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611
7–1.
Ans:
NC=0
612
7–2.
Ans:
NC=0
613
Ans:
7–3.
Two beams are attached to the column such that structural
connections transmit the loads shown. Determine the
internal normal force, shear force, and moment acting in the
column at a section passing horizontally through point A.
Ans:
SOLUTION
S
+
ΣFx=0;
6-6-VA=0
185 mm
23 kN
16 kN
A
6 kN
6 kN
250 mm
40 mm30 mm
614
*7–4.
The beam weighs 280 lb
>
ft. Determine the internal normal
force, shear force, and moment at point C.
SOLUTION
Entire beam :
a
+ΣMA=0;
-2.8 (3) +Bx (8) =0
+
c
Σ
Fy=
0;
A
y
-2.8 =0
Segment AC :
+ Q ΣFx=0;
-NC+1.05 cos 53.13°+2.80 sin 53.13°-0.84 sin 53.13°=0
A
C
B
8 ft
3 ft
7 ft
6 ft
Ans:
NC=2.20 kip
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.
SOLUTION
+ ©MA=0; -20(10) +RB(1.5) =0
A
20 lb
20 lb
10 in. 400.5 in.
1 in.
B
C
Ans:
616
7–6.
617
SOLUTION
Support Reactions. Referring to the FBD of the entire beam shown in Fig. a,
Internal Loadings. Referring to the FBD of the right segment of the beam sectioned
through C, Fig. b,
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
Ans:
618
SOLUTION
Support Reactions. Referring to the FBD of the entire beam shown in Fig. a
Internal Loadings. Referring to the FBD of the left segment of beam sectioned
through C, Fig. b,
*7–8.
Determine the internal shear force and moment acting at
point C in the beam.
AC B
500 lb/ft
6 ft 6 ft
3 ft 3 ft
900 lb ft 900 lb ft
619
7–9.
Determ
i
ne t
h
e norma
l
force,s
h
ear force,an
d
moment at a
s
620
7–10.
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
Ans:
621
7–11.
Determine the internal normal force,shear force,and
moment at points Cand Dof the beam.
SOLUTION
Entire beam:
a
Segment CBD:
Ans.
Segment D:
:
+©Fx=0; -ND-5
13 (690) =0
VC=-648.62 =-649 lb
By=1411.54 lb
+©MA=0; -150 (5) -600 (7.5) +By(15)-12
13 (690) (25) =0
15 ft 10 ft
5ft12 ft
12
13
5
690 lb
40 lb/ft
60 lb/ft
A
CBD
Ans:
NC=265
lb
622
*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.
SOLUTION
Due to symmetry,
+c©Fy=0; Ay+By-w(L-a)
4-wa-w(L-a)
4=0
Ay=By
L
a
w
623
Ans:
SOLUTION
Support Reaction. Referring to the FBD of member AB shown in Fig. a
a
+ΣMB=0;
5(6) +6-A
y
(12) =0
A
y
=3.00 kip
Internal Loading. Referring to the left segment of member AB sectioned through D,
Fig. b,
S
+
ΣFx=0;
ND=0
Ans.
Referring to the left segment of member BC sectioned through E, Fig. c,
S
+
ΣFx=0;
NE=0
Ans.
7–13.
Determine the internal normal force, shear force, and
moment in the beam at sections passing through points D
andE. Point D is located just to the left of the 5-kip load.
6 ft 4 ft 4 ft
BC
DE
6 ft
5 kip
1.5 kip/ft
6 kip ft
A
624
7–14.
Ans:
MC=-15.0 kip #ft
625
SOLUTION
Support Reactions. Referring to the FBD of the entire beam shown in Fig. a,
a
+ΣMA=0;
By(6) -
1
2
(6)(6)(2) =0
B
y
=6.00 kN
S
+
ΣFx=0;
Bx=0
7–15.
Determine the internal normal force, shear force, and
moment at point C.
3 m
3 m
C
AB
6 kN/m
Ans:
NC=0
626
*7–16.
Determine the internal normal force, shear force, and
moment at point Cof the beam.
SOLUTION
Beam:
a
+©MB=0; 600 (2) +1200 (3) -Ay(6) =0
3m 3m
400 N/m
200 N/m
A
C
B
Ans:
NC=0
627
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
Pipe:
Rack:
+c©Fy=0; NBcos 30° -150 =0
6 in.
30
A
B
C
Ans:
628
7–18.
Ans:
629
7–19.
Ans:
630
*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.
SOLUTION
With reference to Fig.a, we obtain
Using this result and referring to Fig.b, we have
a
108.25 cos 30°(1.5) -1
2(15)(1.5)(0.5) -MC=0+©MC=0;
15 lb/ft
60 lb
B
C
A
D30
3 ft 1.5 ft
Ans:
