sd i-module4- rajesh sir

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

Design of Concrete Structures


2

(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 Members: 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.


Columns


Columns

• A ‘compression member’ is a structural element which is subjected (predominantly) to axial compressive forces.

• Compression members 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)].


1) Tied columns: where the main longitudinal bars are enclosed within closely spaced lateral ties

2) Spiral columns: where the main longitudinal bars are enclosedwithin 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

Classification of Columns Based on Type of Reinforcement


6

1. Columns with axial loading (applied concentrically)

2. Columns with uniaxial eccentric loading

3. Columns with biaxial eccentric loading

Classification of Columns Based on Type of Loading

P

ex

ey

axial compression with biaxial bending.

P

e

axial compression with uniaxial bending.

P

axial compression

Dept. of CE, GCE Kannur Dr.RajeshKN7

statically equivalent to

The combination of axial compression (Pu) with bending moment (Mu) at any column section

Load P applied with an eccentricity e = Mu/Pu with respect to the longitudinal centroidal axis of the column section.


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.

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.


• 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.


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 unsupported length l and the boundary conditions at the column ends

Code (Cl. 25.1.3) defines the ‘unsupported length’ l of a column explicitly for various types of constructions.

Unsupported Length


Both ends fixed

One end fixed and the other hinged

One end fixed and the other end free

Both ends hinged

Effective lengthEnd conditions

2l

2l

l

2l

Effective lengths for idealised boundary conditions (Euler’s theory)


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 terms of coefficients β1 and β2 which represent the degrees of rotational freedom at the top and bottom ends of the column.

Code recommendations for idealised boundary conditions (Cl. E–1)


Recommended effective length ratios for normal usage (Table 28, Page 94)

1. columns braced against sideway:

a) both ends ‘fixed’ rotationally : 0.65 b) one end ‘fixed’ and the other ‘pinned : 0.80c) both ends ‘free’ rotationally (‘pinned’) : 1.00

2. columns unbraced against sideway:

a) both ends ‘fixed’ rotationally : 1.20b) one end ‘fixed’ and the other ‘partially fixed’ : 1.50c) one end ‘fixed’ and the other free : 2.00


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


Code requirements on slenderness limits, minimumeccentricities and reinforcement

Slenderness Limits (Cl. 25.3.1)

Ratio of the unsupported 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) .


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:

• unsupported length/500 plus lateral dimension/30, • subject to a minimum of 20 mm.

Minimum eccentricities (Cl. 25.4)


Thus, for a column with a rectangular section:

500 3020

whichever is greatermm,min

xx

l De

+⎧= ⎨⎩

500 30

20whichever is greater

mm,miny

y

l De

+⎧= ⎨⎩


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.


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.


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-to-centre) 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.


• 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

Cover to reinforcement (Cl. 26.4.2.1) :


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.


tie diameter

tie spacing

Lateral Ties:

φlong is the diameter of longitudinal bar to be tied D is the least lateral dimension of the column

4

6mm,maxlong

t

φφ

⎧= ⎨⎩

16

300mm,mint long

Ds φ

⎧⎪= ⎨⎪⎩

where


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

Spirals:


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

Ag = gross area of cross-section = Ac + Asc ;Asc = total area of longitudinal reinforcement Ac = net area of concrete in the section = Ag – Asc

0 c sP C C= + cc c sc scf A f A= +

( )cc g sc sc scf A A f A= − +

( )cc g sc cc scf A f f A= + −


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), members may be designed by a simplified equation:

( )0 4 0 67 0 4. . .u ck g y ck scP f A f f A= + −


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 unsupported 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

slenderness ratios

• 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.

Design problem 1

3000 600 53000 450 6 67.

ex x x x x x x

ey y y y y y y

l D k l D k kl D k l D k k

= = × =⎧= ⎨ = = × =⎩


2. Minimum Eccentricities

500 3020

whichever is greatermm,min

xx

l De

+⎧= ⎨⎩

3000 500 450 3020

whichever is greatermm,minye+⎧

= ⎨⎩

3000 500 600 3020

whichever is greatermm,minxe+⎧

= ⎨⎩

26mm=

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.



