Design Procedure For Column 4b726b
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Design of Concrete Structures
Design of Columns and Footings Dr. Rajesh K. N. Assistant Professor in Civil Engineering Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur
Dr.RajeshKN
(Analysis and design in Module II, III and IV should be based on Limit State Method. Reinforcement detailing shall conform to SP34) MODULE IV (13 hours) Design of Compression : Effective length and classification as per IS 456, short columns subjected to axial compression with and without uniaxial/biaxial eccentricities; slender columns - Use of SP 16 charts. Design of Footings: Wall footings, isolated footings – axial and eccentric loading, rectangular and trapezoidal combined footings.
2 Dept. of CE, GCE Kannur
Dr.RajeshKN
Columns
Dept. of CE, GCE Kannur
Dr.RajeshKN
Columns • A ‘compression member’ is a structural element which is subjected (predominantly) to axial compressive forces. • Compression are most commonly encountered in reinforced concrete buildings as columns • Column is a compression member, the ‘effective length’ of which exceeds three times the least lateral dimension (Cl. 25.1.1) • ‘Pedestal’ is a vertical compression member whose ‘effective length’ is less than three times its least lateral dimension [Cl. 26.5.3.1(h)].
Dept. of CE, GCE Kannur
Dr.RajeshKN
Classification of Columns Based on Type of Reinforcement
1) Tied columns: where the main longitudinal bars are enclosed within closely spaced lateral ties 2) Spiral columns: where the main longitudinal bars are enclosed within closely spaced and continuously wound spiral reinforcement 3) Composite columns: where the reinforcement is in the form of structural steel sections or pipes, with or without longitudinal bars
Dept. of CE, GCE Kannur
Dr.RajeshKN
Classification of Columns Based on Type of Loading 1. Columns with axial loading (applied concentrically) 2. Columns with uniaxial eccentric loading 3. Columns with biaxial eccentric loading
P
P e
axial compression Dept. of CE, GCE Kannur
axial compression with uniaxial bending.
ey ex
P
axial compression with biaxial bending. 6
Dr.RajeshKN
The combination of axial compression (Pu) with bending moment (Mu) at any column section
statically equivalent to
Load P applied with an eccentricity e = Mu/Pu with respect to the longitudinal centroidal axis of the column section.
7 Dept. of CE, GCE Kannur
Dr.RajeshKN
Classification of Columns Based on Slenderness Ratios (Cl. 25.1.2), • ‘Slenderness ratio’ is the ratio of ‘effective length’ to its lateral dimension • It provides a measure of the vulnerability to failure of the column by elastic instability (buckling). Buckling tendency varies with slenderness ratio.
1. Short columns: slenderness ratios with respect to the ‘major principal axis’ (lex/Dx) as well as the ‘minor principal axis’ (ley/Dy) are both less than 12 2. Slender (or long) columns.
Dept. of CE, GCE Kannur
Dr.RajeshKN
• Columns with low slenderness ratios fail under ultimate loads with the material (concrete, steel) reaching its ultimate compressive strength, and not by buckling
• Columns with very high slenderness ratios may undergo buckling (with large lateral deflection) under relatively low compressive loads, and fail suddenly.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Uned Length Code (Cl. 25.1.3) defines the ‘uned length’ l of a column explicitly for various types of constructions.
Effective length of a column The effective length of a column in a given plane is defined as the distance between the points of inflection in the buckled configuration of the column in that plane. The effective length depends on the uned length l and the boundary conditions at the column ends
Dept. of CE, GCE Kannur
Dr.RajeshKN
Effective lengths for idealised boundary conditions (Euler’s theory) End conditions
Effective length
Both ends hinged
l
One end fixed and the other end free One end fixed and the other hinged Both ends fixed
Dept. of CE, GCE Kannur
2l l
2
l 2
Dr.RajeshKN
Code recommendations for idealised boundary conditions (Cl. E–1)
Use of Code Charts Charts are given in Fig. 26 and Fig. 27 of the Code for determining the effective length ratios of braced columns and unbraced columns respectively in of coefficients β1 and β2 which represent the degrees of rotational freedom at the top and bottom ends of the column.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Recommended effective length ratios for normal usage (Table 28, Page 94) 1. columns braced against sideway: a) both ends ‘fixed’ rotationally b) one end ‘fixed’ and the other ‘pinned c) both ends ‘free’ rotationally (‘pinned’)
: 0.65 : 0.80 : 1.00
2. columns unbraced against sideway: a) both ends ‘fixed’ rotationally : 1.20 b) one end ‘fixed’ and the other ‘partially fixed’ : 1.50 c) one end ‘fixed’ and the other free : 2.00
Dept. of CE, GCE Kannur
Dr.RajeshKN
Eccentric loading The general case of loading on a compression member is axial compression combined with biaxial bending. • A state of biaxial eccentric compression • Axial load P acts eccentric to the longitudinal centroidal axis of the column cross section, with eccentricities ex and ey with respect to the major and minor principal axes
Dept. of CE, GCE Kannur
Dr.RajeshKN
Code requirements on slenderness limits, minimum eccentricities and reinforcement Slenderness Limits (Cl. 25.3.1) Ratio of the uned length (l) to the least lateral dimension (d) of a column
Furthermore, in case of cantilevered column, where D is the depth of the cross-section measured in the plane of the cantilever and b is the width (in the perpendicular direction) (Cl. 25.3.2) .
Dept. of CE, GCE Kannur
Dr.RajeshKN
Minimum eccentricities (Cl. 25.4) Very often, indirect eccentricities may arise due to: • lateral loads not considered in design; • live load placements not considered in design; • accidental lateral/eccentric loads; • errors in construction (such as misalignments); and • slenderness effects underestimated in design.
Hence, the Code requires every column to be designed for a minimum eccentricity emin (in any plane) equal to: • uned length/500 plus lateral dimension/30, • subject to a minimum of 20 mm.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Thus, for a column with a rectangular section:
ex ,min
⎧l 500 + Dx 30 =⎨ whichever is greater 20 mm ⎩
e y ,min
⎧l 500 + D y 30 =⎨ whichever is greater 20 mm ⎩
Dept. of CE, GCE Kannur
Dr.RajeshKN
Code Requirements on Reinforcement and Detailing Longitudinal Reinforcement (Cl. 26.5.3.1) Minimum Reinforcement: The longitudinal bars must have a cross sectional area not less than 0.8 percent of the gross area of the column section.
Maximum Reinforcement: The maximum cross-sectional area of longitudinal bars should not exceed 6 percent of the gross area of the column section. But a reduced maximum limit of 4 percent is recommended in general for better placement and compaction of concrete and also at lapped splice locations.
Dept. of CE, GCE Kannur
Dr.RajeshKN
A minimum limit is specified by the Code inorder to: • ensure nominal flexural resistance under unforeseen eccentricities in loading; and • prevent the yielding of the bars due to creep and shrinkage effects, which result in a transfer of load from the concrete to the steel.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Minimum diameter / number of bars and their location: Longitudinal bars in columns (and pedestals) should not be less than 12 mm in diameter and should not be spaced more than 300 mm apart (centre-tocentre) along the periphery of the column At least 4 bars should be provided in a column with rectangular cross-section (one at each corner), and at least 6 bars) in a circular column (equally spaced near the periphery.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Cover to reinforcement (Cl. 26.4.2.1) : • A minimum clear cover of 40 mm or bar diameter (whichever is greater), to the column ties is recommended for columns in general • In small-sized columns (D < 200 mm and whose reinforcing bars do not exceed 12mm) a reduced clear cover of 25 mm is permitted
Dept. of CE, GCE Kannur
Dr.RajeshKN
Transverse Reinforcement (Cl. 26.5.3.2) General: All longitudinal reinforcement in a compression member must be enclosed within transverse reinforcement, comprising either lateral ties (with internal angles not exceeding 135o) or spirals. This is required: • to prevent the premature buckling of individual bars; • to confine the concrete in the ‘core’, thus improving ductility and strength; • to hold the longitudinal bars in position during construction; and • to provide resistance against shear and torsion, if required. Dept. of CE, GCE Kannur
Dr.RajeshKN
Lateral Ties: tie diameter
tie spacing
⎧φlong ,max 4 φt = ⎨ ⎩ 6 mm
⎧ D ⎪ st = ⎨16φlong ,min ⎪ ⎩ 300 mm
where φlong is the diameter of longitudinal bar to be tied D is the least lateral dimension of the column
Dept. of CE, GCE Kannur
Dr.RajeshKN
Spirals: Helical reinforcement provides very good confinement to the concrete in the ‘core’ and enhances ductility of the column at ultimate loads. The diameter and pitch of the spiral may be computed as in the case of ties— except when the column is designed to carry a 5 percent overload (as permitted by the Code), in which case pitch
and
Dept. of CE, GCE Kannur
Dr.RajeshKN
DESIGN OF SHORT COLUMNS UNDER AXIAL COMPRESSION Equilibrium of Axial Loading If fcc and fsc denote respectively the stresses in the concrete and the longitudinal steel, corresponding to the uniform compressive strain εc , then it follows that
P0 = Cc + Cs = f cc Ac + f sc Asc = f cc ( Ag − Asc ) + f sc Asc = f cc Ag + ( f sc − f cc ) Asc Ag = gross area of cross-section = Ac + Asc ; Asc = total area of longitudinal reinforcement Ac = net area of concrete in the section = Ag – Asc Dept. of CE, GCE Kannur
Dr.RajeshKN
Design Strength of Axially Loaded Short Columns (Cl. 39.3) Where the calculated minimum eccentricity (in any plane) does not exceed 0.05 times the lateral dimension (in the plane considered), may be designed by a simplified equation:
Pu = 0.4 f ck Ag + ( 0.67 f y − 0.4 f ck ) Asc
Dept. of CE, GCE Kannur
Dr.RajeshKN
Design problem 1 Design the reinforcement in a column of size 450 mm x 600 mm, subject to an axial load of 2000 kN. The column has an uned length of 3.0m and is braced against sidesway in both directions. Use M 20 concrete and Fe 415 steel. 1. Short Column or Slender Column? Given: lx = ly = 3000 mm, Dy = 450 mm, Dx = 600 mm
⎧ lex Dx = k x l x Dx = k x × 3000 600 = 5k x slenderness ratios = ⎨ ⎩ley D y = k y l y D y = k y × 3000 450 = 6.67k y • Effective length ratios kx and ky are less than unity (since the column is braced against sideway in both directions,) •and hence the slenderness ratios are both less than 12. • Hence, the column may be designed as a short column. Dept. of CE, GCE Kannur
Dr.RajeshKN
2. Minimum Eccentricities
ex ,min ex ,min
e y ,min
⎧l 500 + Dx 30 =⎨ whichever is greater 20 mm ⎩
⎧3000 500 + 600 30 =⎨ whichever is greater 20 mm ⎩ ⎧3000 500 + 450 30 =⎨ whichever is greater 20 mm ⎩
= 26 mm
= 21mm
As 0.05Dx = 0.05 x 600 = 30.0 mm > ex,min = 26.0 mm and 0.05Dy = 0.05 x 450 = 22.5 mm > ey,min = 21.0 mm, the Code formula for axially loaded short columns can be used.
