : Case 7: Case 8: Case 9:

Simply ed Interior Edges of the Long span continuous One edge of Long span, one edge of short span continuous (corner ) Edges of the short span continuous One edge of short span continuous One edge of long span continuous Two edge of long span, one edge of short span continuous Two edge of short span, one edge of long span continuous

Uses of Tables of coeffecients Table 1 - to compute the total negative moment at the continuous edge(column strip) Table 2 - to compute the positive dead load moment at the midspan(middle strip) Table 3- to compute the positive live load moment at the midspan(middle strip) Table 4 - to compute shear at the long and short span Total Positive moment at midspan Sum of positive dead load moment and positive live load moment Negative Moment at Discontinuous egde =1/3 of Positive moment at the midspan Minimum thickness 1/180 times the perimeter of the but not to be less than 75 mm

hmin

2(la  lb )  180

Problem Design the thickness and reinforcement of the cornel shown usin the ACI moment coeffecient method. The slab has beam between all s which are poured monolothically with the slab. fc’ =20 MPa, fy = 350 MPa The slab is to a uniform dead load of 4.5 kPa and uniform live load of 3.9 kPa. Concrete weighs 23.5 kN/m3. B 5.3 m 500 mm

h Section A-A 300 mm

300 mm

h Section B-B 300 mm

4.3 m

for design

A

A 5.3 m

500 mm

5.3 m 300 mm

B

Solution Clear span length short span = la = 4.3 – 0.3 = 4.0m long span = lb = 5.3 – 0.3 = 5.0m Slab thickness:

hmin

2(la  lb )  180

2(4000  5000) h  100mm  75mm 180 Weight of slab Ws = 0.10(23.5) =2.35 kN/m Total Dead load WD = 4.5 + 2.35 = 6.85 kN/m

Factored loads Dead load = 1.4 (6.85) = 9.59 kN/m Live load = 1.7 (3.9) = 6.63 kN/m Total factored load = 16.22 kN/m Effective depth ( using 12 mm bars) d = 100 -1/2(12) – 20 =74 mm

m

la 4   0.80 lb 5

Check for Shear From Table 4 Wa = 0.71, Wb = 0.21, showing that the shear in the short direction is more critical than the long direction. uniform load per m width Wu = 0.71( 16.22) = 11.52 kN/m 11.52 kN/m c.s for shear 0.074 m

4m R=11.52(4)/2=23.04 kN

Shear force at critical section Vu =23.04 -0.074(11.52) = 22.19 kN=22190 N Allowable shear force

0.85 20 (1000)74 Vu   46883 N 6

Design Moments Short Direction Negative Moment at continuous edge: From Table 1 CaNeg =0.071, w = 16.22 Ma,Neg = 0.071(16.22)(4)2=18.43 kN/m Positive Moment at the midspan Dead Load

M apos ,dl  C a ,dl wl a2

From Table 2 ,Ca,dl =0.039 w= 9.59

MaPos,dl=0.039(9.59)(4)2 =5.98 kN.m Live Load

M apos ,ll  C a ,ll wl a2

From Table 3 ,Ca,ll =0.048 w= 6.63

MaPos,ll=0.048(6.63)(4)2 =5.09 kN.m Total Positive Moment at Midspan MaPos = 5.98 + 5.09 = 11.07 kN.m Negative Moment at discontinuous edge MaNeg = -1/3MaPos MaNeg = -1/3(11.07) = -3.69 kN.m

Long Direction Negative Moment at continuous edge: From Table 1 CbNeg =0.029, w = 16.22

M b.neg  CbNeg wlb

2

Mb,Neg = 0.029(16.22)(5)2=11.76 kN/m Positive Moment at the midspan Dead Load

M bpos ,dl  Cb.dl wlb2

From Table 2 ,Cbdl =0.016 w= 9.59

MbPos,dl=0.016(9.59)(5)2 =3.84 kN.m Live Load

M bPos ,ll  Cb,ll wlb2

From Table 3 ,Cb,ll =0.020 w= 6.63

MbPos,ll=0.020(6.63)(5)2 =3.32 kN.m Total Positive Moment at Midspan MbPos = 3.84 + 3.32 = 7.16 kN.m Negative Moment at discontinuous edge MbNeg = -1/3MaPos MbNeg = -1/3(7.16) = -2.39kN.m