3. Factored Load Pu = 2000 x 1.5 = 3000 kN

4. Design of Longitudinal Reinforcement

( )33000 10 0 4 20 450 600 0 67 415 0 4 20. . . scA× = × × × + × − ×

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.



tie diameter

tie spacing

5. Lateral Ties:4

6mm,maxlong

t

φφ

⎧= ⎨⎩

16

300mm,mint long

Ds φ

⎧⎪= ⎨⎪⎩

25 46mm⎧

= ⎨⎩

45016 20 320

300mm

⎧⎪= × =⎨⎪⎩

Hence provide 8 φ ties @ 300 c/c


600

450

4#20ϕ

4#25ϕ8ϕ@300c/c

6. Detailing


DESIGN OF SHORT COLUMNS UNDER COMPRESSION WITHUNIAXIAL BENDING

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

Design problem 2

Arrangement of bars:

SP : 16 Charts for “equal reinforcement on four sides” may be used.(Charts 43-46 )


Assuming an effective cover d‘ = 60 mm d‘/D = 60/600 = 0.1

u

ck

Pf bd

31400 1020 300 600

×=

× ×0 389.=

2u

ck

Mf bd

6

2280 10

20 300 600×

=× × 0 13.=

Referring to Chart 44 of SP : 16 d‘/D = 0.1

preqd = 0.11 × 20 = 2.2

As,reqd = 2.2 × 300 × 600/100 = 3960 mm2

Detailing of longitudinal reinforcement

preqd /20 = 0.11


Detailing

600

300

4#28ϕ at corners4#22ϕ inner rows

8ϕ@200c/c staggered


Code Procedure for Design of Biaxially Loaded Columns (Cl. 39.6)

DESIGN OF SHORT COLUMNS UNDER AXIAL COMPRESSIONWITH BIAXIAL BENDING

The simplified method adopted by the Code is based on Bresler’sformulation

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 as follows:

1 1

1nn

uyux

ux uy

MMM M

αα ⎛ ⎞⎛ ⎞+ ≤⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

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 anaccompanying axial load Pu = PuR.


In biaxial loading, Code (Cl. 39.6) suggests the following equation, applicable for all grades of steel:

0 45 0 75. .uz ck c y scP f A f A= +

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; andαn is assumed to vary linearly for values of Pu/Puz between 0.2 and 0.8

( )0 45 0 75 0 45. . .uz ck g y ck scP f A f f A= + −


1. Given Pu, Mux , Muy , verify that the eccentricities ex = Mux / Puand 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.

Code Procedure


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 unsupported length of the column is 3.5m. Design the reinforcement in the column, assuming M 25 concrete and Fe 415 steel.

Design problem 3

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.


2. Check minimum eccentricities

Applied eccentricities:

ex = Mux / Pu = 190 × 103/1300 = 146 mmey = 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.


Assume longitudinal reinforcement percentage as 3.5%.

Designing for uniaxial eccentricity with Pu = 1300 kN and

3. Trial section: Longitudinal reinforcement

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??.

Uniaxial moment capacities: Mux1, Muy1 [Here, due to symmetry, Mux1= Mux2]

= 0.325u

ck

Pf bd


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

2u

ck

Mf bd

=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


Values of Puz and αn

= (0.45 × 25 × 4002) + (0.75 × 415 – 0.45 × 25) × 5892= (1800 × 103 + 1767.6 × 103) N = 3568 kNPu/Puz = 1300/3568 = 0.364 (which lies between 0.2 and 0.8)

( )0 45 0 75 0 45. . .uz ck g y ck scP f A f f A= + −

αn= 1.273

1 1

nn

uyux

ux uy

MMM M

αα ⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 273 1 273190 110264 264

. .⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 0.986 < 1.0

Hence, the trial section is safe under the applied loading.

4. Check for biaxial bending


5. Transverse reinforcement

The minimum diameter ϕt and maximum spacing st of the lateral ties are specified by the Code

tie diameter

tie spacing

4

6mm,maxlong

t

φφ

⎧= ⎨⎩

16

300mm,mint long

Ds φ

⎧⎪= ⎨⎪⎩

25 46mm⎧

= ⎨⎩

40016 25 400

300mm

⎧⎪= × =⎨⎪⎩

Provide 8 ϕ ties@ 300 c/c


6. Detailing


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)


where d is the least lateral dimension of the column (or diameter of thecore in a spiral column). Alternatively, for more exact calculations,

1 25160 min

. er

lCr

= −

where rmin is the least radius of gyration of the column.