(
)
Pu = 0.4 f ck Ag + 0.67 f y − 0.4 f ck Asc Dept. of CE, GCE Kannur
Dr.RajeshKN
3. Factored Load
Pu = 2000 x 1.5 = 3000 kN
4. Design of Longitudinal Reinforcement
Pu = 0.4 f ck Ag + ( 0.67 f y − 0.4 f ck ) Asc 3000 × 103 = 0.4 × 20 × 450 × 600 + ( 0.67 × 415 − 0.4 × 20 ) Asc Asc = 3111 mm2 Provide 4–25 ϕ at corners : 4 x 491 = 1964 mm2 and 4–20 ϕ additional: 4 x 314 = 1256 mm2 Asc = 3220 mm2 > 3111 mm2 p = (100x3220) / (450x600) = 1.192 > 0.8 (minimum reinf.) — OK. Dept. of CE, GCE Kannur
Dr.RajeshKN
5. Lateral Ties: tie diameter
⎧φlong ,max 4 φt = ⎨ ⎩ 6 mm
tie spacing
⎧ D 450 ⎧ ⎪ st = ⎨16φlong ,min = ⎪⎨16 × 20 = 320 ⎪ ⎪ 300 mm ⎩ 300 mm ⎩
⎧ 25 4 =⎨ ⎩6 mm
Hence provide 8 φ ties @ 300 c/c
Dept. of CE, GCE Kannur
Dr.RajeshKN
6. Detailing
600
450 4#20ϕ 8ϕ@300c/c 4#25ϕ
Dept. of CE, GCE Kannur
Dr.RajeshKN
DESIGN OF SHORT COLUMNS UNDER COMPRESSION WITH UNIAXIAL BENDING Design problem 2 Using the design aids given in SP:16, design the longitudinal reinforcement in a rectangular reinforced concrete column of size 300 mm × 600 mm subjected to a factored load of 1400 kN and a factored moment of 280 kNm with respect to the major axis. Assume M20 concrete and Fe415 steel. Given: b = 300 mm, D = 600 mm, fck = 20 MPa, fy = 415 MPa, Pu = 1400 kN, Mux = 280 kNm Arrangement of bars: SP : 16 Charts for “equal reinforcement on four sides” may be used. (Charts 43-46 ) Dept. of CE, GCE Kannur
Dr.RajeshKN
Assuming an effective cover d‘ = 60 mm
d‘/D = 60/600 = 0.1
Pu 1400 × 103 = 0.389 = f ck bd 20 × 300 × 600 280 × 10 Mu = 2 = 0.13 2 20 300 600 × × f ck bd 6
Referring to Chart 44 of SP : 16 preqd /20 = 0.11
d‘/D = 0.1
preqd = 0.11 × 20 = 2.2
As,reqd = 2.2 × 300 × 600/100 = 3960 mm2 Detailing of longitudinal reinforcement Dept. of CE, GCE Kannur
Dr.RajeshKN
Detailing
600 8ϕ@200c/c staggered 300
4#22ϕ inner rows
Dept. of CE, GCE Kannur
4#28ϕ at corners
Dr.RajeshKN
DESIGN OF SHORT COLUMNS UNDER AXIAL COMPRESSION WITH BIAXIAL BENDING Code Procedure for Design of Biaxially Loaded Columns (Cl. 39.6) The simplified method adopted by the Code is based on Bresler’s formulation An approximate relationship between MuR,x and MuR,y (for a specified Pu = PuR) is established. This relationship is conveniently expressed in a non-dimensional form αn αn as follows:
⎛ M uy ⎞ ⎛ M ux ⎞ ⎟⎟ ≤ 1 ⎜ ⎟ + ⎜⎜ ⎝ M ux1 ⎠ ⎝ M uy1 ⎠
where Mux and Muy denote the factored biaxial moments acting on the column, and Mux1 and Muy1 denote the uniaxial moment capacities with reference to the major and minor axes respectively, all under an accompanying axial load Pu = PuR. Dept. of CE, GCE Kannur
Dr.RajeshKN
In biaxial loading, Code (Cl. 39.6) suggests the following equation, applicable for all grades of steel:
Puz = 0.45 f ck Ac + 0.75 f y Asc Puz = 0.45 f ck Ag + ( 0.75 f y − 0.45 f ck ) Asc where Ag denotes the gross area of the section and Asc the total area of steel in the section. αn = 1 for Pu/Puz < 0.2; αn = 2 for Pu/Puz > 0.8; and αn is assumed to vary linearly for values of Pu/Puz between 0.2 and 0.8
Dept. of CE, GCE Kannur
Dr.RajeshKN
Code Procedure 1. Given Pu, Mux , Muy , that the eccentricities ex = Mux / Pu and ey = Muy / Pu are not less than the corresponding minimum eccentricities. 2. Assume a trial section for the column. 3. Determine Mux1 and Muy1 , corresponding to the given Pu (using SP:16). Ensure that Mux1 and Muy1 are significantly greater than Mux and Muy respectively; otherwise, suitably redesign the section. 4. Determine Puz , and hence αn 5. Check the adequacy of the section; if necessary, redesign the section and check again.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Design problem 3 A column (400 mm × 400 mm), effectively held in position but restrained against rotation at one end, is subjected to factored loads: Pu = 1300 kN, Mux = 190 kNmand Muy = 110 kNm. The uned length of the column is 3.5m. Design the reinforcement in the column, assuming M 25 concrete and Fe 415 steel. Given: Dx = Dy = 400 mm, l = 3500 mm, Pu = 1300 kN, Mux = 190 kNm, Muy = 110 kNm, fck = 25MPa, fy = 415MPa. 1. Slenderness ratios Effective length = 0.8l for the braced column, lex = ley = 0.8 × 3500 = 2800 mm lex/Dx = ley/Dy = 2800/400 = 7.0 < 12 Hence the column may be designed as a short column. Dept. of CE, GCE Kannur
Dr.RajeshKN
2. Check minimum eccentricities Applied eccentricities: ex = Mux / Pu = 190 × 103/1300 = 146 mm ey = Muy / Pu = 110 × 103/1300 = 84.6 mm Minimum eccentricities as per Code: ex, min = ey, min = 3500/500 + 400/30 = 20.3 mm > 20 mm The applied eccentricities are larger than the minimum eccentricities.
Dept. of CE, GCE Kannur
Dr.RajeshKN
3. Trial section: Longitudinal reinforcement Assume longitudinal reinforcement percentage as 3.5%. As,reqd = 3.5 × 4002/100 = 5600 mm2 Provide 12 – 25 dia: As = 491 × 12 = 5892 mm2 > 5600 mm2. The arrangement of bars is shown in Figure??.
Deg for uniaxial eccentricity with Pu = 1300 kN and Uniaxial moment capacities: Mux1, Muy1 [Here, due to symmetry, Mux1 = Mux2]
Pu f ck bd Dept. of CE, GCE Kannur
= 0.325 Dr.RajeshKN
pprovided = 5892 × 100/4002 = 3.68 p/fck = 3.68/25 = 0.147 d’= 40 + 8 + 25/2 = 60.5 mm (assuming a clear cover of 40 mm and 8 mm ties) d’/D = 60.5/400 = 0.151 = 0.15 (say) Referring to Chart 45
Mu f ck bd 2
=0.165
Mux1 = Muy1 = 0.165 × 25 × 4003 = 264 × 106 Nmm = 264 kNm which is significantly greater than Mux = 190 kNm and Muy = 110 kNm
Dept. of CE, GCE Kannur
Dr.RajeshKN
4. Check for biaxial bending Values of Puz and αn
Puz = 0.45 f ck Ag + ( 0.75 f y − 0.45 f ck ) Asc = (0.45 × 25 × 4002) + (0.75 × 415 – 0.45 × 25) × 5892 = (1800 × 103 + 1767.6 × 103) N = 3568 kN Pu/Puz = 1300/3568 = 0.364 (which lies between 0.2 and 0.8)
αn= 1.273 αn
αn
⎛ M uy ⎞ ⎛ M ux ⎞ ⎟⎟ ⎜ ⎟ + ⎜⎜ ⎝ M ux1 ⎠ ⎝ M uy1 ⎠
1.273
⎛ 190 ⎞ =⎜ ⎟ ⎝ 264 ⎠
1.273
⎛ 110 ⎞ +⎜ ⎟ ⎝ 264 ⎠
= 0.986 < 1.0
Hence, the trial section is safe under the applied loading. Dept. of CE, GCE Kannur
Dr.RajeshKN
5. Transverse reinforcement The minimum diameter ϕt and maximum spacing st of the lateral ties are specified by the Code
tie diameter
⎧φlong ,max 4 φt = ⎨ ⎩ 6 mm
tie spacing
⎧ D 400 ⎧ ⎪ st = ⎨16φlong ,min = ⎪⎨16 × 25 = 400 ⎪ ⎪ 300 mm ⎩ 300 mm ⎩
⎧ 25 4 =⎨ ⎩6 mm
Provide 8 ϕ ties@ 300 c/c
Dept. of CE, GCE Kannur
Dr.RajeshKN
6. Detailing
Dept. of CE, GCE Kannur
Dr.RajeshKN
DESIGN OF SLENDER COLUMNS Code Procedures for Design of Slender Columns a. Strength Reduction Coefficient Method (B-3.3 of the Code) • This is a highly simplified procedure, which is given in the Code for the working stress method of design • Although this method has been prescribed for WSM, it can be extended to the limit state method (LSM) for the case of axial loading (without primary bending moments)
Dept. of CE, GCE Kannur
Dr.RajeshKN
According to this procedure the permissible stresses in concrete and steel are reduced by multiplying with a strength reduction coefficient Cr given by:
le Cr = 1.25 − 48d
where d is the least lateral dimension of the column (or diameter of the core in a spiral column). Alternatively, for more exact calculations,
Cr = 1.25 −
le 160rmin
where rmin is the least radius of gyration of the column.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Additional Moment Method (Cl. 39.7.1) The method prescribed by the Code for slender column design by the limit state method is the ‘additional moment method’ According to this method, every slender column should be designed for biaxial eccentricities (which include the P–∆ moment “additional moment”) components eax = Max/Pu and eay = May/Pu :
M ax
P D ⎧l ⎫ = u ⎨ ex ⎬ 2000 ⎩ D ⎭
The total design moments
2
M ay
Pu b ⎧ ley ⎫ = ⎨ ⎬ 2000 ⎩ b ⎭
2
M ux = M ux + M ax = Pu ( ex + eax ) M uy = M uy + M ay = Pu ( e y + eay )
Dept. of CE, GCE Kannur
Dr.RajeshKN
Mux, Muy
are the primary factored moments (obtained from first-order structural analyses)
Max, May
are the additional moments with reference to bending about the major and minor axes respectively.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Code recommends the following modifications for the design of slender columns in general: The additional moments Max, May may be reduced by multiplying factors (Cl. 39.7.1.1) defined as:
Puz − Pu ≤1 kx = Puz − Pb , x
Puz − Pu ≤1 ky = Puz − Pb , y
Pb is the axial load corresponding to the condition of max compressive strain of 0.0035 in concrete and tensile strain of 0.002 in outermost layer of compression steel
Dept. of CE, GCE Kannur
Dr.RajeshKN
• For braced columns subject to unequal primary moments M1, M2 at the two ends, the value of Mu to be considered in the computation of the total moment M u
M u = 0.4 M 1 + 0.6 M 2 ≥ 0.4 M 2 where M2 is the higher column end moment. Also, M u ≥ M 2 • M1 and M2 are considered to be of opposite signs if the column is bent in double curvature.
Dept. of CE, GCE Kannur
Dr.RajeshKN
• For unbraced columns, the lateral drift effect needs to be included. An approximate way for this is by assuming that the additional moment Ma acts at the column end where the maximum primary moment M2 is present. Hence, for design purposes, the total moment may be taken as
M u = M 2 + M a
Dept. of CE, GCE Kannur
for unbraced columns
Dr.RajeshKN
Procedure for design of slender columns 1. Determine the effective depth and slenderness ratio in each direction. 2. a) Determine initial moment Mui from given primary end moments Mu1 and Mu2 in each direction. b) Calculate emin and Mu,min in each direction. c) Compare the moments computed in steps a and b and take the greater of the two as initial moment Mui in each direction. 3. a) Find additional moment Ma in each direction. b) Find total moment Mut in each direction, without considering reduction factor ka . c) Make preliminary design for Pu and Mut and find area of steel. Thus p is known.