D

D

1.25 m

-11.76

A B C

1.0 m 2.0m 1.0 m

E

B

-3.69

+11.07

F

F

E -18.43

1.25 m

2.5 m

A C

-2.39

+7.16

Row A

Row B

Row C

Row D

Row E

-18.43

11.07

-3.69

-11.76

7.16

1000

1000

1000

1000

1000

1000

74

74

74

74

68

74

1.72

Across D, E&F

Mu b

Across D, Across D, Across A, E&F E&F B&C

Across A, B&C

Row F

Across A, B&C

-2.39

d Ru

3.74

2.25

2.39

ρ

0.0122

0.00692

0.0074

0.0052

Use ρ

0.0122

0.00692

0.0074

0.0052

As

902.8

512.08

547.6

384.8

s

125

220

206

293.9

618

150

300

300

12 mm

12 mm

12 mm

12 mm

Top

Top

Use s

110

Bar size

12 mm

Position

Top

Smax = 5t = 500 mm

220 12 mm Bottom

660 220

Bottom

Top

Temperature bars At = 0.0020(1000)100 = 200 S = 392 mm use s = 300 mm Bar size 10 mm

Concrete Design Semi- Final Exam 1.A square tied column 300 mm by 300 mm is reinforced with 6 of 20 mm bars with fc’ = 20.7 MPa and fy = 345 MPa. Determine the following : a) Ultimate axial load capacity of the column. b) spacing of 10 mm lateral ties

 (20) 2 Pu =Ф0.80{0.85fc’(Ag-Ast) + Astfy} Ast  6  1884.95mm 2 4 0.7(0.80){0.85(20.7)[(300)(300)  1884.95]  1884.95(345)} Pu   1232.38kN 1000 Spacing of 10 mm ties

S  16(20)  400mm S  48(10)  480mm S  300mm

Use S = 300 mm

2. Calculate the ultimate axial load capacity of the composite column shown below .fc’=21 MPa, Fy=250 MPa,fy =350MPa. Reinforcing bars consist of 8 of 20 mm diameter bars. 400 mm

400 mm

W10 x 150

Properties of W 10 x 150 Area = 27000 mm2 Depth = 250 mm

8 [(20) 2 ] As   2513mm 2 4 Ac  400(400)  27000  2513  130487 mm 2



Pu   0.85 0.85 fc ' Ac  As f y  AWF Fy



Pu  (0.7)0.85 0.85(21)(130487)  2513(350)  27000(250) Pu  5925452 N  5,925.52kN

3. A cantilever beam 300 mm wide and 450 mm deep is reinforced with 6 of 22 mm straight top bars with fc’ = 20.7 MPa and fy = 415 MPa. Calculate required development length. Solution

Ldb 

Ldb 

25 f y

fc ' 25(415)

20.7

 2280mm

Modification Factor Top bar = 1.3 Required development length

Ld  1.3Ldb  1.3( 2280)  2964mm

4. A rectangular beam 350 mm wide and 600 mm deep is reinforced with 4 of 32 mm compression bars with fc’ = 20.7 MPa and fy = 275 MPa. Calculate the required development length.

0.24d b f y

Ldb 

fc'

0.24(32)275 Ldb   464.2mm 20.7 Ldb  0.04d b f y Ldb  0.04(32)(275)  352mm Use L

db

 464.2mm

No applicable modification factor

Ld  464.2mm

5. A rectangular beam with b = 300 mm, d = 500 mm is provided with 10 mm vertical stirrups with fy = 276 MPa. Assuming fc’ = 21 MPa; a) Determine the required spacing if Vu = 42 kN b) Determine the required spacing if Vu = 95 kN Solution Shear carried by concrete

1 Vc  6

fc'bw d

1 Vc  21(300)500  114,564 N 6

Vc 0.85(114564 )   48689.7 N 2 2

a) b)

