Question and Answers on Terzaghi’s Bearing Capacity Theory (usefulsearch.org) (useful search)

1. A POWERPOINT PRESENTATION ON
PREVIOUS YEAR SOLVED QUISTION PAPER
Question on Bearing Capacity
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2. Q.1 A strip footing 1m wide is laid at a depth of 2m in c-ϕ soil
having the following characteristics. calculate the ultimate
bearing capacity qf by terzaghi analysis, take properties of soil
c=20kN/m²,ϓ=18kN/m³and ϕ=28º give terzaghi bearing
capacity factors (ϕ=28º) as Nc=32,Nq=18,Ny=16.
(june 2010)
sol: given: b=1m
d=2m
ϓ=18kn/m²
ϕ=28º
c=20kn/m²
Nc=32,Nq=18,Ny=16
Since ϕ=28º, footing is likely to fall by general
shear failure.
Hence,ultimate bearing capacity

3. qf=cNc +ϓDNq+0.5ϓBNy
=20X32+18X2X18+0.5X18X1X16
=1432kN/m²
qnf=qf-ϓD = 1432-18*2
=1396kN/m²
if factor of safety F=3
qs=qnf/F+ ϓD
=1396/3+18X2
qs=501.33kN/m²

4. Q2. A strip footing is to be designed to carry a total load of
1000 kN/m at a depth of 1 m in sandy soil. The soil has C=0,
ϕ=36º .
Unit weight of soil below water table = 20kN/m³
Unit weight of soil above water table = 18kN/m³
Groundwater table exist at foundation level. Given for ϕ=36º
Nq=47, Ny=43. Assume unit weight weight of water = 10
kN/m³ factor of safety = 3. ( june 2008 )
Solⁿ: Given : load P = 1000kN/m
Depth D =1m
Cohesion C = 0 ( for sandy soil)
Φ = 36°
Net ultimate stress qnf=cNc+Rw1 (Nq-1)σ +0.5 Rw2 Ny bΎ
Rw1= 0.5(1+Zw1/D)
Rw2= 0.5(1+Zw2/B)

5. here Ground water table is at foundation level
Hence Zw1=D=1m
Zw2=0 B
Then Rw1=1 D Zw1
Rw2=0.5
B Zw2
qnf= 0+(47-1)18+0.5*0.5*43*20* B = 828+215B
qs = qnf/3+ΎD = (828+215B)/3 +18 = 294+71.6B
Safe stress= load/area = 1000/B
Hence 1000/B = 294+71.6B
B = 2.2m

6. Q3. Calculate the depth at which the footing (2mX2m) should
be placed to transfer total load of 200 tons with a factor of
safety3. The soil is sandy having Φ = 30° and unit weight 2
gms/cm³. Ground Water level is too deep. Given Nq=22 and
Ny=20 for Φ = 30° . ( Dec 2008 , June 2011 , June 2012 )
Solⁿ: Given : load P = 200 tons = 200 x 1000 kg
Depth D =?
Cohesion C = 0 ( for sandy soil)
Φ = 30°
Sy =0.8 (for square footing)
Ύ = 2 gms/cm³ = 2000 kg/m³
Net ultimate stress
qnf=cNc+ (Nq-1)σ +0.5 Sy Ny bΎ
= 0+ (22 -1)X 2000XD + 0.5X0.8X20X2X2000
=42000D + 32000

7. safe bearing capacity(kg/m²) = qns = qnf/3 +ΎD
= (42000D +32000)/3 + 2000D
= 16000D +10666.6
Safe stress( kg/m²) = load / area = 200X 1000/ 2X2
= 50000
Hence 50000 = 16000D+10666.6
D = 2.458 m

8. Q.4 A r.c.c. column has a square footing founded at a 3.0 m
depth below the surface of clay soil of average density of18
kN/m³ and a shear strength of 80 kN/m² . The wall cohesion
on the vertical sides of the footing which is acting over the
lower half portion only may be taken as one half of the
shear strenth . The total load applied to the soil is 2000 kN .
Calculate the size of footing for a safety factor of 3.
(june 2007)
sol: given: b=?
d=3m
ϓ=18 kN/m³
ϕ=0
c=(80/2)= 40kN/m²
Nc=5.7, Nq=1, Ny=0 ( for ϕ=0 )
qnf=cNc+ (Nq-1)σ +0.5 Sy Ny bΎ
= 40X5.7+0+0 =228 kN/m²

