--~------------------ --- --- ©Nf1D1 ~~@ lYJ~Tirii.-.,
H-140
·-
.
----=~
~.£w®~
I
J{YDRAULICS and
GEOTECJINICAL ENGINEERING
-~-···_,____-----·-.---,--~-- NOV. 2019
M =Pe
M=700(0.5)
1. CE Board Nov. 2019
A. rectangular footing 3 m x 2 m has a
th 'c~ness of 0.40 m. If P = 700 kN is acting at
d1 . - 1-50 m and d2 = 0.50, find the maximum
soil. pressure, neglecting the weight of the
footing.
d =l.5
I
. A
j d]=().50
--
- - ~' . 5 0 1
2m
i
M=P( 1- ; )
700(0.5) =
100(1- ; )
0.5 = _
3 - x_
3
1.5 =3-x
x= 1.5
f
p=
(3)(1.5)
2
3m
700 = fmax (3)( 1.5)
Solution:
e = ~.5 >
.
2
fmax = 311.11 kPa
~ =0.333
Soil pressure is triangular.
I
x= 1.5
~ ~ =1: .
-X -l .
3
·. I
I
-I•
· CE Board Nov. 2019
A
.
th· r:Ctangular fooling 3 m x 2 mhas 1
k~c. nes~ of 0.4 m. An eccentric load P=!l
18 acting at d1 = 1.0 and d2 =10 detennlt
e
th max. · soi·1 pressure considering· '1he ~-~
of
kN/ 1~bng. Assume concrete to weigh ~
m.
--t9J~ 10.4
2m
3m
H-141
•ci, Enlf•rl~ ucansure Exa111nauons
4· CE Board Nov. 2019
. solution:
:: _!_(3)(2) 3 =2
11
12
A recta_ngular footing 3 m x 2 m carries an
eccen~nc load P acting at d1 =1.5, d2 =0.6 m.
What 18 th e area still in contact with the soil?
I ::~=4.5
1
12
3
d1=1.5 __
e=0.5 = =.0.5
6
'
!
P MC
uset =-A•+ ~
.
I
ll8X
Wt. of footing= 3(2)(0.4)(24) = 57.6 kN
(p + 57.6) M (1.5)
f -~__:_+-.--Ynm
3(2)
IY
f = 800 +57.6 + 800(0.5)(1.5)
nm
3(2)
4.5
fnm
=276.27 kPa
..~I
__
3m
---
Solution:
X
3
'
Arectangular footing 3 m x 2 m carries an
eccentric load P =700 kN acting at d1 =1.5
m, d2 =1.0. Determine the max. soil pressure.
Neglecting the weight of the footing.
I
I
,~-I
I
- --- -
-
,-=--:-1
'\
10,4
Solution:
I
X
M=P(0.4)
2 m .,
1.5
M=P(1·
i)
2m
P(3-X)
0.4 P= - 3 -
f ~p
1.2 =3-X
X =1.8
A
f ~ 700
Area st.ill in contact with soil:
~~-
~ ~ 3(2} =116.67 ~Pa
j
x= 1.8
2
e >6
0.4 > 0.333
3. CE Board Nov. 2019
l.5
,
A= 3(x)
A=3(1.8)
A= 5.4m 2
H -142
Clvll Engineering ucensure bamlnauons
5. CE Board Nov. 2019
Evaluate the kinetic energy of a unit weight of
water in meters, flowing at 4 m/s?
7. CE Board Nov. 2019
KE= 2g
A mercury barometer at the base
mountain reads 620 mm. At the sam:f .the
another barometer at the top of a rn tirne,
reads 450 mm. Assuming w of air isounta·lO
constant at 10 N/m 3, what is the approto_~
height of the mountain?
Pnate
KE= (4)2
2(9.81)
Solution:
Solution:
v2
KE= 0.815
2
,,
\
h
6. CE Board Nov. 2019
The crest gate shown consists of a cylindrical
surface of which AB is the base, supported by
a structural frame hinged at C. The length of
the gate per perpendicular to the paper is 10
m. Compute the horizontal force components
of the total pressure AB.
Solution:
:~i~Jn,3
(_
I
P1 = P2 + wh
P1- P2 = 10h
9810(13.6)(0.62) - 981 0(1:5.6)( 0.45) =10h
h = 2268 m.
B
I
• I
I
(j'
ht
,,,
'?