Vc Stirrups not necessary Vu  42000 N  2 Vu  95000 N Vu Vs   Vc  95000 Vs   114564  2799 N 0.85 2 (10) 2 Av   157.1mm 2 4 S

Av f y d Vs

157.1( 276)500   7745mm 2799

1 3

1 fc 'bw d  21(300)(500)  229129 N 3

1 Vs  fc 'bw d 3 d 500 S   250mm 2 2 Use S = 250 mm

Semi-Final Exam 1. A square concrete column has a dimension of 400 mm by 400 mm. The column length is 4.5 m and is to be used in a frame braced against sidesway ( k =1.0, EI = 40MN.m2). The column is bent in single curvature by ultimate factored moment of 100 kN.m at the top and 120 kN.m at the bottom. The column carries factored axial dead load of 150 kN and factored axial live load of 320 kN. Is slenderness effects be considered in the analysis ? If so, what is the magnified moment capacity of this column ? 2. Solve the previous problem considering the effect of wind loads producing an unfactored axial load of 125 kN an unfactored moment of 90 kN.m.

8 of 28 mm

350 mm 500 mm

3. For the column shown in figure, PD = 250 kN, PL = 410kN, MD = 140 kN.m, and ML = 152 kN.m. The column belongs to a braced frame and is bent in single curvature about the x axis by equal moments at each end of the member . The uned length is 6 m. fc’ = 20.7 MPa, f y = 413.7 MPa . Determine the following : a) Critical load Pc b) Magnified moment 350 mm

Solution to #1

M 2b  120kN .m M 1b  100kN .m M 2s  0 r  0.3h  0.3(400)  120mm Lu  4.5m  4500mm kLu 1(4500)   37.5 r 120 M 1B 100 34  12  34  12  24 M 2B 120 kLu M 1B  34  12 r M 2B

Column is Long

+ since the column is bent in single curvature

M 1b 100 Cm  0.6  0.4  0.6  0.4  0.933 M 2b 120 Pu  150  320  470kN

 2 EI  2 40(10) 6 Pc    19,495,515 N  19,495.5kN 2 2 (kLu ) {(1)(4.5)} Cm 0.933 b    0.967 Pu 470 1 1 0.7(19495.5) Pc M c   b M 2b   c M 2 s  0.967(120)   s (0)  116 .04kN .m

Solution to #2

M 2 s  0.75[1.4M D  1.7 M L  1.7 M W ] M 2 s  0.75[1.4(0)  1.7(0)  1.7(90)]  114 .75kN .m [1.4 PD  1.7 PL ]  150  320  470kN

Specified in the previous problem

Pu  0.75[1.4 PD  1.7 PL  1.7 PW ]  0.75[470  1.7(125)]  511.87 kN Cm 0.933 b    0.97 Pu 511.87 1 1 0.7(19495.5) Pc

 s  1.0  Frame is braced against sidesway M c   b M 2b   s M 2 s  0.97(120)  1.0(114 .75)  231.15kN .m

Solution to # 3

8 (28) 2 As   4926mm 2 4 Pu=1.4PD + 1.7PL = 1047 kN M1b= M2b =1.4MD + 1.7ML = 454.4 kN.m

Ec  4700 fc'  4700 20.7  21383.7 MPa

350(500)3 Ig   3645.83 x106 mm 4 12 I se  As z 2  4926(175) 2  150.86 x106 mm 4

Es  200000 MPa

1.4 M D 1.4(140) d    0.431 Mu 454.4 Ec I g EI 

 E s I se

5 1 d

21383.71(3645.83)106  200000(150.86)10 6 5 EI  1  0.431

EI  4.576 x1013 N .mm 2

 2 EI Pc  (kLu ) 2  2 4.576 x1013 Pc   12545.36kN 2 (6000 ) M 1b 550.54 Cm  0.6  0.4  0.6  0.4  1.0 M 2b 550.54 Cm 1 b    1.135 Pu 1047 1 1 Pc 0.7(12545.36) M c   b M 2b  1.135(454.4)  515.78kN .m