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7–1.
Determine the shear force and moment at points Cand D.
SOLUTION
Support Reactions: FBD (a).
a
Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)],
we have
Ans.
Ans.
a
Ans.
Applying the equations of equilibrium to segment ED [FBD (c)], we have
Ans.
Ans.
aAns.+©M
D
=0; -M
D
-300 (2) =0M
D
=-600 lb
#
ft
+c©F
y
=0; V
D
-300 =0V
D
=300 lb
:
+©F
x
=0; N
D
=0
M
C
=-857 lb
#
ft
+©M
C
=0; M
C
+500(4) -114.29 (10) =0
+c©F
y
=0; 114.29 -500 -V
C
=0V
C
=-386 lb
:
+©F
x
=0N
C
=0
A
y
=114.29 lb
+©M
B
=0; 500(8) -300(8) -A
y
(14) =0
6ft
ACD
E
B
6ft2ft
4ft4 ft
300 lb
200 lb
500 lb
7–2.
SOLUTION
Support Reactions: FBD (a).
a
Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)],
we have
Ans.
Ans.
aAns.
Applying the equations of equilibrium to segment BD [FBD (c)], we have
Ans.
Ans.
a
+©M
D
=0; 1.00(8) +40 -M
D
=0
+c©F
y
=0; V
D
+1.00 =0V
D
=-1.00 kip
:
+©F
x
=0; N
D
=0
+©M
C
=0; M
C
-7.00(8) =0M
C
=56.0 kip
#
ft
+c©F
y
=0; 7.00 -8-V
C
=0V
C
=-1.00 kip
:
+©F
x
=0N
C
=0
:
+©F
x
=0A
x
=0
+c©F
y
=0; A
y
+1.00 -8=0A
y
=7.00 kip
+©M
A
=0;
B
y
(24) +40 -8(8) =0B
y
=1.00 kip
Determine the internal normal force and shear force,and
the bending moment in the beam at points Cand D.
Assume the support at Bis a roller.Point Cis located just to
the right of the 8-kip load.
40 kip ft
8ft8ft8 ft
8 kip
AB
CD
*7–4.
The beam weighs 280 lb
>
ft. Determine the internal normal
force, shear force, and moment at point C.
SOLUTION
Entire beam :
A
C
B
8 ft
3 ft
7 ft
6 ft
615
7–5.
The pliers are used to grip the tube at B. If a force of 20 lb
is applied to the handles,determine the internal shear force
and moment at point C. Assume the jaws of the pliers exert
only normal forces on the tube.
Segment BC:
Ans.
a
Ans.MC=133 lb #in.
+©MC=0; -MC+133.3 (1) =0
VC=-133 lb
+Q©Fy=0; VC+133.3 =0
RB=133.3 lb
A
20 lb
10 in. 400.5 in.
1 in.
B
C
7–6.
SOLUTION
a
a
Ans.a=L
3
2PL aL
3-ab=0
M=2P
A
L
3
-a
B
L-aaL
3b=0+©M=0;
C
y
=2P
A
L
3
-a
B
L-a
-Pa2L
3-ab+C
y
1L-a2+Pa =0+©M
A
=0;
Determine the distance aas a fraction of the beam’s length
Lfor locating the roller support so that the moment in the
beam at Bis zero.
L
AB
C
aL/3
P
SOLUTION
7–7.
Determine the internal shear force and moment acting at
point C in the beam.
6 ft 6 ft
4 kip/ft
AB
C
7–9.
Determ
i
ne t
h
e norma
l
force,s
h
ear force,an
d
moment at a
s
ection passing through point C. Take P=8 kN.
SOLUTION
a
Ans.
Ans.
a
Ans.M
C
=6kN
#
m
+©M
C
=0; -M
C
+8(0.75) =0
V
C
=-8kN
+c©F
y
=0; V
C
+8=0
N
C
=-30 kN
:
+©F
x
=0; -N
C
-30 =0
+c©F
y
=0; A
y
=8kN
:
+©F
x
=0; A
x
=30 kN
T=30 kN
+©M
A
=0; -T(0.6) +8(2.25) =0
0.75 m
C
P
A
B
0.5 m
0.1 m
0.75 m 0.75 m
7–10.