According to this procedure the permissible stresses in concrete and steel are reduced by multiplying with a strength reduction coefficient Crgiven by:

1 2548

. er

lCd

= −


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 :

2

2000u ex

axP D lM

D⎧ ⎫= ⎨ ⎬⎩ ⎭

2

2000eyu

ay

lP bMb

⎧ ⎫= ⎨ ⎬

⎩ ⎭

The total design moments ( )ux ux ax u x axM M M P e e= + = +

( )uy uy ay u y ayM M M P e e= + = +


Mux, Muyare the primary factored moments (obtained from first-order structural analyses)

are the additional moments with reference to bending about the major and minor axes respectively.

Max, May


1,

uz ux

uz b x

P PkP P

−= ≤

−

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

The additional moments Max, May may be reduced by multiplying factors (Cl. 39.7.1.1) defined as:

1,

uz uy

uz b y

P PkP P

−= ≤

−

Code recommends the following modifications for the design of slender columns in general:


• 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 uM

1 2 20 4 0 6 0 4. . .uM M M M= + ≥

2uM M≥

where M2 is the higher column end moment.

Also,

• M1 and M2 are considered to be of opposite signs if the column is bent in double curvature.


Hence, for design purposes, the total moment may be taken as

2u aM M M= + for unbraced columns

• 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.


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.

Procedure for design of slender columns


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.


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 unsupported 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.

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.

Given:

Braced slender column design problem 1


1. Slenderness ratios lex = ley = 0.85 × 7200 = 6120 mm

lex/Dx = 6120/400 = 15.3 > 12ley/Dy = 6120/300 = 20.4 > 12

Hence, the column is slender about both the axes.

2. Minimum eccentricities & Minimum moments

( )7200 500 300 3020

greatermm,minye+⎧

= ⎨⎩

( )7200 500 400 3020

greatermm,minxe+⎧

= ⎨⎩

27 73 mm.=

24 4 mm.=

,min ,min.ux u xM P e= 1200 27 73.= × 33 276kNm.=

,min ,min.uy u yM P e= 1200 24 4.= × 29 28kNm.=


3. Primary moments for design

As the column is braced and bent in double curvature,

0 6 50 0 4 20 22 kNm. .uxM = × − × = 0 4 50 20.≥ × =

0 6 30 0 4 12 13 2 kNm. . .uyM = × − × = 0 4 30 12.≥ × =

Since the minimum moments are larger than these, the primary moments for design are:

33 276kNm.uxM = 29 28kNm.uyM =


Without modification factor, additional eccentricitieseax = Dx (lex/Dx)2/2000 = 400 (15.3)2/2000 = 46.82 mmeay = 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 kNmMay = Pu (kayeay) = 1200(0.5 × 62.42 × 10-3) = 37.452 kNm

4. Additional moments

ux ux axM M M= + 33 276 28 092 61 368kNm. . .= + =

uy uy ayM M M= + 29 28 37 452 66 732 kNm. . .= + = 61 368kNm.uxM> =

5. Total factored moments


• Designing for a resultant uniaxial moment with respect to the minor axis (y-axis),

2 21 15 61 368 66 732. . .uM ≅ + = 104.26 kNm

6. Trial section

combined with Pu = 1200 kNm

Pu/(fckbD) = (1200 × 103/(30 × 400 × 300) = 0.5Mu/(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


→ 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


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.

7. Check additional moments

For d’/Dx = 0.15, k1= 0.196, k2= 0.203,

Puz = 0.45fck Ag + (0.75fy – 0.45fck)As

= (0.45 × 20 × 300 × 400) + (0.75 × 415 – 0.45 × 20) × 4909 = 2563.7 kN

1 2,b x

ck ck

P pk kf bD f

= +

570 03kN, .b xP∴ =

For d’/Dy = 0.2, k1= 0.184, k2= 0.028, 1 2,b y

ck ck

P pk kf bD f

= +

455 34 kN, .b yP∴ =


Modification factors:

,

uz ux

uz b x

P PkP P

−=

−2563 65 1200 0 684 1

2563 65 570 03. .

. .−

= = ≤−

,

uz uy

uz b y

P PkP P

−=

−

Hence, the actual (revised) total moments are:

= 33.276 + 1200(0.684 × 46.82 × 10-3) = 71.706 kNm

= 29.28 + 1200(0.647 × 62.42 × 10-3) = 77.74 kNm

ux

uy

M

M

2563 65 1200 0 647 12563 65 455 34

. .. .

