Simplified Reinforced Concrete Design 2010 Ns 2e2e4c

CHAPTER 1 Introduction Concrete Concrete is a mixture of water, cement, sand, gravel crushed rock, or other aggregates. The aggregates (sand, gravel, crushed rock) are held together in a rocklike mass with a paste of cement and water.

REINFORCED CONCRETE As with most rocklike mass, concrete has very high compressive strength but have a very low tensile strength. As a structural member, concrete can be made to carry tensile stresses (as in beam in flexure). In this regard, it is necessary to provide steel bars to provide the tensile strength lacking in concrete. The composite member is called reinforced concrete.

AGGREGATES Aggregates used in concrete may be fine aggregates (usually sand) and coarse aggregates (usually gravel or crushed stone). Fine aggregates are those that es through a No. 4 sieve (about 6 mm in size). Materials retained are coarse aggregates. The nominal maximum sizes of coarse aggregate are specified in Section 5.3.3 of NS. These are follows: 1/5 the narrowest dimension between sides of forms, 1/3 the depth of slabs, or 3/4 the minimum clear spacing between individual reinforcing bars or wires, bundles of bars, or prestressing tendons or ducts. These limitations may not be applied if, in the judgment the Engineer, workability and methods of consolidation are such that concrete can be placed without honeycomb or voids. WATER According to Section 5.3.4, water used in mixing concrete shall be clean and free from injurious of oils, acids, alkalis, salts organic materials or other substances that may be deleterious to concrete or reinforcement. Mixing water for prestressed concrete or for concrete that will contain aluminum embedment’s, including that portion of mixing water contributed in the form of free moisture on aggregates, shall not be used in concrete unless the following are satisfied: (a) Selection of concrete proportions shall be based on concrete mixes using water from the same source and (b) mortar test cubes made with non-portable mixing water shall have 7-days and 28 day strengths equal to at least 90

MODULUS OF ELASTICITY Unlike steel and other materials, concrete has no definite modulus of elasticity. Its value is dependent on the characteristics of cement and aggregates used, age of concrete and strengths. According to NS (Section 5.8.5), modulus of elasticity Ec for concrete for values of wc, between 1500 and 2500 kg/ m3 may be taken as. Eq. 1-1

√

Where is the day 28-day compressive strength of concrete in MPa is the unit weight on concrete in . For normal weight concrete, √ Modulus of elasticity Es for nonprestressed reinforced may be taken as 200,000 MPa.

DETAILS OF REINFORCEMENT STANDARD HOOKS Standard hooks refer to one of the following: 1. 180-degree bend plus extension but not less than 60 mm at free end of bar. 2. 90-degree bed plus extension at free end of bar. 3. For stirrups and tie hooks: a) 61 mm diameter bar and smaller, 90-degree bend plus extension at free end bar, or b) 20 and 25 mm diameter bar, 90-degree bend, plus extension at free end of bar, or c) 25mm diameter bar and smaller, 135-degree bend d plus extension at free end of bar.

MINIMUM BEND DIAMETERS (SECTION 407.3) Diameter of bend measured on the inside of the bar, other than for stirrups and ties in sizes 10mm through 15 mm, shall not be less than the values in Table 1.1. Inside diameter of bend for stirrups and ties shall not be less than 16 mm bar and smaller. For bars larger than 16 mm, diameter of bend shall be in accordance with Table 1.1

Inside diameter of bend in welded wire fabric /9plain or deformed) for stirrups and ties shall not be less than for deformed wire larger than D56 and for all other wires. Bends with inside diameter of less than 8db shall not be less than from nearest welded intersection. Table 1.1- Minimum Diameters of Bend Bar Size

Minimum Diameter 10 mm to25 mm 28 mm, 32 mm, and 36 mm

PLAIN REINFORCEMENT (407.6) Reinforcement, prestressing tendons, and ducts shall not be accurately placed and adequately before concrete is placed, and shall be secured against displacement within tolerance permitted. Unless otherwise specified by the Engineer, reinforcement prestressing tendons, and prestressing ducts shall be placed within the following tolerances: Tolerance for depth d, and minimum concrete over a flexural walls and compression shall be as follows: Effective depth, d d d

Tolerance on d

Tolerance on minimum concrete cover -10 mm -12 mm

Except that tolerance for the clear distance to formed soffits shall be minus 6 mm and tolerance for cover shall not exceed minus 1/3 the minimum concrete cover required in the design drawings or specifications. Tolerance for longitudinal location of bends and ends of reinforcement shall be mm except at discontinuous ends of where tolerance shall be mm.

50

SPACING LIMITS FOR REINFORCEMENT According for Section 5.7.6 of NS, the minimum clear spacing between parallel bars in a layer should be db but not less than 25 mm. Where parallel reinforcement is placed in two or more layers, bars in the upper layers should be placed directly above bars in the bottom layer with clear distance between layers not less than 25mm. In spirally reinforced or tied reinforced compression , clear distance between longitudinal bars shall be not less than 1.5 db nor 40mm. In walls and slabs other than concrete joist construction, primary flexural reinforced shall be spaced not for farther apart than three times the wall or slab thickness, nor 450 mm. BUNDLED BARS Groups of parallel reinforcing bars bundled in to act as unit shall be limited to four in any one bundle. Bundled bars shall be enclosed within stirrups or ties and bars larger than 32 mm shall not be bundle in beams. The individual bars within the span of flexural should terminate at different points with at least 40 d b stagger. Since spacing limitations and minimum concrete cover of most are based on a single diameter db, bundled bars shall be treated as a single bar of a diameter derived from the equivalent total area.

Figure 1.1 Bundled-bar arrangement

Diameter of single bar equivalent to bundled bars according to NS to be used for spacing limitation and concrete cover.

=

3-25mm

(25)2 x 3

Equivalent diameter, D

D2

CONCRETE PROTECTION FOR REINFORCEMENT (SECTION 407.8.1) Cast-in –place Concrete (nonprestressed). The following minimum concrete cover shall be provided for reinforcement: Minimum cover, mm (a) Concrete cast against permanently exposed to earth 75

(b)

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

50 40

(C) Concrete not exposed to weather or in with ground: slabs, walls, joists: 32 mm bar and smaller Beams, columns Primary reinforcement, ties, stirrups, spirals Shells, folded place : 20 mm bar and larger 16 mm, Wr1 or D31 wire, and smaller

20 40 20 15

Precast concrete (Manufactured Under Plant Conditions).The Following minimum concrete shall be provided for reinforcement Minimum cover, mm (a)

(b)

Concrete exposed to earth or weather: Wall s: 20 32 mm bar and smaller Other : 40 20 mm through 32 mm bars 30 16 mm bar, W31 wire, and smaller Concrete not exposed to weather or in with ground: 15 slabs, walls, joists: db but not less 15, & 32 mm bar and smaller need not exceed 40 Beams, columns 10 Primary reinforcement Ties, stirrups, spirals Shells, folded plate : 20 mm bar and larger 16 mm, Wr1 or D31 wire, and smaller

15 10

Prestressed Concrete The following minimum concrete cover shall be provided for prestressed and nonprestressed reinforcement, ducts and end fittings. Minimum cover, mm (a)

Concrete cast against permanently exposed to earth

75

(b)

Concrete exposed to earth or weather: Wall s, slabs joists other

25 40

(C)

Concrete not exposed to weather or in with ground: 20 slabs, walls, joists: Beams, columns: 40 Primary reinforcement, 25 Ties, stirrups, spirals Shells, folded plate : 10 16 mm, Wr1 or D31 wire, and smaller db but not less than Other Reinforcement 20

Bundled Bars For bundled bars, the minimum concrete cover shall be equal to the equivalent diameter of the bundle, but need to be greater than 50 mm, except for concrete cast against and permanently exposed to earth, the minimum cover shall be 75 mm.

SHRINKAGE AND TEMPERATURE REINFORCEMENT (2010 NS) Shrinkage and temperature reinforcement is required at right angles to the principles reinforcement to minimize cracking and to tie the structure together to ensure its acting as assumed in the design. The provisions of this section are intended for structural slabs only; they are not intended for soil-ed slabs on grade. Reinforcement for shrinkage and temperature stresses normal to flexural reinforcement shall be provided in structural slabs where the flexural reinforcement extends in one direction only.

Shrinkage and temperature reinforcement shall be provided in accordance with either of the following: a) Where shrinkage and temperature movements are significantly restrained, the requirements of 408.3.4 and 408.3.3 shall be considered. b) Deformed reinforcement conforming to 43.6.3 used for shrinkage and temperature reinforcement shall be provided in accordance with the following: Areas of shrinkage and temperature reinforced shall be provided at least the following rations of reinforcement area to gross concrete area, but no less than 0.014: a) Slabs where Grade 280 or 350 deformed bars are used…….0.0020 b) Slabs where Grade 420 deformed bars or welded wire reinforcement are used………………………………………………………………...0.0018 c) Slabs where reinforcement with stress exceeding 420 MPa measured at a yield strain of 0.35 percent is used…………………………………………………….....…..0.0018x415/ Shrinkage and temperature reinforcement shall be spaced not farther apart than five times the slab thickness, nor farther apart than 450 mm.

LOADS

The most important and most critical task of an engineer is the determination of the loads that can be applied to a structure during its life, and the worst possible combination of these loads that might occur simultaneously. Loads on structure may be classified as dead loads or live loads. DEAD LOAD Dead loads are loads of constant magnitude that remain in one position. This consists mainly of the weight of the structure and other permanent attachments to the frame . LIVE LOAD Live loads are loads that may change in magnitude and position. Live loads that move under their own power called moving loads. Other Live loads are those caused by wind, rain, earthquakes, soils, and temperature changes. Wind and earthquake loads are called lateral loads. ARRENGMENTS OF LIVE LOAD Live loads may be applied only to the floor or roof under consideration, and the far ends of columns built integrally with the structure may be considered fixed. It is permitted by the code to assume the following arrangement of live loads: (a) Factored dead load on all spans with full factored live load on two adjacent spans, and (b) Factored dead load on all spans with full factored live load on alternative spans. REQUIRED STRENGHT (FACTIRED LOAD), U Required strength U to resist dead load (D) and live load (L) shall be at least equal to: Eq. 1-2 U=1.4D + 1.7L If resistances to structural effects of a specified wind load W are included in design, the following combination of D, L, and W shall be investigated to determine the greatest required strength U: Eq. 1-3

U=0.75(1.4D + 1.7L + 1.7W)

Where load combinations shall be include both full value and zero value of L to determine the more severe condition, and Eq. 1-4

U=0.9D + 1.3W

But for any combination of D, L, and W, required strength U shall not be less than Eq.12 If resistance to specified earthquake loads of forces E is included in design, the following combinations of D, L and E shall be investigated to determine the greatest required strength U: Eq. 1-5

U=1.1D + 1.3L + 1.1E

Where load combinations shall included both full value and zero value of L to determine the more severe condition, and Eq. 1-6

U=0.9D + 1.1E

But for any combination of D, L, and E, required strength U shall not be less than Eq. 12 If resistance to earth pressure H is included in design, required strength U shall be at least equal to: Eq. 1-7

U=1.4D + 1.7L + 1.7 H

Except where D or L reduces the effect of H, 0.9D shall be substituted for 1.4D and zero value of L shall be used to determine the greatest required strength U. For any combination of D, L and H, required strength U shall not be less than. If resistance to loadings due to weight and pressure of fluids with well defined densities and controllable maximum heights F is included in design, such loading shall have a factor of 1.4 and to be added to all loading combinations that include live load. If resistance to impact effects is taken into in design, such effects shall be included with live load L.