Dept. of CE, GCE Kannur
Dr.RajeshKN
4. a) Get Puz . Also find Pb each direction, for the reinforcement ration p obtained above. b) Find reduction factor ka in each direction. c) Determine the modified design value of moment in each direction Mut = Mui + ka Ma. 5. Redesign the column for Pu and Mut . If the column is slender about both the axes, design the column for biaxial bending, for (Pu , Muxt ) about x-axis and (Pu , Muyt ) about y-axis.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Braced slender column design problem 1 Design the longitudinal reinforcement for a braced column, 300 mm × 400 mm, subject to a factored axial load of 1200 kN and factored moments of 50 kNm and 30 kNm with respect to the major axis and minor axis respectively at the top end. Assume that the column is bent in double curvature (in both directions) with the moments at the bottom end equal to 40 percent of the corresponding moments at top. Assume an uned length of 7.2 m and an effective length ratio of 0.85 in both directions. Use M 20 concrete and Fe 415 steel. Effective cover is 60 mm. Given: Dx = 400 mm, Dy = 300 mm, Pu = 1200 kN; Mux = 50 kNm, Muy = 30 kNm, at top; Mux = 0.4x50 = 20 kNm, Muy = 0.4x30 = 12 kNm, at bottom; l = 7200 mm, kx = ky = 0.85. Dept. of CE, GCE Kannur
Dr.RajeshKN
1. Slenderness ratios
lex = ley = 0.85 × 7200 = 6120 mm lex/Dx = 6120/400 = 15.3 > 12 ley/Dy = 6120/300 = 20.4 > 12
Hence, the column is slender about both the axes. 2. Minimum eccentricities & Minimum moments
ex ,min
⎧7200 500 + 400 30 =⎨ ( greater ) = 27.73 mm 20 mm ⎩
e y ,min
⎧7200 500 + 300 30 =⎨ ( greater ) = 24.4 mm 20 mm ⎩
M ux ,min = Pu .ex ,min = 1200 × 27.73 = 33.276 kNm M uy ,min = Pu .e y ,min = 1200 × 24.4 Dept. of CE, GCE Kannur
= 29.28 kNm Dr.RajeshKN
3. Primary moments for design As the column is braced and bent in double curvature,
M ux = 0.6 × 50 − 0.4 × 20 = 22 kNm ≥ 0.4 × 50 = 20
M uy = 0.6 × 30 − 0.4 × 12 = 13.2 kNm ≥ 0.4 × 30 = 12 Since the minimum moments are larger than these, the primary moments for design are:
M ux = 33.276 kNm
Dept. of CE, GCE Kannur
M uy = 29.28 kNm
Dr.RajeshKN
4. Additional moments Without modification factor, additional eccentricities eax = Dx (lex/Dx)2/2000 = 400 (15.3)2/2000 = 46.82 mm eay = Dy (ley/Dy)2/2000 = 300 (20.4)2/2000 = 62.42 mm Assume modification factors kax = kay ≈ 0.5 (to be verified later), Additional moments: Max = Pu (kaxeax) = 1200(0.5 × 46.82 × 10-3) = 28.092 kNm May = Pu (kayeay) = 1200(0.5 × 62.42 × 10-3) = 37.452 kNm
5. Total factored moments
M ux = M ux + M ax = 33.276 + 28.092 = 61.368 kNm M uy = M uy + M ay = 29.28 + 37.452 = 66.732 kNm > M ux = 61.368 kNm Dept. of CE, GCE Kannur
Dr.RajeshKN
6. Trial section • Deg for a resultant uniaxial moment with respect to the minor axis (y-axis),
M u ≅ 1.15 61.3682 + 66.7322 = 104.26 kNm combined with Pu = 1200 kNm Pu/(fckbD) = (1200 × 103/(30 × 400 × 300) = 0.5 Mu/(fckbD2) = 104.26 × 106/(30 × 400 × 3002) = 0.144 d’/D ≈ 60/300 = 0.20 with “equal reinforcement on all sides”, referring to Chart 46 of SP : 16, p/fck = 0.19 Dept. of CE, GCE Kannur
Dr.RajeshKN
→ preqd = 0.19 × 20 = 3.8 → As,reqd = 3.8 × 300 × 400/100 = 4560 mm2 Provide 10–25 ϕ bars. 4909 > 4560 → pprovided = 4909 × 100/(300 × 400) =4.09 → p/fck = 4.09/20 = 0.2045
Dept. of CE, GCE Kannur
Dr.RajeshKN
7. Check additional moments d’/Dx= 60/400 = 0.15 mm d’/Dy= 60/300 = 0.2 mm From Table 60 of SP :16, the ultimate loads Pb,x and Pb,y at balanced failure can be determined.
Pb , x p = k1 + k2 For d’/Dx = 0.15, k1= 0.196, k2= 0.203, f ck bD f ck ∴ Pb, x = 570.03kN For d’/Dy = 0.2, k1= 0.184, k2= 0.028,
Pb, y
p = k1 + k2 f ck bD f ck ∴ Pb, y = 455.34 kN
Puz = 0.45fck Ag + (0.75fy – 0.45fck)As = (0.45 × 20 × 300 × 400) + (0.75 × 415 – 0.45 × 20) × 4909 = 2563.7 kN Dept. of CE, GCE Kannur
Dr.RajeshKN
Modification factors:
Puz − Pu 2563.65 − 1200 kx = = = 0.684 ≤ 1 Puz − Pb , x 2563.65 − 570.03 2563.65 − 1200 Puz − Pu = = 0.647 ≤ 1 ky = 2563.65 − 455.34 Puz − Pb , y Hence, the actual (revised) total moments are:
M ux = 33.276 + 1200(0.684 × 46.82 × 10-3) = 71.706 kNm -3 M uy
= 29.28 + 1200(0.647 × 62.42 × 10 ) = 77.74 kNm
Dept. of CE, GCE Kannur
Dr.RajeshKN
8. Check safety under axial load with biaxial bending Referring to the design Charts 45 & 46 in SP : 16, uniaxial moment capacities corresponding to Pu/fckbD = 0.5 and p/fck = 0.2045 are obtained as: Mux1/fckbD2 = 0.18 (for d’/Dx = 0.15) Muy1/fckbD2 = 0.152 (for d’/Dy = 0.2)
= 71.706 kNm Mux1 = 0.18 × 30 × 300 × 4002 = 172.8 kNm > M ux = 77.74 kNm Muy1 = 0.152 × 30 × 400 × 3002 = 109.44 kNm > M uy
Pu/Puz = 1200/2563.7 = 0.468 (which lies between 0.2 and 0.8)
αn= 1.447 Dept. of CE, GCE Kannur
Dr.RajeshKN
αn
αn
⎛ M uy ⎞ ⎛ M ux ⎞ ⎟⎟ ⎜ ⎟ + ⎜⎜ ⎝ M ux1 ⎠ ⎝ M uy1 ⎠
1.447
⎛ 71.706 ⎞ =⎜ ⎟ 172 8 . ⎝ ⎠
= 0.89 < 1.0
Dept. of CE, GCE Kannur
1.447
⎛ 77.74 ⎞ +⎜ ⎟ 109 44 . ⎝ ⎠
— Hence, safe
Dr.RajeshKN
Braced slender column design problem 2 Design the longitudinal reinforcement for a braced column, 300 mm × 400 mm, subject to a factored axial load of 1500 kN and factored moments of 60 kNm and 40 kNm with respect to the major axis and minor axis respectively at the top end. Assume that the column is bent in double curvature (in both directions) with the moments at the bottom end equal to 50 percent of the corresponding moments at top. Assume an uned length of 7.0 m and an effective length ratio of 0.85 in both directions. Use M 30 concrete and Fe 415 steel. Given: Dx = 400 mm, Dy = 300 mm, Pu = 1500 kN; Mux = 60 kNm, Muy = 40 kNm, at top; Mux = 30 kNm, Muy = 20 kNm, at bottom; l = 7000 mm, kx = ky = 0.85. Dept. of CE, GCE Kannur
Dr.RajeshKN
1. Slenderness ratios lex = ley = 0.85 × 7000 = 5950 mm lex/Dx = 5950/400 = 14.88 > 12 ley/Dy = 5950/300 = 19.83 > 12 Hence, the column should be designed as a slender column. 2. Minimum eccentricities
ex ,min
⎧7000 500 + 400 30 =⎨ ( greater ) = 27.33 mm 20 mm ⎩
e y ,min
⎧7000 500 + 300 30 =⎨ ( greater ) 20 mm ⎩
Dept. of CE, GCE Kannur
= 24 mm
Dr.RajeshKN
3. Primary moments for design As the column is braced and bent in double curvature,
M ux = 0.6 × 60 − 0.4 × 30 = 24 kNm ≥ 0.4 × 60 = 24
M uy = 0.6 × 40 − 0.4 × 20 = 16 kNm ≥ 0.4 × 40 = 16 Corresponding (primary) eccentricities: ex = 24 × 103/1500 = 16 mm < exmin = 27.33 mm ey = 16 × 103/1500 = 10.67 mm < eymin = 24.00 mm The primary eccentricities should not be less than the minimum eccentricities. Hence primary moments for design are: Mux = 1500 × (27.33 × 10–3) = 41.0 kNm Muy = 1500 × (24.00 × 10–3) = 36.0 kNm Dept. of CE, GCE Kannur
Dr.RajeshKN
4. Additional moments Without modification factor, additional eccentricities eax = Dx (lex/Dx)2/2000 = 400 (14.88)2/2000 = 44.28 mm eay = Dy (ley/Dy)2/2000 = 300 (19.83)2/2000 = 58.98 mm Assume modification factors kax = kay ≈ 0.5 (to be verified later), Additional moments: Max = Pu (kaxeax) = 1500(0.5 × 44.28 × 10-3) = 33.2 kNm May = Pu (kayeay) = 1500(0.5 × 58.98 × 10-3) = 44.2 kNm
5. Total factored moments
M ux = M ux + M ax = 41 + 33.2 = 74.2 kNm M uy = M uy + M ay = 36 + 44.2 = 80.2 kNm > M ux = 74.2 kNm Dept. of CE, GCE Kannur
Dr.RajeshKN
6. Trial section • Deg for a resultant uniaxial moment with respect to the minor axis,
M u ≅ 1.15 74.22 + 80.22 = 126 kNm combined with Pu = 1500 kNm Pu/(fckbD) = (1500 × 103/(30 × 400 × 300) = 0.417 Mu/(fckbD2) = 126 × 106/(30 × 400 × 3002) = 0.117 Assuming 25ϕ main bars, 8ϕ ties and 40mm clear cover, d' = 60.5 mm d’/D ≈ 60.5/300 = 0.201 ≈ 0.20 with “equal reinforcement on all sides”, referring to Chart 46 of SP : 16, p/fck = 0.13 Dept. of CE, GCE Kannur
Dr.RajeshKN
→ preqd = 0.13 × 30 = 3.9 → As,reqd = 3.9 × 300 × 400/100 = 4680 mm2 Provide 8–28 ϕ [As = 8 × 616 mm2] 4928 > 4680 → pprovided = 4928 × 100/(300 × 400) =4.107 → p/fck = 4.107/30 = 0.137
Dept. of CE, GCE Kannur
Dr.RajeshKN
7. Check additional moments • Assuming a clear cover of 40 mm, d’= 40 + 8 + 28/2 = 62 mm d’/Dx= 0.155 mm = 0.15 mm (say) d’/Dy= 0.207 mm = 0.2 mm (say) From Table 60 of SP :16, the ultimate loads Pb,x and Pb,y at balanced failure can be determined.
Pb , x p = k1 + k2 For d’/Dx = 0.15, k1= 0.196, k2= 0.203, f ck bD f ck
∴ Pb, x = 805.71kN For d’/Dy = 0.2, k1= 0.184, k2= 0.028,
Pb, y
p = k1 + k2 f ck bD f ck ∴ Pb, y = 676.21kN
Puz = 0.45fck Ag + (0.75fy – 0.45fck)As = (0.45 × 30 × 300 × 400) + (0.75 × 415 – 0.45 × 30) × 4928 = 3087 kN Dept. of CE, GCE Kannur
Dr.RajeshKN
Modification factors:
3087 − 1500 Puz − Pu = = 0.696 ≤ 1 kx = 3087 − 805.71 Puz − Pb , x
Puz − Pu 3087 − 1500 = = 0.658 ≤ 1 ky = Puz − Pb , y 3087 − 676.21 Hence, the actual (revised) total moments are:
M ux = 41.0 + 1500(0.696 × 44.28 × 10-3) = 87.23 kNm -3 M uy
= 36.0 + 1500(0.658 × 58.98 × 10 ) = 94.21 kNm
Dept. of CE, GCE Kannur
Dr.RajeshKN
8. Check safety under axial load with biaxial bending Referring to the design Charts in SP : 16, uniaxial moment capacities corresponding to Pu/fckbD = 0.417 and p/fck = 0.137 are obtained as: Mux1/fckbD2 = 0.135 (for d’/Dx = 0.15) Muy1/fckbD2 = 0.115 (for d’/Dy = 0.2) Mux1 = 0.135 × 30 × 300 × 4002 = 194.4 × 106 Nmm = 194.4 kNm
> M ux = 87.23kNm Muy1 = 0.115 × 30 × 400 × 3002 =124.2 × 106 Nmm = 124.2 kNm
> M uy = 94.21kNm
Pu/Puz = 1500/3087 = 0.486 (which lies between 0.2 and 0.8)
αn= 1.477 Dept. of CE, GCE Kannur
Dr.RajeshKN
αn
αn
⎛ M uy ⎞ ⎛ M ux ⎞ ⎟⎟ ⎜ ⎟ + ⎜⎜ ⎝ M ux1 ⎠ ⎝ M uy1 ⎠
1.477
⎛ 87.23 ⎞ =⎜ ⎟ 194 4 . ⎝ ⎠
= 0.971 < 1.0
Dept. of CE, GCE Kannur
1.477
⎛ 94.21 ⎞ +⎜ ⎟ 124 2 . ⎝ ⎠
— Hence, safe
Dr.RajeshKN
Footings
Dept. of CE, GCE Kannur
Dr.RajeshKN
The purpose of the foundation is to effectively the superstructure: 1. by transmitting the applied load effects (reactions in the form of vertical and horizontal forces and moments) to the soil below, 2. without exceeding the safe bearing capacity of the soil, 3. and ensuring that the settlement of the structure is within tolerable limits, and as nearly uniform as possible
• ‘Footings’ are shallow foundations used when soil of sufficient strength is available within a relatively short depth below the ground surface. Dept. of CE, GCE Kannur
Dr.RajeshKN
TYPES OF FOOTINGS P
P
P
STEPPED
FLAT
SLOPED
isolated footings P1
P2
P1
P2
property line
individual footing areas overlap
central beam (if required)
combined footings Dept. of CE, GCE Kannur
Dr.RajeshKN
Isolated Footings For ordinary structures located on reasonably firm soil, a separate isolated footing is provided for every column.