SOLUTION
a
Ans.
Ans.
Ans.
a
Ans.M
C
=0.400 kN
#
m
+©M
C
=0; -M
C
+0.533(0.75) =0
V
C
=-0.533 kN
+c©F
y
=0; V
C
-0.533 =0
N
C
=-2kN
:
+©F
x
=0; -N
C
-2=0
+c©F
y
=0; A
y
=0.533 kN
:
+©F
x
=0; A
x
=2kN
P=0.533 kN
+©M
A
=0;
-2(0.6) +P(2.25) =0
T
he cablewill fail when subjectedtoatensionof2kN.
D
etermine the largest vertical load Pthe frame will support
a
nd calculate the internal normal force,shear force,and
m
oment at asection passing through point Cfor this loading.
0.75 m
C
P
A
B
0.5 m
0.1 m
0.75 m 0.75 m
7–11.
Determine the internal normal force,shear force,and
moment at points Cand Dof the beam.
SOLUTION
Segment CBD:
Ans.
Ans.
a
Ans.
Segment D:
Ans.
Ans.
a
Ans.MD=-3.18 kip #ft
+©MD=0; -MD-690 a12
13 b(5) =0
VD=637 lb
+c©Fy=0; VD-12
13 (690) =0
ND=-265 lb
:
+©Fx=0; -ND-5
13 (690) =0
MC=-4231.38 lb #ft =-4.23 kip #ft
-12
13 (690) (13) -MC=0
+©MC=0; -6(1) -120 (1.5) +1411.54 (3)
VC=-648.62 =-649 lb
+c©Fy=0; VC-6-120 +1411.54 -690 a12
13 b=0
NC=-265 lb
:
+©Fx=0; -NC-5
13 (690) =0
By=1411.54 lb
15 ft 10 ft
5ft12 ft
12
13
5
690 lb
40 lb/ft
60 lb/ft
A
CBD
*7–12.
Determine the distance abetween the bearings in terms of
the shaft’s length Lso that the moment in the symmetric
shaft is zero at its center.
a
Since ;
Ans.a=0.366 L
2a2+2aL -L2=0
3a2+(L-a)(L+2a)-3a(L+a)=0
M=0
+©M=0;
-M-wa
2aa
4b-w(La)
4aa
2+L
6-a
6b+w
4(L+a)aa
2b=0
Ay=By=w
4(L+a)
+c©Fy=0; Ay+By-w(L-a)
4-wa-w(L-a)
4=0
L
a
w
623
Ans:
ND=0
VD=3.00 kip
MD=12.0 kip #ft
NE=0
VE=-8.00 kip
ME=-20.0 kip #ft
7–14.
SOLUTION
a
a
Ans.
Ans.
Ans.
a
Ans.
Ans.
Ans.V
D
=1114 lb =1.11 kip
+c©F
y
=0; V
D
-3000 +1886 =0
:
+©F
x
=0; N
D
=0
M
D
=3771 lb
#
ft =3.77 kip
#
ft
+©M
D
=0; -M
D
+1886(2) =0
V
C
=2014 lb =2.01 kip
+c©F
y
=0; -2500 +4514 -V
C
=0
:
+©F
x
=0; N
C
=0
M
C
=-15
000 lb
#
ft =-15.0 kip
#
ft
+©M
C
=0; 2500(6) +M
C
=0
B
y
=1886 lb
+c©F
y
=0; 4514 -2500 -900 -3000 +B
y
=0
:
+©F
x
=0; B
x
=0
A
y
=4514 lb
+©M
B
=0; -A
y
(14) +2500(20) +900(8) +3000(2) =0
The shaft is supported by a journal bearing at Aand a thrust
bearing at B. Determine the normal force, shear force, and
moment at a section passing through (a) point C, which is
just to the right of the bearing at A, and (b) point D, which
is just to the left of the 3000-lb force.
