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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
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
*7–24.
SOLUTION
Since ,
Ans.
a
b=1
4
-a+b=3
4b
-1
6b(a-b)=1
8
-1
6b(2a+b)(a-b)=1
4(2a+b)a1
2b
V
C
=0
+c©F
y
=0; -w
6b(2a+b)(a-b)-w
4aa+b
2b-V
C
=0
:
+©F
x
=0; A
x
=0
A
y
=w
6b(2a+b)(a-b)
2(2a+b)c2
3(2a+b)-(a+b)d+A
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.
7–25.
SOLUTION
Ans.ME=-24.0 kip #ft
©ME=0; ME+6(4) =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
7–26.
SOLUTION
Free body Diagram: The support reactions at Aneed not be computed.
Internal Forces: Applying equations of equilibrium to segment BC, we have
+©F
Determine the internal normal force, shear force, and
bending moment at point C.
A
3m 3m
0.3 m
C
B
8 kN/ m
40 kN
3m
60°
637
Ans:
NC=-1.60 kN
VC=200 N
MC=200 N #m
*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
SOLUTION
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
7–30.
Determine the normal force, shear force, and moment
acting at a section passing through point D.
SOLUTION
a
Ans.
Ans.
a
Ans.M
D
=2612 lb
#
ft =2.61 kip
#
ft
+©M
D
=0; -M
D
-600(1) +927.4(4 cos 30°) =0
V
D
=-203 lb
Q+©F
y
=0; V
D
-600 +927.4 cos 30° =0
N
D
=-464 lb
+a©F
x
=0; N
D
-927.4 sin 30° =0
A
y
=985.1 lb
+c©F
y
=0; A
y
-800 cos 30° -700 -600 cos 30° +927.4 =0
A
x
=100 lb
:
+©F
x
=0; 800 sin 30° -600 sin 30° -A
x
=0
B
y
=927.4 lb
+600 sin 30°(3 sin 30°) +B
y
(6 cos 30° +6 cos 30°) =0
+©M
A
=0; -800(3) -700(6 cos 30°) -600 cos 30°(6 cos 30° +3 cos 30°)
800 lb
700 lb
600 lb
2 ft
3 ft
1.5 ft
1.5 ft
1 ft
3 ft
D
BA
C30 30
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
1.5 kN/m
B
D
E
SOLUTION
7–34.
Member AB:
a
Segment BE:
Ans.
Ans.
Ans.Mg=1.6875 kN #m=1.69 kN #m
+©Mg=0; Mg-2.25 (0.75) =0
VE=0
:
+©Fx=0; VE+2.25 -2.25 =0
NE=1.25 kN
+c©Fy=0; 1.25 -NE=0
By=1.25 kN
+©MA=0;
2.25 (3) -3 (1) -By(3) =0
Cx=2.25 kN
:
+©Fx=0; 2.25 +Cx-4.5 =0
Bx=2.25 kN
Determine the internal normal force, shear force, and
moment at point E.
1.5 m
1.5 m1.5 m
1.5 kN/m
B
D
E
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.
45°45°
3m 3m
0.2 m
dd
AB
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
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
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:
Ans.
Ans.
a
Ans.M=0.0759
w0r2
+©MO=0; M-(0.0759 rw0)(r)=0
N=0.0759 w0r
N=-rw0c1
2ap
4b-1
4sin 90° dcos 45° +a1
2rw0sin245°bsin 45°
+a©Fy=0; -N-FRy cos 45° +FRx sin 45° =0
V=0.278 w0r
V=a1
2rw0sin245°bcos 45° +w0a1
2
p
4-1
4sin 90°bsin 45°
Q+©Fx=0; -V+FRx cos 45° +FRy sin 45° =0
FRy =Lu
0
w0sin u(rdu) sin u=rw0Lu
0
sin2udu=rw0cl
2u-1
4sin 2ud
FRx =Lu
0
w0sin u(rdu) cos u=rw0Lu
0
sin ucos udu=1
2rw0sin2u
w=w0sin u
θ
r
θ
w=w
0
sin
650
*7–40.
Ans.
Ans.
a
Ans.M= - 0.957 r2w0
+ ©MO=0; -M-0.957 rw0(r)=0
V= - 0.907 rw0
+a©Fy¿=0; -V+0.375 rw0sin 30° -1.2637 rw0cos 30° =0
N= - 0.957 rw0
+
b©Fx¿=0; N+0.375 rw0cos 30° +1.2637 rw0sin 30° =0
FRy =rw0c1
2(p)a120°
180° b-1
4sin 240° d=1.2637 rw0
FRx =1
2rw0sin2120° =0.375 rw0
FRy =Lu
0
w0sin u(r du) sin u=rw0Lu
0
sin2udu=rw0c1
2u1
4sin 2udrw0(sin u)20
u
=rw0(sin u)
θ
r
θ
w=w
0
sin
Solve Prob. 7–39 for u = 120°.