8. CE Board Nov. 2019
I
: 60.
6m
P=y hA
~
P =9.81( }h)(10)
A jet of water 25 mm in diameter and having a
velocity of 8 m/s strikes against a plat at right
angles. Determine the force on the plate if the
plate is moving in the same direction as the ~t
with a uniform velocity of 3 mis.
Solution:
h =6 Sin so·
h=S.196
~8.
VI =8
P = 9.81(5.196)2 (10)
5
P =529.74kN
·,- __,.., Ucansure Exa111nauons
~
f pA(V1 - V2 )2
11. CE Board Nov. 2019
2
2
~- CE Board Nov.
2019
f =100o( : } 0.025 ) ( 6 - 3)
r:12.21N
hydraulic jack is used to raise a 10 kN car. A
A of 40 kN is applied on a _25 cn:1 .dia,:net~r
What diameter of Jack
mm ,s
~uired to raise the car?
:er.
1
H-143
m
A tria~g~lar gate or height 1.2 m and base
~- 9 .m is msta11ed in a position that its plane is
mchned 60 degrees with the horizontal with its
vertex at the top and the base Is parallel to the
water surface. The vertex is at a depth of 2 m
vertically below the water surface. Fresh water
is on one side of the gate. If the gate is hinged
at the bottom, evaluate the force normal to the
gate at its vertex that will be required to open it
in kN .
Solution:
Solution:
s-~2 =0.08148 KN/cm2
- {(25)
F
S= I
A
.
10
0.08148 = ~ d2
4
d=125cm= 125mm
10. CE Board Nov. 2019
Hinged
Aclosed cylindrical tank, 1 min diameter, 3 _m
h~h. is full of water. It is rotating about its
vertical axis with a speed of 120 rpm. Find the
-re at the top just under the cover along
Ille circumference.
lotut;on:
Q)::~
60
<o:: 12.57 ,.,.,,.,sec.
au,.
---
I
,,," ... CO2 r2
2g
)h
f---t:I ----)1
•'
,, h
__ ::;-_:__ _
:\
I
I
,
\'
'
,
,
,_
h~h
P. ~ 981
. (2.01) =19.75 l(Pa
p =9.81(2.693)(1)(1.2)(0,9)
p =14.2kN ·.
I
I
-
0.826
3m
\
2(9.81)
.. 2.01 rn
h =2 + 0.8 Sin so·
ti =2.693 m
P=y hA
,, : l,
; ~~
h,
cemer of
pressure
P=l4.3 kN
,
0.374
D= I.Om
,
~
1.2
H-144
,
CIVIi En11111r1nu ucensure bamlnadons
-- + 2
y - O.B Sin 60-
Solution:
y =3.109
C
s =~
1 +e
C
I = 0.9(1.2)3 = 0.0432
9
36
Ss =Ay
s = 8(0.315)
S = (0.9)(1. 2) (3.109)
S =0.241 m.
J
l
0
1. ( 200 + 120_
1+ 1.132 og
200
2
s
I ( po + LlP
og
p
- - - - -- -
Ss = 1.679
. . -· ··· -·
a
13. CE Board Ndtf
I
e=-g
A woman with a glass of water having a height
of 300 mm is inside the elevator with an
upward acceleration of 3 m/s2. Determine the
pressure at the bottom of the glass.
ss
0 0432
=0.0257
1.679
d =e +0.8
e= ·
Solution:
I
d =0.0257 + 0.8
p=yh(1+
a)
w
• g
d =0.8257
Force normal to the gate at its vertex that will
be required to open it in kN:
.p = 9.81(0.3)( 1+ ~)
9.81
LMA=O
p = 3.84kPa
I
14.3(0.374) = F(1.2)
F = 4.45kN
14. CE Board Nov. 2019
12. CE Board Nov. 2019
Evaluate the plastic settlement. in meter(s), on
a layer of plastic clay due to an increase of
pressure caused by loads above it under the
following conditions:
Initial intergranular pressure = 200 kPa
Increase in integranular pressure = 120 kPa
Thickness of the clay layer = m
Coefficient of consolidation = 0.315
Void ratio of the clay= 1. 1°32
a
'
A ship having a displacement of 24000 tons
and a draft of 10.5 m in the ocean enters a
harbor of fresh water. If · horizontal cross·
section of the ship flt the water line i~ 3000
sq.m, what depth of fresh water is req~ired 10
float the ship? Assume a marine tone 1s 1_~
kg and that seawater and fresh water werg
10.1 kN/m3 and 9.81 kN/m3 respectively.
