Combined Footing Vtu Document

7/23/2019 Combined Footing Vtu Document

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DESIGN AND DETAILING OF STEEL IN COMBINED FOOTINGS(For class held from 14-22 MAY 07)

Dr. M. C. Nataraja,

Professor, Civil Engineering Department,

Sri Jayachamarajendra Collge of Engineering, Mysore-5a70 006

Phone: 0821-2343521, 9880447742

e-mail: [email protected]

Combined footing

Whenever two or more columns in a straight line are carried on a single spread footing, it is

called a combined footing. Isolated footings for each column are generally the economical.

Combined footings are provided only when it is absolutely necessary, as

i) When two columns are close together, causing overlap of adjacent isolatedfootings

ii) Where soil bearing capacity is low, causing overlap of adjacent isolated footings

iii) Proximity of building line or existing building or sewer, adjacent to a building

column.

The combined footing may be rectangular, trapezoidal or Tee-shaped in plan. The geometric

proportions and shape are so fixed that the centeroid of the footing area coincides with the

resultant of the column loads. This results in uniform pressure below the entire area of

footing.

Trapezoidal footing is provided when one column load is much more than the other. As a

result, the both projections of footing beyond the faces of the columns will be restricted.Rectangular footing is provided when one of the projections of the footing is restricted or the

width of the footing is restricted.

Rectangular combined footing

Longitudinally, the footing acts as an upward loaded beam spanning between columns and

cantilevering beyond. Using statics, the shear force and bending moment diagrams in the

longitudinal direction are drawn. Moment is checked at the faces of the column. Shear force is

critical at distance d from the faces of columns or at the point of contra flexure. Two-wayshear is checked under the heavier column.

The footing is also subjected to transverse bending and this bending is spread over atransverse strip near the column.

Combined footing may be of slab type or slab and beam type or slab and strap beam type.


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Design:

1. Locate the point of application of the column loads on the footing.

2. Proportion the footing such that the resultant of loads passes through the center of

footing.3. Compute the area of footing such that the allowable soil pressure is not exceeded.

4. Calculate the shear forces and bending moments at the salient points and hence draw

SFD and BMD.

5. Fix the depth of footing from the maximum bending moment.

6. Calculate the transverse bending moment and design the transverse section for depth

and reinforcement. Check for anchorage and shear.

7. Check the footing for longitudinal shear and hence design the longitudinal steel

8. Design the reinforcement for the longitudinal moment and place them in the

appropriate positions.

9. Check the development length for longitudinal steel

10. Curtail the longitudinal bars for economy

11. Draw and detail the reinforcement12. Prepare the bar bending schedule


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P2

l ba

x RCombined footing with loads, (Slab type)

P1

L/2L/2

Types of combined footings (Plan)


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Design of combined footingSlab and Beam type

Two interior columns A and B carry 700 kN and 1000 kN loads respectively. Column A is

350 mm x 350 mm and column B is 400 mm X 400 mm in section. The centre to centre

spacing between columns is 4.6 m. The soil on which the footing rests is capable of providingresistance of 130 kN/m

2. Design a combined footing by providing a central beam joining the

two columns. Use concrete grade M25 and mild steel reinforcement.

Solution: Data

fck= 25 Nlmm2,

fy= 250 N/mm2,

fb (SBC)= l30 kN/m2,

Column A = 350 mm x 350 mm,

Column B = 400 mm x 400 mm,

c/c spacing of columns = 4.6 m ,

PA = 700 kN and PB = 1000kN

Required: To design combined footing with central beam joining the two columns.

Ultimate loads

PuA= 1.5 x 700 = 1050 kN, PuB = 1.5 x 1000 = 1500 kN

Proportioning of base sizeWorking load carried by column A = PA

= 700 kNWorking load carried by column B = PB = 1000 kN

Self weight of footing 10 % x (PA + PB) = 170 kN

Total working load = 1870 kN

Required area of footing = Af= Total load /SBC=1870/130 = 14.38 m2

Let the width of the footing = Bf = 2m

Required length of footing = Lf= Af/Bf = 14.38/2 = 7.19m

Provide footing of size 7.2m X 2m, Af= 7.2 x 2 = 14.4 m2

For uniform pressure distribution the C.G. of the footing should coincide with the C.G. of

column loads. Let x be the distance of C.G. from the centre line of column A


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For M25 and Fe 250, Q u max = 3.71 N/mm2

Required effective depth = (60.2 x 106/(3.71 x 1000)) = 128 mmSince the slab is in contact with the soil, clear cover of 50 mm is assumed.

Using 20 mm diameter bars

Required total depth = 128 + 20/2 + 50 =188 mm say 200 mmProvided effective depth = d = 200-50-20/2 = 140 mm

To find steel

Mu/bd2

=3.073.73, URS

Pt=1.7%

Ast = 2380 mm2

Use 20 mm diameter bar at spacing = 1000 x 314 / 23 84 say 130 mmArea provided =1000 x 314 / 130 = 2415 mm

2

Check the depth for one - way shear considerations

Design shear force=Vu=177x(0.825-0.140)=121kNNominal shear stress=v=Vu/bd=121000/(1000x140)=0.866MPaPermissible shear stress

Pt = 100 x 2415 /(1000 x 140 ) = 1.7 %, uc = 0.772 N/mm2

Value of k for 200 mm thick slab = 1.2

Permissible shear stress = 1.2 x 0.772 = 0.926 N/mm2

uc > v and hence safe

The depth may be reduced uniformly to 150 mm at the edges.

Check for development lengthLdt= [0.87 x 250 / (4 x 1.4)] =39 = 39 x 20 = 780 mm

Available length of bar = 825 - 25 =800mm > 780 mm and hence safe.

Transverse reinforcementRequired A st = 0.15 bD / 100 = 0.15 x 1000 x 200/100 = 300mm

2

Using 8 mm bars, spacing = 1000 x 50 / 300 = 160 mmProvide distribution steel of 8mmat 160 mmc/c

pu=177 kN/m2

1m


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(c) Design of Longitudinal Beam:Two columns are joined by means of a beam monolithic with the footing slab. The load from the slab will

be transferred to the beam. As the width of the footing is 2 m, the net upward soil pressure per meter

length of the beam

= wu = 177 x 2 = 354 kN/m

Shear Force and Bending Moment

VAC = 354 x 0.9 = 318.6 kN, VAB = 1050-318.6 = 731.4 kN

VBD = 354 x 1.7 = 601.8kN, VBA = 1500-601.8 = 898.2 kN

Point of zero shear from left end C

X1 = 1050/354 = 2.97m from C or X2 = 7.2-2.97 = 4.23 m from DMaximum B.M. occurs at a distance of 4.23 m from D

M uE = 354 x 4.232/ 2 - 1500 (4.23 - 1.7) = -628 kN.m

Bending moment under column A = M uA = 354 x 0.92/ 2 = 143.37 kN.m

Bending moment under column B = M uB = 354 x 1.72

= 511.5 kN-m

Let the point of contra flexure be at a distance x from the centre of column A

Then, Mx= I050x - 354 (x + 0.9 )2/ 2 = 0

Therefore x = 0.206 m and 3.92 m from column A i.e. 0.68 m from B.


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Depth of beam from B.M. Considerations

The width of beam is kept equal to the maximum width of the column i.e. 400 mm. Determine the

depth of the beam where T- beam action is not available. The beam acts as a rectangular section in

the cantilever portion, where the maximum positive moment = 511.5 kN/m.

d = (511.5 x 106/ (3.73 x 400)) = 586 mm

V1=318.6 kN

V4=601.8 kNV2=731.4 kN

V3=898.2 kN

SFD

+

-

+

-

2.97 m 4.23 m

E

1050 kN 1500 kN

354 kN/m

0.9 m 1.7 m4.6 m

C

A B

D

E

ME=628 kN-m

MA=143.37 kN-m

MB=511.5 kN-mBMD at Ultimate

+

-

.+

0.206 m0.68m


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Provide total depth of 750mm. Assuming two rows of bars at an effective cover of 70 mm. Effective

depth provided = d= 750-70 = 680 mm (Less than 750mm and hence no side face steel is

needed).

