WORLD FOR CIVIL

Topic 3.82b: Mixed Mechanical/Thermal - Example 2

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 F_x = 0
Sum F_y = F_br - F_st + F_br = 0
Sum Torque_B = -F_st (6 ft) + F_br (12 ft) = 0
The torque equation gives us: F_st = 2 F_br and the sum of y forces also gives us: F_st = 2 F_br
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)(40^oC)(96 in) - F_br (96 in) / (15x10⁶ lbs/in² )(.5 in²)] = [(12x10^-6 / ^oC)(40^oC)(96 in) + F_st (96 in) / (30x10⁶ lbs/in² )(.5 in²)]
or
(0.0768 - 1.28x10^-5 F_br ) = (0.0461 + 0.64x10^-5 F_st )
or
1.28x10^-5 F_br + 0.64x10^-5 F_st = 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 : F_st = 2 F_br Substituting, we obtain:
1.28x10^-5 (F_br ) + 0.64x10^-5 (2 F_br ) = 0.0307 , or 2.56 x 10^-5 (in./lb) F_br = .0307 in.
Then solving F_br = 1,200 lb.; and F_st = 2 F_br = 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)(40^oC)(96 in) + (2,400 lbs)(96 in) / (30x10⁶ lbs/in² )(.5 in²)] = 0.06144 in.