7–41.
Determine the x, y, zcomponents of force and moment at
point Cin the pipe assembly. Neglect the weight of the pipe.
dna ekaT F
2
=5–300j+150k6lb.F
1
=5350i–400j6lb
2ft
1.5fty
z
C
B
3ft
F
1
SOLUTION
7–42.
SOLUTION
Free body Diagram: The support reactions need not be computed.
Internal Forces: Applying the equations of equilibrium to segment BC, we have
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.(M
C
)
z
=-178 lb
#
ft
©M
z
=0; (M
C
)
z
+24 (2) +80 (2) –30 =0
©M
y
=0;
(M
C
)
y
–24 (3) =0(M
C
)
y
=72.0 lb
#
ft
©M
x
=0; (M
C
)
x
–10(2) =0(M
C
)
x
=20.0 lb
#
ft
©F
z
=0; (V
C
)
z
–10 =0(V
C
)
z
=10.0 lb
©F
y
=0; N
C
=0
©F
x
=0; (V
C
)
x
–24–80 =0(V
C
)
x
=104 lb
F
1
F
2
2ft
x
z
y
3ft
C
B
A
M
1.5 ft
Determine the x, y, z components of force and moment at
point C in the pipe assembly. Neglect the weight of the pipe.
The load acting at (0, 3.5 ft, 3 ft) is F
1
= 5–24i – 10k6 lb and
M = {–30k} lb
#
ft and at point (0, 3.5 ft, 0) F
2
= {–80i
} lb.
655
7–45.
SOLUTION
For
Ans.
c
Ans.
For
Ans.
c
M=Px –aa
L
bPx –Px +Pa
+©M=0;
a1–a
LbPx –P(x–a)–M=0
V=-aa
LbP
+c©Fy=0; a1–a
LbP–P–V=0
a6x6L
M=a1–a
LbPx
+©M=0;
a1–a
LbPx –M=0
:
+©Fx=0; A=0
V=a1–a
LbP
+c©Fy=0; a1–a
LbP–V=0
0…x…a
:
Draw the shear and moment diagrams for the shaft
(
a
)
in
terms of the parameters shown; (b) set
There is a thrust bearing at Aand a journal
bearing at B.
L=6m.
a=2m,P=9 kN,
P
a
AB
L
657
7–46.
Draw the shear and moment diagrams for the beam (a) in
terms of the parameters shown; (b) set
L=12 ft.
a=5 ft,P=800 lb,
Ans.
aAns.
For
Ans.
a
Ans.
For
Ans.
a
Ans.
M=P(L–x)
+©M=0; –M+P(L–x)=0
+c©F
y
=0;
V=-P
L–a6x…L
M=Pa
+©M=0; –Px +P(x–a)+M=0
+c©F
y
=0;
V=0
a6x6L–a
+©M=0;
M=Px
+c©F
y
=0;
V=P
aa
L
PP
658
7–46. Continued
Ans:
For 0 …x6a
,
V=P
,
M=Px
For a6x6L–a
,
V=0
,
M=Pa
For L–a6x…L
,
V=–P
,
M=P(L–x)
For 0 …x65
ft
,
V=800
lb
M=800x
lb #ft
For 5
ft 6x67
ft
,
V=0
M=4000
lb #ft
For 7
ft 6x…12
ft
,
V=–800
lb
M=(9600 –800x)
lb #ft
(b) Set
For
Ans.
aAns.
For
Ans.
a
Ans.
For
Ans.
a
Ans.M=(9600 –800x)lb
#
ft
+©M=0; –M+800(12 –x)=0
+c©F
y
=0;
V=-800 lb
7ft6x…12 ft
M=4000 lb
#
ft
+©M=0; –800x+800(x–5) +M=0
+c©F
y
=0;
V=0
5ft6x67ft
+©M=0;
M=800xlb
#
ft
+c©F
y
=0;
V=800 lb
0…x65ft
P=800 lb, a=5ft, L=12 ft
659
7–47.
SOLUTION
(a) For
Ans.
a
Ans.
For
Ans.
a
(b) For b=7fta=5 ft,P=600 lb,
–Pb
a+bx+P1x–a2+M=0+©M=0;
V=-
Pa
a+b
Pb
a+b–P–V=0+c©F
y
=0;
a6x…1a+b2
M=Pb
a+bx
M–Pb
a+bx=0+©M=0;
V=Pb
a+b
Pb
a+b–V=0+c©F
y
=0;
0…x6a
Draw the shear and moment diagrams for the beam (a) in
terms of the parameters shown; (b) set
b=7 ft.
a=5 ft,P=600 lb,
AB
P
ab
7–47. Continued
*7–48.
SOLUTION
Draw the shear and moment diagrams for the cantilevered
C
B
5ft
100 lb
800 lb ft
5ft
A
beam.
7–49.
Ans.
aAns.
For
Ans.
aAns.
For
Ans.
aAns.
(b)
Set ,
For
Ans.
cAns.
For
Ans.