Check the depth for Two-way Shear

The column B can punch through the footing only if it shears against the depth of the beam along its

two opposite edges, and along the depth of the slab on the remaining two edges. The critical section

for two-way shear is taken at distance d/2 (i.e. 680/2 mm) from the face of the column. Therefore, the

critical section will be taken at a distance half the effective depth of the slab (ds/2) on the other side as

shown in Fig.

In this case b = D = 400 mm, db = 680 mm, ds = 140 mm

Area resisting two - way shear

= 2(b x d b + d s x d s) + 2 (D + d b)ds

= 2 (40 0 x 6 80 + 140 x 140) + 2(400+680) 140

= 885600 mm2

B=400 x 400 mm

2000B

D

D+db

D+ds

D+db/2

350 x 350 400 x 400400

7200 mm

2000

mm+

0.8m

2.7m 1.9m1.5m

0.825m

A B

a=0.9m b=1.7m4.6m


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Design shear=Pud= column loadW u x area at critical section= 1500 - 177 x(b + d s) x (D + d b)

=1500-177 x (0.400+0.140) x (0.400+ 0.680)

=1377.65kN

v=Pud/bod= 1377.65x1000/885600=1.56 MPa

Shear stress resisted by concrete = uc = uc x K swhere, uc = 0.25 fck= 0.25 25 = 1.25 N/mm

2

K s = 0.5 + d / D = 0.5 + 400/400 = 1.5 1 Hence Ks = 1uc =1 x 1.25 = 1.25 N/mm

2

Therefore unsafe and the depth of slab need to be increased. However the same depth is taken.

Area of Reinforcement

Cantilever portion BD

Length of cantilever from the face of column = 1.7 - 0.4 / 2 = 1.5 m.

Ultimate moment at the face of column = 354 x 1.52/ 2 = 398.25 kN-m

Mumax = 3.71 x 400 x 6802 x 106 = 686 kN.m >398.25 kN-mTherefore Section is singly reinforced.

Mu/bd2

=398.25x106/(400x680

2)=2.15 3.73, URS

Pt=1.114%

Ast =3030 mm2

Provide 3 - 32 mm + 4 - 16 mm at bottom face, Area provided = 3217 mm2

Ld = 39 x 32 =1248 mm

Curtailment

All bottom bars will be continued up to the end of cantilever. The bottom bars of 3 - 32 will becurtailed at a distance d (= 680 mm) from the point of contra flexure ( = 680 mm) in the portionBE with its distance from the centre of support equal to 1360 mm from B.

Cantilever portion ACLength of cantilever from the face of column = 900 - 350 / 2 = 725 mm

Ultimate moment = 354 x 0.7252/2 = 93 kN-m

Mu/bd2

=93x106/(400x680

2) =0.52 3.73, URS

Pt=0.245% (Greater than minimum steel)

A st =660 mm2

Provide 4 - 16 mm at bottom face, Area provided = 804 mm2

Continue all 4 bars of 16 mm diameter through out at bottom.

Region AB between points of contra flexuresThe beam acts as an isolated T- beam.

bf = [L o /( L o/ b +4)] + b w, where,

L o = 4.6 - 0.206 - 0.68 = 3.714 m = 3714 mm

b= actual width of flange = 2000 mm, b w = 400 mm

bf = [3714 / (3714 / 2000 + 4) + 400] =1034mm < 2000mm


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D f = 200 mm, Mu = 628 kN-m

Moment of resistance Mufof a beam for x u = D f is :

(Muf) = [0.36 x 25 x 1034 x 200 (680 - 0.42x200)] x10-6

= 1109 kN.m > Mu ( = 628 kN-m)

Therefore Xu 4542 mm

2

pt= 100 x 4624/(400x680) = 1.7 %

Curtailment

Consider that 2 - 32 mm are to be curtailedNo. of bars to be continued = 3 - 16 + 3 - 32 giving area = Ast =3016 mm

