995
*9–116. Continued
Ans:
FR=27.0 kN
x=0.778 m
y=0.833 m
Thus,
x=
L
A
xp(x, y)dA
LA
p(x, y)dA
=
21.0 kN #m
27.0 kN =
7
9 m =0.778 m Ans.
y=
L
A
yp(x, y)dA
LA
p(x, y)dA
=
22.5 kN #m
27.0 kN =0.833 m Ans.
9–117.
SOLUTION
Ans.y=
L
FR
L
FR
dF
R
=
32.0
24.0 =1.33 m
T
h
e
l
oa
d
over t
h
e p
l
ate var
i
es
li
near
l
y a
l
ong t
h
e s
id
es of t
h
e
plate such that Determine the resultant
force and its position on the plate.1x,y2
p=2
3[x14–y2] kPa.
p
3m
4m
y
x
8kPa
9–118.
The rectangular plate is subjected to a distributed load over
its entire . The load is defined by the expression
where represents the
pressure acting at the center of the plate. Determine the
magnitude and location of the resultant force acting on
the plate.
p0
p=p0sin (px>a)
x
a
p
0
y
p
SOLUTION
Ans.
Since the loading is symmetric, the location of the resultant force is at the center of
the plate. Hence,
Ans.x=
a
2y=
b
2
=
4ab
p2p0
=p0
B
a–
a
p
cos
px
ab2a
0a–
b
p
cos
px
bb2b
0
R
FR=LFR
dFR=p0La
0asin
px
adxbLb
0asin
py
bdyb
sin (py>b),
surface
9–119.
A wind loading creates a positive pressure on one side of
the chimney and a negative (suction) pressure on the other
side, as shown. If this pressure loading acts uniformly along
the chimney’s length, determine the magnitude of the
resultant force created by the wind.
SOLUTION
2
2
p p0 cos u
p
u
*9–120.
When the tide water Asubsides, the tide gate automatically
swings open to drain the marsh B.For the condition of high
tide shown, determine the horizontal reactions developed
at the hinge Cand stop block D.The length of the gate is
6mand its height is 4 m. rw=1.0 Mg/m3.
SOLUTION
Thus,
Resultant Forces:
Equations of Equilibrium:
Ans.
Ans.Cx=46.6 kN
264.87 –117.72 –100.55 –Cx=0:
+©F
x=0;
Dx=100.55 kN =101 kN
264.87132–117.7213.3332–Dx142=0a+©MC=0;
F
R2=1
21117.722122=117.72 kN
F
R1=1
21176.582132=264.87 kN
wE=29.43162=176.58 kN>m
wD=19.62162=117.72 kN>m
pE=1.01103219.812132=29 430 N>m2=29.43 kN>m2
pD=1.01103219.812122=19 620 N>m2=19.62 kN>m2
A
B
C
D
3m
2m
4m
9–121.
The tank is filled with water to a depth of
Determine the resultant force the water exerts on side Aand
side Bof the tank. If oil instead of water is placed in the tank,
to what depth dshould it reach so that it creates the same
resultant forces? and rw=1000 kg>m3.ro=900 kg>m3
d
=4m.
SOLUTION
Ans.
At side B:
Ans.
For oil
At side A:
Ans.d=4.22 m
F
RA
=
1
2(17 658 d)(d)=156 960 N
=17 658 d
=2(900)(9.81)d
w
A=brogd
F
RB
=
1
2(117 720)(4) =235 440 N =235 kN
=117 720 N>m
=3(1000)(9.81)(4)
w
B=brwgd
F
RA
=
1
2(78 480)(4) =156 960 N =157 kN
=78 480 N/m
=2(1000)(9.81) (4)
d
BA
3m 2m
9–122.
The concrete “gravity” dam is held in place by its own
weight. If the density of concrete is rc
=
2.5 Mg
>
m
3
, and
water has a density of rw
=
1.0 Mg
>
m
3
, determine the
smallest dimension d that will prevent the dam from
overturning about its end A.
A
d
1 m
6 m
d – 1
SOLUTION
b=1 m
2
(
FR
)
h=
1
2
(58.86)(6) =176.58 kN
x1=0.5
x2=1+
1
3
(d–1) =
1
3
(d+2) x3=1+
2
3
(d–1) =
1
3
(2d+1)
y=
1
3
(6) =2 m
Equation of Equilibrium. Write the moment equation of equilibrium about A by
referring to the FBD of the dam, Fig. a,
a+
ΣMA=0;
147.15(0.5) +[73.575(d–1)]
c1
3
(d+2)
d
+[29.43(d–1)]
c1
3
(2d+1)
d
–176.58(2) =0
44.145d2+14.715d–338.445 =0
Solving and chose the positive root
d=2.607 m =2.61 m
Ans.
1002
9–123.