The soil bearing pressures from below tend to make the base slab of the footing bend upwards, hence the footing needs to be suitably reinforced by a mesh provided at the bottom
Dept. of CE, GCE Kannur
Dr.RajeshKN
Combined Footings If isolated footings are attempted in the following cases, it results in an overlap of areas • When two or more columns are close to each other and/or • if they are relatively heavily loaded and/or • if they rest on soil with low safe bearing capacity Combined Footings are provided in the above cases.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Soil pressures under isolated footings The plan area of a footing base slab is selected so as to limit the maximum soil bearing pressure induced below the footing to within a safe limit. • safe soil bearing capacity (‘allowable soil pressure’), qa, given to the structural designer by the geotechnical consultant is applicable for service load conditions, as qa includes the factor of safety. • Hence, the calculation for the required area of a footing must be based on qa and the service load effects. • The ‘partial load factors’ to be used for different load combinations (DL, LL, WL/EL) should, therefore, be those applicable for the serviceability limit state and not the ‘ultimate limit state’ when used in association with qa. Dept. of CE, GCE Kannur
Dr.RajeshKN
• the prescribed allowable soil pressure qa at a given depth is generally the gross pressure, which includes the pressure due to the existing overburden (soil up to the founding depth), and not the net pressure (in excess of the existing overburden pressure). • Hence, the total load to be considered in calculating the maximum soil pressure q (≤ qa) must include the weight of the footing itself and that of the backfill. • Often, these weights are initially taken as 10 – 15 percent of the axial load on the column
Dept. of CE, GCE Kannur
Dr.RajeshKN
Distribution of Base Pressure • The distribution of the soil reaction acting at the base of the footing depends on the rigidity of the footing as well as the properties of the soil. • The distribution of soil pressure is generally non-uniform. However, for convenience, a linear distribution of soil pressure is assumed in normal design practice. Concentrically Loaded Footings • In a symmetrically loaded footing, where the resultant vertical (service) load P + ΔP (where P is the load from the column and ΔP the weight of footing plus backfill) es through the centroid of the footing, the soil pressure is assumed to be uniformly distributed, and its magnitude q is given by
q= Dept. of CE, GCE Kannur
P + ΔP A
where A is the base area of the footing. Dr.RajeshKN
P GROUND LEVEL
backfill ∆ P footing FOUNDING LEVEL
GROSS SOIL PRESSURE
q = (P+∆P)/A L
B
Dept. of CE, GCE Kannur
area A = BL
Dr.RajeshKN
• Limiting q to the allowable soil pressure qa will give the minimum required area of footing: P + ΔP
Areqd =
qa
Eccentrically Loaded Footings • The load P acting on a footing may act eccentrically with respect to the centroid of the footing base due to: • the column transmitting a moment M in addition to the vertical load • the column carrying a vertical load offset with respect to the centroid of the footing • the column transmitting a lateral force located above the foundation level, in addition to the vertical load Dept. of CE, GCE Kannur
Dr.RajeshKN
P M e = M/P
P
e
ΔP
ΔP C FOOTING
C FOOTING
(a)
(b) P H e H
resultant thrust P
C FOOTING (c) Dept. of CE, GCE Kannur
Dr.RajeshKN
General design considerations • The major design considerations in the structural design of a footing relate to flexure, shear (both one-way and two-way action), bearing and bond (development length). • Deflection control is not a consideration in the design of footings which are buried underground (and hence not visible). • However, control of crack-width and protection of reinforcement by adequate cover are important serviceability considerations, particularly in aggressive environments • minimum cover prescribed in the Code (Cl. 26.4.2.2) is 50 mm • In order to compute the factored moments, shears, etc., acting at critical sections of the footing, a fictitious factored soil pressure qu, corresponding to the factored loads, should be considered. Dept. of CE, GCE Kannur
Dr.RajeshKN
Thickness of Footing Base Slab • The thickness of a footing base slab is generally based on considerations of shear and flexure, which are critical near the column location. • Generally, shear considerations predominate, and the thickness is based on shear criteria.
Both one-way shear and two-way shear (‘punching shear’) need to be considered [Cl. 34.2.4.1].
Dept. of CE, GCE Kannur
Dr.RajeshKN
critical sections for moment
L a B
a
b
d/2
b
b +d
d/2 Mu
Pu d
critical section for one–way shear
a +d
Vu1 d qu
Dept. of CE, GCE Kannur
critical section (all around) for two– way shear Vu2
(b)
(a)
Dr.RajeshKN
t/4
critical sections for moment
masonry wall
t
Pu
Pu Mu
Mu critical section for moment d
qu (c)
Dept. of CE, GCE Kannur
q u
(d)
Dr.RajeshKN
DESIGN EXAMPLE OF ISOLATED FOOTING
Design Problem 1: Square isolated footing, Concentrically loaded Design an isolated footing for a square column, 450 mm × 450 mm, reinforced with 8–25 φ bars, and carrying a service load of 2300 kN. Assume soil with a safe bearing capacity of 300 kN/m2 at a depth of 1.5 m below ground. Assume M 20 grade concrete and Fe 415 grade steel for the footing, and M 25 concrete and Fe 415 steel for the column.
Dept. of CE, GCE Kannur
Dr.RajeshKN
1. Size of footing Given: P = 2300 kN, qa = 300 kN/m2 (at h = 1.5 m below ground surface) Assuming the weight of the footing + backfill to be 10 % of the load P = 2300 kN, base area required =
2300 × 11 . 300
Minimum size of square footing =
= 8.43 m2
8.43
= 2.904 m
Assume a 3 m × 3 m footing base
Dept. of CE, GCE Kannur
Dr.RajeshKN
2. Thickness of footing slab based on shear Net soil pressure at ultimate loads (assuming a load factor of 1.5)
qu =
2300 × 1.5 = 383 kN/m2 3.0 × 3.0
(a) One-way shear
3000
The critical section is at a distance d from the column face Factored shear force
d 1275-d
450 3000
450
Vu1 = 0.383 N/mm2 × 3000 mm × (1275 – d) = (1464975 – 1149d) N. critical section for shear Dept. of CE, GCE Kannur
Dr.RajeshKN
Assuming τ c = 0.36 MPa (for M 20 concrete with, say, pt = 0.25) [Table 19], One-way shear resistance Vc1 = 0.36 × 3000 × d = (1080d) N Vu1 ≤ Vc1 1464975 – 1149d ≤ 1080d d ≥ 658 mm
Dept. of CE, GCE Kannur
Dr.RajeshKN
(b) Two-way shear The critical section is at d/2 from the periphery of the column ⇒ Factored shear force Vu2 = 0.383 × [30002 – (450 + d)2] Two-way shear resistance
where ks = 1.0 for a square column
τ c = 0.25 20 = 1.118 MPa (Cl. 31.6.3.1)
Dept. of CE, GCE Kannur
450 450
Vc2 = ksτ c × ⎡⎣4 × ( 450 + d ) d ⎤⎦
d/2
450+d
d/2 450+d critical section (all around) for two– way shear Vu2 Dr.RajeshKN
Vc2 = 1.0 × 1.118 × 4d (450 + d) = (2012.4d + 4.472d2) N
Vu2 ≤ Vc2 ⇒ 0.383 × [30002 – (450 + d)2] ≤ 2012.4d + 4.472d2 3369442.5 - 0.383 d2 - 900d ≤ 2012.4d + 4.472d2 4.855 d2 + 2912.4 d - 3369442.5 ≥ 0 d ≥ 585.5 Evidently, in this problem, one-way shear governs the thickness. Assuming a clear cover of 75 mm and 16 φ bars in both directions, with an average d = 658 mm, thickness D ≥ 658 + 75 + 16 = 749 mm Dept. of CE, GCE Kannur
Dr.RajeshKN
Provide D = 750 mm. The effective depths in the two directions will differ by one bar diameter, which is not significant in relatively deep square footings. For the purpose of flexural reinforcement calculations, an average value of d may be assumed as: d = 750 – 75 – 16 = 659 mm
Assuming unit weights of concrete and soil as 24 kN/m3 and 18kN/m3 respectively, actual gross pressure at footing base (under service loads) q = 2300/(3.0 × 3.0) + (24 × 0.75) + (18 × 0.75) = 287 kN/m2 < 300 kN/m2
Dept. of CE, GCE Kannur
— OK.
Dr.RajeshKN
3. Design of flexural reinforcement Factored moment at column face (in either direction): Mu = 0.383 × 3000 × 12752/2 = 933.9 × 106 Nmm 3000
=0.717 MPa
( pt ) reqd = 0.207 (Table 2, SP:16, Page 48)
3000
Mu 933.9 × 106 = 2 3000 × 6592 Bd
critical section for moment
450 450
1275
Ast,min = 0.0012BD = 0.0012 × 3000 × 750 = 2700 mm2 ⇒ pt,min = 100 × 2700/(3000 × 659) = 0.137 < 0.207 Dept. of CE, GCE Kannur
Dr.RajeshKN
But this reinforcement is less than the 0.25% assumed for one-way shear. Hence Ast,reqd = 0.25 × 3000 × 659/100 = 4943 mm2 Using 16 mm φ bars, number of bars required = 4943/201 = 25 [corresponding spacing s = {3000 – (75 × 2) – 16}/(25 –1) = 118 mm — is acceptable.] Provide 25 nos 16 φ bars both ways Required development length
Ld =
For M 20 concrete and Fe 415 steel,
φ ( 0.87 f y ) 4τ bd
Ld =
Cl. 26.2.1
0.87 × 415 φ = 47φ 4 × 1.2 × 1.6
For 16 φ bars in footing, Ld = 47.0 × 16 = 752 mm Length available = 1275 – 75 = 1200 mm > 752 mm — Hence, OK. Dept. of CE, GCE Kannur
Dr.RajeshKN
450
25 nos 16 φ both ways
750
dav = 659
75
75 3000 SECTION ‘XX’
section for moment section for one-way shear section for two-way shear
d/2 X 450
d
3000
d/2
X
d/2 450
(3000 – 450)/2 = 1275
PLAN
Dept. of CE, GCE Kannur
Dr.RajeshKN
4. Transfer of force at column base
Dept. of CE, GCE Kannur
Dr.RajeshKN
Design Problem 2: Isolated footing, Eccentrically loaded
Dept. of CE, GCE Kannur
Dr.RajeshKN
DESIGN OF COMBINED FOOTINGS
Dept. of CE, GCE Kannur
Dr.RajeshKN
Summary
Design of Compression : Effective length and classification as per IS 456, short columns subjected to axial compression with and without uniaxial/biaxial eccentricities; slender columns - Use of SP 16 charts. Design of Footings: Wall footings, isolated footings – axial and eccentric loading, rectangular and trapezoidal combined footings.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Design of Columns and Footings Dr. Rajesh K. N. Assistant Professor in Civil Engineering Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur
Dr.RajeshKN
(Analysis and design in Module II, III and IV should be based on Limit State Method. Reinforcement detailing shall conform to SP34) MODULE IV (13 hours) Design of Compression : Effective length and classification as per IS 456, short columns subjected to axial compression with and without uniaxial/biaxial eccentricities; slender columns - Use of SP 16 charts. Design of Footings: Wall footings, isolated footings – axial and eccentric loading, rectangular and trapezoidal combined footings.
