QUESTION ONE
The cantilever retaining wall shown in fig.1. Supports a granular material of saturated density dry density 2000kg/m3 and allowable bearing pressure is 110KN/m. design the wall by using angle of internal friction φ=30 coefficient of friction u=0.45, ɤconcrete 24KN/m3 high yield steel fy=460N/mm2 and grade 35 concrete.
200
4500 4900
400
400 800 350 2250
Solution
Given: ɤ= 200Kg/m3=2000*10=20000N/m3
ɤ=20KN/m3
BC= 110KN/m2
φ=300
u=0.45
ɣc=24KN/m3
fy =460N/mm2
fcu =35N/mm2
Coefficient of earth pressure
Wall stability.
Consider 1m length of wall
Stability calculation table
Load type
Horizontal load
Distance from A
Moment about A
Active pressure
Pa= 2
= 2
=80.034KN
130.723 (-)
Total load=80.034
130.723 (-)= total
Vertical load(KN)
wall
W1=0.2*4.5*24=21.6
W2=
1.05
0.9
22.68=21.6*1.05
7.29=8.1*0.9
base
Vb=3.4*0.4*24=32.64
1.7
32.64*1.7=55.488
backfill
Vs=2.25*4.5*20=202.5
2.275
202.5*2.275=460.687
key
Vk=0.4*0.35*25=3.5
0.975
3.5*0.975=3.4125
TOTAL=268.34
TOTAL=549.55
Distance of centre of gravity (c.g) of vertical force from the face of toe (ptA).
c.g. = =1.56
Eccentricity (e)=
Base properties
Area = 3.4*1=3.4m2
Modulus (Z) = 2*1=1.92
Maximum soil pressure at A
Pmax= = =98.49KNm
Pmax<BC; 98.49KNm < 110KNm: safe
Pmin= =68.18KN/m2
Pmin>0; →safe
Factor of safety against overturning
FS greater than1.55
FS=
FS>1.55→4.20>1.55→ok;→safe
F.S against sliding
F.S=
Pp= 2=19.2KN
FS=
FS>1.55→1.74>1.55→safe
D. STRUCTURAL DESIGN
A) Steem
 Pressure at the base= ka ɤ h= *20*4.5=30KN/m2
 Shear (V) = 2= 2=67.5KN
Vu=67.5KN*1.4=94.5 KN
 Moment== 3=101.25KNm
Mu=101.25*1.4=141.75KNm
 d=h─cover─φ/2
Assume φ=20mm
Cover=75mm
d=350-75-10=265mm
MRC= 0.156*fcu*bd2=0.156*35*100*2652=383.42KNm
K= = *106=0.0576
Z=d [0.5+(0.25- 1/2] = d[0.5+(0.25- )1/2] = 246.78mm
Z<0.95d→246.78<251.75
As= 6=1435.27mm2/mm
As=1435.27mm2/m
Provide T16 @ 140mm, As=1436mm2/m
ɣ= 3=0.254
Ɣc= √( ) √( ) √
√( ) ( )1/3=0.541<3→ok
(400/d) 1/4 = (400/265)1/4=1.1>1→ok
(35/25)1/3= 1.11
Ɣc= = 0.42
ɣ< Ɣc→0.254<0.42→no compression bars required (safe)
Anchorage length = 40* φ of bar=40*16=640mm
B) heel
=
Pressure=68.18+20.05=880.23KN/m2
Total downward pressure = weight of earth + self-weight of heel
Weight of earth=20*4.5=90KN/m2
Self-weight of heel=24*0.4=9.6KN/m2
Total=90KN/m2+9.6KN/m2=99.6KN/m2
+ =
 shear=11.37*2.25*1+ =
=25.58+22.55=48.138KN
Vu=1.4*48.138=67.39KN
 moment=25.58* +22.55+ *2.25=62.61KNm
Mu=68.61*1.4=87.65KNm
 d=350-75- =265mm
 k=Mu/fcu bd2= *106=0.0356
k<0.156→No compression bars required
 Z=d[0.5+(0.25- 1/2] = d[0.5+(0.25- )1/2] = 254.25mm
Z>0.95d=251.75
Z>0.95d→let’s take 0.95d=251.75
As=87.65*106/0.87*460*251.75=869.97mm2/m
Provided T16 @ 220mm→As=913mm2/m
ɣ= 3=0.181
Ɣc= √( ) √( ) √
√( ) ( )1/3=0.7<3→ok
(400/d) 1/4 = (400/265)1/4=1.1>1→ok
(35/25)1/3= 1.11
Ɣc= = 0.54
ɣ< Ɣc→0.181<0.54→no compression bars required (safe)
 Distribution bars= =520mm2
 Provide T10 @ 150 mm→As=523mm2
C) KEY
As min=520mm2=0.13%*1000*400
Provided=T10@150mm, As=523mm2/m
D) TOE
= → x=23.178
Pressure=23.178+68.18=91.358KN/m2
Total pressure= self-weight of toe=24*0.4=96KN/m2
+ =
 Shear=58.58*0.8*1+ 46.864+9.368=56.135KN
Vu=56.135*14=78.58KN
 Moment=46.264*
Mu=23.68*1.4=33.16KNm
 D=350-75-10=265mm
 K= *106=0.0134
K<0.156→no compression bars
 Z=d[0.5+(0.25- 1/2] = d[0.5+(0.25- )1/2] = 260.96mm
Z>(0.95d=251.75)
Z>0.95d→ok
 As= 6=317.51mm2/mm
Provided T10@240mm→As=327mm2/m
Anchorage length=40*10=400mm
ɣ= 3=0.2118
Ɣc= √( ) √( ) √
√( ) ( )1/3=0.497<3→ok
(400/d) 1/4 = (400/265)1/4=1.1>1→ok
(35/25)1/3= 1.11
Ɣc= = 0.366
ɣ< Ɣc→0.2118<0.366→no compression bars required (safe)
E) Sketch of reinforcement
T10@150mm T16@140mm
T16@220mm
T10@150mm
T10@240mm T10@150mm