*7–92.
Draw the shear and moment diagrams for the beam.
SOLUTION
1.5 m
6 kN/m6 kN/m
1.5 m
A B C
7–93.
Draw theshear and moment diagramsfor thebeam.
SOLUTION
Ans.
a
Ans.M={x
2
>2–x
3
>45} N
#
m
+©M=0;
M+(x
2
>15) ax
3b–1x(x>2) =0
V={x–x
2
>15} N
+c©F
y
=0; 1x–x
2
>15 –V=0
15 ft
1 kip/ft
2kip/ft
A
SOLUTION
7–94.
The cable supports the three loads shown. Determine the
sags yB and yD of B and D. Take P1 = 800 N, P2 = 500 N. 1 m
3 m 6 m 6 m 3 m
A
E
B
C
D
yByD
4 m
P2P2
P1
6
715
SOLUTION
7–95.
The cable supports the three loads shown. Determine the
magnitude of P1 if P2 = 600 N and yB = 3 m. Also find sag yD. 1 m
3 m 6 m 6 m 3 m
A
E
B
C
D
yByD
4 m
P2P2
P1
6
18
*7–96.
Determine the tension in each segment of the cable and the
cable’s total length.
A
B
7ft
4ft
D
SOLUTION
7–97.
The cable supports the loading shown. Determine the
distance
xB
the force at B acts from A. Set P = 800 N.
4 m
1 m
600 N
D
C
B
A
xB
6 m
P
SOLUTION
7–99.
The cable supports the three loads shown. Determine the
sags and of points Band D.Take
P2=250 lb.
P1=400 lb,yD
yB
4ft
A
E
B
C
D
y
B
y
D
14 ft
P
2
P
2
721
*7–100.
The cable supports the three loads shown. Determine the
magnitude
of if and Also find the
sag
yD.
yB=8ft.P2=300 lbP1
SOLUTION
At
B
At
C
(1)
(2)
At
D
:
+©F
x=0; 12
2(4 +yD)2+144 T
DE –15
2(14 –yD)2+225 T
CD =0
+c©F
y=0; 6
2436 (854.2) +14 –yD
2(14 –yD)2+225 T
CD –P
1=0
:
+©F
x=0; –20
2436 (854.2) +15
2(14 –yD)2+225T
CD =0
T
BC =854.2 lb
T
AB =983.3 lb
+c©F
y=0; –6
2436 T
BC +8
2208 T
AB –300 =0
:
+©F
x=0; 20
2436 T
BC –12
2208 T
AB =0
4ft
12 ft 20 ft 15 ft 12 ft
A
E
B
C
D
y
B
y
D
14 ft
P
2
P
2
P
1
P1=658 lb
P1=658 lb
+
c
ΣF
y
=0;
4+y
D
2
(4
+
y
D
)2
+
144
TDE –
14 –y
D
2
(14
–
y
D
)2
+
225
TCD –300 =0 (4)
From Eq. 1,
T
CD
2
(14
–
y
D
)2
+
225
=54.545 lb
and then from Eq. 3,
T
DE
2
(4
+
y
D
)2
+
144
=68.182 lb
Now Eq. 2 simplifies to
P1+54.545yD=1009.09
(5)
and Eq. 4 simplifies to
68.182(4 +yD)–54.545(14 –yD)=300
yD=6.444 =6.44 ft
Ans.
7–101.
SOLUTION
(1)
Joint C:
(2)
(3)
Combining Eqs. (1) and (2):
(4)
Joint D:
(5)
From Eqs. (1) and (3):
From Eqs. (4) and (5): Ans.
Ans.
T
CD =4.60 kN
T
BC =4.53 kN
T
AB =6.05 kN
P=0.8 kN
yB=3.53 m
3yBP–16 yB+48 =0
15 –2yB
2(yB–3)2+9T
CD =12
+c©F
y=0; 3
213 T
DE –yB–3
2(yB–3)2+9T
CD –6=0
:
+©F
x=0; 2
213 T
DE –3
2(yB–3)2+9T
CD =0
3
2(yB–3)2+9T
CD =16
yB
(yB–3)T
BC =3P
+c©F
y=0; yB–3
2(yB–3)2+9T
CD –P=0
:
+©F
x=0; 3
2(yB–3)2+9T
CD –T
BC =0
yBT
BC =16
+c©F
y=0; yB
2y2
B+16 T
AB –4=0
2y2
B+16 T
Determine the force Pneeded to hold the cable in the
position shown, i.e., so segment BC remains horizontal.Also,
compute the sag and the maximum tension in the cable.yB
4kNP
6kN
y
B
3m
A
BC
D
E
7–102.
SOLUTION
At ,
At ,
At ,
Ans.w=51.9 lb/ft
F
H
=2705 lb
T
max
=F
H
cos u
max
=3000
u
max
=tan
–1
(0.48) =25.64°
dy
dx
2
max
=tan
u
max
=w
F
H
x
2
x=25 ft
y=6 ft
F
H
=52.08 wx=25 ft
y=w
2F
H
x
2
C
1
=C
2
=0
y=0x=0
dy
dx =0x=0
y=1
F
H
LaLwdxbdx
Determine the maximum uniform loading measured in
that the cable can support if it is capable of sustaining
a maximum tension of 3000 lb before it will break.
lb>ft,
w,
50 ft
6ft
w
7–103.