2

Moment of resistance of continuing bars

Mur= (0.87 x 250 x 3016 ( 680((250 x 3016) / (25 x 400) x 10-6

= 396.6 kN-m

Let the theoretical point of curtailment be at a distance X from the free end C, thenMuc= Mur Therefore -354 x

2/ 2 + 1050 (x- 0.9) = 396.6

x2-5.93x + 7.58 =0, Therefore x = 4.06m or 1.86m from C

Actual point of curtailment = 4.06 + 0.68 = 4.74 m from C or 1.86 - 0.68 = 1.18 m from C

Terminate 2 - 32 mm bars at a distance of 280 mm (= 1180 - 900) from the column A and760mm (= 5500 - 4740) from column B. Remaining bars 3 - 32 shall be continued beyondthe point of inflection for a distance of 680 mm i.e. 460 mm from column A and up to the

outer face of column B. Remaining bars of 3 - 16 continued in the cantilever portion.

Design of shear reinforcement

Portion between column i.e. ABIn this case the crack due to diagonal tension will occur at the point of contra flexure because

the distance of the point of contra flexure from the column is less than the effective depth d(=

680mm)

(i) Maximum shear force at B = Vumax = 898.2 kN

Shear at the point of contra flexure = VuD - 898.2-354 x 0.68 = 657.48 kN

v=657000/(400x680) =2.42 MPa c,max.Area of steel available 3 - 16 + 3 - 32 , Ast = 3016 mm

2

pt = 100 x 3016 / (400 x 680) = 1.1%

c=0.664MPav > cDesign shear reinforcement is required.

Using 12 mm diameter 4 - legged stirrups,Spacing = 0.87 x 250 x (4 x 113) /(2.42-0.664)x400 =139 mm say 120 mm c/c

Zone of shear reinforcements between v to c= m from support B towards A

(ii) Maximum shear force at A = Vu max = 731.4 kN.

Shear at the point contra flexure = VuD = 731.4 - 0.206 x 354 = 658.5 kN


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v=658500/(400x680) =2.42MPa c,max.Area of steel available = 4624 mm

2, pt= 100 x 4624 / (400 * 680) = 1.7 %

uc = 0.772 N/ mm2

v > cDesign shear reinforcement is required.

Using 12 mm diameter 4 - legged stirrups,Spacing = 0.87 x 250 x (4 x 113) /(2.42-0.774)x400 =149 mm say 140 mm c/c

Zone of shear reinforcement.

From A to B for a distance as shown in figure

For the remaining central portion of 1.88 m provide minimum shear reinforcement using 12

mm diameter 2 - legged stirrups at

Spacing , s = 0.87 x 250 x (2 x 113) / (0.4 x 400) = 300 mm c/c< 0.75d

Cantilever portion BDVumax = 601.8kN, VuD = 601.8 - 354(0.400 / 2 + 0.680) = 290.28kN.

v=290280/(400x680) =1.067MPa c,max.

Ast = 3217 mm2

Therefore pt = 100 x 3217/(400 x 680) = 1.18%

c =0.683N/mm2

(Table IS:456-2000)

v > c and v - c0.4. Provide minimum steel.

Using 12 mm diameter 2- legged stirrups,

Spacing = 0.87 x 250 x (2 x 113) /(0.4x400) =307.2 mm say 300 mm c/c

Cantilever portion AC

Minimum shear reinforcement of 12 mm diameters 2 - legged stirrups at 300mm c/c will besufficient in the cantilever portions of the beam as the shear is very less.


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7200 mm

2000mm

Plan of footing slab

20@130 8@160

5-323- 16

C/S at Centre C/S at the junctionRi ht of B

400400

750750

200

2000

3-16

3-324-16

4-16

350x350 400x400

12@300,2L Stp

12@120,4L Stp

12@300,2L Stp

12@140,4L Stp

12@300,2L Stp

5-32 + 3- 16 3- 16

3-32+

4-16

0.9 m 4.6 m 1.7 m

3-32 + 3- 163- 16 Side face2- 12


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Design of combined footingSlab and Beam type-2

Design a rectangular combined footing with a central beam for supporting two columns400x400 mm in size to carry a load of 1000kN each. Center to center distance between the

columns is 3.5m. The projection of the footing on either side of the column with respect to

center is 1m. Safe bearing capacity of the soil can be taken as 190kN/m2. Use M20 concrete

and Fe-415 steel.