2 Dept. of CE, GCE Kannur
Dr.RajeshKN
Columns
Dept. of CE, GCE Kannur
Dr.RajeshKN
Columns • A ‘compression member’ is a structural element which is subjected (predominantly) to axial compressive forces. • Compression are most commonly encountered in reinforced concrete buildings as columns • Column is a compression member, the ‘effective length’ of which exceeds three times the least lateral dimension (Cl. 25.1.1) • ‘Pedestal’ is a vertical compression member whose ‘effective length’ is less than three times its least lateral dimension [Cl. 26.5.3.1(h)].
Dept. of CE, GCE Kannur
Dr.RajeshKN
Classification of Columns Based on Type of Reinforcement
1) Tied columns: where the main longitudinal bars are enclosed within closely spaced lateral ties 2) Spiral columns: where the main longitudinal bars are enclosed within closely spaced and continuously wound spiral reinforcement 3) Composite columns: where the reinforcement is in the form of structural steel sections or pipes, with or without longitudinal bars
Dept. of CE, GCE Kannur
Dr.RajeshKN
Classification of Columns Based on Type of Loading 1. Columns with axial loading (applied concentrically) 2. Columns with uniaxial eccentric loading 3. Columns with biaxial eccentric loading
P
P e
axial compression Dept. of CE, GCE Kannur
axial compression with uniaxial bending.
ey ex
P
axial compression with biaxial bending. 6
Dr.RajeshKN
The combination of axial compression (Pu) with bending moment (Mu) at any column section
statically equivalent to
Load P applied with an eccentricity e = Mu/Pu with respect to the longitudinal centroidal axis of the column section.
7 Dept. of CE, GCE Kannur
Dr.RajeshKN
Classification of Columns Based on Slenderness Ratios (Cl. 25.1.2), • ‘Slenderness ratio’ is the ratio of ‘effective length’ to its lateral dimension • It provides a measure of the vulnerability to failure of the column by elastic instability (buckling). Buckling tendency varies with slenderness ratio.
1. Short columns: slenderness ratios with respect to the ‘major principal axis’ (lex/Dx) as well as the ‘minor principal axis’ (ley/Dy) are both less than 12 2. Slender (or long) columns.
Dept. of CE, GCE Kannur
Dr.RajeshKN
• Columns with low slenderness ratios fail under ultimate loads with the material (concrete, steel) reaching its ultimate compressive strength, and not by buckling
• Columns with very high slenderness ratios may undergo buckling (with large lateral deflection) under relatively low compressive loads, and fail suddenly.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Uned Length Code (Cl. 25.1.3) defines the ‘uned length’ l of a column explicitly for various types of constructions.
Effective length of a column The effective length of a column in a given plane is defined as the distance between the points of inflection in the buckled configuration of the column in that plane. The effective length depends on the uned length l and the boundary conditions at the column ends
Dept. of CE, GCE Kannur
Dr.RajeshKN
Effective lengths for idealised boundary conditions (Euler’s theory) End conditions
Effective length
Both ends hinged
l
One end fixed and the other end free One end fixed and the other hinged Both ends fixed
Dept. of CE, GCE Kannur
2l l
2
l 2
Dr.RajeshKN
Code recommendations for idealised boundary conditions (Cl. E–1)
Use of Code Charts Charts are given in Fig. 26 and Fig. 27 of the Code for determining the effective length ratios of braced columns and unbraced columns respectively in of coefficients β1 and β2 which represent the degrees of rotational freedom at the top and bottom ends of the column.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Recommended effective length ratios for normal usage (Table 28, Page 94) 1. columns braced against sideway: a) both ends ‘fixed’ rotationally b) one end ‘fixed’ and the other ‘pinned c) both ends ‘free’ rotationally (‘pinned’)
: 0.65 : 0.80 : 1.00
2. columns unbraced against sideway: a) both ends ‘fixed’ rotationally : 1.20 b) one end ‘fixed’ and the other ‘partially fixed’ : 1.50 c) one end ‘fixed’ and the other free : 2.00
Dept. of CE, GCE Kannur
Dr.RajeshKN
Eccentric loading The general case of loading on a compression member is axial compression combined with biaxial bending. • A state of biaxial eccentric compression • Axial load P acts eccentric to the longitudinal centroidal axis of the column cross section, with eccentricities ex and ey with respect to the major and minor principal axes
Dept. of CE, GCE Kannur
Dr.RajeshKN
Code requirements on slenderness limits, minimum eccentricities and reinforcement Slenderness Limits (Cl. 25.3.1) Ratio of the uned length (l) to the least lateral dimension (d) of a column
Furthermore, in case of cantilevered column, where D is the depth of the cross-section measured in the plane of the cantilever and b is the width (in the perpendicular direction) (Cl. 25.3.2) .
Dept. of CE, GCE Kannur
Dr.RajeshKN
Minimum eccentricities (Cl. 25.4) Very often, indirect eccentricities may arise due to: • lateral loads not considered in design; • live load placements not considered in design; • accidental lateral/eccentric loads; • errors in construction (such as misalignments); and • slenderness effects underestimated in design.
Hence, the Code requires every column to be designed for a minimum eccentricity emin (in any plane) equal to: • uned length/500 plus lateral dimension/30, • subject to a minimum of 20 mm.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Thus, for a column with a rectangular section:
ex ,min
⎧l 500 + Dx 30 =⎨ whichever is greater 20 mm ⎩
e y ,min
⎧l 500 + D y 30 =⎨ whichever is greater 20 mm ⎩
Dept. of CE, GCE Kannur
Dr.RajeshKN
Code Requirements on Reinforcement and Detailing Longitudinal Reinforcement (Cl. 26.5.3.1) Minimum Reinforcement: The longitudinal bars must have a cross sectional area not less than 0.8 percent of the gross area of the column section.
Maximum Reinforcement: The maximum cross-sectional area of longitudinal bars should not exceed 6 percent of the gross area of the column section. But a reduced maximum limit of 4 percent is recommended in general for better placement and compaction of concrete and also at lapped splice locations.
Dept. of CE, GCE Kannur
Dr.RajeshKN
A minimum limit is specified by the Code inorder to: • ensure nominal flexural resistance under unforeseen eccentricities in loading; and • prevent the yielding of the bars due to creep and shrinkage effects, which result in a transfer of load from the concrete to the steel.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Minimum diameter / number of bars and their location: Longitudinal bars in columns (and pedestals) should not be less than 12 mm in diameter and should not be spaced more than 300 mm apart (centre-tocentre) along the periphery of the column At least 4 bars should be provided in a column with rectangular cross-section (one at each corner), and at least 6 bars) in a circular column (equally spaced near the periphery.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Cover to reinforcement (Cl. 26.4.2.1) : • A minimum clear cover of 40 mm or bar diameter (whichever is greater), to the column ties is recommended for columns in general • In small-sized columns (D < 200 mm and whose reinforcing bars do not exceed 12mm) a reduced clear cover of 25 mm is permitted
Dept. of CE, GCE Kannur
Dr.RajeshKN
Transverse Reinforcement (Cl. 26.5.3.2) General: All longitudinal reinforcement in a compression member must be enclosed within transverse reinforcement, comprising either lateral ties (with internal angles not exceeding 135o) or spirals. This is required: • to prevent the premature buckling of individual bars; • to confine the concrete in the ‘core’, thus improving ductility and strength; • to hold the longitudinal bars in position during construction; and • to provide resistance against shear and torsion, if required. Dept. of CE, GCE Kannur
Dr.RajeshKN
Lateral Ties: tie diameter
tie spacing
⎧φlong ,max 4 φt = ⎨ ⎩ 6 mm
⎧ D ⎪ st = ⎨16φlong ,min ⎪ ⎩ 300 mm
where φlong is the diameter of longitudinal bar to be tied D is the least lateral dimension of the column
Dept. of CE, GCE Kannur
Dr.RajeshKN
Spirals: Helical reinforcement provides very good confinement to the concrete in the ‘core’ and enhances ductility of the column at ultimate loads. The diameter and pitch of the spiral may be computed as in the case of ties— except when the column is designed to carry a 5 percent overload (as permitted by the Code), in which case pitch
and
Dept. of CE, GCE Kannur
Dr.RajeshKN
DESIGN OF SHORT COLUMNS UNDER AXIAL COMPRESSION Equilibrium of Axial Loading If fcc and fsc denote respectively the stresses in the concrete and the longitudinal steel, corresponding to the uniform compressive strain εc , then it follows that
P0 = Cc + Cs = f cc Ac + f sc Asc = f cc ( Ag − Asc ) + f sc Asc = f cc Ag + ( f sc − f cc ) Asc Ag = gross area of cross-section = Ac + Asc ; Asc = total area of longitudinal reinforcement Ac = net area of concrete in the section = Ag – Asc Dept. of CE, GCE Kannur
Dr.RajeshKN
Design Strength of Axially Loaded Short Columns (Cl. 39.3) Where the calculated minimum eccentricity (in any plane) does not exceed 0.05 times the lateral dimension (in the plane considered), may be designed by a simplified equation:
Pu = 0.4 f ck Ag + ( 0.67 f y − 0.4 f ck ) Asc
Dept. of CE, GCE Kannur
Dr.RajeshKN
Design problem 1 Design the reinforcement in a column of size 450 mm x 600 mm, subject to an axial load of 2000 kN. The column has an uned length of 3.0m and is braced against sidesway in both directions. Use M 20 concrete and Fe 415 steel. 1. Short Column or Slender Column? Given: lx = ly = 3000 mm, Dy = 450 mm, Dx = 600 mm
⎧ lex Dx = k x l x Dx = k x × 3000 600 = 5k x slenderness ratios = ⎨ ⎩ley D y = k y l y D y = k y × 3000 450 = 6.67k y • Effective length ratios kx and ky are less than unity (since the column is braced against sideway in both directions,) •and hence the slenderness ratios are both less than 12. • Hence, the column may be designed as a short column. Dept. of CE, GCE Kannur
Dr.RajeshKN
2. Minimum Eccentricities
ex ,min ex ,min
e y ,min
⎧l 500 + Dx 30 =⎨ whichever is greater 20 mm ⎩
⎧3000 500 + 600 30 =⎨ whichever is greater 20 mm ⎩ ⎧3000 500 + 450 30 =⎨ whichever is greater 20 mm ⎩
= 26 mm
= 21mm
As 0.05Dx = 0.05 x 600 = 30.0 mm > ex,min = 26.0 mm and 0.05Dy = 0.05 x 450 = 22.5 mm > ey,min = 21.0 mm, the Code formula for axially loaded short columns can be used.