2500 lb
A
CDB
3000 lb
75 lb/ft
6ft12ft2ft
2
626
*7–16.
Determine the internal normal force, shear force, and
moment at point Cof the beam.
SOLUTION
Segment AC:
Ans.
Ans.
a
Ans.MC=1350 N #m=1.35 kN #m
+©MC=0; -800 (3) +600 (1.5) +150 (1) +MC=0
VC=50 N
+c©Fy=0; 800 -600 -150 -VC=0
:
+©Fx=0; NC=0
:
+©Fx=0; Ax=0
Ay=800 N
3m 3m
400 N/m
200 N/m
A
C
B
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–17.
The cantilevered rack is used to support each end of a
smooth pipe that has a total weight of 300 lb. Determine the
normal force, shear force, and moment that act in the arm at
its fixed support Aalong a vertical section.
SOLUTION
Rack:
Ans.
Ans.
a
Ans.MA=1.80 lb #in.
+©MA=0; MA-173.205(10.3923) =0
VA=150 lb
+c©Fy=0; VA-173.205 cos 30° =0
NA=86.6 lb
:
+©Fx=0; -NA+173.205 sin 30° =0
NB=173.205 lb
6 in.
30
A
B
C
7–18.
Determine the internal normal force, shear force, and the
moment at points Cand D.
SOLUTION
Support Reactions: FBD (a).
a
Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)],
we have
Ans.
Ans.
a
Ans.
Applying the equations of equilibrium to segment BD [FBD (c)], we have
Ans.
Ans.
a
8.485132-611.52-M
D
=0+©M
D
=0;
V
D
+8.485 -6.00 =0V
D
=-2.49 kN+c©F
y
=0;
N
D
=0:
+©F
x
=0;
M
C
=4.97 kN
#
m
M
C
-3.515 cos 45°122=0+©M
C
=0;
3.515 sin 45° -N
C
=0N
C
=2.49 kNa+©F
y¿
=0;
3.515 cos 45° -V
C
=0V
C
=2.49 kNQ+ ©F
x¿
=0;
:
+©F
x
=0A
x
=0
A
y
+8.485 -12.0 =0A
y
=3.515 kN+c©F
y
=0;
B
y
=8.485 kN
B
y
16+6 cos 45°2-12.013+6 cos 45°2=0+©M
A
=0;
2 kN/m
3m 3m
B
D
C
A
6m
2m
45˚
7–19.
SOLUTION
Entire beam:
a
Segment AC:
Ans.
Ans.
a
Ans.MC=4.72 kip #ft
+©MC=0; -1.8 (3) +0.45 (1.5) +MC=0
VC=1.35 kip
+c©Fy=0; 1.8 -0.45 -VC=0
NC=-4.32 kip
:
+©Fx=0; 4.32 +NC=0
Ay=1.8 kip
+c©Fy=0; Ay-1.8 =0
Ax=4.32 kip
:
+©Fx=0; Ax-4.32 =0
T=4.32 kip
+©MA=0;
-1.8 (6) +T(2.5) =0
Determine the internal normal force, shear force, and
moment at point C.
8ft
3ft
4ft
150lb/ft
2ft
0.5ft
A
C
B
*7–20.
Rod AB is fixed to a smooth collar D, which slides freely
along the vertical guide. Determine the internal normal
force, shear force, and moment at point C. which is located
just to the left of the 60-lb concentrated load.
Using this result and referring to Fig.b, we have
Ans.
Ans.
a
Ans.
The negative signs indicates that NCand VCact in the opposite sense to that shown
on the free-body diagram.
MC=135 lb #ft
108.25 cos 30°(1.5) -1
2(15)(1.5)(0.5) -MC=0+©MC=0;
V
C=-22.5 lb
V
C-60 -1
2(15)(1.5) +108.25 cos 30° =0+c©F
y=0;
N
C=-54.1 lb-N
C-108.25 sin 30° =0©F
x=0;
:
+
2(15)(3) -60 -1
2(15)(1.5) =0+c©F
15 lb/ft
60 lb
B
C
A
D30
3 ft 1.5 ft
