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20.1104 Intro to Eng. Analysis, Studio version
Fall 1998 – Test #4 Friday Dec. 4, 1998
Instructions: 60-min time limit. You can use Maple to do the problems and check your
answers, but clearly show your setup and solution for each problem on the test pages – read each
problem to see what we want specifically.
1. (30 points) A heavy-weight lift as shown in the sketch below is used at a construction site. Assume
that it is a pin-jointed truss. Find the loads in members PQ, RP, RQ and RS.
Solution:
Use the method of sections (using section a-a’) to solve for PQ, RQ, and RS:
MR = (1000 lb)(64 ft) – FPQ (8 ft) = 0
FPQ = 8000 lb (T)
FY = FRQ sin 45 – 1000 lb = 0
FRQ = 1412 lb (T)
FX = FRS + FRQ cos 45 + FPQ = 0
FRS = -9000 lb
or
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FRS = 9000 lb (C)
Next, use the method of joints to get PR:
FY = FPR + FRQ sin 45 = 0
FPR = -1000 lb
or
FPR = 1000 lb (C)
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2. (10 points – this should take no more than 5 minutes to answer) For the same truss that is pictured
in problem 1, the left end of the truss is enlarged below. Two members AB and AC are added (dotted
lines). Explain why members AB and AC would not make any difference in the loading of the rest of
the truss under the 1000 lb loading shown.
Solution:
Members AB and BC are zero force members and do not carry any of the load that is being
placed on the truss.
This can be shown using equilibrium equations at joint A:
FX = FAB = 0
FY = FAC = 0
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3. (30 points)
(a) For the simple pin-jointed truss shown below, set up a solution that would yield the values of the
forces in all of the members. In your solution, identify the number of unknowns in this problem, and
the number of available equations. DO NOT SOLVE FOR THE NUMERICAL VALUES OF THE
UNKNOWNS AT THIS STAGE.
Solution:
 There are 6 joints in this equation which will provide us with 12 equations (FX = 0 and
FY = 0 for each joint).
 There are also 12 unknowns in this problem: 3 ground reactions (AY, DX, and DY), as well
as 9 member loads.
 Therefore, we can solve for all 12 unknowns with the 12 equations that are available.
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(b) Using whatever method is fastest for you, solve for the forces in members CD and DE in the truss
from part (a), indicating whether the forces are tensile or compressive.
Solution:
First, treat the entire truss as a rigid body to solve for ground reactions at D:
MA = -(1000 lb)(15 ft) + DY (4.5 ft) = 0
DY = 333.3 lb
FX = DX = 0
Next, use the method of joints to get CD and DE:
FY = DY + FCD sin 45 = 0
FCD = -471.5 lb
or
FCD = 471.5 lb (C)
FX = -FCD cos 45 – FDE + DX = 0
FDE = 333.3 lb (T)
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4. (15 points) Draw the free body diagram of each part of the smooth piston and link mechanism
shown below, which is used to crush cans (C). Explain (in terms of procedure) how to solve for the
force exerted on the can but do not solve the problem numerically.
Solution:
 Note that on the FBDs above:
NW is the normal force from contact with the wall, and P is the compressive force
caused by the can (this is what we want to solve for).
Also, since B and D are pin joints, each will have two reaction forces (which are equal
and opposite on the two FBDs).
However, member AB is a two force member, which simplifies the FBD that we
would need to draw (only one force acts through this member, with its direction along
a line that connects the two endpoints of member AB).
 To solve for the desired unknown force P:
Set up equilibrium equations (FX = 0, FY = 0 and MD = 0) for the FBD of the
handle member. This will allow you to solve for the unknowns DX and DY.
 Then move on to the FBD of the can...
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Set up equilibrium equations (FX = 0 and FY = 0) and solve for the unknown P.
5. (15 points) Determine the required force P that must be applied at the handles of the pruning shears
so that the blade applies a normal force of 20 lb on the twig at E. Note that the lower handle is one
piece and extends only as far as pin A. The upper handle is one piece all the way out to the lower
cutting blade (note cross-hatching on upper handle); some of the upper handle passes behind the lower
handle near pin A.
Solution:
From FBD I:
MA = -DX (0.5 in) – P (5.5 in) + 20 lb (1 in) = 0
FX = DX + AX = 0
FY = -20 lb + DY + AY – P = 0
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From FBD II:
where  = tan-1(0.5/0.75) = 33.7
MB = AX (0.5 in) + AY (0.75 in) + P (4.75 in) = 0
FX = -FBC cos 33.7 - AX = 0
FY = -FBC sin 33.7 – AY + P = 0
For extra points, solve these 6 equations and 6 unknowns in Maple to get:
AX = -13.3 lb
AY = -6.46 lb
DX = 13.3 lb
DY = 28.9 lb
P = 2.42 lb
FCB = 16.0 lb
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