9. safe bearing capacity(kN/m²) = qns = qnf/3 +ΎD
= 228/3 + 18X 3
= 120(kN/m²)
Safe stress( kg/m²) = load / area = 2000/B²
hence 2000/B² = 120
B = 4.08 m

10. Q.5 A strip footing 1.5 m wide at its base is located at a depth of
1 m below the ground surface. The foundation level are
Ύ = 18 kN/m³, C = 30 kN/m² , Φ =20°. Determine the safe
bearing capacity using a factor of safety of 2.5. Use Terzaghi
analysis and assume soil fails by local shear.[ Given for
Φ =20°. bearig capacity factors Nc’ = 11.8, Nq’ = 3.9,
Ny’ = 1.7 ]. (June 2009 )
sol:given: b=1.5m
d=1m
ϓ=18kn/m²
ϕ=20º
c=30kn/m²
Nc’=11.8, Nq’=3.9, Ny’=1.7
Since ϕ=28º, footing is likely to fail by local
shear failure.
Hence,ultimate bearing capacity

11. qnf=(2/3)cNc’+ (Nq’-1)σ +0.5 Ny’ bΎ
= (2/3) 30X11.8 + (3.9-1) X 18 X1 + 0.5X 1.7 X 1.5 X18
= 311.15 kn/m²
qs = qnf /2.5 + ΎD
= 311.15/2.5 +18X1
qs = 142.46 kn/m²

12. Q6. Calculate the net ultimate bearing capacity of a rectangular
footing
2 m X 4m in plan founded at a depth of 1.5m below the ground
surface. The load on the footing acts at an angle of 15º to the
vertical and is eccentric in the direction of width by 15 cm. The
saturated unit weight of soil is18 kN/m³ . The rate of loading
is slow and hence the effective stress shear strength parameters
can be used in the analysis . C’ = 15 kN/m² and ϕ’=25º .
Natural water table is at a depth of 2 m below the ground
surface. ( for ϕ’=25º ,Nc’=20.7, Nq’=10.7, Ny’=10.9)
Use IS : 6403 – 1981 recommendations.
sol:given: b X L=2m x 4m
d=1.5m
ϓ= ϓsat= 18 kN/m³
ϕ =ϕ’=25º
c=c’=15 kN/m²

13. Nc’=20.7, Nq’=10.7, Ny’=10.9
Since ϕ=28º, footing is likely to fail by local
shear failure.
Hence,ultimate bearing capacity
qnf = c Nc Sc dc ic + σ ( Nq -1) Sq dq iq + 0.5 Ύ B Ny Sy dy iy
W’
for D’w/B = 0.25 (D’w=2 – 1.5 = 0.5m), W’ = 0.625
eₓ = 0.15m ; effective width B’ = B -2 eₓ = 2-0.3 = 1.7m
Sc = Sq = 1 + 0.2B’/L = 1+ 0.2(1.7/4) = 1.025
Sy = 1 - 0.4B’/L = 1 - 0.4(1.7/4) = 0.83
dc = 1 + 0.2 (Df/B’)tan(45º+ ϕ/2) = 1.28
dq = dy = 1 + 0.1 (Df/B’)tan(45º+ ϕ/2) = 1.14

14. ic = iq = ( 1 – α/90 )² = ( 1- 15/90 )² = 0.69
iy = ( 1 – α/ ϕ )² = ( 1- 15/25 )² = 0.16
substituting these values
qnf = 15 X 20.7 X 1.085 X 1.28 X0.69 + 27 X 9.7 X 1.085 X 1.14X
0.69
+ 0.5 X 18 X 1.7X 10.9 X 0.83 X 1.14 X0.16 X0.625
qnf = 536.8 kN/m²