The factor of safety for tipping of the concrete dam is defined as the
ratio of the stabilizing moment due to the dam’s weight divided by
the overturning moment about O due to the water pressure.
Determine this factor if the concrete has a density of
r
conc =
2.5 Mg
>
m
3
and for water rw
=
1 Mg
>
m
3
.
y
1 m
x
4 m
6 m
O
SOLUTION
Ans:
F.S. =2.71
1003
*9–124.
The concrete dam in the shape of a quarter circle.
Determine the magnitude of the resultant hydrostatic force
that acts on the dam per meter of length. The density of
water is
SOLUTION
Resultant Force Component:The vertical component consists of the weight
ofwater contained in the shaded area shown in Fig. a.For a 1–m length of dam,
we have
The horizontal component consists of the horizontal hydrostatic pressure.
Since the width of the dam is constant (1 m),this loading can be represented
by a triangular distributed loading with an intensity of
at point C,Fig.a.
Thus,the magnitude of the resultant hydrostatic force acting on the dam is
Ans.F
R=3Fh
2+Fv2=344.145
2+18.95
2=48.0 kN
F
h=1
2 (29.43)(3) =44.145 kN
1000(9.81)(3)(1) =29.43 kN>m
wC=rghCb=
Fh
F
v=rgAABCb=(1000)(9.81)
B
(3)(3) –p
4 (32)
R
(1) =18947.20 N =18.95 kN
Fv
rw=1 Mg>m3.
3 m
Ans:
FR=48.0 kN
1004
9–125.
The tank is used to store a liquid having a density of
80 lb
>
ft3. If it is filled to the top, determine the magnitude of
force the liquid exerts on each of its two sides ABDC
and BDFE.
2
2
Ans:
F1=9.60 kip
F2=40.3 kip
4 ft
6 ft
12 ft
3 ft
B
A
C
D
F
E
1005
9–126.
The parabolic plate is subjected to a fluid pressure that
varies linearly from 0 at its top to 100 lb
>
ft at its bottom B.
Determine the magnitude of the resultant force and its
location on the plate.
SOLUTION
Ans:
FR=427 lb
y
=
1.71 ft
x=0
2 ft 2 ft
4 ft
y x2
y
x
9–127.
SOLUTION
F
2=(58 860)(3) =176 580
F
1=1
2(3)(58 860) =88 290
w2=1000(9.81)(3)(2) =58 860 N/m
w1=1000(9.81)(3)(2) =58 860 N/m
The 2-m-wide rectangular gate is pinned at its center Aand
is prevented from rotating by the block at B. Determine the
reactions at these supports due to hydrostatic pressure.
rw
=1.0 Mg>m3.
6m
1.5m
A
B
1.5m
1007
*9–128.
SOLUTION
Resultant Force and its Location: Here,.The volume of the
differential element is .the
Evaluating integrals using Simpson’s rule,we have
Ans.
Ans.y=
L
FR
y
‘dF
R
L
FR
dF
R
=
–866.7
6934.2 =-0.125 m
=-866.7 N #m
L
FR
ydF
R=17658 L0.5 m
–0.5 m
y(0.5 –y)(21–4y2)dy
=6934.2 N =6.93 kN
F
R=LFR
dF
R=17658 L0.5 m
–0.5 m
(0.5 –y)(21–4y2)dy
dV =dF
R=p(2xdy)=8829(0.5 –y)[221–4y2]dy
x=21–4y2
T
h
e tan
k
i
s f
ill
e
d
w
i
t
h
a
li
qu
id
w
hi
c
h
h
as a
d
ens
i
ty of
Determine the resultant force that it exerts on
the elliptical end plate, and the location of the center of
pressure, measured from the xaxis.
900 kg>m3.
0.5 m
1m
0.5 m
1m
x
y
4y
2
x
2
1
Ans:
FR=6.93 kN
y=–0.125 m
1008
9–129.
Determine the magnitude of the resultant force acting on
the gate ABC due to hydrostatic pressure. The gate has a
width of 1.5 m. .rw
=1.0 Mg>m3
B
C
1.25 m
1.5 m
Ans.F=2F2
x+F2
y=2(73.58)2+(70.069)2=102 kN
Fy=F1+F2=70.069 kN
F2=1
2(1.5)(2)a2
tan 60° b(1000)(9.81) =16.99 kN
F1=c(22.072)a1.25 +2
tan 60° bd =53.078 kN
Fx=1
2(29.43)(2) +(22.0725)(2) =73.58 kN
w2=1000(9.81)(2)(1.5) =29.43 kN>m
Ans:
F=102
kN
1009
9–130.
The semicircular drainage pipe is filled with water.
Determine the resultant horizontal and vertical force
components that the water exerts on the side AB of the
pipe per foot of pipe length; gw=62.4 lb>ft3.
SOLUTION
Thus,
p=62.4(2) =124.8 lb>ft2
B
2ft
A
F
Rh
=125
lb