Solution: Data

fck= 20Nlmm2,

fy= 415mm2,

fb (SBC)= l90 kN/m2,

Column A = 400 mm x 400 mm,

Column B = 400 mm x 400 mm,

c/c spacing of columns = 3.5,PA = 1000 kN and PB = 1000 kN

Required: To design combined footing with central beam joining the two columns.

Ultimate loads

PuA= 1.5 x 1000 = 1500 kN, PuB = 1.5 x 1000 = 1500 kN

Proportioning of base sizeWorking load carried by column A = PA

= 1000 kN

Working load carried by column B = PB = 1000 kN

Self weight of footing 10 % x (PA + PB) = 200 kN

Total working load = 2200 kN

Required area of footing = Af= Total load /SBC=2200/190 = 11.57 m2

Length of the footing Lf= 3.5 +1 +1 =5.5m

Required width of footing = b= Af/Lf = 11.57/5.5 = 2.1m

Provide footing of size 5.5 x 2.1 m

For uniform pressure distribution the C.G. of the footing should coincide with the C.G. of

column loads. As the footing and columns loads are symmetrical, this condition is satisfied.


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The details are shown in Figure

Total load from columns = P = (1000 + 1000) = 2000 kN.

Upward intensity of soil pressure=Column loads, P/Af= 2000/5.5 x2.1 =173.16 kN/m2< SBC

Design of slab:Intensity of Upward pressure = p =173.16 kN/m

2

Consider one meter width of the slab (b=1m)

Load per m run of slab at ultimate = 173.16 x 1 = 173.16 kN/m

Cantilever projection of the slab (For smaller column) =1050 - 400/2 = 850mm

Maximum ultimate moment = 173.16 x 0.8502/2 = 62.55 kN-m. (Working condition)

For M20 and Fe 415, Q u max = 2.76 N/mm2

Required effective depth = (62.15 x1.5 x 106

/(2.76 x 1000)) = 184.28 mmSince the slab is in contact with the soil clear cover of 50mm is assumed.

Using 20 mm diameter bars, effective cover =20/2 +50 say 75 mm

Required total depth = 184.28 + 75 = 259.4 mm. However provide 300 mm from shear

consideration as well. Provided effective depth = d = 300-75 = 225 mm

To find steel

Mu/bd2

=1.5 x 62.15 x106/1000 x 225

2= 1.84 2.76, URS

Pt=0.584 %

p=173.162

1m

1000 kN 1000 kN

3.5 m 1 m1 m

x

R p=173.16 kN/m2

(safe)


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Ast = 1314 mm2

Use #20mm diameter bar at spacing = 1000 x 314 / 1314 =238.96 mm say 230mm c/c

Area provided =1000 x 314 / 230 = 1365 mm2. Hence safe. This steel is required for the entire

length of the footing.

Check the depth from one - way shear considerationsDesign shear force=Vu=1.5 x173.16 x (0.850-0.225) = 162.33 kN

Nominal shear stress = v= Vu/bd =162330/(1000x225) =0.72 MPaPermissible shear stress

Pt = 100 x 1365 /(1000 x 225 ) = 0.607 %, uc = 0.51 N/mm2

Value of k for 300 mm thick slab = 1

Permissible shear stress = 1 x 0.51= 0.51 N/mm2

uc < v and hence unsafe.

The depth may be increased to 400 mm so that d = 325mm

Mu/bd2

=1.5 x 62.15 x106/1000 x 325

2= 0.883 2.76, URS

Pt=0.26 %, Ast = 845 mm2

Use #16mm diameter bar at spacing = 1000 x 201 / 845 =237.8 mm, say 230 mm c/c

Area provided =1000 x 201 / 230 = 874 mm2.