Solution:
W = 24000(1000)(9.81)
1000
W =235440kN
I, Cflll EnglnNrlRD Ucansura Examinauons
1
H-145
w
Solution:
11',S ,
d1=10.5
__
·_ _ _ •_
!~;
CJ
·
•inal w il •"'face
H = 12 m
-- -- - - ~
JS
Salt water
y=IO.I kN/m3
W=VD
w
w
v=-o
e -- e- -1 e
3
I
2
3
e. = -· e
I
V=Ad
~e)
s - -·H!e\ _ ,_,1_
d= ~
A
1+e
Fresh water
y=9 .81 kN/m 3
S::: 12[ e . 2/3eJ·
'
A1
=A2 =3000 m2
d -d =
2
1
v2 v1 ~.
S = 12[ 0.90 - i(0.90)]
1+ 0.90
S =1.89m
,;·,;:
2
1
d2 -d = ~ - ~
1
d2 -10.5=
3000
v- ~
3000
, d2 :: 10.73 m.
15· cEBoard Nov.
2019
Alayer of soft cla h .
. . I "d t·
0.9Q i
Y avmg an .m1t1a
vo1 ra 10
~ ap \ 12 m. thick. Under a compressive
. decreas! led above it, the void ratio is
reduction ; by one-third. Evaluate the
·
in th8 thickness of the clay layers.
. Of
1+ e
16. CE Board Nov. 2019
In accordance with the Boussinesq theory, the
stress at depth Z, a semi-infinite,
homogeneous. isotropic soil mass below the
corner of a flexible area B x L in plan due to a
uniform load "q" per unit area may be
estimated from the expression P = q lz. The
value of lz may be obtained from the table
attached. A square footing 4 m. on a side
transmits a load of 1.6 MN onto the soil where
a layer of clay is formed at a depth of 8 m. to
12 m. below the base of the footing.
CD Deter~ine the pressure exerted by the
footing onto the surface.
® Estimate the stress in kPa, below t~e
c~nter of the footing the load at the m~dheight of the clay layer in accordance with
Boussinesq theory·
H-146
Clvll En11naer111 ucansure Examlnadons
@
If the stress induced on the underlying soil.
formation is just to spread on an angle of
2 vertical to 1 horizontal, evaluate the
stress at the mid height of the clay below
the center of the footing with this
assumption.
.~;
n- 2
-10 =0.2
P = 100(0.018)(4)
P= 7.2kPa
NOTE: There are four smaller e
qua/ areas
that shares comer A.
q = 1600
4(4)
.. .. .
q = 1001cPa
'
4m
. ..
.
@
I
4m
@
m=
z
lz = 0.018 from the table 3 _ 1,
,Page H-349;
P = qlz
➔- - -
➔---1< 4m
n= ~
m = n =0.20
Solution:
CD Pressure exerted by the footing onto the
I J kN
suiface
tP=l6CI
q=:
A
2
-2 ·=·0.20
10
m=-
Stress at mid height of clay using 2'1
·
method
p
Stress below the center of the footing the
lo~d at the mid-height of the clay layer
usmg
•·
&ussinerhoo~
r-_,.g-f:,;;J.-,,
,
~
I
,
:
I
·1
l-- ,
!
;z=lOm
:.
I
8 m!
I
,
:
I
• m_
I
,
,.;'
l
.s• _·: · • , ... •
., I
.· I
j
2m
A'
I
I
I
q= p
. A
q = 1600
14(14)
I
The4 m .x 4 mcon · t,
equal areas of 2 ,;, x 2 s,s s of 4 SIJ)aller
m. and each of the
corner of these ~ .
our areas shears comer
A.
