651
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
F
2
2ft
1.5fty
z
x
C
B
3ft
F
1
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.
1V
C
2
y
–400 –300 =01V
C
2
y
=700 lb©F
y
=0;
N
C
+350 =0N
C
=-350 lb©F
x
=0;
Ans:
NC=–350 lb
(V
C
)
y
=700 lb
(M
7–42.
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.
653
Ans:
Nx=–500 N
SOLUTION
Internal Loadings. Referring to the FBD of the free end segment of the pipe
assembly sectioned through B, Fig. a,
ΣFx=0;
Nx+300 +200 =0
Nx=–500 N
Ans.
ΣF
y
=0;
V
y
–100 =0
V
y
=100 N
Ans.
ΣMx=0;
ΣM
M
M
The negative signs indicate that
Nx
and
M
y act in the opposite sense to those shown
in FBD.
7–43.
Determine the x, y, z components of internal loading at a
section passing through point B in the pipe assembly. Neglect
the weight of the pipe. Take F1 =
5
200i – 100j – 400k
6
N
and F2 =
5
300i – 500k
6
N.
x
z
y
B
A
1 m
1.5 m F1
F2
1 m
654
*7–44.
Determine the x, y, z components of internal loading
at a section passing through point B in the pipe
assembly. Neglect the weight of the pipe. Take
F1 =
5
100i – 200j – 300k
6
N and F2 =
5
100i + 500j
6
N.
x
z
y
B
A
1 m
1.5 m F1
F2
1 m
SOLUTION
Internal Loadings. Referring to the FBD of the free end segment of the pipe
assembly sectioned through B, Fig. a
ΣFx=0;
Nx+100 +100 =0
Nx=–200 N
Ans.
ΣF
y
=0;
V
y
+500 –200 =0
V
y
=–300 N
Ans.
ΣMx=0;
ΣM
M
M
The negative signs indicates that
Nx
,
V
y,
M
y and
Mz
act in the senses opposite to
those shown in FBD.
655
7–45.
SOLUTION
(a)
For
Ans.
V=a1–a
LbP
+c©Fy=0; a1–a
LbP–V=0
0…x…a
(
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
656
7–45. Continued
(b)
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,
SOLUTION
(a) For
0…x6a
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
For 0 …x65
V=800
(b) Set
For
0…x65ft
P=800 lb, a=5ft, L=12 ft
659
7–47.
660
7–47. Continued
(b)
c
For
0 … x…5 ft
By=250 lb+ c ©F
y=0;
Ay=350 lb
Ay(12) –600(7) =0+©MB=0;
Ans:
For 0 …x6a, V=
Pb
a+b
, M=
Pb
a+b
x
*7–48.
SOLUTION
Draw the shear and moment diagrams for the cantilevered
C
B
5ft
100 lb
800 lb ft
5ft
A
beam.
662
7–49.
SOLUTION
(a)
(b)
Set ,
For
0…x68
3m
L=8mM
0
=500 N
#
m
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
Ans.
h
e s
h
ear an
d
moment
di
agrams for t
h
e
b
eam.
A
BC
a a
w
Ans:
0…x6a:
V=–wx, M=–
x2
665
*7–52.
SOLUTION
Support Reactions: From FBD (a),
a
Shear and Moment Functions: For [FBD (b)],
Ans.
For [FBD (c)],
V+3wL
8–w1L–x2=0+c©F
y=0;
L
2
<x◊L
wL
8–V=0V=wL
8
+c©F
y=0;
0◊x<
L
2
Cy1L2–wL
2a3L
4b=0Cy=3wL
8
+ ©MA=0;
Draw the shear and moment diagrams for the beam.
C
w
A
B
L
L
––
2
666
7–53.
Draw the shear and bending-moment diagrams for the beam.
SOLUTION
Support Reactions:
a
Shear and Moment Functions: For [FBD (a)],
0…x<20 ft
10001102–200 –A
y
1202=0A
y
=490 lb+©M
B
=0;
C
A
B
20 ft 10 ft
50 lb/ft
200lb·ft
Ans:
For 0 …x620
ft
667
7–54.
SOLUTION
(a) For
a
(b) Set w= 500 lb/ft, L= 10 ft
For
+c©Fy=0; 2500 –500x–V=0
0…x…10 ft
+©M=0;
–wL
2x+wxax
2b+M=0
+c©Fy=0; wL
2–wx–V=0
0…x…L
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
+c©F
y
=0; V–20 =0
86x…11
+c©F
y
=0; 133.75 –40x–V=0
0…x68
40 kN/m
20 kN
150 kN m
A
BC
8m 3m
Ans:
For 0 …x68
M
=
(
133.75x
–
20x
2
)
kN
#
m
For 8
m6x…11
m
669
*7–56.
SOLUTION
:
:
+c©Fy=0; 0.75 –1.5 (x–2) –V=0
2m 6x64m
0…x…2m
Draw the shear and moment diagrams for the beam.
2m
4m
1.5 kN/m
ABC
Ans:
V=0.75 kN
V=3.75 –1.5 x kN
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
––
3
L
––
3
L
––
3
L
SOLUTION
Support Reactions: From FBD (b),
a
From FBD (a),
a
From FBD (c),
Shear and Moment Functions: For [FBD (d)],
7wL
18 –wx–V=0+c©F
y
=0;
0◊x<L
D
y
1L2+wL
6aL
3b–4wL
3aL
3b=0D
y
=7wL
18
+©M
C
=0;
F
y
aL
3b–wL
3aL
6b=0F
y
=wL
6
+©M
E
=0;