Where structural effects T of differential settlement, creep, and shrinkage expansion of shrinkage-compensating concrete or temperature change may be significant in design, required strength U shall be equal to

Eq. 1-8

U=1.75(1.4D +1.4T + 1.7L)

But required strength U shall not be less than Eq. 1-9

U=1.4(D + T)

Estimations of differential settlement, creep, and shrinkage expansion of shrinkage compensating concrete or temperature change shall be based on a realistic assessment of such effects occurring in service.

STRENGTH REDUCTIONS FACTORS, φ (PHI) The design strength provided by a concrete member, its connections to other , and its cross sections, in of flexure, axial load, shear, and torsion shall be taken as the nominal strength multiplied by a strength reduction factor φ having following values. (a) Flexure without axial load………………………………… 0.90 (b) Axial tension, and axial tension with flexure…………… 0.90 (c)Axial tension and axial tension with flexure: 1. Spiral reinforcement…………………………………….. 0.75 2. The reinforcement & other reinforced ……... 0.75 (d) Shear and torsion…………………………………………….. 0.85 (e) Bearing on concrete………………………………………….. 0.70 (f) Post-tensioned anchorage zones…………………………… 0.85

ACI-318-05 (NS C101-10-210) Notations gross of concrete sections . For a hollow section, concrete only and does not include the area of the void(s)

is the area of the

area of shear reinforcement spacing, web width, or diameter of circular section, mm D = dead loads, or related internal moments and forces d = distance from extreme compression fiber to centroid of longitudinal tension reinforcement, mm E

= load effects of earthquake, or related internal moments and forces specified yield strength

of transverse reinforcement, MPa

F = loads due to weight and pressures of fluids with well-defined densities and controllable maximum heights, or related internal moments and forces. h

= overall thickness or height of member, mm

H = loads due to weight and pressure of soil water in soil, or other materials, or related internal moments and forces. L = live loads or related internal moments and forces. roof live loads or related internal moments and forces. factored moment at section, N-mm factored axial force normal to cross section occurring simultaneously with to be taken as positive for compression and negative for tension, N R = rain load, or related internal moments and forces.

or

;

T = cumulative effect of temperature, creep, shrinkage , differential settlement, and shrinkage-compensating concrete. U

= required strength to resist factored loads or related internal moments and forces, = nominal shear strength provided by concrete, N = nominal shear strength. = nominal shear strength provided by shear reinforcement N = factored shear force at section, N

W

= wind load, related internal moments and forces

= net tensile strain in extreme layer of longitudinal tension steel at nominal strength, excluding strains due to effective prestress, creep, shrinkage, and temperature = strength reduction factor = ratio of

to

CHAPTER 9 – STRENGTH AND SERVVICEABILITY REQUIREMENTS 9.1- GENERAL 9.1.1 Structures and structural shall be designed to have design strengths at all sections at least equal to the required strengths calculated for the factored loads and forces in such combinations as are stipulated in this code. 9.1.2 also shall meet all other requirements of this code to ensure adequate performance at service load levels. 9.1.3 Design of structures and structural using the load factor combinations and strength reduction factors of Appendix C shall be permitted. Use of load factor combinations from this chapter in conjunction with strength reduction factors of appendix C shall be permitted.

9.2 Required strength 9.2.1 Required strength U shall be at least to the effects of factored loads in Eq. (9-1) through (9-7). The effect of one or more loads not acting simultaneously shall be investigated.

U = 1.4 (D+F) U = 1.2(D+F+T) + 1.6(L+H) + 0.5( or R) U = 1.2D + 1.6( or R) + (1.0L or 0.8W) U = 1.2D + 1.6W + 1.0L + 0.5( or R) U = 1.2D + 1.0E+ 1.0L U = 0.9D + 1.6W+ 1.6H U = 0.9D + 1.0E+ 1.6H

(9-1) (9-2) (9-3) (9-4) (9-5) (9-6) (9-8)

Except as follows: a) The load factor on the live load L in Eq. (9-3) to (9-5) shall be permitted to be reduced to 0.5 except for garages, areas occupied as places of public assembly, and all where L is greater than 4.8N/ . b) Where wind load W has not been reduced by a directionality factor, it shall be permitted to use 1.3 W in Eq. (9-4) and (9-6). c) Where E, the load effects of earthquake, is based on service-level seismic forces, 1.4E shall be used in place of 1.0E Eq. (9-5) and (9-7). d) The load factor on H, loads due to weight and pressure of soil, water in soil or other materials, shall be set equal to zero in Eq. (9-6) and (9-7) if the structural action due to H counteracts that due to W or E. Where lateral earth pressure provides resistance to structural actions from other forces, it shall be not be included in H but shall be included in the design resistance. 9.2.2 If resistance to impact effects is taken into id design, such effects shall be included with L.

9.2.3 Estimations of differential settlement, creep, shrinkage, expansion of shrinkage-compensating concrete. or temperature change shall be based on a realistic assessment of such effects occurring in service. 9.2.4 If structure is in a flood zone, or is subjected to forces from atmospheric ice loads, the flood or ice loads and the appropriate load combinations of SEI/ASCE7 shall be used. 9.2.5 For post-tensioned anchorage zone design, a load factor of 1.2 shall be applied to the maximum prestressing steel jacking force.

9.3 Design strength 9.3.1 Design strength provided by a member, its connections to other , and its cross sections, in of flexure, axial load, shear and torsion, shall be taken as the nominal strength calculated in accordance with requirements and assumptions of this code, multiplied by the strength reduction factors in 9.3.2,9.3.4, and 9.3.5. 9.3.2 Strength reduction factor shall be as given in 9.3.2.1 through 9.3.2.7: 9.3.2.1 Tension-controlled sections as defined in 10.3.4………….0.90 (See also 9.3.2.7) 9.3.2.2 Compression-controlled sections, as defined 10.3.3: a) with spiral reinforcement conforming to 10.9.3…..0.70 b) Other reinforced ……………………………………..0.65 For sections in which the net tensile strain in the extreme tension steel at nominal strength is between the limits for compression-controlled and tension-controlled sections, shall be permitted to be linearly increase from that for compression-limit to 0.005.

Alternatively, when Appendix B is used, for in which does not exceed 415 MPa, with symmetric reinforcement, and with (d-d’)/h not less than 0.70, shall be permitted to be increased linearly to 0.90 as decreases from 0.10 to zero. For other reinforced , shall be permitted to be increased from 0.10 or , whichever is smaller, to zero. 9.3.2.3 –Shear and torsion…………………………………………0.75 9.3.2.4 – Bearing on concrete (except for post-tensioned and anchorage zones and struct-and-tie models)…………………….……………0.65

CHAPTER 1 Analysis and Design of Beam NOTAIONS AND SYMBOLS USED = depth of equivalent stress block, mm = area of tension reinforcement, mm2 = area of skin reinforcement per unit height in one side face, mm2/ m   = width of compression face of member, mm  = distance from extreme compression fiber to neutral axis, mm = distance from extreme compression fiber to centroid of tension reinforcement, mm =distance from extreme compression fiber to centroid of compression reinforcement, mm = thickness of concrete cover measured from extreme tension fiber to center of  bar or wire, mm =modulus of elasticity of concrete, MPa  = modulus of elasticity of steel 200,000 MPa  =specified compressive stress of concrete, MPa   =calculated stress in reinforcement at service loads, MPa  =specified yield strength of steel, MPa   =overall thickness of member, mm =moment of inertia of gross concrete section about centroidal axis, neglecting  reinforcement =moment of inertia of reinforcement about centroidal axis of member cross section

=nominal moment, N-mm =factored moment at section, N-mm =factor defined in Section 410.4 in Page 16 =strain in concrete (maximum = 0.003) =strain in steel below yield point = =strain in steel at yield point =ration of tension reinforcement =balance steel ratio =strength reduction factor



Ø

ASSUMPTION IN STRENGTH DESIGN IN FLEXURE (CODE SECTION 5.10.2) 1. Strain in reinforcement and concrete shall be based assumed directly proportional to the distance from the neutral axis. Expect for deep flexural with overall depth to clear span to ratio, h/L> 2/5 for continuous spans and h/L >4/5 for simple spans, a nonlinear distribution of strain shall be considered (See Sec. 5.10.7). 2. Maximum usable strain at extreme concrete compression fiber, equal to 0.003 3. For

below shall

,

be taken as

x

for

>

,

=

shall be assumed

.

4. Tensile strength of concrete shall be neglected in axial and flexural calculations. 5. Relationships between compressive stress distribution and concrete strain may be assumed rectangular, trapezoidal, parabolic, or any other from that result in prediction of strength in substantial agreement with results of comprehensive tests. 6. For rectangular distribution of stress: a) Concrete stress of 0.85 shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross-section and a straight line located parallel to the maximum compressive strain. b) Distance c from fiber of maximum strain to the neutral axis hall is measured in the direction perpendicular to N.A.

c) Factor shall be taken as 0.85 for ≤ 30 MPa and shall be reduced continuously at rate of 0.008 for each 1 MPa of strength in excess of 30 MPa, but shall not be taken less than 0.65. i.e i. ii.

For For

≤ 30 MPa, = 0.85 > 30 MPa, =0.85-0.008( -30) but not shall be less than 0.65

RECTANGULAR BEAM REINFORCED FOR TENSION ONLY (SINGLY REINFORCED) b

0.85

c

0.003

a

d

c

d-a/2

NA T= Stress Diagram

As

Strain Diagram

Figure 2.1: Stress and strain diagram for singly reinforced and rectangular beam Eq. 2-1 For For

[

Eq. 2-2

(

]

C=T 0.85

but shall not be less than 0.65

Multiplying Eq. 2-2 by d/d:

The term

is called the ratio of steel reinforcement and is denoted as .

