Retaining Wall Design Calculation Example

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’=H₁+H₂+H₃=6tan10⁰+18+2.75=21.81ft   ,                             Pa=1/2y₁KaH’2   ;

For Ø₁’=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		ƩMR=203.17

Note:y_concrete=150lb/ft³

The overturning moment ,M_O, is:

M_O=P_hH’/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(k₁Ø’₂)+Bk₂c’₂+Pp)/(Pvcosα);    where we consider k₁=k₂=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-ƩM₀)/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/ft² ;

The ultimate bearing capacity of the soil can be determinate from the following equation:

q_u=c’₂N_cF_cdF_el+qN_qF_qdF_qi+1/2B’N_yF_ydF_yi          where:

q=y₂D=(4)(0.107)=0.428 kip/ft²;

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/ft²

FS(bearing capacity)=q_u/q_toe=10.21/2.44=4.18    >2 => OK