SOLUTION
Equations of Equilibrium:
a
+ΣMA=0;
NB(24) –2(12)(6) –
1
2
(2)(12)(16) =0
NB=14.0
k
a
+ΣMB=0;
1
2
(2)(12)(8) +2(12)(18) –Ay(24) =0
Ay=22.0
k
+
SΣFx=0;
Ax=0
12 ft 12 ft
B
A
2 k
>
ft
*2–24. Determine the reactions on the beam. The support at
B can be assumed to be a roller.
SOLUTION
A
B
C
12 kN
>
m
60 kN
?
m
608
6 m
2–25. Determine the horizontal and vertical components of
reaction at the pins A and C.
5 1
AB
48 ft
600 lb>ft 400 lb>ft
48 ft
20 ft
SOLUTION
a
+ ΣMA=0;
By(96) +
a12
13 b
20.8(72) –
a5
13 b
20.8(10
)
–
a12
13 b
31.2(24) –
a5
13 b
31.2(10) =
0
By=5.117 k =5.12 k
+
c
ΣFy=0;
Ay–5.117 +
a12
13 b
20.8 –
a12
13 b
31.2 =
0
Ay=14.7 k
+
SΣFx=0;
–Bx+
a5
13 b
31.2 +
a5
13 b
20.8 =0
Bx=20.0 k
2–26. Determine the reactions at the truss supports A
and B. The distributed loading is caused by wind.
Ans.
By=5.12 k
Ay=14.7 k
Bx=20.0 k
Ans.
Ans.
Ans.
5 2
2–27. The compound beam is fixed at E and supported by
rockers at A and B. There are hinges (pins) at C and D.
Determine the reactions at the supports. The 4-kN load is
applied just to the right of the pin at D.
2 m 2 m 2 m
AD
B E
C
6 kN
6 kN>m
4 kN
3 m3 m
SOLUTION
Ans.
NB=7.50 kN
NA=150 kN
Ex=0
E
y
=10.0 kN
ME=30.0 kN #m
5 3
SOLUTION
a+ΣMA=0;
–20 kN(3 m) –20 kN(6 m) –20 kN(9 m) –20 kN(12 m)
–8 kN(sin
60°)(15 m) +By(9 m) =0
By=78.2 kN
+
SΣFx=0;
–Ax+8 kN(cos
60°)=0
Ax=4 kN
+
c
ΣFy=0;
–20 kN –20 kN –20 kN –20 kN –8 kN(sin 60°)
+78.2 kN +Ay=0
Ay=8.71 kN
A
8 kN
608
B
3 m3 m3 m3 m3 m
20 kN 20 kN 20 kN 20 kN
*2–28. Determine the reactions on the beam.
Ans.
By=78.2 kN
Ax=4 kN
Ay=8.71 kN
Ans.
Ans.
Ans.
SOLUTION
12 kN>m
B
C
A
3
m6
m
2–30. Determine the reactions at the supports A and C of the
compound beam. Assume C is a roller, B is a pin, and A is
fixed.
5 6
P2P
w2
w1
L
__
3
L
__
3
L
__
3
2–31. The beam is subjected to the two concentrated loads as
shown. Assuming that the foundation exerts a linearly varying
load distribution on its bottom, determine the load intensities
w1
and
w2
for equilibrium (a) in terms of the parameters
shown; (b) set
P=500
lb,
L=12
ft.
w
2P
L
w
12
w
12
Ans.
w
1=
2P
L
w
2=
4P
L
w1=83.3 lb>ft
w2=167 lb>ft
SOLUTION
4 m 2 m
0.3
m
3 m
B
A
C
3 kN
>
m
6 kN
*2–32. Determine the horizontal and vertical components of
reaction at the supports A and C. Assume the members are pin
connected at A, B, and C.
SOLUTION
5b
AC
B
4
m
6 kN>m
3 m
2–33. Determine the horizontal and vertical components of
reaction at the supports A and C.