Eq. 2-3 and

Eq. 2-4 Let Eq. 2-5

Nominal Moment Capacity: From the stress diagram in Figure 2.1:

Eq.2-6

Ultimate Moment Capacity (Design Strength):

Eq.2-7

Coefficient of Resistance Eq.2-8 Eq.2-9

Solving for an

in Eq. 2-8 and replacing it with,

, yields the following formula the

steel ratio : Eq.2-10

[

√

]

BALANCE DESIGN Balance design refers to a design so proportioned that the maximum stresses in concrete (with strain of 0.003) and steel and (with strain of

⁄ ) are reached simultaneously once he

ultimate load is reached, causing them to fail simultaneously. UNDERREINFORCED DESIGN Underreinforced design is a design in which the steel reinforced is lesser than what is required for balance condition . If the ultimate load is approached, the steel will begin to yield although the compression concrete is still understressed. If the load is further increased, the steel will continue to elongate, resulting in appreciable deflections and large visible crack in the tensile concrete. Failure under this condition is ductile and will give warning to the of the structure to decrease the load.

OVERREINFORCED DESIGN Overreinforced design is a design in which the steel reinforcement is more than what is required for balanced condition. If the beam is overreinforced, the steel will not 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. Overreinforced as well as balanced design should be avoided in concrete because of its brittle property, that is why the Code limits the tensile steel percentage (P max=0.75pb) to ensure underreinforced beam with ductile type of failure to give occupants warning before occurs.

BALANCED STEEL RATIO : In balanced condition, the concrete and steel yield simultaneously, In this condition, the strain in concrete reached is maximum usable value of and the strain in steel is =200,000 MPa.

By ratio and proportion in the triangle shown in Figure2.2:

⁄

Note:

⁄

where

Eq.2-11

But a =

c=c Eq. 2-12

Note: Eq. 2-12 is for singly reinforced rectangular sections only. Eq. 2-11 is applicable to nay shape. MAXIMUM STEEL REINFORCEMENT Section 410.4.3: For flexural and for subject to combined flexure and compressive axial load when the design axial load strength is less than the smaller of or , the ratio of reinforcement that would produce balance strain condition for the section under flexure without axial; load. For with compression reinforcement, the portion of equalized by compression reinforcement need not be reduced by the0.75 factor. Eq. 2-13 and Eq. 2-14

This limitation is to ensure that the steel reinforcement will yield first to ensure ductile failure.

MINIMUM REINFORCEMENT OF FLEXURAL 410.61 At very section of flexural where tensile reinforcement is required by analysis, the area provided shall not be less than that given by: Eq. 2-15 Eq.2-16

√

and not less than

410.62 For statically determinate T-section with flange in tension, the area be equal to or greater than the smaller value given either by: Eq. 2-17 or Eq. 2-15 with

shall

√

set equal to the width of the flange.

410.6.3 The requirements of Sections 410.6.1 and 410.6.2 need to be applied if at every section the area of the tensile reinforcement is at least one-third greater than that required by analysis. 410.6.4 For structural slabs and footings of uniform thickness, the minimum area of tensile reinforcement in the direction of span shall be the same as that required by Section 407.13 (Shrinking and Temperature Reinforcement). Maximum spacing of this reinforcement shall not exceed three times the thickness and 450 mm. The provision for minimum amount of reinforcement applies to beams, which for architectural and other reasons are much larger in cross-section than required by strength consideration. With a very small amount of tensile reinforcement, the computed moment strength as a reinforced concrete section computed from its modulus of rapture. Failure in such a case can be quite sudden.

STEPS IN DEG A SINGLY REINFORCED RECTANGULAR BEAM FOR FLEXURE: Note: The assumptions made in steps II, V,and VIII are the author’s recommendation based on his experience. I. II. III. IV. V.

Identify the values of the dead load and live load to be carried by the beam. (DL & LL) Approximate the weight of beam (DL) between 20% to 25% of (DL+LL).This weight is added to the de load. Compute the factored load and factored moment: Ex: factored Load =1.4 DL+1.7L Compute the factored moment to be resisted by the beam, Try a value of steel ratio from 0.5 but must not be less than . This value will provided enough alloance in the actual value of due to rounding-off of the number of bars to be used, for it not to exceed the maximum of 0.05ρb.

(

)

VI.

Compute the value of

VII.

Solve for

VIII.

Try ratio ( from d=15b to d=2b), and solve for d, (round-off this value to reasonable dimension). Check also the minimum thickness of beam required by the Code a given in Table 2.1 in page 36. After solving for d, substitute its value to Step VII, and solve for b. Compute the weight of the beam and compare it to the assumption made in Step II.

IX.

:

Solve for the required steel area and number of bars. Number of bars(diameter = D)

x number of bars =

STEPS IN COMPUTING THE REQUIRED TENSION STEEL AREA WITH KNOWN MOMENT NT AND OTHER PROPERTIES: I.

if if II.

OF A BEM

Solve for

(1-0.59 design as singly reinforced (Step II) design as doubly reinforced (Step III) Solve for :

[

√

]

III. Compression reinforcement is necessary. (See Chapter 3)

STEPS IN COPUTING OF A BEAM WITH KNOWN TENSION STEEL AREA AND OTHER BEAM PROPERTIES:

I.

Solve for :

II.

Check if steel yields by computing

III.

`

if ,steel yields, proceed to III if ,steel does not yield, proceed to step IV. Note: if ,the given is not adequate for the beam dimension.

IV.

Solve for

from the strain diagram: [Note:

=200,000MPa]

⁄

Eq. 2-18

[

]

T=C

but a=

Solve c by quadratic formula and solve for

and a:

(

)

(

)

or (

)

(

)

MINIMUM THICKNESS OF FLEXURAL According to Section 5.9.5 of NA, minimum thickness stipulated in Table 2.1 shall apply for one-way construction not ing are attached to portions or other construction likely to be damaged by large deflections, unless computation of deflection indicates a lesser thickness can be used without adverse effects.

Table 2.1 MINIMUM THICKNESS OF NON-PRESTRESSED BEAMS OR ONE-WAY SLABS UNLESS DEFLECTIONS ARE COMPUTED *

Member

Minimum thickness, h Simply One end Both ends Cantilever ed continuous continuous not ing or attached to partitions or other construction likely to be damaged by large deflections

Solid one-way L/20 slabs

L/24

L/28

L/10

Beams or L/16 ribbed one-way slabs

L/18.5

L/21

L/8

Span length L is in millimeters Values given shall be used directly for with normal density concrete ( ) and grade 415 reinforcement. For other conditions, the values shall be modified as follows: (a)

(b)

For structural lightweight concrete having weights in the range 1500-2000 values shall be multiplied by (1.65-0.005 ) but not less than 1.09, where the unit mass in . For other than 415 MPa, the values shall be multiplied by (0.4 +

is

BEAM DEFLECTION (SECTION 5.9.5 Sect. 5.9.5.2.2 Where deflections are to be computed, deflections that occur immediately on application of load shall be computed by usual methods or formulas for elastic deflections, considering effects of cracking and reinforcement on member stiffness. Sect. 5.9.5.2.3 Unless stiffness values are obtained by a more comprehensive analysis, immediate deflection shall be computed with the modulus of elasticity for concrete and with the effective moment of inertia as follows, but not greater than .

Eq.2-19

*

+

[

*

+3]

Where = = modulus of rapture of concrete, MPa, for normal weight √

Concrete

= maximum moment in member at stage deflections is computed. = moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement. = moment of inertia of cracked section transformed to concrete = distance from centroidal axis of gross section, neglecting reinforcement, to extreme fiber in tension. When Lightweight aggregate is used, one of the following modifications shall apply: (a)

When

is specified and concrete is proportioned in accordance with Sec. 5.5.2,

shall be modified by substituting 1.8 exceed√

for √

but the value of 1.8

shall not

.

(b) When is not specified, shall not be multiplied by 0.75 for “all lightweight” concrete, and 0.85 for “ sand-lightweight” concrete. Linear interpolation is permitted if partial sand replacement is used. Sect. 5.9.5.2.4: For continuous , effective moment of inertia may be taken as the average of values obtained from Eq. 2-19 for the critical positive and negative moment sections. For prismatic , effective moment of inertia may be taken as the value obtained from Eq. 2-19 at midspan for simple and continuous spans, and at the cantilevers. Sect.5.9.5.2.5: Unless values are obtained by a more comprehensive analysis, additional long-term deflection resulting from creep and shrinkage of flexural (normal weight or lightweight concrete) shall be determined by multiplying the immediate caused by the sustained load considered, by the factor. Eq. 2-10

Where shall be taken the value of reinforcement ratio for non-prestressed compression reinforcement at midspan for simple and continuous spans,a nd at for cantilevers. It is permitted to assume the time-dependent factor for sustained loads to be equal to: 5 years or more……………………2.0 12 months…………………………...1.4 6 months……………………………..1.2 3 months………………………………1.0

Deflection computed in accordance with Sec. 5.9.5.2.2 through Sec.5.9.5.2.5 shall not exceed limits stipulated in Table 2.2.

Table 2.2: Maximum Permissible Computed Deflections

Type of member

Deflection to be considered

Deflection limitation

Flat roofs not ing or attached to nonstructural elements likely to be damage by large deflections Floors not ing or attached to nonstructural elements likely to be damaged by large deflections Roof or floor construction ing, or attached to nonstructural elements not likely to be damaged by large deflections Roof or floor construction ing, or attached to nonstructural elements not likely to be damaged by large deflections

Immediate deflection due to L/180* live load LL

Immediate deflection due to L/360* live load LL

That part of the total L/480** deflection occurring after attachment of non structural elements (sum of the longtime deflection due to all sustained loads and the L/20**** immediate deflection due to any additional live load)****



 



Limit not intended to safeguard against ponding. Ponding should be cheated by suitable calculations of deflections, including added deflections due to ponded water and considering long-term effects of all sustained loads, camber, construction tolerances, and reliability of provisions for damage. Limit may be exceeded if adequate measures are taken to prevent damage to ed or attached elements. Long=time deflections shall be determined in accordance with Sec.5.9.5.2.5 or Sec. Attachment of nonstructural elements. This amount shall be determined on basis of accepted engineering, data relating to time-deflection characteristics of similar to those being considered. But not greater than tolerance provided for nonstructural elements. Limit may be exceeded if camber is provided so that deflection minus camber does not exceeded limit.

NS COEFFICICIENTS FOR CONTINUOUS BEAMS AND SLASBS Section 5.8.3.3 of NS states that in lieu of frame analysis, the following approximate moment and shears are permitted for design of continuous beams and one-way slabs (slabs reinforced to resist flexural stresses in only one direction), provided: a) There are two or more spans, b) Spans are approximately equal, with the larger of two adjacent spans not greater than uniformly than the shorter by more than 20 percent, c) Loads are uniformly distributed, d) Unit live does not exceeded three times unit dead load, and e) are prismatic. Positive moment End spans Discontinuous end unrestrained…………………… Discontinuous end integral with ………….. Interior spans………………………………… Negative moment at exterior face of first interior Two spans ………………………………………………………….....

More than two spans………………………………………………..... Negative moment at other faces of interior s………….…… Negative moment at face of all s for: Slabs with spans not exceeding 3 m; and beams Where ratio of sum of column stiffness to beams Stiffness exceeds eight at each end of the span……………………

Negative moment at interior face of exterior built integrally with Where is a spandrel beam…………………………… When is a column…………………………….……..…. Shear in end at face of First interior ……………………………..................................... Shear at face of all other s………………………………………..