(
)
Pu = 0.4 f ck Ag + 0.67 f y − 0.4 f ck Asc Dept. of CE, GCE Kannur
Dr.RajeshKN
3. Factored Load
Pu = 2000 x 1.5 = 3000 kN
4. Design of Longitudinal Reinforcement
Pu = 0.4 f ck Ag + ( 0.67 f y − 0.4 f ck ) Asc 3000 × 103 = 0.4 × 20 × 450 × 600 + ( 0.67 × 415 − 0.4 × 20 ) Asc Asc = 3111 mm2 Provide 4–25 ϕ at corners : 4 x 491 = 1964 mm2 and 4–20 ϕ additional: 4 x 314 = 1256 mm2 Asc = 3220 mm2 > 3111 mm2 p = (100x3220) / (450x600) = 1.192 > 0.8 (minimum reinf.) — OK. Dept. of CE, GCE Kannur
Dr.RajeshKN
5. Lateral Ties: tie diameter
⎧φlong ,max 4 φt = ⎨ ⎩ 6 mm
tie spacing
⎧ D 450 ⎧ ⎪ st = ⎨16φlong ,min = ⎪⎨16 × 20 = 320 ⎪ ⎪ 300 mm ⎩ 300 mm ⎩
⎧ 25 4 =⎨ ⎩6 mm
Hence provide 8 φ ties @ 300 c/c
Dept. of CE, GCE Kannur
Dr.RajeshKN
6. Detailing
600
450 4#20ϕ 8ϕ@300c/c 4#25ϕ
Dept. of CE, GCE Kannur
Dr.RajeshKN
DESIGN OF SHORT COLUMNS UNDER COMPRESSION WITH UNIAXIAL BENDING Design problem 2 Using the design aids given in SP:16, design the longitudinal reinforcement in a rectangular reinforced concrete column of size 300 mm × 600 mm subjected to a factored load of 1400 kN and a factored moment of 280 kNm with respect to the major axis. Assume M20 concrete and Fe415 steel. Given: b = 300 mm, D = 600 mm, fck = 20 MPa, fy = 415 MPa, Pu = 1400 kN, Mux = 280 kNm Arrangement of bars: SP : 16 Charts for “equal reinforcement on four sides” may be used. (Charts 43-46 ) Dept. of CE, GCE Kannur
Dr.RajeshKN
Assuming an effective cover d‘ = 60 mm
d‘/D = 60/600 = 0.1
Pu 1400 × 103 = 0.389 = f ck bd 20 × 300 × 600 280 × 10 Mu = 2 = 0.13 2 20 300 600 × × f ck bd 6
Referring to Chart 44 of SP : 16 preqd /20 = 0.11
d‘/D = 0.1
preqd = 0.11 × 20 = 2.2
As,reqd = 2.2 × 300 × 600/100 = 3960 mm2 Detailing of longitudinal reinforcement Dept. of CE, GCE Kannur
Dr.RajeshKN
Detailing
600 8ϕ@200c/c staggered 300
4#22ϕ inner rows
Dept. of CE, GCE Kannur
4#28ϕ at corners
Dr.RajeshKN
DESIGN OF SHORT COLUMNS UNDER AXIAL COMPRESSION WITH BIAXIAL BENDING Code Procedure for Design of Biaxially Loaded Columns (Cl. 39.6) The simplified method adopted by the Code is based on Bresler’s formulation An approximate relationship between MuR,x and MuR,y (for a specified Pu = PuR) is established. This relationship is conveniently expressed in a non-dimensional form αn αn as follows:
⎛ M uy ⎞ ⎛ M ux ⎞ ⎟⎟ ≤ 1 ⎜ ⎟ + ⎜⎜ ⎝ M ux1 ⎠ ⎝ M uy1 ⎠
where Mux and Muy denote the factored biaxial moments acting on the column, and Mux1 and Muy1 denote the uniaxial moment capacities with reference to the major and minor axes respectively, all under an accompanying axial load Pu = PuR. Dept. of CE, GCE Kannur
Dr.RajeshKN
In biaxial loading, Code (Cl. 39.6) suggests the following equation, applicable for all grades of steel:
Puz = 0.45 f ck Ac + 0.75 f y Asc Puz = 0.45 f ck Ag + ( 0.75 f y − 0.45 f ck ) Asc where Ag denotes the gross area of the section and Asc the total area of steel in the section. αn = 1 for Pu/Puz < 0.2; αn = 2 for Pu/Puz > 0.8; and αn is assumed to vary linearly for values of Pu/Puz between 0.2 and 0.8
Dept. of CE, GCE Kannur
Dr.RajeshKN
Code Procedure 1. Given Pu, Mux , Muy , that the eccentricities ex = Mux / Pu and ey = Muy / Pu are not less than the corresponding minimum eccentricities. 2. Assume a trial section for the column. 3. Determine Mux1 and Muy1 , corresponding to the given Pu (using SP:16). Ensure that Mux1 and Muy1 are significantly greater than Mux and Muy respectively; otherwise, suitably redesign the section. 4. Determine Puz , and hence αn 5. Check the adequacy of the section; if necessary, redesign the section and check again.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Design problem 3 A column (400 mm × 400 mm), effectively held in position but restrained against rotation at one end, is subjected to factored loads: Pu = 1300 kN, Mux = 190 kNmand Muy = 110 kNm. The uned length of the column is 3.5m. Design the reinforcement in the column, assuming M 25 concrete and Fe 415 steel. Given: Dx = Dy = 400 mm, l = 3500 mm, Pu = 1300 kN, Mux = 190 kNm, Muy = 110 kNm, fck = 25MPa, fy = 415MPa. 1. Slenderness ratios Effective length = 0.8l for the braced column, lex = ley = 0.8 × 3500 = 2800 mm lex/Dx = ley/Dy = 2800/400 = 7.0 < 12 Hence the column may be designed as a short column. Dept. of CE, GCE Kannur
Dr.RajeshKN
2. Check minimum eccentricities Applied eccentricities: ex = Mux / Pu = 190 × 103/1300 = 146 mm ey = Muy / Pu = 110 × 103/1300 = 84.6 mm Minimum eccentricities as per Code: ex, min = ey, min = 3500/500 + 400/30 = 20.3 mm > 20 mm The applied eccentricities are larger than the minimum eccentricities.
Dept. of CE, GCE Kannur
Dr.RajeshKN
3. Trial section: Longitudinal reinforcement Assume longitudinal reinforcement percentage as 3.5%. As,reqd = 3.5 × 4002/100 = 5600 mm2 Provide 12 – 25 dia: As = 491 × 12 = 5892 mm2 > 5600 mm2. The arrangement of bars is shown in Figure??.
Deg for uniaxial eccentricity with Pu = 1300 kN and Uniaxial moment capacities: Mux1, Muy1 [Here, due to symmetry, Mux1 = Mux2]
Pu f ck bd Dept. of CE, GCE Kannur
= 0.325 Dr.RajeshKN
pprovided = 5892 × 100/4002 = 3.68 p/fck = 3.68/25 = 0.147 d’= 40 + 8 + 25/2 = 60.5 mm (assuming a clear cover of 40 mm and 8 mm ties) d’/D = 60.5/400 = 0.151 = 0.15 (say) Referring to Chart 45
Mu f ck bd 2
=0.165
Mux1 = Muy1 = 0.165 × 25 × 4003 = 264 × 106 Nmm = 264 kNm which is significantly greater than Mux = 190 kNm and Muy = 110 kNm
Dept. of CE, GCE Kannur
Dr.RajeshKN
4. Check for biaxial bending Values of Puz and αn
Puz = 0.45 f ck Ag + ( 0.75 f y − 0.45 f ck ) Asc = (0.45 × 25 × 4002) + (0.75 × 415 – 0.45 × 25) × 5892 = (1800 × 103 + 1767.6 × 103) N = 3568 kN Pu/Puz = 1300/3568 = 0.364 (which lies between 0.2 and 0.8)
αn= 1.273 αn
αn
⎛ M uy ⎞ ⎛ M ux ⎞ ⎟⎟ ⎜ ⎟ + ⎜⎜ ⎝ M ux1 ⎠ ⎝ M uy1 ⎠
1.273
⎛ 190 ⎞ =⎜ ⎟ ⎝ 264 ⎠
1.273
⎛ 110 ⎞ +⎜ ⎟ ⎝ 264 ⎠
= 0.986 < 1.0
Hence, the trial section is safe under the applied loading. Dept. of CE, GCE Kannur
Dr.RajeshKN
5. Transverse reinforcement The minimum diameter ϕt and maximum spacing st of the lateral ties are specified by the Code
tie diameter
⎧φlong ,max 4 φt = ⎨ ⎩ 6 mm
tie spacing
⎧ D 400 ⎧ ⎪ st = ⎨16φlong ,min = ⎪⎨16 × 25 = 400 ⎪ ⎪ 300 mm ⎩ 300 mm ⎩
⎧ 25 4 =⎨ ⎩6 mm
Provide 8 ϕ ties@ 300 c/c
Dept. of CE, GCE Kannur
Dr.RajeshKN
6. Detailing
Dept. of CE, GCE Kannur
Dr.RajeshKN
DESIGN OF SLENDER COLUMNS Code Procedures for Design of Slender Columns a. Strength Reduction Coefficient Method (B-3.3 of the Code) • This is a highly simplified procedure, which is given in the Code for the working stress method of design • Although this method has been prescribed for WSM, it can be extended to the limit state method (LSM) for the case of axial loading (without primary bending moments)
Dept. of CE, GCE Kannur
Dr.RajeshKN
According to this procedure the permissible stresses in concrete and steel are reduced by multiplying with a strength reduction coefficient Cr given by:
le Cr = 1.25 − 48d
where d is the least lateral dimension of the column (or diameter of the core in a spiral column). Alternatively, for more exact calculations,
Cr = 1.25 −
le 160rmin
where rmin is the least radius of gyration of the column.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Additional Moment Method (Cl. 39.7.1) The method prescribed by the Code for slender column design by the limit state method is the ‘additional moment method’ According to this method, every slender column should be designed for biaxial eccentricities (which include the P–∆ moment “additional moment”) components eax = Max/Pu and eay = May/Pu :
M ax
P D ⎧l ⎫ = u ⎨ ex ⎬ 2000 ⎩ D ⎭
The total design moments
2
M ay
Pu b ⎧ ley ⎫ = ⎨ ⎬ 2000 ⎩ b ⎭
2
M ux = M ux + M ax = Pu ( ex + eax ) M uy = M uy + M ay = Pu ( e y + eay )
Dept. of CE, GCE Kannur
Dr.RajeshKN
Mux, Muy
are the primary factored moments (obtained from first-order structural analyses)
Max, May
are the additional moments with reference to bending about the major and minor axes respectively.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Code recommends the following modifications for the design of slender columns in general: The additional moments Max, May may be reduced by multiplying factors (Cl. 39.7.1.1) defined as:
Puz − Pu ≤1 kx = Puz − Pb , x
Puz − Pu ≤1 ky = Puz − Pb , y
Pb is the axial load corresponding to the condition of max compressive strain of 0.0035 in concrete and tensile strain of 0.002 in outermost layer of compression steel
Dept. of CE, GCE Kannur
Dr.RajeshKN
• For braced columns subject to unequal primary moments M1, M2 at the two ends, the value of Mu to be considered in the computation of the total moment M u
M u = 0.4 M 1 + 0.6 M 2 ≥ 0.4 M 2 where M2 is the higher column end moment. Also, M u ≥ M 2 • M1 and M2 are considered to be of opposite signs if the column is bent in double curvature.
Dept. of CE, GCE Kannur
Dr.RajeshKN
• For unbraced columns, the lateral drift effect needs to be included. An approximate way for this is by assuming that the additional moment Ma acts at the column end where the maximum primary moment M2 is present. Hence, for design purposes, the total moment may be taken as
M u = M 2 + M a
Dept. of CE, GCE Kannur
for unbraced columns
Dr.RajeshKN
Procedure for design of slender columns 1. Determine the effective depth and slenderness ratio in each direction. 2. a) Determine initial moment Mui from given primary end moments Mu1 and Mu2 in each direction. b) Calculate emin and Mu,min in each direction. c) Compare the moments computed in steps a and b and take the greater of the two as initial moment Mui in each direction. 3. a) Find additional moment Ma in each direction. b) Find total moment Mut in each direction, without considering reduction factor ka . c) Make preliminary design for Pu and Mut and find area of steel. Thus p is known.
