631
Ans:
NE=720 N
VE=1.12 kN
NF=0
SOLUTION
Support Reactions. Referring to the FBD of member BC shown in Fig. a,
a
+ΣMB=0;
Cy(3) –1200
a4
5b
(2) =0
C
y
=640 N
B
S
+
ΣFx=0;
Internal Loadings. Referring to the right segment of member AB sectioned through E,
Fig. b
S
+
ΣFx=0;
720 –NE=0
NE=720 N
Ans.
VE=1120 N =1.12 kN
+ΣME=0;
Referring to the left segment of member CD sectioned through F, Fig. c,
S
+
ΣFx=0;
NF=0
+ΣMF=0;
7–21.
Determine the internal normal force, shear force, and
moment at points E and F of the compound beam. Point E
is located just to the left of 800 N force.
A
1 m
400 N/m
800 N 1200 N
2 m 1 m
1.5 m 1.5 m
D
EF
BC
5
4
3
1.5 m
632
7–22.
Determine the internal normal force, shear force, and
moment at points Dand Ein the overhang beam. Point Dis
located just to the left of the roller support at B, where the
couple moment acts.
2kN/m
5kN
3m 1.5 m
3
4
5
ADBE
C
6 kN m
1.5 m
SOLUTION
The intensity of the triangular distributed load at Ecan be found using the similar
triangles in Fig. b.With reference to Fig. a,
Using this result and referring to Fig. c,
Ans.
Also, by referring to Fig. d, we can write
Ans.
:
+©Fx=0; 5a4
5b–NE=0NE=4kN
:
+©Fx=0; 5a4
5b–ND=0ND=4kN
The negative sign indicates that VD,MD, and MEact in the opposite sense to that
shown on the free-body diagram.
VD=–9
MD=–18
ME=–4.875
kN #m
633
7–23.
Determine the internal normal force,shear force,and
moment at point C.
3m 2m
1.5 m
1m
0.2 m 400 N
A
C
B
SOLUTION
Beam:
Segment AC:
:
+©Fx=0; NC–400 =0
:
+©Fx=0; –Ax+400 =0
Ans:
634
*7–24.
SOLUTION
a
Since ,
V
C
=0
+©M
B
=0; –w
2(2a+b)c2
3(2a+b)–(a+b)d+A
y
(b)=0
BCA
ab/2b/2
w
a
ABC
Determine the ratio of a
b for which the shear force will be
zero at the midpoint C of the beam.
635
7–25.
SOLUTION
a
a
©MD=0; MD+1
2(0.75)(6) (2) –3(6) =0
+©MB=0; 1
2(1.5)(12)(4) –Ay(12) =0
Determ
i
ne t
h
e norma
l
force,s
h
ear force,an
d
moment
i
n
the beam at sections passing through points Dand E.Point
Eis just to the right of the 3-kip load.
6ft4ft
A
4ft
BC
DE
6ft
3 kip
1.5 kip/ft
Ans:
ND=0
NE=0
636
7–26.
Ans:
Ans:
NC=–1.60 kN
SOLUTION
Support Reactions. Referring to the FBD of the entire assembly shown in Fig. a,
Internal Loading. Referring to the FBD of the left segment of the assembly
sectioned through C, Fig. b,
7–27.
Determine the internal normal force, shear force, and
moment at point C.
A
C
E
D
B
1 m 1 m 2 m
1 m
800 N m
200 N
638
*7–28.
Determine the internal normal force,shear force,and
moment at points Cand Din the simply supported beam.
Point Dis located just to the left of the 10-kN
concentrated load.
A
CD
B
1.5 m
6kN/m
10 kN
1.5 m 1.5 m 1.5 m
Using these results and referring to Fig. c,
Also, by referring to Fig. d,
Ans.
:
+©F
x=0; N
D=0
SOLUTION
The intensity of the triangular distributed loading at Ccan be computed using the
similar triangles shown in Fig. b,
With reference to Fig. a,
1.5 =6
3or wC=3kN>m
Ans:
NC=0
639
7–29.
SOLUTION
Determine the normal force, shear force, and moment
acting at a section passing through point C.
800 lb
700 lb
600 lb
2ft
3ft
1.5 ft
1.5 ft
1ft
3ft
D
AB
C30 30
a
A
y
=985.1 lb
+c©F
y
=0; A
y
–800 cos 30° –700 –600 cos 30° +927.4 =0
+©M
A
=0;
–800 (3) –700(6 cos 30°) –600 cos 30°(6 cos 30° +3cos 30°)
640
7–30.