When =clear span positive moment or shear and average of adjacent clear spans for negative moment.

Figure 2.3: Shear and moment for continuous beam or slab with spans and discontinuous end integral with

Figure 2.5 Shear and moment for continuous beam or slab with more than two spans and discontinuous end unrestrained

ACI-318-05 (NS C101-10-2010) 10.2 Design assumptions (410.3) 10.2.1 Strength design of for flexure and axial loads shall be based on assumptions given in 10.2.2 through 10.2.7, and on satisfaction of applicable conditions of equilibrium and compatibility of strains. 10.2.2 Strain in reinforcement and concrete shall be assumed directly proportional to the distance from the neutral axis, except that, for deep beams as defined in 10.7.1, an analysis that considers a nonlinear distribution of strain shall be used alternatively, it shall be permitted to use a struct-and tie model. See 10.7,118, and Appendix A. 10.2.3 Maximum usable strain at extreme concrete compression fiber shall be assumed equal to 0.003. 10.2.4 Stress in reinforcement below shall be taken as times steel strain. For strains greater than that corresponding to , stress in reinforcement shall be considered independent of strain and equal to . 10.2.5 Tensile strength of concrete shall be neglected in axial and flexural calculations of reinforced concrete, except when meeting requirements of 18.4. 10.2.6 The relationship between concrete compressive stress distribution and concrete strain shall be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction of strength in substantial agreement with results of comprehensive tests. 10.2.7 Requirements of 10.2.6 are satisfied by an equivalent rectangular concrete stress distribution defined by the following: 10.2.7.1 Concrete stress of 0.85 shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross section and a

straight line located parallel to the neutral axis at distance a= maximum compressive strain.

form the fiber of

10.2.7.2 Distance from the fiber of maximum strain to the neutral axis, c , shall be measured in direction perpendicular to the neutral axis.

10.2.7.3 For between 17 and 18 MPa, above 28 MPa, shall not be taken less than 0.65

shall be taken as 0.85. For

10.3 General principles and requirements (410.4) 10.3.1 Design of cross sections subject to flexure or axial loads, or to combined flexure and axial loads, shall be based on stress and strain compatibility using assumptions in10.2.

10.3.2 Balanced strain conditions exist at a cross section when tension reinforcement reaches the strain corresponding to just as concrete in compression reaches its assumed ultimate strain of 0.003. 10.3.3 Sections are compression-controlled if the next tensile strain in the extreme tension steel, , is equal to or less than the compression-controlled strain limit when the concrete in reaches its assumed strain limit of 0.003. The compressioncontrolled strain limit is the net tensile strain in the reinforcement at balanced strain conditions. For Grade 415 reinforcement, and for all prestressed reinforcement, it shall be permitted to set the compression-controlled strain limit equal to 0.002. 10.3.4 Sections are tension-controlled if the net tensile strain in the extreme tension steel is equal to greater than 0.005 when the concrete in compression reaches its assumed strain limit of 0.003. Sections with between the compressioncontrolled strain limit and 0.005 constitute a transition region between compressioncontrolled and tension-controlled sections. Derivation: for E = 200 GPa The beam is tension-controlled When = 0.005 (or =1000MPa)

Eq. 2-21

For rectangular beam: ( )

Eq. 2-22

10.3.5 For nonprestressed flexural and nonprestressed with factored axial compressive load less than 0.10 steel strain at nominal strength shall not be less than 0.004. 10.3.5.1 Use of compression reinforcement shall be permitted in conjunction with additional tension reinforcement to increase the strength of flexural . Derivation: for E =200 GPa

Maximum steel area and

when beam is singly reinforced:

Eq. 2-23

For rectangular section: T= )b

Eq. 2-14

(

Eq. 2-25

)

)

Eq. 2-26

10.3.6 Design axial strength of compression shall not be taken greater than computed by Eq. (10-1) or (10-2). 10.3.6.1 For nonprestressed with spiral reinforcement conforming to 7.10.4 or composite conforming to 10.16: [ ] (10-1) 10.3.6.2 For non nonprestressed with spiral reinforcement conforming to 7.10.5: [ ] (10-2)

10.3.6.3 For prestressed , design axial shall not be taken greater than 0.85 (for with spiral reinforcement) or 0.80 (for with tie reinforcement) of the design axial strength at zero capacity .

10.3.7 subject to compressive axial load shall be designed for the maximum moment that can accompany the axial load. The factored axial force at given eccentricity shall not exceed that given in 10.3.6. The maximum factored moment shall be magnified for slenderness effects in accordance with 10.10 10.4 Distance between lateral s of flexural

10.4.1 Spacing of lateral s for a beam shall not exceed 50 times b, the least width of compression flange or face. 10.4.2 Effects of lateral eccentricity of load shall be taken into in determining spacing of lateral s.

10.5.1 Minimum reinforcement of flexural 10.5.1 At every section of flexural where tensile reinforcement is required by analysis, except as provided in 10.5.2, 10.5.3, and 10.5.4, as provided shall not be less than that given by √

√

(10-3)

and not less than (10-3)

10.5.2 –For statically determinate with a flange in tension, shall not be less than the value given by eq. (10-3), except that is replaced by either or the width of the flange, whichever is smaller. STEPS IN THE DESIGN OF SINGLY REINFORCED RECTANGULAR BEAM FOR FLEXURE Note: The assumption made in steps II, V, and VIII are the authors recommendation based on his experience. I. Determine the values of loads, Dl, LL and other loads II. Approximate the weight of beam (DL) as follows: Small beams: 2kN/m Medium-sized beams: 3.5kN/m Large-sixed beams: 7kN/m or Weight of beam in kN/m=24kN/ x beam area in III. Compute the factored load on different load combinations Example: Factored Load =1.2 DL + 1.6 LL

IV. V.

Compute the factored moment to be resisted by the beam, Try a value of steel ratio from 0.7 to 0.8 but must not be less than . This value of will provided enough allowance in the actual value of due to rounding-off the numbers bars to be used so that it will not exceed the maximum .

( (

) )

√ VI.

Compute the value of

VII.

Solve for the reduction factor Solve for c: Note: For singly reinforced rectangular beam, proportional to c:

is directly

c=(assumed factor) x The assumed factor may range from 0.7 to 0.8 as suggested in step V.

if

if VIII.

IX.

Solve for

:

Try a ratio d/b (from d= 1.5b to d=2b), and solve for d. (round-off this value to reasonable dimension) Check also the minimum thickness of beam required by the code as given in Table 2.1 in Page 26. After solving for d, substitute its value to Step VII, and solve for b. Compute the weight of the beam and it to the assumption made in Step II.

X.

Solve for the required steel area and number of bars. Number of bars (diameter=D) x number of bars =

STEPS IN FINDING THE REQUIRED TENSION STEEL AREA OF A BEAM WITH KNOW REQUIRE MOMENT AND OTHER BEAM PROPERTIES Given b, d,

I.

Solve for

and

: and

.

(

)

if if

design as Singly Reinforced (Step II) design as Doubly Reinforced (Step V)

II.

Determine if the section in tension-controlled or transition From Eq. 2-11: if ,proceed to step III if region, proceed to step IV

III. √ √

IV. √

Solve for c and

*

√

+

if

V. Compression reinforcement is necessary.(See chapter 2)

STEPS IN FINDING OF A BEAM WITH KNOWN TENSION STEEL AREA AND OTHER BEAM PROPERTIES: Given: b, d, I. II.

,

,

:

Solve for Check if steel yields by computing ( ) if steel yields, proceed to step III if steel dos not yield, proceed to step IV. Note: if the given is not adequate for the beam dimension. √

III. Solve for : ;c=

=_________

if if

IV. Compression-controlled b

0.85 a

.

c=

d d-a/2

T=C but a=

c=__________ ; a=

T=

=__________

(

)

(

)

or

ILLUSTRATIVE PROBLEMS DESIGN PROBLEMS PROBLEM 2.1 A reinforced concrete rectangular beam 300 mm wide has an effective depth of 460 mm and is reinforced for tension only. Assuming and , determine the balance steel area in sq.mm.

SOLUTION

PROBLEM 2.2 A rectangular beam has b = 300 mm and d =490 mm. Concrete compressive strength and steel yield strength . Calculate the required tension steel area if the factored moment is (a) 20 kN-m,(b)140 kN-m,(c) 485 kN-m, and (d)620 kN-m.

SOLUTION Solve for

[

]

a) 20 x

*

√

*

√

+

+

√

b)

(singly reinforced) 140 x 1

*

√

+

*

c)

√

+

(singly reinforced) 485 x

[

√

[

√

] ]

d) The beam will be doubly reinforced. See Chapter 3.

PROBLEM 2.3 (CE MAY 2012)

A reinforced concrete beam has a width of 300 mm and an overall depth of 480 mm. The beam is simply ed over span of 5 m. Steel strength MPa and concrete . Concrete cover is 70 mm from the centroid of the steel area. Unit weight concrete is 23.5kN/ .Other than the weight of the beam , the beam carries a superimposed dead of 18 kN/m and a live load of 14 kN/m. Use the strength design method. a) Determine the maximum factored moment on the beam. b) If the design ultimate moment capacity of the beam is 280 kN-m, determine the required number of 20 mm tension bars. c) If the beam will carry a factored load of 240 kN at midsoan, determine the required number of 20 mm tension bars.

SOLUTION Given: b=300m d=480-70=410 mm

Bar diameter , Weight of beam, a) Maximum factored moment on the beam. Factored load, Factored load, Maximum factored moment:

b) Solve for

to determine whether compression steel is needed

Required

<

(singly reinforced)

[ *

√

N=6.9 say 7 bars

3. (weight of beam)

] +

√

*

*

+

√

+

2498 =

N

PROBLEM 2.4 (CE MAY 1993) A reinforced concrete beam has a width of 300 mm and an effective depth to tension bars of 600 mm. compression reinforcement if needed will be placed at a depth of 60 mm below the top. If and , determine the tension steel area if the beam is to resist an ultimate moment of 650 kN-m. SOLUTION Solve for

and

:

[1-0.59(0.309) Since

, the beam may be designed as singly reinforced. 650 x 1

Solve for : √

(

*

√

)

+

PROBLEM 2.5 (CE November 2000) A rectangular concrete beam has a width of 300 mm and an effective depth of 550 mm. The beam is simply ed over a span 6 m and is used to carry a uniform dead load of 25 kN/m and a uniform live load of 40 kN/m. Assume and . Compression reinforcement if necessary shall be placed at a depth 80 mm from the outermost compression concrete. a) Determine 80 mm from the outermost compression concrete. b) Determine the required tension steel area.

c) Determine the required number of 25-mm tension bars. SOLUTION a) Maximum steel area:

9

b) Required tension steel area: Factored load:

Required strength:

=463.5kN-m Solve for

[

]

√

(

*

√

)

+

c) Number of 25 mm bars: Number of 25-mm bars= Number of 25-mm bars= PROBLEM 2.6 (CE MAY 2009) A reinforced concrete beam has a width of 300 mm and total depth of 600 mm. The beam will be design to carry a factored moment of 540kN-m. Concrete strength and steel yield strength . Solve using the strength design method. a) Determine the balanced steel ratio in percent. b) Determine the minimum effective depth of the beam using a steel ratio equal to 0.5 of balanced steel ratio. c) Determine the minimum effective depth of the beam using the maximum allowable steel ratio.