Check the depth from one - way shear considerationsDesign shear force=Vu=1.5 x 173.16 x (0.850-0.325)= 136.36 kN

Nominal shear stress = v = Vu/bd =136360/(1000x325) = 0.42 MPaPermissible shear stress

pt = 100 x 874/(1000 x 325 ) = 0.269 %, uc = 0.38 N/mm2

Value of k for 400 mm thick slab = 1

Permissible shear stress = 1 x 0.38= 0.38 N/mm2

Again uc < v and hence slightly unsafe.However provide steel at closure spacing, #16 @150 mm c/c.

Ast=201 x 1000/150 =1340 mm2 and pt =0.41% and hence uc=0.45 MPa and safe.

Check for development lengthLdt= 47 times diameter = 47x16=768 mm

Available length of bar = 850 - 25 =825mm > 768 mm and hence safe.

Transverse reinforcementRequired A st = 0.12 bD / 100 = 0.12 x 1000 x 400/100 = 480mm

2

Using 10 mm bars, spacing = 1000 x 79 / 480 = 164.58 mm

Provide distributionsteel of#10 mm at 160 mm c/c

Design of Longitudinal BeamTwo columns are joined by means of a beam monolithic with the footing slab. The load fromthe slab will

be transferred to the beam. As the width of the footing is 2.1 m, the net upward soil pressure per meter

length of the beam under service.

= w = 173.16 x 2.1 = 363.64 kN/m

Shear Force and Bending Moment at service conditionVAC = 363.64 x 1 = 363.14 kN, VAB = 1000-363.14 = 636.36 kN

VBD = 363.14 kN, VBA = 636.36 kN


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Point of zero shear is at the center of footing at L/2, i.e., at E

Maximum B.M. occurs at E

ME = 363.64 x 2.752/ 2 - 1000 (2.75 - 1) = -375 kN.m

Bending moment under column A = MA = 363.64 x 12

/ 2 = 181.82 kN.mSimilarly bending moment under column B = MB = 181.82 kN-m

Let the point of contra flexure be at a distance x fromC

Then, Mx= 363.63x2/21000(x-1) = 0

Therefore x = 1.30 m and 4.2m from C

Depth of beam from B.M. Considerations:The width of beam is kept equal to the maximum width of the column i.e. 400 mm. Determine the

depth of the beam from absolute maximum BM. This is in the central part where T- beam action is

available. Assume the beam as rectangular at the center of span where the moment is maximum,

we have,

d = (375 x 1.5 x 106/ (2.76 x 400)) = 713.8 mmProvide total depth of 800 mm. Assuming two rows of bars at an effective cover of 75 mm, the

effective depth provided = d= 800-75 = 725 mm.


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Check the depth for Two-way Shear:

The column B can punch through the footing only if it shears against the depth of the beam along its

two opposite edges, and along the depth of the slab on the remaining two edges. The critical section

for two-way shear is taken at distance d/2 (i.e. 680/2 mm) from the face of the column. Therefore, the

critical section will be taken at a distance half the effective depth of the slab (ds/2) on the other side asshown in Fig.

MC=375 kN-m

MA=181.82 kN-m MB=181.82 kN-m

V2=636.36 kN

V3=636.36 kN

BMD at

+

-

V1=363.63 kN

V4=363.63 kN

SFD

+

_+

-

1000 kN 1000 kN

363.64 kN/m

0.3 0.3


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In this case b = D = 400 mm, d b = 725 mm, d s = 325 mm

Area resisting two - way shear

= 2(b x d b + d s x d s) + 2 (D + d b)ds

= 2 (4 00 x 72 5+ 325 x 325) + 2(4 00+725) 325 = 1522500 mm

2

Design shear=Pud= column loadW u x area at critical section= 1500173.16 x1.5 x(b + d s) x (D + d b)

=1500-173.16 x 1.5 x (0.400+0.325) x (0.400+

0.725)

=1288.14 kN

v=Pud/bod= 1288.14x1000/1522500=0.845 MPa

Shear stress resisted by concrete = uc = uc x K swhere, uc = 0.25 fck= 0.25 20 = 1.11 N/mm

2

K s = 0.5 + d / D = 0.5 + 400/400 = 1.5 1 Hence K s = 1uc = 1 x 1.11 = 1.11 N/mm

2, Therefore safe

Area of ReinforcementCantilever portion BD and AC

Length of cantilever from the face of column = 0.8 m.