The cable is subjected to a uniform loading of
Determine the maximum and minimum tension in the cable.
w=250 lb
>
ft.
SOLUTION
Ans.
The minimum tension occurs at .
Ans.T
min
=F
H
=13.0 kip
u=0°
T
max
=F
H
cos u
max
=13 021
cos 25.64° =14.4 kip
u
max
=tan
–1
aw
0
L
2F
H
b=tan
–1
a250 (50)
2(13 021) b=25.64°
50 ft
6ft
w
*7–104.
The cable AB is subjected to a uniform loading of 200 N/m.
If the weight of the cable is neglected and the slope angles
at points Aand Bare 30° and 60°, respectively, determine
the curve that defines the cable shape and the maximum
tension developed in the cable.
SOLUTION
y=1
¢
≤
dx
y
B
60°
7–105.
If x=2 ft and the crate weighs 300 lb, which cable segment
AB,BC, or CD has the greatest tension? What is this force
and what is the sag yB?
SOLUTION
Referring to Fig. b, we have
Using these results and analyzing the equilibrium of joint C,Fig. c, we obtain
Ans.
Solving,
Using these results to analyze the equilibrium of joint B,Fig. d, we have
Solving,
Thus, both cables AB and CD are subjected to maximum tension. The sag yBis
given by
Ans.yB=2 ft
yB
2= tan f= tan 45°
T
AB =212.13 lb =212 lb (max)
f=45°
T
AB sin f–100 –158.11 sin 18.43° =0+c©F
y=0;
158.11 cos 18.43° –T
AB cos f=0©F
x=0;
:
+
u=18.43°T
BC =158.11 lb
T
AB =T
CD =212 lb (max)
T
BC sin u+212.13 sin 45° –200 =0+c©F
y=0;
212.13 cos 45° –T
BC cos u=0©F
x=0;
:
+
T
CD =212.13 lb =212 lb (max)
T
CD sin 45°(8) –200(5) –100(2) =0+©MA=0;
F
B=200 lb300(1) –F
B(3) =0+©MF=0;
3 ft 3 ft
3 ft
2 ft
AD
B
C
y
B
7–106.
If y
B
=1.5 ft, determine the largest weight of the crate and
its placement xso that neither cable segment AB,BC,or
CD is subjected to a tension that exceeds 200 lb.
SOLUTION
a
Since the horizontal component of tensile force developed in each cable is
constant, cable CD, which has the greatest angle with the horizontal, will be
subjected to the greatest tension. Thus, we will set TCD =200 lb.
First, we will analyze the equilibrium of joint C,Fig. b.
Using the result of TBC to analyze the equilibrium of joint B,Fig. c, we have
Solving Eqs. (1) and (2)
Ans.W=247 lbx=2.57 ft
(2)
w
3(3 –x)=35.36
176.78a3
5b–158.11 sin 26.57° –w
3(3 –x)=0+c©F
y=0;
T
AB =176.78 lb158.11 cos 26.57° –T
AB a4
5b=0©F
x=0;
:
+
(1)
wx
3=212.13
200 sin 45° +158.11 sin 26.57° –wx
3=0+c©F
y=0;
T
BC =158.11 lb200 cos 45° –T
BC cos 26.57° =0©F
x=0;
:
+
F
B=w
3(3 –x)w(3 –x)–F
B(3) =0+©MF=0;
3
3 ft 3 ft
3 ft
2 ft
AD
B
C
y
B
728
7–107.
The cable supportsagirder which weighs 850
Determine the tension in the cable at points A, B, and C.
lb>ft.
SOLUTION
At ,
At ,
At ,
At ,
At A,
Ans.T
A=
61.7 kip
TA=FH
cos uA
=36 459
cos 53.79° =61 714 lb
uA=53.79°
dy
dx =tan uA=2(425)x
FH
`
x=-58.58 ft
=1.366
FH=36 459 lb
x¿=–200 <22002+4(100)2
2=41.42 ft
(x¿)2+200x¿-1002=0
2(x¿)2=(x¿)2–200x¿+1002
40 =425(100 –x¿)2
FH
x=(100–x¿)y=40 ft
20 =425(x¿)2
FH
x=x¿y=20 ft
y=425
FH
x2
C2=0y=0x=0
C1=0
dy
dx =0x=0
dy
dx =850
FH
x+C1
FH
y=1
FH
(425x2+C1x+C2)
y=1
FHL(Lw0dx)dx
100ft
A
C
B
40 ft
20 ft
7–107. Continued
At B,
Ans.
At C,
Ans.T
C=
50.7 kip
TC=FH
cos uC
=36 459
cos 44.0°
=50 683 lb
uC=44.0°
dy
dx =tan uC=2(425)x
FH
`
x=41.42 ft
=0.9657
TB=FH=36.5 kip