The cross section of a cantilever retaining wall is shown in figure 8.10.Calculate the factors of safety in regards to overturning,sliding and bearing capacity.
Solution:
From the figure ,
H’=H_{1}+H_{2}+H_{3}=6tan10⁰+18+2.75=21.81ft , Pa=1/2y_{1}KaH’2 ;
For Ø_{1}’=34⁰ and α=10⁰, the value of Ka=0.294,so Pa=1/2(117)(0.294)(21.81)^{2}/1000=8.18kip/ft;
Pv=Pasin10⁰=1.42kip/ft; Ph=Pacos10⁰=8.06kip/ft;
Factor of safety against Overturning
The following table should be prepared now for determination of the resisting moment.
Section |
Weight(kip/ft) |
Moment arm from C(ft) |
Moment about C(kipft/ft) |
1 | (1.5)(18)(0.15)=4.05 | 5.75 | 23.29 |
2 | ½(1)(18)(0.15)=1.35 | 4+2/3(1)=4.67 | 6.3 |
3 | (12.5)(2.75)(0.15)=5.15 | 6.25 | 32.23 |
4 | (18+19.06)/2(6)(0.117)=13.01 | ~4+2.5+6/2=9.5 | 123.6 |
Pv=1.42 | 12.5 | 17.75 | |
ƩV=24.986 | ƩM_{R}=203.17 |
Note:y_{concrete}=150lb/ft^{3}
The overturning moment ,M_{O}, is:
M_{O}=P_{h}H’/3=(8.06)(21.81/3)=58.6 kip/ft
FS(overturning)=ƩM_{R}/M_{O}=203.17/58.6=3.47 >2 => OK
Factor of Safety against Sliding
FS_{(sliding)}=((ƩV)tan(k_{1}Ø’_{2})+Bk_{2}c’_{2}+Pp)/(Pvcosα); where we consider k_{1}=k_{2}=2/3 and Pp=0. So
FS_{(sliding)}=((ƩV)tan(2/3Ø’_{2})+B(2/3c’_{2}))/(Pvcosα)=((24.98)tan(2/3(18))+(12.5)2/3(0.9))/8.06=
=1.59 >1.3 =>OK
Factor of Safety against Bearing Capacity Failure
e=B/2-(ƩM_{R}-ƩM_{0})/ƩV=6.25-(203.17-58.6)24.98=0.464 < B/6=12.5/6=2.08ft;
q_{toe}=ƩV/B(1+6e/B)=24.98/12.5(1+6(0.464)/12.5)=2.44 kip/ft^{2} ;
The ultimate bearing capacity of the soil can be determinate from the following equation:
q_{u}=c’_{2}N_{c}F_{cd}F_{el}+qN_{q}F_{qd}F_{qi}+1/2B’N_{y}F_{yd}F_{yi }where:
q=y_{2}D=(4)(0.107)=0.428 kip/ft^{2};
B’=B-2e=12.5-(2)(0.464)=11.572ft;
F_{cd}=1+0.4(D/B’)=1+0.4(4/(11.572)=1.138;
F_{qd}=1+2tanØ’_{2}(1-sinØ’_{2})^{2}(D/B’)=1+0.31(4/11.57)=1.107;
F_{yd}=1;
Fel=Fqi=(1-ψ⁰/90⁰)^{2}
Ψ=tan^{-1}(Pacosδ/ƩV)=(8.06/24.98)=17.88⁰ so
F_{el}=F_{qi}=(1-17.88/90)^{2}=0.642;
F_{yi}=(1-ψ/Ø’_{2})=(1-17.88/18)^{2}=0 ; thus :
q_{u}=(0.9)(13.1)(1.138)(0.642)+(0.428)(5.26)(1.107)(0.642)+1/2(0.107)(11.572)(4.07)(1)(0)=10.21 kip/ft^{2}
FS(bearing capacity)=q_{u}/q_{toe}=10.21/2.44=4.18 >2 => OK