Download ME 236 Engineering Mechanics I Test #2 Solution

ME 236 Engineering Mechanics I  Test #2 Solution
Date: Friday, March 5, 2004
Time: 9:30-10:30 (60 minutes)
Instructions: Covering Chapters 5-6 of the textbook, closed-book test, calculators allowed.
1
(30%) Determine the tension in the cable and the horizontal component of the support force
at A. The pulley at D is frictionless and the cylinder load at C weighs 80 lb.
Solution:
Since the pulley does not have friction, the two segments of the cable are subject to the same
tension force T. Based on the free-body diagram shown above, we have
ΣM A = 0 ⇒ T × 5 +
Thus, we get
T=
13W
(5 + 20 / 5 )
=
2
5
T × 10 − W × 13 = 0
13 × 80
(5 + 20 / 5 )
= 74.583 (lb)
To find the horizontal support force, we have
ΣF x = 0
⇒
Ax −
1
5
T =0
Thus,
Ax =
1
5
T=
1
5
× 74.583 = 33.354 (lb)
2
The Howe bridge truss is subject to the loading shown in the figure.
2.1 (20%) Determine the force in members HG and CD. Indicate it is tension or compression.
2.2 (20%) Identify all the zero-force members. (Note: if you identified a wrong member, you
will loose the same amount of credits as missing one).
Solution:
2.1 First, considering the entire truss as a rigid body, we have a free-body diagram as above.
Based on the diagram, we have
ΣM A = 0 ⇒ − 20 × 4 − 20 × 8 − 40 × 12 + E y × 16 = 0
Thus,
E y = (20 × 4 + 20 × 8 + 40 × 12) / 16 = 45 (KN)
Second, applying the method of section, we consider the truss as shown
in the figure on the right. In order to get FHG in one step, we write the
moment equilibrium equation about D.
ΣM D = 0 ⇒ E y × 4 − FHG × 4 = 0
from which we solve
FHG = E y = 45 (KN)
 Compression
Next, write the moment equation about point H,
ΣM H = 0 ⇒ − 40 × 4 + E y × 8 − FCD × 4 = 0
Thus,
FCD = (−40 × 4 + E y × 8) / 4 = (−40 × 4 + 45 × 8) / 4 = 50 (KN)
 Tension
2.2 Zero-force members:
− Members GF and FE because the two forces in these two members are perpendicular to
each other and they cannot be balanced at joint F if they are nonzero.
− Member HC because it is perpendicular to the other two connecting members at joint C
and no external force at the joint to balance any nonzero force in member HC.
− Member JI because the force in this member is perpendicular to the only other two forces
at joint J.
3
(30%) The derrick is pin-connected to the pivot at A. Determine the largest mass that can be
supported by the derrick if maximum force that can be sustained by the pin at A is 18 kN?
Solution:
(a)
(b)
Take two different sections out of the original system and draw their free-body diagrams as
shown in figures (a) and (b).
Because link AB is a two-force member, the reaction force from pin A to link AB must be along
the link. Assume this force has reached the maximum, i.e.,
FA = 18 (kN)
From the free-body diagram of the crate (b), we have
ΣFy = 0 ⇒
2T2 − W = 0 ⇒ T2 =
W
2
From the free-body diagram (a), we have
ΣFy = 0 ⇒
FA sin 60 o − T2 sin 60 o − W = 0
i.e.,
18 × sin 60 o −
W
sin 60o − W = 0
2
Thus, we get
W = 18 × sin 60 o /(0.5 sin 60 o + 1) = 10.878 (kN)
The maximum mass of the crate is:
m = W / g = 10.878 × 1000 / 9.81 = 1109 (kg)