SOLUTION Given: b=300 mm h=600 mm

a) Balanced steel ratio:

(

)

b) Effective depth using

[

540 x 1

]

PROBLEM 2.7 A concrete one-way slab has a total thickness of 120 mm. The slab will be reinforced with 12-mm-diameter bars with .Concrete strength . Determine the required spacing 12 mm main bar if the total factored moment acting on 1-m width of slab is 23 kN-m width of slab is 23 kN-m. Clear concrete cover is 20 mm. SOLUTION Note: Slabs are practically singly reinforced because of its small depths.

.

12mm bars

.

s

d

s

cover=20 mm

b = 1000 mm

Effective depth, d= 120 -20-1/2(12)=94 mm Width, b = 1000 mm 23 x

(

√

)

h=120 mm

(

√

)

Spacing of bars (for walls and slabs using unit width):

Eq. 2-17

PROBLEM 2.8 A 2.8 m square column fooring has a total thickness of 47 mm. The factored moment at critical section for moment is 640 kN-m. Assume and . Clear concrete cover is 75 mm. Determine the required number of 20 mm tension bars.

SOLUTION Effective depth, d=470-75-1/2(20)=385 mm Width, b =2800 mm Design strength, Maximum and minimum requirements:

(Procedure is not shown anymore see Problem 2.2)

Singly reinforced:

(

√

Number of 20 mm bars:

)

(

√

)

PROBLEM 2.9 Design a rectangular beam reinforced for tension only to carry a dead load moment of 60 kN-m (including its own weight) and a live load moment of 48 kN- m. Use and SOLUTION Required strength:

(Note: this already includes the weight of beam)

Try

Note: this is the author’s suggestion

[

165.6 x

]

Try d = 1.75 b

b=228 mm say 230 mm d=399 say 30 mm

Summary: b = 230 mm d = 400 mm

PROBLEM 2.10 Design a singly reinforced rectangular beam for a 6-m simple span to a superimposed dead load of 29 kN/m and a live load of 44 kN/m. Assume normal weigth oncrete with . Use

SOLUTION Weight of beam: (this is the author’s assumption) Assuming a 300 mm x 600 mm, .7(44) .

√

[

Assume d = 1.75 b

]

(this is the author’s assumption) 546.516 x

Use b = 280 mm, d = 490 mm Minimum beam the thickness (Section 409.6.2.1) (

)

Using 32 mm bars (#100):

(

)

280 mm

.

h

. 6 - #10

Beam weight = 24 (0.28)(0.5545) Beam weight = 3.73 kN/m < 4.32(OK)

PROBLEM 2.11 A propped cantilever beam shown in Figure 2.6 is made of reinforced concrete having a width of 290 mm overall depth of 490 mm. The beam is loaded with uniform dead load of 35 kN/m (including its own weight), and a uniform live load of 55 kN/m. Given Concrete cover is 60 mm from the centroid of the bars. Determine the required tension steel area for maximum positive moment. Assume EI=constant.

290mm

490 mm A

6m

B

2m

C

Figure 2.6 SOLUTION Given:

D

O

A

B

2m C

x R

Moment Diagram

Solve for moment reactions using the three-moment equation:

Mo Lo +

̅̅̅̅

̅̅̅

-489.75 = R(6)- 142.5(8)(4) R=676.875 kN

Maximum positive moment:

142.5(2 + x) - 676.875 = 0 x = 2.75 m

Solve for

:

[

]

At a point of maximum positive moment: (Singly reinforced) 253.828 x

√

*

*

√

+

+

ANALYSIS OF RECTANGULAR STEEL YIELDS (

BEAMS

WHERE

PROBLEM 2.12(CE MAY 1999) A reinforced concrete rectangular beam with b = 400 mm and d= 720 mm is reinforced for tension only with 6-25 mm diameter bars. If and a) The coefficient of resistance of the beam. b) The ultimate moment capacity of the beam. SOLUTION

[

] Answer

Answer

PROBLEM 2.13 A rectangular beam reinforced for tension only has b= 300 m, d = 490 mm. The tension steel area provided is 4,500 sq. mm. Determine the ultimate moment capcity of the beam in kN-m. Assume ,

SOLUTION

[

PROBLEM 2.14 A rectangular beam has b = 300 mm, d = 500 mm, grade 60 reinforcement ( Calculate the design moment

]

SOLUTION

Check if the beam satisfies the minimum requirement: √

[

]

PROBLEM 2.15 A 130-mm-thick-one-way slab is reinforced with 12-mm-diameter tension bars spaced at 110 on centers. Concrete cover is 20 mm, concrete strength MPa and steel yield strength . Unit weight of concrete is 23.5 kN/ . a) What is the ultimate moment capacity of the slab? b) If the slab is simply ed over a span of 4 m, what safe uniform live load pressure can the slab carry?

SOLUTION a) Consider 1 m width of slab, b = 1000 mm Effective depth: d = h – cover- 1/2 d = 130-20-1/2(12)=104 mm

Check if the beam satisfies the minimum steel requirement on flexures:

]

b)

Dead load pressure, Dead load pressure,

x thickness of concrete.

PROBLEM 2.16 A rectangular beam with b = 250 mm and d = 460 m is reinforced for tension only with 3-25 mm bars. The beam is simply ed over a span of 6 m and carries a uniform dead load of 680 N/m including its own weight. Calculate the uniform live load that the beam can carry. Assume and . SOLUTION

)

Check if the beam satisfies the minimum steel requirement on flexure:

[

]

PROBLEM 2.17 (CE JANUARY 2008) A reinforced concrete rectangular beam has a width of 300 mm and an effective depth of 55 mm. The beam is reinforced with six 25-mm-diameter tension bars. Steel yield is 415 MPa and concrete strength is 28 MPa. a) What is the balanced steel ratio? b) What is the maximum steel area for singly reinforced? c) What is the nominal moment capacity of the beam?

SOLUTION

a) Balanced steel ratio:

b) Maximum steel area

c) Nominal moment capacity Using 6-25 mm bars:

PROBLEM 2.18 A 350 mm x 500 mm rectangular is reinforced for tension only with 5-28 mm 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 concentrated dead load of P and 2P as shown in Figure 2.7. Assume . Calculate the following: a) The ultimate moment capacity of the section in kN-m, and b) The maximum value of P in kN.

2P

2m

Figure 2.7

SOLUTION

P

2m

2m

Check if the beam satisfies the minimum requirement: √

[

1.4(2P)

1.4P

]

A

B 2m

C 2m

D 2m

Figure 2.8 – Beam with factored loads For the given loads, the maximum moment can occur at B or C: At point C: Set

440.18 = 1.4P(2) + 11.4(2)(1)

At point B: (First solve for ∑

∑ Set

Thus the maximum value of P such that

will not exceed 440.18 kN-m is 149 kN.

ANALYSIS OF RECTANGULAR BEAMS STEEL DOES NOT YIELDS ( )

WHERE

PROBLEM 2.19 A rectangular beam has b = 300 mm, d = 500 mm, grade 60 reinforcement ( Calculate the ultimate moment capacity of the beam. SOLUTION

From Eq. 2-18

∑

(

)

PROBLEM 2.20

A rectangular beam reinforced for tension only has b=300 mm, d = 490 mm. The tension steel area provided is 7-25 mm diameter bars with . Calculate the ultimate moment capacity of the beam.

SOLUTION

From Eq.2-18:

∑

(

)

(

)

ANALYSIS & DESIGN OF SINGLY REINFORCED NON-RECTANGULAR BEAMS PROBLEM 2.21 Compute the ultimate moment capacity of the beam shown in Figure 2.9. Assume and .

SOLUTION Note: This is not a rectangular beam. Some formulas derived above (such as , ) may not be applicable. The moment can be computed using the assumptions in the Code and the conditions of equilibrium.

Solve for the balanced

to determine whether the given steel yield or not.

From Eq. 2-11

Since

<

, tension steel yields.

(

)

PROBLEM 2.22 Compute the ultimate moment capacity of the beam shown in Figure 2.10. Assume and .

SOLUTION

Solve for

Since solve for c:

, tension steel does not yield (

0.85

(

)

(

) *

+

PROBLEM 2.23 A hallow beam is shown in Figure 2.11. Assume and . a) Calculate the required tension steel area when . b) What is the balanced moment capacity of the beam? c) What is the maximum steel area under singly reinforced condition? d) What is the maximum design moment strength under singly reinforced condition? e) Calculate the required tension steel area when .

Figure 2.11-Hallow beam

SOLUTION To guide us whether “a: will exceed 150 mm or not, let us solve the design moment when a=150 mm. d = 800 – 75 = 725 mm (

)

a) Since the required Assuming tension steel yields:

Check is steel yields:

(

)

b) Balanced condition (See Figure 2.12)

) [

]

Figure 2.12 c) Maximum steel area,

d) Maximum moment , Refer to Figure 2.12: [

[

e) Refer to Figure 2.12

]

]

[

]

BEAM DEFLECTION PROBLEM PROBLEM 2.24 A reinforced concrete beam is 350 mm wide and 600 mm deep. The beam is simply ed over a span of 8 m and carries a uniform dead load of 11 kN/m including its own weight and a uniform live load of 15 kN/m. The beam is reinforced tension bars of 530 mm. Modulus of elasticity of concrete and a) Calculate the maximum instantaneous deflection due to service loads. b) Calculate the deflection for the same loads after five years assuming that 40% of the live load is sustained. SOLUTION

Figure 2.13

Effective moment of inertia,

(

)

*

(

Eq. 2-19

) +

Moment of inertia of cracked section with steel transformed to concrete From Figure 2.13: Modular ratio,

Solve for c: Moment of area above N.A. = Moment of area below N.A. 350 x c x c/2 = 27,208(350-c) c = 219.7 mm

(

(

)

*

)

(

) +

*

(

) +

a) Instantaneous Deflection:

b) Long-term Deflection Since only 40% of the live load was sustained: w = 11 + 0.4(15) = 17 kN/m

Instantaneous deflection

Note: Since deflections are directly proportional to the load, the instantaneous deflection due to sustained load can be found by ratio and proportion using the result in Part”a”.