q =8.16kPa
I 'm=Alz
0. 7 I 0.8 I 0.9 11.0 11.2 11.5 12.0 12.5 13.0 15.0 110.0
n=Blz or m=A/z
1
oo
-
5"
~
-
0.1
•
1
'
l
1
r
I
1
0.005 0.009 0.013 0.017 0.020 0.022 0.024 0.026 0.027 0.028 0.029 0.030 0.031 0.031 0.032 0.032 0.032 0.032
~
~
0.026 0.033 0.039 0.043 0.047 0.050 0.053 0.055 0.057 0.059 · 0.061 0.062 0.062 0.062 0.062 0.062
0.009 •
•
::::,
0.013 0.026 0.037 0.047 0.056 0.063 0.069 0.073 0.077 0.079 0.083 0.086 0.089 0.90 0.090 0.090 0.090 0.090
0.3
=ti
C
0.011 o.033 o.047 0.060 0.011 o.oao a.oar o.093 o.o9a 0.101 0.10s 0.110 0.113 0.115 0.115 0.115 0.115 0.115
o.4
CD
::::,
0.5 0.020 0.039 0.056 0.071 0.084 0.095 0.103 0.110 0.116 0.1'20 0.126 0.131 0.135 0.137 0.137 0.137 0.137 0.137
· C')
CD
0.6 0.022 0.043 0.063 0.080 0.095 0.107 0.117 0.125 0.131 0.136 j0,143 , 0.149 10.153 0.155 0.156 0.156 "0.156 0.156
C')
0
0.024 o.047 o.069 0.081 0.103 0.111 0.12a 0.137 o.1°44 o.149 10.151 /0.154 /o.rns 10.110 0.111 0.112 0.112 0.112
0.1
CD
3\ .
o.a 0.026 o.o5o 0.013 0.093 0.110 0.125 o.137 o.146 0.154 a.mo ,o.rna :o.176 ;o.rn1 lo.1a3 0.181 o.185 o.185 o.185
Q.
CD
o.9 0.021 o.053 0.011 0.098 0.116 0.131 0.144 0.154 o.1s2 !0.1oa 10.1,s :o.1a6 1o.rn2 10.194 /o.19s 0.196 0.196 0.196
=>
1.0 0.028 0.055 0.079 0.101 0.120 0.136 0.149 0.160 0.168 !0.175 i0,185 : 0.19.) ; 0.200 /0.202 / 0203 0.204 I 0.205 0.205
en
o'
1.2 0.029 0.051 0.083 0.10s 0.126 0.143 0.159 o.1ss o.11a Io.1ss ;o.rna ,0.20s ! 0.212 /o.21s i 0.21s 0.211 0.21a 0.218
1.5 o.o3o 0.059 0.086 0.110 0.131 o.149 0.164 0.116 o.186 ! a.193 :o.2os ! 0.215 ; 0.223 10.22s 0.228 / 0.229 0.230 0.230
en
2.0 . 0.031 0.061 o.089 0.113 0.135 0.153 o.169 o.1a1 0.192 I0.200 0.212 0.223 ! 0.232 i0.236 0.238 0.239 0.240 0.240
=>
2.5 0.031 0.062 0.090 0.115 0.137 0.155 0.170 0.183 0.194 0.202 i0.215 !0.226 / 0.236 i0.240 0.242 10.244 0.244 0.244
<O
C:
3.0 0.032 0.062 0.090 0.115 0.137 0.156 0.171 0.184 0.195 0.203 lo.216 0.228 l 0.238 /0.242 0.244 0.246 0.247 0.247
ti>
-,
s.o 0.032 0.062 0.090 0.115 o.137 o.156 0.112 o.185 0.196 0.204 0.217 0.229 /0.239 '0.244 0.246 0.249 0.249 o.249
Q)
coQ) 10.0 10.032 0.062 0.090 0.115 0.137 0.156 0.172 0.185 0.196 0.205 0.218 0.230 I 0.240 0.244 0.247 0.249 0.250 0250
(,/)
\ 0.032 0.062 0.090 0.115 0.137 0.156 0.172 0.185 0.196 0.205 0.218 0.230 / 0.240 0.244 0.247 0.249 0.250 0.250
oo
~ 11 ~FYz/ o., 1•1 o.3 10.41 o.s Io.6
~
..
I
I
~
~
....~
--
-
~
H-148
CMI Enlllnaerlnu ucansurt EX1mlnat1ons
'
Solution:
17. CE Board Nov. 2019
60
Sin 0 =
· A cohesionless specimen of soil under triaxial
shear test was consolidated under a chamber
pressure of 60 kPa. The axial stress on the
specimen was then increased and failure
stress occurred when the axial stress reached
120 kPa. Estimate the angle of internal friction
of the soil in degrees.
Solution:
120
0 =30•
19. CE Board Nov.·1\ •19
W~at should be the discharged velocity of
a
rectangular canal having :.t depth of 1 2
a width of 5.4 m if it has a slope of o:oo~ aM
nd
a
n = 0.013.