641
Ans:
NE=2.20 kN
VE=0
ME=0
VD=600 N
SOLUTION
Support Reactions. Notice that member AB is a two force member.
Referring to the FBD of member BC,
FAB =2200 N
Internal Loadings. Referring to the left segment of member AB sectioned through E,
Fig. b,
NE=2200 N =2.20 kN
VE=0
ME=0
Referring to the left segment of member BC sectioned through D, Fig. c
VD=600 N
7–31.
Determine the internal normal force, shear force, and
moment acting at points D and E of the frame.
2 m
900 N m
600 N
D
E
B
A
4 m
C
1.5 m
.
642
SOLUTION
Support Reactions. Notice that member BC is a two force member. Referring to the
FBD of member ABE shown in Fig. a,
a
+ΣMA=0;
FBC
a3
5b
(4) –6(7) =0
FBC =17.5 kN
Ax=4.50 kN
Internal Loadings. Referring to the FBD of the lower segment of member ABE
sectioned through D, Fig. b,
+ΣMD=0;
*7–32.
Determine the internal normal force, shear force, and
moment at point D.
A
D
E
C
B
6 kN
3 m
3 m
1 m
3 m
643
7–33.
Determine the internal normal force, shear force, and
moment at point Dof the two-member frame.
1.5 m
1.5 m1.5 m
1.5m
1.5 kN/m
B
D
E
SOLUTION
Member BC:
Member AB:
a
Segment DB:
:
+©Fx=0; –ND–2.25 =0
+©MA=0; 2.25 (3) –3 (1) –By(3) =0
Ans:
644
7–34.
SOLUTION
Member BC:
Member AB:
Segment BE:
Determine the internal normal force, shear force, and
moment at point E.
1.5 m
1.5 m1.5 m
1.5m
1.5 kN/m
B
D
E
Ans:
645
7–35.
The strongback or lifting beam is used for materials
handling. If the suspended load has a weight of 2 kN and a
center of gravity of G, determine the placement dof the
padeyes on the top of the beam so that there is no moment
developed within the length AB of the beam.The lifting
bridle has two legs that are positioned at 45°, as shown.
SOLUTION
Support Reactions: From FBD (a),
a
F
F162–2132=0F
E=1.00 kN+©ME=0;
45°45°
3m 3m
0.2 m
0.2 m
dd
E
AB
F
G
646
Ans:
NC=184 N
VC=–78.6 N
SOLUTION
Support Reactions. Not required
Internal Loadings. Referring to the FBD of bottom segment of the curved rod
sectioned through C, Fig. a
Referring to the FBD of bottom segment of the curved rod sectioned through B,
Fig. b
*7–36.
Determine the internal normal force, shear force, and
moment acting at points B and C on the curved rod.
45
30
0.5 m
B
C
A
200 N
3
4
5
647
7–37.
Determine the internal normal force,shear force,and
moment at point Dof the two-member frame.
2m
1.5 m
250 N/m
300 N/m
4m
A
C
D
E
B
SOLUTION
Member AB:
Member BC:
Segment DB:
648
7–38.
Determine the internal normal force, shear force, and
moment at point Eof the two-member frame.
2m
1.5 m
250 N/m
300 N/m
4m
A
C
D
E
B
SOLUTION
Member AB:
Member BC:
Segment EB:
:
+©Fx=0; –NE–1258.33 –225 =0
Ans:
649
7–39.
The distributed loading sin ,measured per unit
length, acts on the curved rod.Determine the internal normal
force,shear force,and moment in the rod at .u=45°
uw
=
w0
SOLUTION
Resultants of distributed loading:
+a©Fy=0; –N–FRy cos 45° +FRx sin 45° =0
Q+©Fx=0; –V+FRx cos 45° +FRy sin 45° =0
w=w0sin u
θ
r
θ
w=w
0
sin
Ans:
V=0.278 w0
r
650
*7–40.
SOLUTION
Resultants of distributed load:
FRx =Lu
0
w0sin u(r du) cos u=rw0Lu
0
sin ucos u=1
2rw0sin2u
θ
r
θ
w=w
0
sin
Solve Prob. 7–39 for u = 120°.
Ans:
N=–0.957
r w0