Long-term deflection =

Long-term deflection = 16.36 + 2(10.7) Long-term deflection = 37.76 mm

PROBLEM 2.25 (CE NOVEMBER 2002) The continuous reinforced concrete beam shown in Figure 2.14 is subjected to a uniform service dead load of 16 k/m and a service live load of 32 kN/m,resulting in the bending moment diagram shown. Twenty percent of the live load will be sustained in nature, while 80% will be applied only intermittently. The concrete strength The modulus of elasticity of concrete is given by the expression and the modulus of rapture is given by the expression . Determine the following: a) The effective moment of inertia at the s (maximum negative moment). b) The effective moment of inertia for the continuous member. c) The additional deflection (in addition to the initial deflection) after 5 years, under the sustained loading if the instantaneous deflection due to the combined service dead and live load is 5 mm.

Figure 2.14

SOLUTION √ √

√ √

a) Effective moment of inertia at the s Maximum moment, Distance from NA of gross section to extreme tension fiber,

Moment of inertia of gross section, Moment of inertia of cracked section,

(

)

(

*

)

(

) +

*

(

) +

b) Effective moment of inertia for the continuous member

At maximum negative moment (at )

Solving for

at maximum positive moment (at midspan)

(

)

*

(

) +

c) Additional long term deflection= long term deflection x

Solving for the instantaneous deflection under sustained loading: Instantaneous deflection = 5mm (given) Instantaneous loading = 16 kN/m + 32 kN/m Instantaneous loading = 48 kN/m Sustained loading = 16 + 20%(32)

Sustained loading = 22.4 kN/m Sine deflection is directly proportional to the load:

Additional long term deflection = 2.333 x =2.333 x 2 Additional long term deflection = 4.67 mm

ONE-WAY SLAB Reinforced concrete design slabs are large flat plates that are ed at its sides by reinforced concrete beams, walls, columns, steel beams, or by the ground. If a slab is ed on two opposite sides only, they are referred to a one-way slabs since the bending occurs in one direction only. If the slab is ed on all four sides, it is called two-way slab since the bending occurs in both direction. If a rectangular slab is ed in all four sides but the long is two or more times the short side, the slab will, for all practical purposes, act as one way slab, with bending occurring in the short direction.

A one-way slab is considered as a wide, swallow, rectangular beam. The reinforcing steel is usually spaced uniformly over its width. One way-way slabs are analyzed by considering one-meter strip, which is assumed independent of the adjacent strips. This method of analysis is somewhat conservative because we neglect the lateral restraint provided by the adjacent strips. MAXIMUM SPACING OF REINFORCEMENT According to Section 407.7.5, the flexural reinforcement shall not be spaced farther apart than 3 times the slab thickness, nor 450 mm. SHRINKAGE AND TEMPERATURE REINFORCEMENT, Concrete shrinks as it hardens. In addition, temperature changes occur that causes expansion and construction of concrete. In this effect, the code (407.13) requires that one-way slab, where flexural reinforcement extends in one direction only, should be reinforced for shrinkage and temperature stresses perpendicular to flexural reinforcement. According to Section 407.132.2.1, the area of shrinkage reinforcement shall provide at least the following ratios of gross concrete area bh, (where h is the slab thickness) but not less than 0.0014. a) Where Grades 230 & 275 deformed bars are used………………..0.0020 b) Where Grade 415 deformed bars or welded wire fabric (plain or deformed ) are used…………………………………..0.0018 c) Where reinforcement with measured at yield strain of 0.35% are used………………………………………….

Shrinkage and temperature reinforcement may not be spaced not farther apart than 5 times the slab thickness, nor 450 mm (Section 407.13.2.2).

STEPS IN THE DESIGN OF ONE-WAY SLABS (FLEXURE) I. Identify the uniform floor pressure (Pa) to be carried by the slab. This load may consist of: 1) Live load pressure 2) Dead load pressure 3) Ceiling load and other attachments below the slab II. Determine the minimum slab thickness “h” from Table 2.1. If necessary adjust this value depending on your judgment. III. Compute the weight of slab (Pa) Weight = IV. Calculate the factored moment ( to be carried by the slab. Uniform load, . V. Compute the effective depth, d: d=h-covering (usually 20 mm)-1/2 (main bar diameter) VI. Compute the required steel ratio : Solve for from *

VII.

√

+

Solve for If is less than and greater than , use If is greater than , increase the depth of slab to ensure ductile failure If is less than Compute the required main bar spacing.

Spacing, Use the smallest of the following for the main bar spacing: a) b) c) 450 mm

VIII.

Temperature bars: See Page 81 for the required steel ratio,

Use the smallest of the following for temperature bar spacing: a) b) c) 450 mm

ILLUSTRATIVE PROBLEMS Problem 2.36 Design a one-way slab having a simple span 3 m. The slab is to carry a uniform live load of 7,500 Pa. Assume and for main and temperature bars. The slab is not exposed to earth or weather. Use unit weight of concrete . SOLUTION Consider 1 m strip of slab, b= 1000 m Uniform live load,

Minimum slab thickness from Table 2.1: (

)

(

)

Effective depth:

d = 120-20 mm (covering)-1/2 bar diameter (12mm) d=94 mm

Weight of slab:

Factored floor pressure load:

*

+

√

Check for

and

*

√

:

per meter width of slab Using 12-mm main bars: Spacing s =

+

Maximum spacing required by the Code: a) b)

Thus, use 12 mm main bars at 135 mm o.c. Temperature bars: (Grade 275)

Spacing =

Maximum spacing required by the Code: a) b) 450 mm OK

Thus, use 10 mm temperature bars at 325 mm o.c.

PROBLEM 2.27 Design a one-way slab to carry a service live load of 4000 Pa. The slab has a length of 4m with both ends continuous. Assume and for main bars and for temperature bars. Steel cover is 20 mm. Unit weight of concrete is 23.5 kN/ .

SOLUTION Consider 1 m strip, b = 1000 mm Uniform live load,

Minimum slab thickness from Table 2.1:

Weight of beam (DL):

Maximum factored moment, Section 408.4 (See Page 29) LL < 3 DL

Effective depth, d = 1.50 – 20 – 1/2 (12) Effective depth, d = 124 mm At midspan:

*

√

+

*

√

+

Use

Spacing, s =

Maximum spacing required by the Code: a) b) 450 mm

Thus, use 12 mm bottom bars at 270 mm o.c. at midspan

At :

*

√

+

*

√

+

Use

Spacing,

Thus, use 12 mm top bars @ 265 mm o.c. at Temperature bars (10 mm): (

Spacing, s =

Maximum spacing required by the Code: a) b) 450 mm

Thus, use 10 mm temperature bars @ 260 mm o.c.

PROBLEM 2.28

A one-way slab having a simple span of 3 m is 160 mm thick. The slab is reinforced with 12 mm tension bars spaced at 140 mm o.c. Steel covering is 20 mm. Calculate the uniform live load pressure that a slab can carry. Use . Unit weight of concrete is 23.5 kN/ . SOLUTION

Consider 1 m strip of slab, b = 1000 m

Dead load:

Effective depth:

d = 160 – 20 – 1/2(12) d = 134 mm

Steel area,

[

]

[

]

= 22.696 kN/m

10.25 = Uniform pressure x 1 Uniform live load pressure = 10.25 kPa

Solved Problems Using 2010 NS

PROBLEM 2.29 A reinforced concrete beam has width of 310 mm and an effective depth of 490 mm. , . Determine the following: a) The balanced steel area b) The maximum steel area for singly reinforced condition c) The maximum design strength if the beam is singly reinforced d) The required steel area if the beam is subjected to dead load moment of 120 kNm and live load moment of 170 kN-m.

SOLUTION Since (

)

a) Balanced steel area:

(

)

b) Maximum steel area when beam is singly reinforced: From Eq. 2-24:

c) Maximum design strength, From Eq. 2-25 :

From Eq. 2-26:

d)

Thus, the beam is singly reinforced. Determine if the beam is tension-controlled: From Eq. 2-22:

Since the required

Check if it is really tension-controlled:

PROBLEM 2.30

the section is tension controlled.

Given the following data for a rectangular beam: width , , . Dead load moment load moment SOLUTION

Solve for

to determine if compression steel area is required. (

) (

)

=0.8237

Solve for

to determine if the section is tension-controlled. (

Since

)

, the section is within “transition region’, i.e 0.65 <

, effective depth Live

(

)

PROBLEM 2.31 Given the following properties of a rectangular concrete beam: b = 280 mm, d = 480 mm, , . The beam is reinforced for tension only. Determine the design strength under the following conditions. a) When the beam is reinforced with three 25 mm diameter bars. b) When the beam is reinforced with four 25 mm diameter bars. c) When the beam is reinforced with seven 25 mm diameter bars.

SOLUTION

(

)

a)

The section is tension-controlled,

b)

)

The section within” transition region”, i. e 0.65 <

c)

The section is compression-controlled,

PROBLEM 2.32

A hallow beam is shown in Figure 2.16. Assume and . a) Calculate the required tension steel area when b) What is the balanced moment capacity of the beam? c) What is the maximum steel area under singly reinforced condition? d) What is the maximum design moment strength under singly reinforced condition? e) Calculate the required tension steel area when .

SOLUTION This problem is the same as Problem 2.23.

To guide us whether “a” will exceed 150 mm or not, let us solve the design moment when a =150 mm.

a) Since the required

Stress in steel

b) Balanced condition:

241.27) = 454.37

[

]

c) Maximum steel area,

(114.11)=517.95

[

d) Maximum moment,

]

:

[

]

e) Refer to Figure 2.17

=575-1/2(0.85c-150)

(

)

[

] ]

PROBLEM 2.33 Design a singly reinforced rectangular beam to carry dead load moment of 110 kN-m (including self weight) and live load moment of180 kN-m. Use steel ratio and take Assume and

SOLUTION

]

Note: For singly reinforced rectangular beam,

is directly proportional to c.

Thus,

PROBLEM 2.34 Repeat Problem 2.33 using a steel ratio

SOLUTION

SUPPLEMENTARY PROBLEMS PROBLEM 2.35 A rectangular beam has . Determine (a) the maximum design moment if the beam is singly reinforced and (b) the required steel area if the beam is required to carry a dead load moment of 50 kN-m and a live load moment of 30 kN-m. Use the 2001 NS.

PROBLEM 2.36 Repeat Problem 2.35 using the 2010 NS.

PROBLEM 2.37 Design a rectangular beam reinforced for tension only carry dead load moment of 85 kN-m (including its estimated weight) and a live load of 102 kN-m. Use and use d= 1.75b. Assume and . Use the 2001 NS

PROBLEM 2.38 Repeat Problem 2.37 using the 2010 NS.

PROBLEM 2.39 A reinforced concrete beam has the following properties: Use 2001 NS) beam with, effective depth, concrete strength, reinforcing steel, reinforcing steel modulus, service dead load moment a) If the beam is to be designed for a balanced condition, find the required area of steel area reinforcement, design balanced moment, and the corresponding service live load moment. b) Find the maximum steel area, the maximum design moment, and the

corresponding service live load moment if the beam is to be designed as singly reinforced.

