CHAPTER 7
PROBLEM 7.1
Determine the internal forces (axial force, shearing force,
and bending moment) at Point J of the structure indicated.
Frame and loading of Problem 6.76.
SOLUTION
From Problem 6.76:
720 lb
x=C
140 lb
y
=C
PROBLEM 7.2
Determine the internal forces (axial force, shearing force, and bending
moment) at Point J of the structure indicated.
Frame and loading of Problem 6.78.
SOLUTION
From Problem 6.78:
30 lb=C
FBD of AE:
30 lbV=
PROBLEM 7.3
Determine the internal forces at Point J when
α
= 90°.
SOLUTION
Reactions
( 90 )
α
= °
0: 0
Ay
MΣ= =B
12
0: 780 N 0
13
y
FA
Σ= − =
PROBLEM 7.4
Determine the internal forces at Point J when
α
= 0.
SOLUTION
Reactions
( 0)
α
=
0: (780 N)(0.720 m) (0.3 m) 0
Ay
MBΣ= + =
1872 N
y
B= −
1872 N
y
=B
12
0: 1872 N 0
13
y
FA
Σ= − =
PROBLEM 7.5
For the frame and loading shown, determine the internal forces at
the point indicated:
Point J.
SOLUTION
From free body diagram, all reactions = 0
PROBLEM 7.6
For the frame and loading shown, determine the internal
forces at the point indicated:
Point K.
SOLUTION
From free body diagram, all reaction = 0
Free body KC:
PROBLEM 7.7
An archer aiming at a target is pulling with a 45–lb force on the
bowstring. Assuming that the shape of the bow can be approximated
by a parabola, determine the internal forces at Point J.
SOLUTION
FBD Point A:
By symmetry
12
TT=
PROBLEM 7.7 (Continued)
0: (37.5 lb)sin (39.094 ) 0
y
FF
′
Σ = + °=
23.647F= −
23.6 lb=F
76.0°
PROBLEM 7.8
For the bow of Problem 7.7, determine the magnitude and location of
the maximum (a) axial force, (b) shearing force, (c) bending moment.
PROBLEM 7.7 An archer aiming at a target is pulling with a 45–lb
force on the bowstring. Assuming that the shape of the bow can be
approximated by a parabola, determine the internal forces at Point J.
SOLUTION
Free body: Point A
3
0: 2 45 lb 0
5
x
FT
Σ= − =
37.5 lbT=
PROBLEM 7.8 (Continued)
(a) Maximum axial force
0: (30 lb) cos (22.5 lb)sin 0
V
FF
θθ
Σ = −+ − =
Free body: Portion bow CK
30cos 22.5sinF
θθ
= −
F is largest at
( 0)C
θ
=
30.0 lb at
m
FC=
(b) Maximum shearing force
0: (30 lb)sin (22.5 lb) cos 0
V
FV
θθ
Σ = −+ + =
30sin 22.5cosV
θθ
= +
m
PROBLEM 7.9
A semicircular rod is loaded as shown. Determine the internal forces at Point J.
SOLUTION
FBD Rod:
0: (2 ) 0
Bx
M ArΣ= =
0
x
=A
PROBLEM 7.10
A semicircular rod is loaded as shown. Determine the internal forces at Point K.
SOLUTION
FBD Rod:
0: 120 N 0 120 N
yy y
FBΣ= − = =B
0: 2 0 0
A xx
M rBΣ= = =B
PROBLEM 7.11
A semicircular rod is loaded as shown. Determine the internal
forces at Point Jknowing that
θ
= 30°.
SOLUTION
FBD AB:
43
0: 2 (280 N) 0
55
A
M rCrC r
Σ= + − =
400 N=C
4
0: (400 N) 0
5
xx
FAΣ= − + =
320 N
x=A
PROBLEM 7.12
A semicircular rod is loaded as shown. Determine the
magnitude and location of the maximum bending moment in
the rod.
SOLUTION
Free body: Rod ACB
43
0: (0.16 m) (0.16 m)
55
(280 N)(0.32 m) 0
A CD CD
MF F
Σ= +
−=
400 N
CD =F
PROBLEM 7.12 (Continued)
Free body: (For 90°)BJ
θ
>
0: (280 N)(0.16 m)(1 cos ) 0
J
MM
φ
Σ= − − =
(44.8 N m)(1 cos )M
φ
= ⋅−
PROBLEM 7.13
The axis of the curved member AB is a parabola with vertex at A. If a
verticalload P of magnitude 450 lb is applied at A, determine the
internal forces at J when h= 12 in., L= 40 in., and a= 24 in.
SOLUTION
Free body AB
0: 450 lb 0
yy
FBΣ= − + =
450 lb
y
=B
0: (12 in.) (450 lb)(40 in.) 0
Ax
MBΣ= − =
1500 lb
x=B
0: 1500 lb
x
FΣ= =A
PROBLEM 7.13 (Continued)
0: (450 lb)sin19.8 (1500 lb)cos19.8 0FFΣ = − °− °=
1563.8 lbF= +
1564 lb=F
19.8°
PROBLEM 7.14
Knowing that the axis of the curved member AB is a parabola with
vertex at A, determine the magnitude and location of the maximum
bending moment.
SOLUTION
Parabola
2
y kx=
At B:
2
h kL=
2
/k hL=
Equation of parabola
22
/y hx L=
PROBLEM 7.15
Knowing that the radius of each pulley is 200 mm and
neglecting friction, determine the internal forces at Point J of
the frame shown.
SOLUTION
FBD Frame with pulley and cord:
0: (1.8 m) (2.6 m)(360 N)
(0.2 m)(360 N) 0
Ax
MBΣ= −
−=
560 N
x=B