1.050 Solid Mechanics, Fall 2004
Problem Set#2 Solution
Problem 2.1
First, we need to find the reaction forces at point A and B, we
get that
C
ΣFy = 0; R A + R B = P
ΣM A = 0; R B (2 L )− P (L + x p ) = 0
L
2L
P
D
E
RB =
P  L + xp

2  L



RA =
P  L − xp

2  L



y
A
60o
B
x
RA
x
RB
p
Figure 1.1
Now consider the horizontal member, the only
possible way to have equilibrium condition is that the
forces action at point D and E are both horizontal forces
(i.e., no vertical forces acting on the points). You can simply take a moment at either point D or E and you will find
that if there is any vertical forces at either D or E, the member DE cannot be in the equilibrium condition (summation
of moment is not equal to zero).
FD
FE
D
E
Figure 1.2
CY
CX
Figure 1.1 shows you the member FBD of the member BC. From the
FBD, we get that
L
60o
FE
ΣFy = 0; C y + R B = P
P
ΣFx = 0; C x = FE
y
L
B
x
ΣM c = 0; P (x p )+ FE (L sin 60 ) = RB (L )
FE =
x
p
Figure 1.3
RB
Cy =
RB (L )− P (x p )
Lsin60
x 
P
1 − p 
L 
2
=
x 
P 
1 − p  = C x = FD
L 
3
Problem 2.2
From figure 2.1 (a) and (b), we can find the equivalent force by
ΣM f = 3W (x ) = W (L )+ W (4 L )+ W (5 L ) = 10WL
f
l
L
L
L
2L
W
W
x =
L
10
L
3
ΣM f = 0; Rl (6 L ) = 10WL
W
Rl =
f
l
5
W
3
ΣFy = 0; R f + Rl = 3W
x=?
3W
2L
a
2L
c
R f = 3W − Rl =
e
W
W
f
h
j
2L
2L
4
W
3
αL
W
l
2L
Rf
Rl
Figure 2.1 (a), (b), and (c) for top, middle, and bottom, respectively
To find the forces in the member ch with a single FBD, we use method of section and cut the truss as
shown in the figure 2.2. We get that
ΣM h = 0; Fac (αL )+ R f (2 L ) = W (L )
Fac = −
5W
3α
ΣFy = 0; R f + Fch (sin θ ) = W
Fch = −
W
3(sin θ )
2L
a
αL
Fch
W
f
h
2L
Rf
ΣFx = 0; Fac + Fch + Fhj = 0
Fhj =
5W
W
W 5
1 
+
=
+

3α 3(sin θ )
3 α sinθ 
Fac
c
Figure 2.2
Fhj
. The negative sign for Fch indicates that the direction that I assumed in the figure above is not correct. (If the angle θ is 60 degree, the Fch is equal to F ch
2W
= –---------- )And the correct direction is opposite of
3 3
that in the figure 2.2. It is important to note the correct Fch in this case is compression.
Problem 2.3
First, we need to find the reaction at point A and B. Assuming each member length is L (except the
member CF which is longer than other members), we get that
L
ΣM A = 0; Rby   + Rbx (L sin 60 ) = P(3L )
2
ΣFy = 0; Rby = Ray + P
F
Ax
A
30o
C
E
G
60o
Rby
B
60o
D
P
60o
Rbx
Figure 3.1
However, since AC is a two force member, Ray must be 0 to maintain the equilibrium condition for
that member.. Therefore, we get that Rby = P and .
5P
R bx = ------3
Now consider cutting the joint B, we get that
ΣFx = 0; Rbx0 = Fbc (cos 60 )+ Fbd + Rbx
FCB
ΣFy = 0; 0R=byP =+ Fbc (sin 60 )
B
o
60
Rbx
P
FDB
2P
F bc = –------3
Fbd = -Rbx − Fbc (cos 60 ) =
- 5P + 2 P
- 4P
−
=
3
3 (2 )
3
Now consider the joint D, we get that {This needs to be checked}
∑ F y = 0:
∑ F x = 0:
F cd ⋅ sin 60 + F de ⋅ sin 60 = 0
F cd = –F de
– F cd ⋅ cos 60 + F de ⋅ cos 60 – F bd = 0
F bd
F cd = –------------------
2 cos60
Fcd =
Fbd
 4 P  2 
4P
= 
  =
2(cos 60 )  3  2 
3 in tension
FCD
Problem 2.4
FED
60o
FDB
The reactions acting at point A and B are equivalent to PL/
h. They are equal but opposite in the direction. Also, the total vertical reaction at the point A and B are equal to P (acting upward).
D
Sy
A
Ax
h
Bx
L
By
P