−= = ≤

−


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)

8. Check safety under axial load with biaxial bending

Mux1 = 0.18 × 30 × 300 × 4002 = 172.8 kNm

77 74 kNm.uyM> =Muy1 = 0.152 × 30 × 400 × 3002 = 109.44 kNm

Pu/Puz = 1200/2563.7 = 0.468 (which lies between 0.2 and 0.8)

αn= 1.447

71 706kNm.uxM> =


1 1

nn

uyux

ux uy

MMM M

αα ⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 447 1 44771 706 77 74172 8 109 44

. .. .. .

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 0.89 < 1.0 — Hence, safe


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 unsupported length of 7.0 m and an effective length ratio of 0.85 in both directions. Use M 30 concrete and Fe 415 steel.

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.

Given:

Braced slender column design problem 2


1. Slenderness ratios

lex = ley = 0.85 × 7000 = 5950 mm

lex/Dx = 5950/400 = 14.88 > 12ley/Dy = 5950/300 = 19.83 > 12

Hence, the column should be designed as a slender column.

2. Minimum eccentricities

( )7000 500 300 3020

greatermm,minye+⎧

= ⎨⎩

( )7000 500 400 3020

greatermm,minxe+⎧

= ⎨⎩

27 33 mm.=

24 mm=


3. Primary moments for design

As the column is braced and bent in double curvature,

0 6 60 0 4 30 24 kNm. .uxM = × − × = 0 4 60 24.≥ × =

0 6 40 0 4 20 16kNm. .uyM = × − × = 0 4 40 16.≥ × =

Corresponding (primary) eccentricities:ex = 24 × 103/1500 = 16 mm < exmin = 27.33 mmey = 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 kNmMuy = 1500 × (24.00 × 10–3) = 36.0 kNm


Without modification factor, additional eccentricitieseax = Dx (lex/Dx)2/2000 = 400 (14.88)2/2000 = 44.28 mmeay = 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 kNmMay = Pu (kayeay) = 1500(0.5 × 58.98 × 10-3) = 44.2 kNm

4. Additional moments

ux ux axM M M= + 41 33 2 74 2 kNm. .= + =

uy uy ayM M M= + 36 44 2 80 2 kNm. .= + = 74 2 kNm.uxM> =

5. Total factored moments


• Designing for a resultant uniaxial moment with respect to the minor axis,

2 21 15 74 2 80 2. . .uM ≅ + = 126 kNm

6. Trial section

combined with Pu = 1500 kNm

Pu/(fckbD) = (1500 × 103/(30 × 400 × 300) = 0.417Mu/(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


→ 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


From Table 60 of SP :16, the ultimate loads Pb,x and Pb,y at balanced failure can be determined.

7. Check additional moments

For d’/Dx = 0.15, k1= 0.196, k2= 0.203, 1 2,b x

ck ck

P pk kf bD f

= +

805 71kN, .b xP∴ =

For d’/Dy = 0.2, k1= 0.184, k2= 0.028, 1 2,b y

ck ck

P pk kf bD f

= +

676 21kN, .b yP∴ =

d’/Dx= 0.155 mm = 0.15 mm (say)d’/Dy= 0.207 mm = 0.2 mm (say)

• Assuming a clear cover of 40 mm, d’= 40 + 8 + 28/2 = 62 mm

Puz = 0.45fck Ag + (0.75fy – 0.45fck)As

= (0.45 × 30 × 300 × 400) + (0.75 × 415 – 0.45 × 30) × 4928 = 3087 kN


Modification factors:

,

uz ux

uz b x

P PkP P

−=

−3087 1500 0 696 1

3087 805 71.

.−

= = ≤−

,

uz uy

uz b y

P PkP P

−=

−3087 1500 0 658 1

3087 676 21.

.−

= = ≤−

Hence, the actual (revised) total moments are:

= 41.0 + 1500(0.696 × 44.28 × 10-3) = 87.23 kNm

= 36.0 + 1500(0.658 × 58.98 × 10-3) = 94.21 kNm

ux

uy

M

M


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)

8. Check safety under axial load with biaxial bending

Mux1 = 0.135 × 30 × 300 × 4002 = 194.4 × 106 Nmm = 194.4 kNm

94 21kNm.uyM> =Muy1 = 0.115 × 30 × 400 × 3002 =124.2 × 106 Nmm = 124.2 kNm

Pu/Puz = 1500/3087 = 0.486 (which lies between 0.2 and 0.8)

αn= 1.477

87 23kNm.uxM> =


1 1

nn

uyux

ux uy

MMM M

αα ⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 477 1 47787 23 94 21194 4 124 2

. .. .. .

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 0.971 < 1.0 — Hence, safe


Footings


The purpose of the foundation is to effectively support 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.