+c©F
y
=0; V=0
8
3m6x616
3m
+©M=0;
M=0
+c©F
y
=0; V=0
0…x68
3m
L=8mM
0
=500 N
#
m
+©M=0;
M=0
+c©F
y
=0; V=0
2L
36x…L
+©M=0;
M=M
0
+c©F
y
=0; V=0
L
36x62L
3
+©M=0;
M=0
+c©F
y
=0; V=0
Draw the shear and moment diagrams for the beam (a) in
terms of the parameters shown; (b) set
L=8m.
M
0
=500 N
#
m,
L/3L/3L/3
M
0
M
0
7–50.
If the beam will fail when the maximum shear
force is or the maximum bending moment is
Determine the magnitude of the
largest couple moments it will support.
M
0
M
max
=2kN
#
m.
V
max
=5kN
L
=9m,
SOLUTION
L/3L/3L/3
M
0
M
0
664
7–51.
SOLUTION
+c©Fy=0; –V–wx=0
0…x6a
Draw t
h
e s
h
ear an
d
moment
di
agrams for t
h
e
b
eam.
A
BC
a a
w
2
*7–52.
Shear and Moment Functions: For [FBD (b)],
Ans.
aAns.
For [FBD (c)],
Ans.
a
Ans.M=w
81–L2+5Lx –4x22
3wL
81L–x2–w1L–x2aL–x
2b–M=0+ ©M=0;
V=w
815L–8x2
V+3wL
8–w1L–x2=0+c©F
y=0;
L
2
<x◊L
M–wL
81x2=0M=wL
8x+ ©M=0;
wL
8–V=0V=wL
8
+c©F
y=0;
0◊x<
L
2
Ay+3wL
8–wL
2=0Ay=wL
8
+c©F
y=0;
2a3L
4b=0Cy=3wL
8
Draw the shear and moment diagrams for the beam.
C
w
A
B
L
L
––
2
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
667
7–54.
Ans.
a
Ans.
(b) Set w= 500 lb/ft, L= 10 ft
For
Ans.
a
Ans.M=(2500x–250x2)lb#ft
+©M=0;
–2500x+500 x2
2+M=0
V=(2500 –500x)lb
+c©Fy=0; 2500 –500x–V=0
0…x…10 ft
M=w
2aLx –x2b
M=wL
2x–wx2
2
+©M=0;
–wL
2x+wxax
2b+M=0
V=w
2(L–2x)
V=-wx +wL
2
+c©Fy=0; wL
2–wx–V=0
Theshaft is supportedbyathrust bearing at Aand a
journal bearing at B. Draw the shear and moment diagrams
for the shaft (a) in terms of the parameters shown; (b) set
L=10 ft.w=500 lb>ft,
L
AB
w
7–55.
Draw t
h
es
h
ear an
d
moment
di
agrams for t
h
e
b
eam.
SOLUTION
Ans.
a
Ans.
Ans.
c
Ans.M=20x–370
+©M=0;
M+20(11 –x)+150 =0
V=20
+c©F
y
=0; V–20 =0
86x…11
M=133.75x–20x
2
+©M=0;
M+40xax
2b–133.75x=0
V=133.75 –40x
+c©F
y
=0; 133.75 –40x–V=0
0…x68
40 kN/m
20 kN
150 kN m
A
BC
8m 3m
*7–56.
SOLUTION
Ans.
a
Ans.
:
Ans.
a
Ans.M=-0.75 x2+3.75 x–3kN#m
+©M=0;
M+1.5
2(x–2)2–0.75 x=0
V=3.75 –1.5 xkN
+c©Fy=0; 0.75 –1.5 (x–2) –V=0
2m 6x64m
M=0.75 xkN #m
+©M=0;
M–0.75 x=0
V=0.75 kN
+c©Fy=0; 0.75 –V=0
Draw the shear and moment diagrams for the beam.
2m
4m
1.5 kN/m
ABC
670
7–57.
Draw the shear and moment diagrams for the compound
beam. The beam is pin-connected at Eand F.
A
L
w
BEFC
D
L
––
L
––
L
––
L
SOLUTION
From FBD (a),
a
From FBD (c),
a
Shear and Moment Functions: For [FBD (d)],
Ans.
a
Ans.
M=w
18 17Lx –9x
2
2
M+wxax
2b–7wL
18 x=0+©M=0;
V=w
18
17L–18x2
7wL
18 –wx–V=0+c©F
y
=0;
0◊x<L
B
y
+7wL
18 –4wL
3–wL
6=0B
y
=10wL
9
+c©F
y
=0;
4wL
3aL
3b–wL
6aL
3b–A
y
1L2=0A
y
=7wL
18
+©M
B
=0;
D
y
1L2+wL
6aL
3b–4wL
3aL
3b=0D
y
=7wL
18
+©M
C
=0;
E
y
+wL
6–wL
3=0E
y
=wL
6
+c©F
y
=0;
3b–wL
3aL
6b=0F
6