In the structure shown below member ABC is
assumed to be a solid rigid member. Member CD is a cable. For
this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the value of all the support forces acting on the
structure.
C. Determine the force (tension ) in member CD.
Unless otherwise indicated, all joints and support points are
assumed to be pinned or hinged joints.
Solution:
PARTS A
& B
STEP 1:
Draw a free body
diagram showing and labeling all load forces and support
(reaction) forces, as well as any needed angles and dimensions.
STEP 2: Break any forces not already in x and y
direction into their x and y components.
STEP 3: Apply the equilibrium conditions.
Sum Fx = -D cos (37o) + Ax = 0
Sum Fy = D sin (37o) + Ay -
10,000 lbs - 8,000 lbs = 0
Sum TA = (-10,000 lbs)(6.93 ft) - (8,000 lbs)(10.4 ft)
+ D sin (37o)(5.61 ft) + D cos (37o)(9.61
ft) = 0
Solving for the unknowns:
D = 13,800
lbs; Ax = 11,100 lbs; Ay = 9,710 lbs
PART C - Now find internal force in member DC.
In this problem member DC is a single axial member connected to
the wall at point D. Therefore, the force in member DC is equal
and opposite the force exerted on DC by the wall. From parts A
and B, the force on DC due to the wall is 13,800 lbs. Therefore,
force in DC is also 13,800 lbs (in tension since DC is a cable).
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