TYPES OF FOOTINGS

P

FLAT

P

STEPPED

P

SLOPED

isolated footings

combined footings

P1 P2

central beam(if required)

individual footingareas overlap

P1 P2

property line


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 suitablyreinforced by a mesh provided at the bottom


Combined Footings

• 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

If isolated footings are attempted in the following cases, it results in an overlap of areas

Combined Footings are provided in the above cases.


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.


• 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 netpressure (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


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.

• In a symmetrically loaded footing, where the resultant vertical (service) load P + ΔP (where P is the load from the column and ΔPthe weight of footing plus backfill) passes through the centroid of the footing, the soil pressure is assumed to be uniformly distributed, and its magnitude q is given by

Concentrically Loaded Footings

q P PA

=+ Δ

where A is the base area of the footing.


L

Barea A = BL

P

GROUND LEVEL

backfillfooting

∆P

FOUNDING LEVEL GROSS SOIL PRESSURE

q = (P+∆P)/A


• Limiting q to the allowable soil pressure qa will give the minimum required area of footing:

A P Pqreqda

=+ Δ

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


P

ΔP

FOOTING

Pe

C

(b)

P

ΔP

FOOTING

M

e = M/P

C

(a)

C FOOTING

e

P

H

H

resultant thrust

(c)


General design considerations

• 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.

• 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 footingswhich 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


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].


a

bd/2

d/2

critical section (allaround) for two–way shear Vu2

a +d

(b)

b +d

L

B b

a

critical sectionsfor moment

critical section forone–way shear

PuMu d

d

qu(a)

Vu1


(d)

PuMu

t/4

critical section formoment

qu

masonrywallt

d

critical sections formoment

PuMu

qu(c)


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.


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× . = 8.43 m2

Minimum size of square footing = 8.43 = 2.904 m

Assume a 3 m × 3 m footing base


2. Thickness of footing slab based on shear

Net soil pressure at ultimate loads (assuming a load factor of 1.5)

qu =××

2300 1 53 0 3 0

.. . = 383 kN/m2

(a) One-way shear

The critical section is at a distance d from the column face

Factored shear force Vu1 = 0.383 N/mm2 × 3000 mm × (1275 – d)

= (1464975 – 1149d) N.

3000

3000

critical section for shear

450450

d 1275-d


Assuming 0 36 MPa.cτ =

(for M 20 concrete with, say, pt = 0.25) [Table 19],

One-way shear resistance Vc1 = 0.36 × 3000 × d = (1080d) N

Vu1 ≤ Vc11464975 – 1149d ≤ 1080d

d ≥ 658 mm


(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

( )2 4 450c s cV k d dτ ⎡ ⎤= × × +⎣ ⎦

where ks = 1.0 for a square column

0 25 20 .cτ =

= 1.118 MPa (Cl. 31.6.3.1)

450

450 d/2

d/2

critical section (allaround) for two–way shear Vu2

450+d

450+d


Vc2 = 1.0 × 1.118 × 4d (450 + d)= (2012.4d + 4.472d2) N

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

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


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 — OK.


3. Design of flexural reinforcement

Factored moment at column face (in either direction):Mu = 0.383 × 3000 × 12752/2 = 933.9 × 106 Nmm

6

2 2933 9 103000 659

.uMBd

×=

×=0.717 MPa

0 02 7( ) .t reqdp =

Ast,min = 0.0012BD = 0.0012 × 3000 × 750 = 2700 mm2

⇒ pt,min = 100 × 2700/(3000 × 659) = 0.137 < 0.207

3000

3000 450

450

critical sectionfor moment

1275

(Table 2, SP:16, Page 48)


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 ( )0 87

4. y

dbd

fL

φτ

= Cl. 26.2.1

For M 20 concrete and Fe 415 steel, 0 87 415 474 1 2 1 6

.. .dL φ φ×

= =× ×

For 16 φ bars in footing, Ld = 47.0 × 16 = 752 mmLength available = 1275 – 75 = 1200 mm > 752 mm — Hence, OK.


dav = 659

450

750

3000

75

3000450

d/2

d/2

d/2d

450

section for moment

section forone-way shear

section fortwo-way shear

25 nos 16 φ both ways

(3000 – 450)/2= 1275

75

XX

PLAN

SECTION ‘XX’


4. Transfer of force at column base


Design Problem 2: Isolated footing, Eccentrically loaded


DESIGN OF COMBINED FOOTINGS


Summary

Design of Compression Members: 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.

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