PROBLEM 2.40 Repeat Problem 2.39 using the 2010 NS.

PROBLEM 2.41 Calculate the ultimate moment capacity of a rectangular beam with , , . Assume . . Use 2001 NS

PROBLEM 2.42 Repeat Problem 2.41 using the 2010 NS.

PROBLEM 2.43 Calculate the ultimate moment capacity of a rectangular beam with , , . Assume , . Use 2010 NS.

PROBLEM 2.44

Repeat Problem 2.43 using the 2010 NS.

PROBLEM 2.45 Calculate the ultimate moment capacity of a rectangular beam with , , . Assume , . Use 2010 NS

PROBLEM 2.46 Repeat Problem 2.45 using the 2010 NS.

CHAPTER 3

Analysis and Design of T-Beams and Doubly Reinforced Beams

T-Beams Reinforced concrete floors usually consist of slab and beams, which are placed or poured monolithically. In this effect, the beam will have extra width on top (which is usually under compression) called flangers, and the resulting section is called a Tbeam. The beam may also be L-shaped if it is located at the end of slab. ANALYSIS AND DESIGN OF T-BEAMS WITH FLANGE IN COMPRESSION Because of the huge amount of compression concrete when the flange of a T-beams is compression, the section is usually tension-controlled (extreme tension yields).

The compression block of T-beam may fall within the flange only or partly in the web. If it falls within the flange as shown in Figure 3.1 (a), the rectangular beam formulas in Chapter 2 applies since the concrete below neutral axis is assumed to be cracked and its shape has no effect on the flexure calculations. If however it cover part of the web as shown in Figure 3.1 (b), the compression concrete no longer consist of a single rectangle and thus the rectangular formulas do not apply.

BALANCED AND MAXIMUM STEEL AREA AND MOMENT The balanced value of “c” for any beam shape, as discussed in Chapter 2 is given by:

and If ”a” is less than the slab thickness, the formulas for rectangular beam may be used, or

However, if “a” is greater than the slab thickness, the following formula will be used.

Eq. 3-1 Eq. 3-2

[

]

DESIGN OF T-BEAMS WITH NEGATIVE MOMENTS

When T-beams are resisting negative moments so that far their flangers are in tension and the bottom of their stems in compression, the formulas for rectangular beams can be applied. The following code requirements shall be applied for this case: 410.7.6: Where flangers of T-beam construction are in tension, part of the flexural tension reinforcement shall be distributed over an effective flange width as defined in Sec. 408.11, or width equal to 1/10 the span, whichever is smaller. If the effective flange width exceeds 1/10 the span, some longitudinal reinforcement shall be provided in the outer portions of the flange. The intention of this section is to minimize the possibilities of flexural cracks that will occur at the top face of the flange due to negative moments.

MINIMUM STEEL RATIO For statically determinate T-section with flange in tension, the minimum steel area is equal to or greater than the smaller value of Eq. 3-3 and Eq. 3-4:

Eq. 3-3

Eq. 3-4

√

√

CODE REQUIREMENTS ON T-BEAMS (SECTION 408.11) NOTE: THESE REQUIREMENTS ARE THE SAME WITH 2010 NS 1.

In T-beam construction, the flange and web shall be built integrally or otherwise effectively bonded together.

2.

The width of slab effective as T-beam shall not exceed 1/4 of the span of the beam, and the effective overhanging flange on each side of th web shall not exceed: a) 8 times the slab thickness, and b) 1/2 the clear distance 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, and c) 1/2 the clear distance to the next web.

For Interior Beam is the smallest of: 1. L/4 2. ⁄ ⁄ 3. For End Beam is the smallest of : 1. L/12 + 2. 3. For symmetrical interior beam ( is the smallest of: 1. L/4 2. 3. center-to-center spacing of beams 4. Isolated beams in which T-shape are used to provide a flange for additional compression area shall have a flange thickness not less than 1/2 the width of the web and an effective flange width not more than four times the width of the web.

5.

Where primary flexural reinforcement in a slab that is considered as a T-beam flange is parallel to the beam, reinforcement perpendicular to the beam shall be provided in the top of the slab in accordance with the following: a) Transverse reinforcement shall be designed to carry the factored load on the overhanging slab with assumed to act as a cantilever. For isolated beam, the full width of the overhanging flange shall be considered. For other T-beams, only the effective overhanging slab needs to be considered. b) Transverse reinforcement shall be spaced not further apart than five times the slab thickness, nor 450 mm.

STEPS IN FINDING THE TENSION STEEL AREA OF SINGLY REINFORCED T-BEAMS WITH GIVEN AND OTHER BEAM PROPERTIES: I.

Solve for

max to determine of compression steel is necessary.

Follow the procedure in Page 105. If If

II.

Solve for

, the beam is singly reinforced, proceed to Step II. , the beam is doubly reinforced

when a = t

Compressive force in concrete, ⁄ (

⁄ )

if if III.

Solve for a: ⁄ ⁄

is the smaller value of:

√

√

IV.

is the smallest value of: √

√

STEPS IN FINDING OF SINGLY REINFORCED T-BEAMS WITH GIVEN AND OTHER BEAM PROPERTIES: I.

Solve for balanced steel area to determine if tension steel yiel. Follow the procedure in Page 105.

If If

,tension steel yields. Proceed to step II , tension does not yield

. II.

Tension steel yields,

Compare

. Compute the area of compression concrete,

with the area of compression flange,

If If

proceed to Step III

III. Solve for a:

⁄

IV.

:

Solve for :

[

]

ILLUSTRATIVE PROBLEMS

SOLVED PROBLEMS IN T-BEAMS USING 2001 NS PROBLEM 3.1 Determine the effective flange with for symmetrical T-beam with a span of 6 m. The beam width of web is 250 mm, the slab thickness is 120 mm, and the clear distance to adjacent beams is 3m. SOLUTION For symmetrical T-beam, the effective flange width is the smallest of: 1. 1/4 span = 6000/4 = 1500 mm 2. 3. clear spacing of beams + Therefore

PROBLEM 3.2 Given the following elements of a T-beam: Flange width, Flange thickness, Width of web, Effective depth, If the beam is reinforced for tension only, determine the ultimate moment capacity when the depth of compression concrete flange equals the flange thickness or .

SOLUTION ⁄

Eq. 3-5

⁄

PROBLEM 3.3 Given the following elements of a T-beam: Flange width, Flange thickness, Width of web, Effective depth, If the beam is reinforced for tension only, determine the following: a) The balanced steel area b) The nominal and ultimate balanced moment capacity c) The maximum steel area d) The nominal and ultimate maximum moment capacity

SOLUTION a) Balanced condition

Figure 3.3

⁄ ⁄

[

]

b) Maximum steel area and moment. Refer to Figure 3.3.

102,466=99,000 + 310(z)

⁄ [

PROBLEM 3.4 A T-beam has the following properties:

Concrete compressive strength load conditions: a) b)

SOLUTION

(

)

,

]

, and steel area for the following

Balanced condition:

FIGURE 3.4

Maximum condition:

⁄

a)

Since

⁄ ⁄

Minimum √

Thus,

is the smaller of: √

b) Since

is more than

is more than t.

⁄

PROBLEM 3.5 Design a T-beam for a floor system for which and . The beams are 4.5 m long and spaced at 3 mo.c. The slab thickness is 100 mm.

SOLUTION

Solve for bf: 1. 2. 3.

L/4 = 1.125 m center-to center spacing of beams = 3 m

Thus, Solve for ⁄

Solve for

to determine if compression steel is needed.

z=23.25 mm

Minimum √

is the smaller value of: √

Thus,

PROBLEM 3.6 Determine the ultimate moment capacity of reinforced concrete T-beam with the following properties: Flange width b = 1500 mm, web width , effective depth d = 600 mm, slab thickness t = 100 mm. Assume and . The beam is reinforced with six 28 mm bars. SOLUTION Solve for balanced

Steel area provided, Therefore, 0.85 0.85(20.7)

72,441 = a (1500)

⁄ ⁄

PROBLEM 3.7 Given the following properties of T-beam: Flange width, Flange thickness, t=1200 Width of web, Effective depth, d = 580 mm Service deal load, Determine the safe service live load if the beam is reinforced for tension only with twelve (12) 28-mm-diameter bars. SOLUTION

Solve for balance

:

Steel area provided is less than the balanced steel area. Steel yields.

142,813=108,000+

34,813 = 400z ⁄

[

]

PROBLEM 3.8 The section of a reinforced concrete T-beam is shown in Figure 3.5. The beam is reinforced with 10 32-mm-diameter tension bars with . Concrete strength . If the total service dead load moment on the beam is 330 kN-m, determine the safe service live load moment.

SOLUTION

Solve for balance

Since

, tension steel does not yield

[

]

⁄

mm

⁄

[

]

SOLVED PROBLEMS IN T-BEAMS USING 2010 NS PROBLEM 3.9 Repeat Problem 3.3 using the 2010 NS.

SOLUTION Given:

a) Balanced condition,

Figure 3.6

⁄

⁄

[

b) Maximum steel area and moment. Refer to Figure 3.6.

]

⁄

[

PROBLEM 3.10 Repeat Problem 3.2 using the 2010 NS.

SOLUTION Given:

⁄ ⁄

Solving for :

]

(

)

⁄

PROBLEM 3.11 Repeat Problem 3.4 using the 2010 NS. Additional questions: c) Find the required steel area if and . d) Find the maximum design moment so that section is tension-controlled if it is reinforced for tension only.

SOLUTION Given:

Solve for ⁄

Solve for

when

⁄

⁄

a)

Since

is less than

,”a” is less than t.

Assume ⁄ ⁄

⁄

Minimum

is the smaller value of: √

Thus, b)

Since

is less than t.

Assume ⁄

⁄

⁄

c)

Since

is more than t.

Assume

⁄ ⁄

d)

Note: If

is less than or equal to

, the beam is tension-controlled.

PROBLEM 3.12 Repeat Problem 3.6 using the 2010 NS. SOLUTION Given:

Solve for balanced

Steel area provided, therefore ,

:

is less than t

Solve for : ⁄

therefore ⁄ ⁄

PROBLEM 3.13 Repeat Problem 3.7 using 2010 NS.

SOLUTION Given the following properties of a T-beam: Flange width, Flange thickness, Width of web, Effective depth, Service deal load,

Solve for balance

:

Steel area provided is less than the balanced steel area. Steel yields.

0.85 “a” >t

⁄ ⁄

[

]

Solve for : ⁄

Since

Transition region

PROBLEM 3.14 Repeat Problem 3.8 using 2010 NS.

