
MECHANICS  CASE STUDY SOLUTION

Problem Diagram
Bottom Shelf
Top Shelf


Since the shelves are not symmetrically loaded, each wire will likely have a
different load. This will cause each of the wire to deflect differently and thus
the shelves to rotate slightly. Complicating the situation is there are two types
of wire used, steel and copper, and lengths of the wires vary.




Load in Wires


The first step is to determine the load in each wire using static equilibrium
equations. Draw a freebody diagram for both shelves so that the load
in each wire can be found. Since there are only two known cable forces for the
shelf on the bottom shelf, it should
be analyzed first. Due to symmetry, it is easy to notice that the
tensions in both cables AB and CD will equal half the weight of the
books.
T_{AB} = T_{CD} = 140 lb
Now, the top shelf can be analyzed since the wire load T_{CD} is
known which only leaves two unknowns, T_{GH} T_{EF}. The
tension in
the
cable
EF
can be calculated by
summing
the
moments about point G,
ΣM_{G}
= 3T_{EF}  1.5 (50)  1 (140) = 0
T_{EF} = 71.67 lb
Summing forces in the vertical direction gives the force in cable GH,
ΣF_{y}
= T_{GH}  140  50 + T_{EF} = 0
T_{GH} = 118.3 lb






Wire Elongation and Shelf Rotation

Shelf Deflection and Rotation 

The tensions in the cables will cause each wire to stretch a small
amount. The elongation of each wire can be found from the equation
where P is the force, L is the length, A is the crosssectional area,
and E is the Young's modulus of each wire. The elongation of each cable
will cause each of the points A, C, D, E, and G to be displaced differently,
as shown. The angle of rotation of the bottom shelf is given
by
Therefore, the deflection of points A and C need to be calculated. The
deflection at point A is simply the elongation of the wire AB,
= 0.06837 in




Top Shelf


The deflection at point C is more complex to determine since it is related
to the deflection of the top shelf. However, the total deflection is
simply the sum of the elongation of CD and the deflection of point D,
The displacement at point D is related to the displacement at points
E and G as
δ_{D} = δ_{G} 
(δ_{G}  δ_{E})/3
However, this requires the deflection of points E and G to be determined.
The deflection at point E is equal to the elongation of the wire EF,
= 0.00768 in
The deflection at point G is
= 0.01268 in
Substituting these results back into the the δ_{C} deflection
equation gives,
= 0.03056 in






Now the rotation angle can be determined,
= tan^{1}(0.0007877)
= 0.04513^{o}
The negative sign inside the arctan term simply means the angle is opposite
of the assumed direction in the diagram above.