Dept. of CE, GCE Kannur
Dr.RajeshKN
4. a) Get Puz . Also find Pb each direction, for the reinforcement ration p obtained above. b) Find reduction factor ka in each direction. c) Determine the modified design value of moment in each direction Mut = Mui + ka Ma. 5. Redesign the column for Pu and Mut . If the column is slender about both the axes, design the column for biaxial bending, for (Pu , Muxt ) about x-axis and (Pu , Muyt ) about y-axis.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Braced slender column design problem 1 Design the longitudinal reinforcement for a braced column, 300 mm × 400 mm, subject to a factored axial load of 1200 kN and factored moments of 50 kNm and 30 kNm with respect to the major axis and minor axis respectively at the top end. Assume that the column is bent in double curvature (in both directions) with the moments at the bottom end equal to 40 percent of the corresponding moments at top. Assume an uned length of 7.2 m and an effective length ratio of 0.85 in both directions. Use M 20 concrete and Fe 415 steel. Effective cover is 60 mm. Given: Dx = 400 mm, Dy = 300 mm, Pu = 1200 kN; Mux = 50 kNm, Muy = 30 kNm, at top; Mux = 0.4x50 = 20 kNm, Muy = 0.4x30 = 12 kNm, at bottom; l = 7200 mm, kx = ky = 0.85. Dept. of CE, GCE Kannur
Dr.RajeshKN
1. Slenderness ratios
lex = ley = 0.85 × 7200 = 6120 mm lex/Dx = 6120/400 = 15.3 > 12 ley/Dy = 6120/300 = 20.4 > 12
Hence, the column is slender about both the axes. 2. Minimum eccentricities & Minimum moments
ex ,min
⎧7200 500 + 400 30 =⎨ ( greater ) = 27.73 mm 20 mm ⎩
e y ,min
⎧7200 500 + 300 30 =⎨ ( greater ) = 24.4 mm 20 mm ⎩
M ux ,min = Pu .ex ,min = 1200 × 27.73 = 33.276 kNm M uy ,min = Pu .e y ,min = 1200 × 24.4 Dept. of CE, GCE Kannur
= 29.28 kNm Dr.RajeshKN
3. Primary moments for design As the column is braced and bent in double curvature,
M ux = 0.6 × 50 − 0.4 × 20 = 22 kNm ≥ 0.4 × 50 = 20
M uy = 0.6 × 30 − 0.4 × 12 = 13.2 kNm ≥ 0.4 × 30 = 12 Since the minimum moments are larger than these, the primary moments for design are:
M ux = 33.276 kNm
Dept. of CE, GCE Kannur
M uy = 29.28 kNm
Dr.RajeshKN
4. Additional moments Without modification factor, additional eccentricities eax = Dx (lex/Dx)2/2000 = 400 (15.3)2/2000 = 46.82 mm eay = Dy (ley/Dy)2/2000 = 300 (20.4)2/2000 = 62.42 mm Assume modification factors kax = kay ≈ 0.5 (to be verified later), Additional moments: Max = Pu (kaxeax) = 1200(0.5 × 46.82 × 10-3) = 28.092 kNm May = Pu (kayeay) = 1200(0.5 × 62.42 × 10-3) = 37.452 kNm
5. Total factored moments
M ux = M ux + M ax = 33.276 + 28.092 = 61.368 kNm M uy = M uy + M ay = 29.28 + 37.452 = 66.732 kNm > M ux = 61.368 kNm Dept. of CE, GCE Kannur
Dr.RajeshKN
6. Trial section • Deg for a resultant uniaxial moment with respect to the minor axis (y-axis),
M u ≅ 1.15 61.3682 + 66.7322 = 104.26 kNm combined with Pu = 1200 kNm Pu/(fckbD) = (1200 × 103/(30 × 400 × 300) = 0.5 Mu/(fckbD2) = 104.26 × 106/(30 × 400 × 3002) = 0.144 d’/D ≈ 60/300 = 0.20 with “equal reinforcement on all sides”, referring to Chart 46 of SP : 16, p/fck = 0.19 Dept. of CE, GCE Kannur
Dr.RajeshKN
→ preqd = 0.19 × 20 = 3.8 → As,reqd = 3.8 × 300 × 400/100 = 4560 mm2 Provide 10–25 ϕ bars. 4909 > 4560 → pprovided = 4909 × 100/(300 × 400) =4.09 → p/fck = 4.09/20 = 0.2045
Dept. of CE, GCE Kannur
Dr.RajeshKN
7. Check additional moments d’/Dx= 60/400 = 0.15 mm d’/Dy= 60/300 = 0.2 mm From Table 60 of SP :16, the ultimate loads Pb,x and Pb,y at balanced failure can be determined.
Pb , x p = k1 + k2 For d’/Dx = 0.15, k1= 0.196, k2= 0.203, f ck bD f ck ∴ Pb, x = 570.03kN For d’/Dy = 0.2, k1= 0.184, k2= 0.028,
Pb, y
p = k1 + k2 f ck bD f ck ∴ Pb, y = 455.34 kN
Puz = 0.45fck Ag + (0.75fy – 0.45fck)As = (0.45 × 20 × 300 × 400) + (0.75 × 415 – 0.45 × 20) × 4909 = 2563.7 kN Dept. of CE, GCE Kannur
Dr.RajeshKN
Modification factors:
Puz − Pu 2563.65 − 1200 kx = = = 0.684 ≤ 1 Puz − Pb , x 2563.65 − 570.03 2563.65 − 1200 Puz − Pu = = 0.647 ≤ 1 ky = 2563.65 − 455.34 Puz − Pb , y Hence, the actual (revised) total moments are:
M ux = 33.276 + 1200(0.684 × 46.82 × 10-3) = 71.706 kNm -3 M uy
= 29.28 + 1200(0.647 × 62.42 × 10 ) = 77.74 kNm
Dept. of CE, GCE Kannur
Dr.RajeshKN
8. Check safety under axial load with biaxial bending Referring to the design Charts 45 & 46 in SP : 16, uniaxial moment capacities corresponding to Pu/fckbD = 0.5 and p/fck = 0.2045 are obtained as: Mux1/fckbD2 = 0.18 (for d’/Dx = 0.15) Muy1/fckbD2 = 0.152 (for d’/Dy = 0.2)
= 71.706 kNm Mux1 = 0.18 × 30 × 300 × 4002 = 172.8 kNm > M ux = 77.74 kNm Muy1 = 0.152 × 30 × 400 × 3002 = 109.44 kNm > M uy
Pu/Puz = 1200/2563.7 = 0.468 (which lies between 0.2 and 0.8)
αn= 1.447 Dept. of CE, GCE Kannur
Dr.RajeshKN
αn
αn
⎛ M uy ⎞ ⎛ M ux ⎞ ⎟⎟ ⎜ ⎟ + ⎜⎜ ⎝ M ux1 ⎠ ⎝ M uy1 ⎠
1.447
⎛ 71.706 ⎞ =⎜ ⎟ 172 8 . ⎝ ⎠
= 0.89 < 1.0
Dept. of CE, GCE Kannur
1.447
⎛ 77.74 ⎞ +⎜ ⎟ 109 44 . ⎝ ⎠
— Hence, safe
Dr.RajeshKN
Braced slender column design problem 2 Design the longitudinal reinforcement for a braced column, 300 mm × 400 mm, subject to a factored axial load of 1500 kN and factored moments of 60 kNm and 40 kNm with respect to the major axis and minor axis respectively at the top end. Assume that the column is bent in double curvature (in both directions) with the moments at the bottom end equal to 50 percent of the corresponding moments at top. Assume an uned length of 7.0 m and an effective length ratio of 0.85 in both directions. Use M 30 concrete and Fe 415 steel. Given: Dx = 400 mm, Dy = 300 mm, Pu = 1500 kN; Mux = 60 kNm, Muy = 40 kNm, at top; Mux = 30 kNm, Muy = 20 kNm, at bottom; l = 7000 mm, kx = ky = 0.85. Dept. of CE, GCE Kannur
Dr.RajeshKN
1. Slenderness ratios lex = ley = 0.85 × 7000 = 5950 mm lex/Dx = 5950/400 = 14.88 > 12 ley/Dy = 5950/300 = 19.83 > 12 Hence, the column should be designed as a slender column. 2. Minimum eccentricities
ex ,min
⎧7000 500 + 400 30 =⎨ ( greater ) = 27.33 mm 20 mm ⎩
e y ,min
⎧7000 500 + 300 30 =⎨ ( greater ) 20 mm ⎩
Dept. of CE, GCE Kannur
= 24 mm
Dr.RajeshKN
3. Primary moments for design As the column is braced and bent in double curvature,
M ux = 0.6 × 60 − 0.4 × 30 = 24 kNm ≥ 0.4 × 60 = 24
M uy = 0.6 × 40 − 0.4 × 20 = 16 kNm ≥ 0.4 × 40 = 16 Corresponding (primary) eccentricities: ex = 24 × 103/1500 = 16 mm < exmin = 27.33 mm ey = 16 × 103/1500 = 10.67 mm < eymin = 24.00 mm The primary eccentricities should not be less than the minimum eccentricities. Hence primary moments for design are: Mux = 1500 × (27.33 × 10–3) = 41.0 kNm Muy = 1500 × (24.00 × 10–3) = 36.0 kNm Dept. of CE, GCE Kannur
Dr.RajeshKN
4. Additional moments Without modification factor, additional eccentricities eax = Dx (lex/Dx)2/2000 = 400 (14.88)2/2000 = 44.28 mm eay = Dy (ley/Dy)2/2000 = 300 (19.83)2/2000 = 58.98 mm Assume modification factors kax = kay ≈ 0.5 (to be verified later), Additional moments: Max = Pu (kaxeax) = 1500(0.5 × 44.28 × 10-3) = 33.2 kNm May = Pu (kayeay) = 1500(0.5 × 58.98 × 10-3) = 44.2 kNm
5. Total factored moments
M ux = M ux + M ax = 41 + 33.2 = 74.2 kNm M uy = M uy + M ay = 36 + 44.2 = 80.2 kNm > M ux = 74.2 kNm Dept. of CE, GCE Kannur
Dr.RajeshKN
6. Trial section • Deg for a resultant uniaxial moment with respect to the minor axis,
M u ≅ 1.15 74.22 + 80.22 = 126 kNm combined with Pu = 1500 kNm Pu/(fckbD) = (1500 × 103/(30 × 400 × 300) = 0.417 Mu/(fckbD2) = 126 × 106/(30 × 400 × 3002) = 0.117 Assuming 25ϕ main bars, 8ϕ ties and 40mm clear cover, d' = 60.5 mm d’/D ≈ 60.5/300 = 0.201 ≈ 0.20 with “equal reinforcement on all sides”, referring to Chart 46 of SP : 16, p/fck = 0.13 Dept. of CE, GCE Kannur
Dr.RajeshKN
→ preqd = 0.13 × 30 = 3.9 → As,reqd = 3.9 × 300 × 400/100 = 4680 mm2 Provide 8–28 ϕ [As = 8 × 616 mm2] 4928 > 4680 → pprovided = 4928 × 100/(300 × 400) =4.107 → p/fck = 4.107/30 = 0.137
Dept. of CE, GCE Kannur
Dr.RajeshKN
7. Check additional moments • Assuming a clear cover of 40 mm, d’= 40 + 8 + 28/2 = 62 mm d’/Dx= 0.155 mm = 0.15 mm (say) d’/Dy= 0.207 mm = 0.2 mm (say) From Table 60 of SP :16, the ultimate loads Pb,x and Pb,y at balanced failure can be determined.
Pb , x p = k1 + k2 For d’/Dx = 0.15, k1= 0.196, k2= 0.203, f ck bD f ck
∴ Pb, x = 805.71kN For d’/Dy = 0.2, k1= 0.184, k2= 0.028,
Pb, y
p = k1 + k2 f ck bD f ck ∴ Pb, y = 676.21kN
Puz = 0.45fck Ag + (0.75fy – 0.45fck)As = (0.45 × 30 × 300 × 400) + (0.75 × 415 – 0.45 × 30) × 4928 = 3087 kN Dept. of CE, GCE Kannur
Dr.RajeshKN
Modification factors:
3087 − 1500 Puz − Pu = = 0.696 ≤ 1 kx = 3087 − 805.71 Puz − Pb , x
Puz − Pu 3087 − 1500 = = 0.658 ≤ 1 ky = Puz − Pb , y 3087 − 676.21 Hence, the actual (revised) total moments are:
M ux = 41.0 + 1500(0.696 × 44.28 × 10-3) = 87.23 kNm -3 M uy
= 36.0 + 1500(0.658 × 58.98 × 10 ) = 94.21 kNm
Dept. of CE, GCE Kannur
Dr.RajeshKN
8. Check safety under axial load with biaxial bending Referring to the design Charts in SP : 16, uniaxial moment capacities corresponding to Pu/fckbD = 0.417 and p/fck = 0.137 are obtained as: Mux1/fckbD2 = 0.135 (for d’/Dx = 0.15) Muy1/fckbD2 = 0.115 (for d’/Dy = 0.2) Mux1 = 0.135 × 30 × 300 × 4002 = 194.4 × 106 Nmm = 194.4 kNm
> M ux = 87.23kNm Muy1 = 0.115 × 30 × 400 × 3002 =124.2 × 106 Nmm = 124.2 kNm
> M uy = 94.21kNm
Pu/Puz = 1500/3087 = 0.486 (which lies between 0.2 and 0.8)
αn= 1.477 Dept. of CE, GCE Kannur
Dr.RajeshKN
αn
αn
⎛ M uy ⎞ ⎛ M ux ⎞ ⎟⎟ ⎜ ⎟ + ⎜⎜ ⎝ M ux1 ⎠ ⎝ M uy1 ⎠
1.477
⎛ 87.23 ⎞ =⎜ ⎟ 194 4 . ⎝ ⎠
= 0.971 < 1.0
Dept. of CE, GCE Kannur
1.477
⎛ 94.21 ⎞ +⎜ ⎟ 124 2 . ⎝ ⎠
— Hence, safe
Dr.RajeshKN
Footings
Dept. of CE, GCE Kannur
Dr.RajeshKN
The purpose of the foundation is to effectively the superstructure: 1. by transmitting the applied load effects (reactions in the form of vertical and horizontal forces and moments) to the soil below, 2. without exceeding the safe bearing capacity of the soil, 3. and ensuring that the settlement of the structure is within tolerable limits, and as nearly uniform as possible
• ‘Footings’ are shallow foundations used when soil of sufficient strength is available within a relatively short depth below the ground surface. Dept. of CE, GCE Kannur
Dr.RajeshKN
TYPES OF FOOTINGS P
P
P
STEPPED
FLAT
SLOPED
isolated footings P1
P2
P1
P2
property line
individual footing areas overlap
central beam (if required)
combined footings Dept. of CE, GCE Kannur
Dr.RajeshKN
Isolated Footings For ordinary structures located on reasonably firm soil, a separate isolated footing is provided for every column.
