MECHANICS  CASE STUDY SOLUTION

Initial Conditions


This problem involves both temperature and mechanical loads. Furthermore, the
ceiling will restrict the overall deflection of the pole as the temperature increases.
The ceiling restriction makes this an indeterminate problem to the first degree
(one redundant support or restraint). As such, in addition to the standard equilibrium
equations, one compatibility relationship is needed to solve this problem.




Compatibility Relationship

Compatibility Relationship


When all loads, including
thermal, are applied, the structure will have a set deflection,
which can be used to define the compatibility condition.
Using the concept of superposition, each type of deflection can be isolated.
The first deflection, is due to the two shelf loads, F_{1} and F_{2}, labeled δ_{F}. These
will compress the pole.
Next, the temperature change will cause the pole
to increase by δ_{T}. This deflection assumes
there is no ceiling. But there is a ceiling that will restrict the deflection
and cause a compression force and deflection, δ_{R}.
The total deflection cannot exceed the initial gap of 1 mm. Thus, the compatibility
relationship is
δ_{F} + δ_{T}  δ_{R}
= 1 mm




Pipe CrossSection 

Before finding the pole deflections, the crosssectional area of the pipe needs
to be calculated.
A = π (2.5^{2} 
2.25^{2}) cm^{2}
=
3.731^{} cm^{2} = 0.0003731^{} m^{2}





Deflections

Load Deflection Using Superposition


Load Deflection
The two loads, F_{1} and F_{2}, each cause the pole to decrease
in length. The deflection from each load can be added together using the principle
of superposition, giving the total deflection from the applied loads as
δ_{F} = δ_{F1} + δ_{F2}
δ_{F} = 0.0004842
m = 0.4842 mm




Thermal Deflection without Ceiling 

Thermal Deflection
The expansion of the pole due to the temperature change (assuming no ceiling) is
δ_{F} =
L α ΔT
= (2.4 m  0.001
m) (23 × 10^{6}/^{o}C) (50^{o}C  20^{o}C)
= 0.0016553 m = 1.6553 mm
This total thermal deflection may not be possible due to the ceiling. If it
is too great, the ceiling will induce a reaction force into the pole that must
be considered. This is calculated in the next paragraph.




Reaction with Ceiling 

Ceiling Reaction Deflection
It is assumed that the pole will expand enough for the pole to press against
the ceiling. This will cause a reaction force on the pole that will compress
the pole. The total compression load is not known but is related to the deflection,
=
8.8081 × 10^{8} R_{2} m =
0.000088081 R_{2} mm






Solving for the Reactions



Now that all the deflections have been established, they can be substituted
into the compatibility relationship,
δ_{F} + δ_{T}  δ_{R} =
1 mm
0.4842 + 1.6553  0.000088081 R_{2}
= 1.0
R_{2} = 1,943 N




Applied and Reaction Loads 

The reaction is positive, thus the initial assumption that the ceiling will
cause compression was correct.
The floor reaction can be found by using the static equilibrium equation,
ΣF_{y} =
0
R_{1}  F_{1}  F_{2} 
R_{2} = 0
R_{1} = (5.494  5.494  1.943) kN
R_{1} =
12.93 kN



