5 questions and answers based on Terzaghi’s Bearing Capacity Theory (usefulsearch.org) (useful search)

1. Questions on Bearing Capacity (Terzaghi’s Theory)
[5 Questions ]
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2. Q .1 A square footing 2.5m by 2.5m is built in homogenous bed of
sand of unit weight 20 kN/m2.abd having an angle of shearing
resistance of 36˚.the depth of base of footing is 1.5 m below
ground surface. Calculate the safe load that can be carried by a
footing with FOS 3against complete shear failure . Use Terzaghi’s
analysis.
SOL: given: b=2.5 m (width of foundation)
γ=20kn/m3 ( unit weight of soil)
D = 1.5 m ( depth of foundation )
Ф =36˚ ( angle of shearing resistance)
c = 0 ( cohesion )

3. • Since the soil is dense and Ф=36˚ footing is likely to fail by
general shear failure.
For sq. footing
qu = 1.3cNc + q Nq + 0.4 γBNγ
Nc = 65.4 , Nq =49.4 ,Ny =54 (Bearing capaity factors)
On Putting the values in above equation
qu = 0 + yDf Nq + 0.4yBNy
= 20*1.5*49.4 + 0.4*20*2.5*54.0
= 2562kn/m2

4. qsafe = ( qu – yD)/ F + yD
= (2562-1.5*20)/3 + 1.5*20
= 874kN/m2
Safe load = qsafe * area of cross-section
= 874*2.5*2.5
= 5462.5 kN

5. Q.2 What will be the max. safe load if in Q.1 the soil is loose sand
of unit wt. 13kN/m2 and ф= 250 ?
Solution: Nq =5.6 and Ny = 3.2 ( Bearing capacity factors
corresponding to ф = 250 )
The soil is loose and ф < 250 , footing is likely to fail by local shear failure .
For sq. footing
qu = 1.3cNc + qNq +0.4yBNy
= 0+16*1.5*5.6+ 0.4*16*2.5*3.2
= 185.6kN/m2
qsafe = qnet /F + yD
= (qu – yD)/F + yD
= 77.87 kN/m2
Safe load = qsafe * Area of cross section
= 77.84*2.5*2.5
= 486.7 kN

6. Q.3 A strip footing 1m wide at its base and is located at a
depth of 0.8m below the ground surface. The properties of
foundation of soil are given below. Determine safe bearing
capacity by Terzaghi’s analysis. Assume that soil fails in local
shear.
Solutin: Given y= 18kN/m2 ( unit wt. of soil)
c= 30kN/m2 (cohesion)
ф=200 ( angle of shearing resistance)
F= 3 (factor of safety)
For ф= 200 Nc = 11.8 , Nq= 3.9, Ny=1.7 ( bearing capacity factors)
The ultimate bearing capacity in Local Shear Failure :
qu = 2/3cNc +yDNq +0.5yBNy
= 2/3*30*11.8 + 0.18*0.8*3.9 + 0.5*18*1*1.7
= 307.46kN/m2

7. qnet = qu – yD
= 307.46 - 18*0.8
= 293.06kN/m2
qsafe = qnet /F +yD
= 97.7+14.4
= 112.1 kN/m2

8. Q.4 In Q.3 if watertable is located at the base of footing . Assume
ysat = 19.5kN/m2 .
Solution : qu = 2/3cNc + yDNqRw1 + 0.5ysatBNyRw2
Since Water table is at the base of footing.
Rw2 = 0.5 and Rw1 = 1
Therefore,
qu = 2/3*30*11.8 + 18*0.8*3.9*1 + 0.5*19.5*1*1.7*0.5
=300.45 kn/m2
qnet= qu – yD
=286.0475kN/m2

9. qsafe = qnet /F +yD
= 109.75kN/m2

10. Q.5 Design a strip footing to carry a load of 750kN/m2 at a
depth of 1.6m in a soil having a unit wt. of 18kN/m3 and
shearing strength parameters as c= 20kN/m2 and ф= 250
.Determine the width of footing .F = 3. Use Terzaghi’s
analysis.
Solution: Given:
D = 1.6m ( Depth of foundation )
c = 20kN/m2 ( cohesion )
ф= 25 0 ( angle of shearing resistance)
y=18 kN/m3 ( unit wt. of soil)
F = 3 ( Factor of safety)
Assuming general shear failure
qu = cNc + yDNq + 0.5yBNy
Nc = 25.1 Nq=12.7 Ny= 9.7 ( Bearing capacity factors corresponding
to ф = 25 0 )

11. qu = 20*25.1 + 18*1.6*12.7 + 0.5*78*B*9.7
= 867.8 + 87.3 B
Intensity of pressure . When F = 3
= qu /3
= (867.8 + 87.3B)/3 [ Eqn 1]
Applied load intensity = 750/B*1 [ Eqn 2]
Equating Eqn 1 and Eqn 2
B = 2.134m