The soil bearing pressures from below tend to make the base slab of the footing bend upwards, hence the footing needs to be suitably reinforced by a mesh provided at the bottom
Dept. of CE, GCE Kannur
Dr.RajeshKN
Combined Footings If isolated footings are attempted in the following cases, it results in an overlap of areas • When two or more columns are close to each other and/or • if they are relatively heavily loaded and/or • if they rest on soil with low safe bearing capacity Combined Footings are provided in the above cases.
Dept. of CE, GCE Kannur
Dr.RajeshKN
Soil pressures under isolated footings The plan area of a footing base slab is selected so as to limit the maximum soil bearing pressure induced below the footing to within a safe limit. • safe soil bearing capacity (‘allowable soil pressure’), qa, given to the structural designer by the geotechnical consultant is applicable for service load conditions, as qa includes the factor of safety. • Hence, the calculation for the required area of a footing must be based on qa and the service load effects. • The ‘partial load factors’ to be used for different load combinations (DL, LL, WL/EL) should, therefore, be those applicable for the serviceability limit state and not the ‘ultimate limit state’ when used in association with qa. Dept. of CE, GCE Kannur
Dr.RajeshKN
• the prescribed allowable soil pressure qa at a given depth is generally the gross pressure, which includes the pressure due to the existing overburden (soil up to the founding depth), and not the net pressure (in excess of the existing overburden pressure). • Hence, the total load to be considered in calculating the maximum soil pressure q (≤ qa) must include the weight of the footing itself and that of the backfill. • Often, these weights are initially taken as 10 – 15 percent of the axial load on the column
Dept. of CE, GCE Kannur
Dr.RajeshKN
Distribution of Base Pressure • The distribution of the soil reaction acting at the base of the footing depends on the rigidity of the footing as well as the properties of the soil. • The distribution of soil pressure is generally non-uniform. However, for convenience, a linear distribution of soil pressure is assumed in normal design practice. Concentrically Loaded Footings • In a symmetrically loaded footing, where the resultant vertical (service) load P + ΔP (where P is the load from the column and ΔP the weight of footing plus backfill) es through the centroid of the footing, the soil pressure is assumed to be uniformly distributed, and its magnitude q is given by
q= Dept. of CE, GCE Kannur
P + ΔP A
where A is the base area of the footing. Dr.RajeshKN
P GROUND LEVEL
backfill ∆ P footing FOUNDING LEVEL
GROSS SOIL PRESSURE
q = (P+∆P)/A L
B
Dept. of CE, GCE Kannur
area A = BL
Dr.RajeshKN
• Limiting q to the allowable soil pressure qa will give the minimum required area of footing: P + ΔP
Areqd =
qa
Eccentrically Loaded Footings • The load P acting on a footing may act eccentrically with respect to the centroid of the footing base due to: • the column transmitting a moment M in addition to the vertical load • the column carrying a vertical load offset with respect to the centroid of the footing • the column transmitting a lateral force located above the foundation level, in addition to the vertical load Dept. of CE, GCE Kannur
Dr.RajeshKN
P M e = M/P
P
e
ΔP
ΔP C FOOTING
C FOOTING
(a)
(b) P H e H
resultant thrust P
C FOOTING (c) Dept. of CE, GCE Kannur
Dr.RajeshKN
General design considerations • The major design considerations in the structural design of a footing relate to flexure, shear (both one-way and two-way action), bearing and bond (development length). • Deflection control is not a consideration in the design of footings which are buried underground (and hence not visible). • However, control of crack-width and protection of reinforcement by adequate cover are important serviceability considerations, particularly in aggressive environments • minimum cover prescribed in the Code (Cl. 26.4.2.2) is 50 mm • In order to compute the factored moments, shears, etc., acting at critical sections of the footing, a fictitious factored soil pressure qu, corresponding to the factored loads, should be considered. Dept. of CE, GCE Kannur
Dr.RajeshKN
Thickness of Footing Base Slab • The thickness of a footing base slab is generally based on considerations of shear and flexure, which are critical near the column location. • Generally, shear considerations predominate, and the thickness is based on shear criteria.
Both one-way shear and two-way shear (‘punching shear’) need to be considered [Cl. 34.2.4.1].
Dept. of CE, GCE Kannur
Dr.RajeshKN
critical sections for moment
L a B
a
b
d/2
b
b +d
d/2 Mu
Pu d
critical section for one–way shear
a +d
Vu1 d qu
Dept. of CE, GCE Kannur
critical section (all around) for two– way shear Vu2
(b)
(a)
Dr.RajeshKN
t/4
critical sections for moment
masonry wall
t
Pu
Pu Mu
Mu critical section for moment d
qu (c)
Dept. of CE, GCE Kannur
q u
(d)
Dr.RajeshKN
DESIGN EXAMPLE OF ISOLATED FOOTING
Design Problem 1: Square isolated footing, Concentrically loaded Design an isolated footing for a square column, 450 mm × 450 mm, reinforced with 8–25 φ bars, and carrying a service load of 2300 kN. Assume soil with a safe bearing capacity of 300 kN/m2 at a depth of 1.5 m below ground. Assume M 20 grade concrete and Fe 415 grade steel for the footing, and M 25 concrete and Fe 415 steel for the column.
Dept. of CE, GCE Kannur
Dr.RajeshKN
1. Size of footing Given: P = 2300 kN, qa = 300 kN/m2 (at h = 1.5 m below ground surface) Assuming the weight of the footing + backfill to be 10 % of the load P = 2300 kN, base area required =
2300 × 11 . 300
Minimum size of square footing =
= 8.43 m2
8.43
= 2.904 m
Assume a 3 m × 3 m footing base
Dept. of CE, GCE Kannur
Dr.RajeshKN
2. Thickness of footing slab based on shear Net soil pressure at ultimate loads (assuming a load factor of 1.5)
qu =
2300 × 1.5 = 383 kN/m2 3.0 × 3.0
(a) One-way shear
3000
The critical section is at a distance d from the column face Factored shear force
d 1275-d
450 3000
450
Vu1 = 0.383 N/mm2 × 3000 mm × (1275 – d) = (1464975 – 1149d) N. critical section for shear Dept. of CE, GCE Kannur
Dr.RajeshKN
Assuming τ c = 0.36 MPa (for M 20 concrete with, say, pt = 0.25) [Table 19], One-way shear resistance Vc1 = 0.36 × 3000 × d = (1080d) N Vu1 ≤ Vc1 1464975 – 1149d ≤ 1080d d ≥ 658 mm
Dept. of CE, GCE Kannur
Dr.RajeshKN
(b) Two-way shear The critical section is at d/2 from the periphery of the column ⇒ Factored shear force Vu2 = 0.383 × [30002 – (450 + d)2] Two-way shear resistance
where ks = 1.0 for a square column
τ c = 0.25 20 = 1.118 MPa (Cl. 31.6.3.1)
Dept. of CE, GCE Kannur
450 450
Vc2 = ksτ c × ⎡⎣4 × ( 450 + d ) d ⎤⎦
d/2
450+d
d/2 450+d critical section (all around) for two– way shear Vu2 Dr.RajeshKN
Vc2 = 1.0 × 1.118 × 4d (450 + d) = (2012.4d + 4.472d2) N
Vu2 ≤ Vc2 ⇒ 0.383 × [30002 – (450 + d)2] ≤ 2012.4d + 4.472d2 3369442.5 - 0.383 d2 - 900d ≤ 2012.4d + 4.472d2 4.855 d2 + 2912.4 d - 3369442.5 ≥ 0 d ≥ 585.5 Evidently, in this problem, one-way shear governs the thickness. Assuming a clear cover of 75 mm and 16 φ bars in both directions, with an average d = 658 mm, thickness D ≥ 658 + 75 + 16 = 749 mm Dept. of CE, GCE Kannur
Dr.RajeshKN
Provide D = 750 mm. The effective depths in the two directions will differ by one bar diameter, which is not significant in relatively deep square footings. For the purpose of flexural reinforcement calculations, an average value of d may be assumed as: d = 750 – 75 – 16 = 659 mm
Assuming unit weights of concrete and soil as 24 kN/m3 and 18kN/m3 respectively, actual gross pressure at footing base (under service loads) q = 2300/(3.0 × 3.0) + (24 × 0.75) + (18 × 0.75) = 287 kN/m2 < 300 kN/m2
Dept. of CE, GCE Kannur
— OK.
Dr.RajeshKN
3. Design of flexural reinforcement Factored moment at column face (in either direction): Mu = 0.383 × 3000 × 12752/2 = 933.9 × 106 Nmm 3000
=0.717 MPa
( pt ) reqd = 0.207 (Table 2, SP:16, Page 48)
3000
Mu 933.9 × 106 = 2 3000 × 6592 Bd
critical section for moment
450 450
1275
Ast,min = 0.0012BD = 0.0012 × 3000 × 750 = 2700 mm2 ⇒ pt,min = 100 × 2700/(3000 × 659) = 0.137 < 0.207 Dept. of CE, GCE Kannur
Dr.RajeshKN
But this reinforcement is less than the 0.25% assumed for one-way shear. Hence Ast,reqd = 0.25 × 3000 × 659/100 = 4943 mm2 Using 16 mm φ bars, number of bars required = 4943/201 = 25 [corresponding spacing s = {3000 – (75 × 2) – 16}/(25 –1) = 118 mm — is acceptable.] Provide 25 nos 16 φ bars both ways Required development length
Ld =
For M 20 concrete and Fe 415 steel,
φ ( 0.87 f y ) 4τ bd
Ld =
Cl. 26.2.1
0.87 × 415 φ = 47φ 4 × 1.2 × 1.6
For 16 φ bars in footing, Ld = 47.0 × 16 = 752 mm Length available = 1275 – 75 = 1200 mm > 752 mm — Hence, OK. Dept. of CE, GCE Kannur
Dr.RajeshKN
450
25 nos 16 φ both ways
750
dav = 659
75
75 3000 SECTION ‘XX’
section for moment section for one-way shear section for two-way shear
d/2 X 450
d
3000
d/2
X
d/2 450
(3000 – 450)/2 = 1275
PLAN
Dept. of CE, GCE Kannur
Dr.RajeshKN
4. Transfer of force at column base
Dept. of CE, GCE Kannur
Dr.RajeshKN
Design Problem 2: Isolated footing, Eccentrically loaded
Dept. of CE, GCE Kannur
Dr.RajeshKN
DESIGN OF COMBINED FOOTINGS
Dept. of CE, GCE Kannur
Dr.RajeshKN
Summary
Design of Compression : Effective length and classification as per IS 456, short columns subjected to axial compression with and without uniaxial/biaxial eccentricities; slender columns - Use of SP 16 charts. Design of Footings: Wall footings, isolated footings – axial and eccentric loading, rectangular and trapezoidal combined footings.
Dept. of CE, GCE Kannur
Dr.RajeshKN
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