SOLUTION

Solve for balanced

Since

:

[

]

⁄

[

DOUBLY REINFORCED BEAM

]

Occasionally, beams are restricted in small sizes by space or aesthetic requirements to such extent that the compression concrete should be reinforced with steel to carry compression. Compression reinforcement is needed to increase the moment capacity of a beam beyond that of a tensilely reinforced makes beams tough and ductile and reduces long-time deflection of beams. Compression steel also helps the beam withstand stress reversals that might occur during earthquakes. Continuous compression bars are also helpful for positioning stirrups and keeping them in place during concrete placement and vibration. Various tests show that compression reinforcement also prevents the beam to collapse even if the compression concrete crushes especially if it is enclosed by stirrups. According to Section 407.12 of NS, compression steel in beams must be enclosed by lateral ties, at least 10 mm in size for longitudinal bars 32 mm or smaller, and at least 12 mm in size for 36 mm and bundled bars. Deformed wire or welded wire fabric of equivalent area is allowed. The spacing of these ties shall not exceed 16 longitudinal bar diameters, 48 tie bar or wire diameters, or least dimension of the compression member. ANALYSIS OF DOUBLY REINFORCED BEAM Doubly reinforced beam is analyzed by dividing the beam into two couples and as shown in Figure 3.8. is the couple due to compression concrete and the part of the tension steel is the couple due to the compression steel and the other part of the tension steel area

Figure 3.8

Compression reinforcement is provided to ensure ductile failure (i.e. tension steel must yield). For the reason, therefore, the stress in tension ( is always equal for . On the other hand, stress of compression steel ( may either be or below . This stress must always be checked. If the compression steel yields, then , otherwise , where the stress of compression steel is given and is given by the following equation. (see derivation in page 137)

is

Eq. 3-6 According to Section 410.4.3 of NS, for with compression reinforcement, the portion of equalized by compression reinforcement need not be reduced by the 0.75 factor. Thus, the maximum permissible is: Eq. 3-7

The expression 0.75

.

STEPS TO DETERMINE AND RECTANGULAR BEAM, GIVEN I.

Solve for

and

OF ADOUBLY REINFORCED AND OTHER BEAM PROPERTIES

If

design Singly Reinforced (See Chapter 2) design as Doubly Reinforced(proceed to step II)

If II.

Figure 3.9 Solve for Solve for

III.

,

and

Solve for the stress of compression steel Solve for a and c: 0.85

Eq. 3-8

If If

proceed to IV proceed to V

IV.

then use

(compression steel yields)

V.

then use

(compression steel will not yield)

STEPS IN FINDING OF A DOUBLY REINFORCED RECTANGULAR BEAM WITH GIVEN , AND OTHER BEAM PROPERTIES There are three possible cases in doubly reinforced beams. Case 1: Both tension and compression yields ( Case 2: Tension steel yields and compression steel does not ( Case 3: Tension steel does not yield compression steel yields. (

Note: For doubly reinforced beams with effective depth d=250 mm or more, it is not possible for both steels not to yield.

I.

Assume compression steel yields (

)

II.

Solve for a and c (assuming tension steel yields): 0.85

Check: If III.

If tension steel; yields, proceed to step III tension steel does not yield, proceed to step IV

Solve for the stress in compression steel

If If IV.

proceed to step IV proceed to step V

Since

(

) (

V.

If

)

compression steel does not yield.

From the stress diagram:

Solve for

by equation formula.

Solve for Solve for a,a= Solve for (

)

(

)

VI.

______

(

) (

⁄ )

DEEP BEAMS According to Section 410.8 of the Code, beams with overall depth to clear span ratios greater than 2/3 for continuous spans, or 4/5 for simple spans, shall be designed as deep flexural taking into nonlinear distribution of strain and lateral buckling.

Beams with web depth that exceed 900 mm have a tendency to develop excessive wide cracks in the upper parts of their tension zones. According to Section 410..7 of NS, if the depth of web exceeds 900 mm, longitudinal skin reinforcement shall be uniformly distributed along both side faces of the member for distance d/2 nearest the flexural tension. The area of skin reinforcement per meter of height on each side face shall be Eq. 3-9 The maximum spacing of the skin reinforcement shall not exceed the lesser of d/b and 300 mm. Such reinforcement may be included in strength computations if a strain compatibility analysis is made to determine stresses in the individual bars or wires. The total area of longitudinal skin reinforcement in both faces need not exceed one-half of the required flexural tensile reinforcement.

ILLUSATIVE PROBLEMS DESIGN PROBLEMS PROBLEM 3.15 a .305-mm wide rectangular beam has an overall depth of 560 mm. The beam is reinforced with four 25-mm-diameter compression bars. The centroid fiber. Assume and . Determine the following: a) The balanced tension steel area and the nominal and ultimate balanced moment. b) The maximum tension steel area and the nominal and ultimate maximum moment. SOLUTION

a) Balanced condition

yield

Balanced steel area, ⁄ (

) ⁄

b) Maximum tension steel area: According to Section 410.4.3, for with compression reinforcement, the portion of equalized by compression reinforcement need not be reduced by the 0.75 factor.

(

⁄ ) (

⁄ ) ⁄

PROBLEM 3.16 (CE NOVEMBER 2009) A reinforced concrete beam has width of 300 mm and effective depth of 460 mm. The beam is reinforced with 2-28 mm compression bars placed 70 mm from extreme concrete. Concrete strength and steel strength . a) What is the balanced steel area considering the contribution of the compression steel? b) What is the maximum tension steel area allowed by the code?

SOLUTION

a) Balanced condition considering compression steel:

b) Maximum steel area: For rectangular beams:

PROBLEM 3.17 A rectangular beam has b=300 mm and d= 490 mm. Concrete compressive strength and steel yield strength . Compressive steel if required shall have its centroid 60 mm from extreme concrete fiber. Calculate the required tension steel area if the factored moment is 620 kN-m. SOLUTION This is the same problem in Chapter 2.

Solve for

:

⁄

Since

the beam must be doubly reinforced.

Note:

Solve for ⁄

Compression steel yields

Tension steel area, Compression steel:

PROBLEM 3.18 A rectangular beam has b=310 mm and d=460 mm. The beam will be designed to carry a service dead load of 230 kN-m and service live load of190 kn-m. Compression reinforcement if necessary will have its centtoid 70 mm from extreme concrete fiber. Determine the required steel area. Use and SOLUTION

Solve for : Note : For rectangular beams,

⁄

Since

Tension Steel:

205.088 x

Compression steel:

compression steel does nt yield,

PROBLEM 3.19 A floor system consists of a 100-mm concrete slab ed by continuous T beam with 9 m span, 1.2 m on centers as shown in Figure 3.10. Web dimensions, as determined by negative-moment requirements, are and Concrete cover is 70 mm from the centroid of the bars. The beam is subjected to a maximum positive factored moment of 1080 kN-m. Use . Unit weight of concrete is 23.5 kN/ a) Calculate the required tension steel area at the point of maximum positive moment. b) Using the tributary area method, what is the uniform service dead load acting on the beam? c) Calculate the uniform service live load acting on the beam.

SOLUTION

Maximum factored moment, Effective flange width, 1. L/4=9/4=2.25 m 2. 16t+ 3.

Use Solve for to determine if compression steel is required. Balanced condition:

Maximum condition:

Since

the compression reinforcement must be provided.

Tension steel area, Compression steel area:

b)

Dead load=weight of concrete: Area=1.2(0.1)+0.28(0.47)=0.2516 ⁄

c) Uniform live load

Maximum positive moment (at midspan)

⁄

INVESTIGATION (ANALYSIS) PROBLEMS PROBLEM 3.20 The beam shown in Figure 3.11 is subjected to a maximum service dead load moment of 230 kN-m. Determine the service live load that the beam can carry. Use

SOLUTION

Assume all steel yield:

compression steel yields

Assumption is correct, all steel yield.

(

) (

) (

)

PROBLEM 3.21 A rectangular beam has the following properties: Width, b=400 mm Effective depth, d=620 mm Tension bars, 3 pcs 25-mm-diameter d’=70 mm Determine the design strength of the beam and the safe service live load if the service dead load is 320 kN-m. SOLUTION

Assume all steel yields:

⁄

All steel yields. Assumption is correct

(

⁄ ) ( ⁄ ) ⁄

PROBLEM 3.22 A 12-m long rectangular reinforced concrete beam is simply ed at its ends. The beam is provided with an addition at the mid span. Width of beam is 300 mm and the overall depth is 450 mm. The beam is reinforced with 25-mm-diameter bars, four bars at the tension side and 2 bars at the compression side .Concrete protective

coverings is 70 mm form the centroid of the bars. Concrete strength and steel yield . Use 0.75 a) Determine the depth of the compression block. b) Determine the nominal moment capacity of the beam. c) Determine the factored uniform load, including its own weight, the beam can carry. SOLUTION

Assuming all steel yields:

⁄ compression steel does not yield Assuming tension steel yields and compression steel does not.

(

⁄ )

c) Maximum factored uniform load:

By there-moment equation: ̅

̅

̅ ̅

PROBLEM 3.23 (CE NOVEMBER 2010) A 6 meter long simply ed reinforced concrete beam has a width of 350mm and

an overall depth of 470 mm. The beam is reinforced with 2-28 mm compression bars on top and 4-28 tension bars at the bottom, each located 70 mm from the extreme concrete fiber. Concrete strength , and steel yield strength Determine the following: a) Depth of compression blocks assuming both tension and compression steel yields. b) What is the ultimate moment capacity of the beam in kN-m? c) Determine the additional concentrated live load that can be applied at midspan if the dead load including the weight of the beam is 20 kN/m.

SOLUTION Given :

Tension steel area Compression steel area,

Assuming tension & compression steel yields:

Thus, compression steel does not yield.

Since compression steel does not yield, Assuming tension steel yields:

(

) (

) ⁄

Ultimate moment capacity= Ultimate moment capacity=

PROBLEM 3.24 A beam section is shown in Figure 3.12. The beam will be subjected to a maximum service dead load of 215 kN-m. What is the safe service live load moment for this beam? Use

SOLUTION Tension steel, Compression steel,

Effective depth to extreme tension bar: Effective depth (to centroid of tension bar)

Assume all steel yields:

Compression steel does not yield.

(

) (

)

PROBLEM 3.25 A beam section is shown in Figure 3.13. The beam will be subjected to a maximum service dead load of 360 kN-m. What is the safe service live load moment for this beam? Use

Tension steel,

Compression steel,

⁄ Effective depth (to centroid of tension bars)

Assume all steel yields:

Tension steel does not yield.

Solve for d:

(

) (

) ⁄

PROBLEM 3.26 Calculate the design flexural strength of the T-beam shown in Figure 3.14. Use

.

SOLUTION

Flange area, Assume all steel yields: ⁄

Area of compression concrete:

⁄

⁄

if the upper layer of tension steel yields ⁄

⁄

All steel yields, assumption is correct: (

) (

)

PROBLEM 3.27 Calculate the design flexural strength of the T-beam shown in Figure 3.15. Use

SOLUTION

Flange area, Assume all steel yields: ⁄

Area of compression concrete:

⁄

⁄