Solution:
r=30
r=30
1.2 m
90
30
.
S1n0=90
0 = 19.47"
5.4 m
A= 5.4(1.2)
A= 6.48
P = 1.2(2) + 5.4
P= 7.8
18. CE Board Nov. 2019
Th_e result of a consolidated drained tri-axial
soil test conducted on a consolidated I
c ay are
-as follows:
Cha~ber confining pressure: 0'3 = 60 kPa
Devrator stress at failure, ~D = 120 kPa
Compute the angle of internal friction in
degrees.
A
R=-p
6.48
R = 7.8 =0.831
V = ;_
R213
n
8112
V = 0.~ (0.831)~3 (0.001)"
13
, .
V =2.15 mis .
O=AV
r=60
,1D =120
r=60
Q = 6.48(2.15)
Q = 13.93 m3/sec
f
CfVII En11neer1na Ucensure Exami
----;;;;;;;
;;;;;---==
:na~t1~o
n~s:__
___
_
' ,. .,..,..,..,r.1:e~.,~, ~.-,., ,~.,-.mi,~,..,-,
. -~·--Asoil sample has a. water content ~f SO% and
degree of saturation of 90%. Find the v0.d
~tio of nit has a sp.gr. = 2.67.
'
I
I
solution:
ro Gs
&MD~x,sw;,Mwe
~
The wet unit w • h
kN/m3. The soil erg t of a soil sample ,s 14.9
the Sp.gr. of theh!~t water content of 12%. If
Weight.
IS 2-67 , find the dry unit
Solution:
Is=-e
oo=
H-149
"(wet ='¥dry (1 + co)
80(2.67)
14 ,9 ='¥dry (1 + 0.12)'
e
Yjry
e=2.37
21. CE Board Nov. 2019
Afireman has to put out a fire but is blocked
by a firewall. To reach over the wail, he
directed the water jet from the nozzle at an
angle of 30· to the horizontal. Evaluate the
velocity of the water in meters per second,
leaving the nozzle of his hose to reach over
the wall if he stands 30 m. away from the wall
and the wall js standing 2 m. higher than the
nozzle of the hose. Neglect friction in the jet.
= 13.30 kN/m3
:lS:;)t~_'Soard Nov. 2019
A 450 kN is trans~itted by a column footing
onto the surface through a square footing 1.5
m. on a side. Assuming that the force exerted
on the underlying soil formation spreads on 2
vertical to 1 horizontal, evaluate the pressure
exerted in footing on a soil 2.7 m. below it.
Solution: .
450kN
;olutiori:
2.7m
~::xtane.
-
gx2
J .35
2v 2 Cos 2 e
h30tan 30• _
15,32:: ~
v2
Ii ...
V .._
19
,6m/s
·-·
1.5
4.2
2
9.81(30)
2(V) 2 Cos 2 30°
450
p = (4.2) 2
p =25.51 kPa
H-150
Clvll En11neer1n1 ucensura EXamlnado~s
•
V2-V1 = h(A)
24. CE Board Nov. 2019
20000 - 19417.48 =·h(3000)
A ship having a displacement of 20000 metr!c
tones enters a harbor of fresh water. The ship
captain recorded a draft of 8.4 m. while the
ship was still in seawater (sp.gr. = 1.03).
Obtain the draft in meters of the ship in fresh
water if the horizontal section of the ship
below the water line is 3000 · m3 in both
instances.
h=0.194m
Draft in fresh water = 8.4 + 0.1 94 .
Draft in fresh water = 8.594 m.
Solution:
-
~hat i~ th~ frictio~ angle of the soil when the
coefficient of passive ff:·sistance is 5?
Solution:
K = 1+ Sin 0
P
'
5
BF1
0
Sea water (sp. gr. = 1.03)
1- Sin 0
= 1_+ Sin 8
1-Sin 0
5 - 5 Sin 8 =1 + Sin 0
6Sin8 =4
0 = 41.8"
26. CE Board Nov. 2019
What
A mercury barometer reads 660. mm.
water
·
would be the corresponding reading in
BF2
Fresh water
barometer?