Ultimate moment at the face of column = 363.64x1.5 x 0.82/ 2 = 177.53 kN-m

Mumax = 2.76 x 400 x 7252

x 10-6

= 580.29 kN.m >177.53 kN-m

Therefore Section is singly reinforced.

Mu/bd2

=177.53x106/(400x725

2)=0.8442.76, URS

400

2100B

D

D+db

D+ds

db/2

400

4000 x 400 400 x 400400

5500 mm

2100

mm+

0.85m

1.75m 1.75m 0.8m

A B

a=1 m b=13.5m


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Pt=0.248%, A st =719.2 mm2

Provide 4 - 16 mm at bottom face, Area provided = 804 mm2, pt=0.278%

Ld = 47x16 =752 mm

Curtailment: All bottom bars will be continued up to the end of cantilever for both columns. If

required two bottom bars of 2-16mm will be curtailed at a distance d (= 725 mm) from the pointof contra flexure in the portion BE as shown in figure.

Region AB between points of contra flexuresThe beam acts as an isolated T- beam.

bf = [ L o/( L o/b + 4 )] + b w, where,

L o = 4.2-1.3=2.9 m = 2900 mm

b= actual width of flange = 2100 mm, b w = 400 mm

bf = [2900 / (2900 / 2100) + 4] + 400 =938.9mm < 2100mm

D f = 400 mm, Mu = 1.5 x 375=562.5 kN-m

Moment of resistance Mufof a beam for x u = D f is :

(Muf) = [0.36 x 20 x 938.9 x 400 (725 - 0.42x400)] x10-6

= 1506 kN.m > Mu ( = 562.5 kN-m)Therefore Xu 2334 mm2

pt = 100 x 2334/(400x725) = 0.805 %

Curtailment: Curtailment can be done as explained in the previous problem. However extendall bars up to a distance d from the point of contra flexure i.e up to 225 mm from the outerfaces of the columns. Extend 2-16mm only up to the end of the footing.

Design of shear reinforcement

Portion between column i.e. ABIn this case the crack due to diagonal tension will occur at the point of contra flexure because

the distance of the point of contra flexure from the column is less than the effective depth d(=

725mm)

(i) Maximum shear force at A or B = Vumax = 1.5 x 636.36 =954.54 kN

Shear at the point of contra flexure = 954.54-1.5x 363.64x0.3 = 790.9 kN

v=790900/(400x725) =2.73 MPa c,max.(2.8 MPa)Area of steel available = 2354 mm

2, 0.805 %

c=0.59MPa, v > cDesign shear reinforcement is required.

Using 12 mm diameter 4 - legged stirrups,Spacing = 0.87 x 415 x (4 x 113) /(2.73-0.59)x400 =190.6 mm say 190 mm c/c

Zone of shear reinforcements is between v to c= m from support B towards A

Cantilever portion BD and ACVumax = 363.64 x 1.5 = 545.45 kN,

Shear from face at distance d = VuD = 545.45-363.64 x1.5(0.400 / 2 + 0.725) = 40.90 kN

v=40900/(400x 725) =0.14 MPa c,max. This is very small


22/22

Steel at this section is 416 mm, Area provided = 804 mm2, pt=0.278%c =0.38N/mm

2(Table IS:456-2000). No shear steel is needed.

Provide minimum steel.

Using 12 mm diameter 2- legged stirrups,

Spacing = 0.87 x 415 x (2 x 113) /(0.4x400) =509.9 mm say 300 mm c/c

400x400 400x400

#12@300,

2L Stp#12@190,

4L Stp

#12@300,

2L Stp#12@190,

4L Stp

#12@300,

2L Stp

4-#25 + 2-#16 2-#16

4-#16

1 m 3.5 m 1 m

2-#16Side face

2-#12

2-#16

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