Partially Constrained Thermal Deformation:
In a structure, rather than being completely constrained or completely free, members more
often are partially constrained. This means a member may expand (or contract) but not as
much as it would if unconstrained.
Example 2
In the structure shown below horizontal member BDF is supported by two brass members, AB
and EF, and a steel member CD. Both AB and EF have cross sectional areas of .5 in2. Member
CD has a cross sectional area of .5 in2.
The structure is initially unstressed and then experiences a temperature increase of 40
degrees celsius. For this structure, determine the axial stress that develops in steel
member CD, and the resulting movement of point D. (At this point we will assume that the
horizontal member BDF does not bend due to forces acting on it. We will consider beam
bending at a later point.) The linear coefficient of expansion, and Young's modulus for
brass and steel are :
br = 20 x
10-6 /oC; st = 12 x 10-6 /oC; Ebr = 15 x 10 6 psi; Est = 30 x 10 6 psi
Solution:
PART I: STATICS
In this problem we first consider how much the brass and steel member
would expand, if free to do so, due to the temperature increase. As the coefficient of
expansion of the brass is larger than the coefficient of expansion of the steel, the brass
would expand more if free to do so. (See Diagram 2)
As the brass expands it pulls on the steel placing the steel in tension. The steel pulls
back on the brass placing the brass in compression. The support reactions, reflecting
these forces, are shown in diagram 2. Notice that the brass and steel members have forces
acting on them at only two points, so they are axial members. This also means that the
external forces on the members due to the floor are equal to the forces in the members.
Additionally we note that the forces in the brass member are equal from symmetry of the
structure. (or if we mentally sum torque about point D, the center of the member BDF, we
see that forces in the brass members would produce opposing torque which would need to be
equal and opposite for equilibrium. Since the distances are equal, the forces in the brass
members need to be equal to produce equal amounts of torque.)
The result of the brass and steel members working against each other is that the
horizontal member BDF moves to an intermediate position, as shown in Diagram 2. We are now
ready to proceed with the Statics.
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. (Diagram 2)
We now write the equilibrium equations:
Sum Fx = 0
Sum Fy = Fbr - Fst + Fbr = 0
Sum TorqueB = -Fst (6 ft) + Fbr (12 ft) = 0
The torque equation gives us: Fst =
2 Fbr and the sum of y
forces also gives us: Fst = 2 Fbr
We do not have enough equations at this point to solve the problem. We need an additional
equation, which we will obtain from the deformation relationships.
PART II: DEFORMATION
From the geometry of the problem, we see that the final net deformation of the
brass and steel members will be equal, or 
br = st Notice that both deformations are positive since
the members expand. Next using our expression for combined thermal and mechanical
deformations: d total = [a DTL
+ FL / EA ], we substitute into the general deformation relationship and
obtain:
[
- FL / EA ]br = [ + FL / EA ]st
The sign of the mechanical deformation term for the brass is negative (-) since the brass
is in compression. The sign of the mechanical deformation term for the steel is positive
(+) since the steel is in tension. We now substitute values from our problem into the
equation and obtain:
[(20x10-6 / oC)(40oC)(96 in) - Fbr (96
in) / (15x106 lbs/in2 )(.5 in2)] = [(12x10-6 /
oC)(40oC)(96 in) + Fst (96 in) / (30x106
lbs/in2 )(.5 in2)]
or
(0.0768 - 1.28x10-5 Fbr ) = (0.0461 + 0.64x10-5 Fst
)
or
1.28x10-5 Fbr + 0.64x10-5 Fst
= 0.0307
This is our additional equation. We can now substitute our relationship between the
brass and steel force from our static equilibrium equations in part I, which gave us : Fst = 2 Fbr Substituting, we obtain:
1.28x10-5 (Fbr ) + 0.64x10-5 (2 Fbr ) =
0.0307 , or 2.56 x 10-5 (in./lb) Fbr = .0307 in.
Then solving Fbr = 1,200 lb.; and Fst = 2 Fbr
= 2,400 lbs, and so the Stress in steel = F/A = 2,400
lbs/.5 in2 = 4,800 lbs/in2
PART III: MOVEMENT
Movement of point D. Point D is connected to steel member CD and so moves the amount CD
deforms, or Movement of D = [ + FL /
EA ]st , or
Movement of D = [(12x10-6 / oC)(40oC)(96
in) + (2,400 lbs)(96 in) / (30x106 lbs/in2 )(.5 in2)] =
0.06144 in.
Return to:
Topic 3.81: Thermal Stress, Strain &
Deformation II
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