Vol. displaced in seawater:
Solution:
W=V (sp.gr.) 'Yw
h = 0.66(t3.6)
. 20000(10Q0) =V1(1.03)(1000)
V1 =19417.48 m3
h =8.98 m. of water
J
Vol. displaced in fresh water:
20000(1000) =v2(1000)
V2 =20000 m3
'
-• EnUlnearlna Ucansura Examlnauons
f11il=MU·14·lff4D
A rectangular tank ~f internal width . 7 m.
partitioned as shown, 1t contains oil and water.
ASSume unit wt. of water is 9.79 kN/m3.
H-151
tH&·MitWM♦Wfl
A rectangular tank has
..
contains oil and water. If~h~a:~,i°~atown. It
of 0.84, find its depth h.
a sp.gr.
If the oil has a sp.gr. of ·o.84, find its depth
h.
@ If a 900 N block of wood is floated in oil,
what is the volume of wood submerged .
@ If a 900 N block of wood is floated in oil.
what is the rise in free surface of the water
in contact with air?
(D
Solution:
Solution:
9.81(h)(0.80) - 1(9.81) = 0
h = 1.19 m of oil
·29~?CE Board Nov. 2019
l-osm ---1m --900 N
h1
I
- --~ '
, Hi()
l
4m
\11.!:::o.sm-::=_ :::;::_
_ -===~J
G)
A tank contains two liquids having a sp.gr. of
0.6 and 1.0 respectively. An object having a
weight of 0.50 kN (sp.gr. = 0.8) is placed in the
container and is submerged s~ch that a
portion of the object _
is ~bove the interface of
the two liquids. Determine the volu~e ~f the
object above the interface of the two hqu1d~.
Solution:
Depth "h":
0+9.79 (0.84) h- 9.79 (1) = 0
h: 1.19 m.
~ Volume of wood:
0.900 =9.79 (Vs)(0.84)
Vs=0.10944 m3. (vol. of wood
submerged)
~ .
, R,_se in free surface of the water in contact
Wffh air
.
s· .
ince the volume of oil does not change:
119
· (0.5)(7) -t1O.1O944 = h1 (O.5)(7)
hi =1.221 m.
w
V =0.0637 m3
V =V1 + V2
O.O637 =V1 + V2
BF1 =V1 (0.6)(9.81)
BF 1 =5.886 V1
'
BF2 =V2(1)(9.8
1)
BF2 =9.81 V2
sp.g1: :::c I·O
H-JS1
CIVIi
Enllnlll'lnt ucensure Ell■lnlll•
W= BF,+ BF2
0.50 :: 5.886 V1 + 9.81 V2
V2=0.0637 - .V1
0.50 =5.886 V1 + 9.81(0.0637 • V1)
0.50 =5.886 V1 + 0.624897 - 9.81 V1
3.924 V1 =0.624897 - 0.50
V1 =0.0318 m3
V = __J_(0.75f3(0 oo.1r2
.
0.012
V =2.175
Q=AV
a =4.8(2.175)
Q = 10.44 m'/sec.
32. CE Board Nov. 2019
30. CE Board Nov. 2019
Whe at any instant, the number of particles
passing every section of the stream are
always equal, the flow is said to be _ _
a)
b}
c}
d}
Continuous t"
Steady flow
Turbulent flow
laminar flow
Determine the disc 1.J~ge flowing in a pi
having a diameter r..: 1 51') mmand a head:
ss
of 3 m/km. Friction f2.: :or :: 0.025.
Solution:
hf= f L v2
02g
3
= 0.025(1000)V~
0.15(2)(9.81)
V =0.594m's
31. CE Board Nov. 2019
Q=AV
Ar~ctangular canal has a depth of 1'.2 m and
a width of 4 m. It has a slope of 0.001 and
n = 0.0 12· Compute the discharge flowing in
Q=:
Q
(0.15) 2 (0.594)
= 0.0105 m3 I sec.
the canal.
33. CE Board Nov. 2019
Solution:
1.2m l
V = -1 R213 5v2
n
A= 4(1.2) =4.8 m2
P =4 + 2( f .2) =6.4 m
R= A
p
R
_ 4.8
-6A =Q.75
4m
In accordance with Rankine's Theory, what
angle e in degrees, results in a coe~t of
lateral pressure equal to 4.00 of a honzontal
backfill?
Solution:
K = ,1+Sin 8
1-Sin 8
= 1+ Sin 8
4
1- Sin 8"
4 - 4 Sin a=1 + Sin e
5 Sin 8 =3
8 =36.9"