WORLD FOR CIVIL

Topic 3.82a: Mixed Mechanical/Thermal - Example 1

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 1
In the structure shown in Diagram 1 horizontal member ABC is pinned to the wall at point A, and supported by a Aluminum member, BE, and by a Brass member, CD. Member BE has a .75 in² cross sectional area and Member CD has .5 in² cross sectional area.
The structure is initially unstressed and then experiences a temperature increase of 40 degrees Celsius. For this structure we wish to determine the stress which develops in the aluminum and brass members. We would also like to determine the deformation of the brass member. (At this point we will assume that the horizontal member ABC 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 aluminum are :
br = 20 x 10-6 /oC; al = 23 x 10-6 /oC; Ebr = 15 x 10 6 psi; Eal = 10 x 10 6 psi

Solution:
The first part of the solution will be to consider the static equilibrium conditions for the structure. However to do this effectively we first need to consider the physical effects of the temperature change to determine the directions of the forces acting on the structure. We do this by first considering how much the brass and aluminum members would expand if they were free to expand, due to the temperature increase.

Both the brass and aluminum members try to expand. In Diagram 2 we have shown the amount each member would expand if free to do so. However both can not expand - one will "win" causing the other to contract. We will assume the brass "wins" forcing the aluminum to compress. The horizontal member, ABC, will rotate upward about point A as shown in Diagram 2. That is, the brass will expand, but not as much as it would if unconstrained - since it is compressing the aluminum which is in turn pushing back on the brass, putting it into compression also. Thus both the brass and aluminum members are in compression. We have shown in Diagram 2 the direction the external support forces will act on the structure. Notice the forces at E and D are along the direction of the members. This is due to the fact that the brass and aluminum are axial members, and thus simply in tension or compression. At point A where the structure is pinned to the wall, we have put in horizontal and vertical support forces A_x and A_y. We are now ready to apply static equilibrium conditions.

PART I: 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. (Already done in Diagram 2) Apply equilibrium conditions:
Sum F_x = A_x = 0 (There is no external horizontal force acting at point A)
Sum F_y = F_br - F_al + A_y = 0
Sum Torque_A = -F_br (14 ft) + F_al (8 ft) = 0
At this point we see we have too many unknowns (F_br, F_al, A_y) and not enough equations to solve for our forces. We need an additional equation to solve for the unknowns. We will obtain the additional equation from the deformation is the problem.

PART II: DEFORMATION
We first write a general relationship between the deformations, which we find from the geometry of the problem. We see (from Diagram 2) that: _br / 14 ft = - _al / 8 ft or simplifying:
_br = -1.75 _al
The negative sign in front of the deformation of aluminum term comes from the fact that, in our assumption, the aluminum gets shorter, and in problems involving mixed thermal and mechanical deformations (as in this problem) we need to keep signs associated with deformations, with elongation being positive, and contractions being negative.
Once we have this general deformation relationship, we now substitute in our deformation expressions for mechanical plus thermal deformation. That is, the net deformation is the sum of the thermal deformation () due to the temperature change, and the mechanical deformation (FL / EA) due to the forces which develop in the members. So the total deformation of a member may be written as _total = [⁺ FL / EA ] where the mechanical deformation term is positive (use + sign) if the member is in tension, and the mechanical deformation term is negative (use - sign) if the member is in compression. If we substitute this expression for the deformations into our general relationship ( _br = -1.75 _a) we obtain:

[- FL / EA ]_br = -1.75[ - FL / EA ]_al (Notice that we use a negative sign for each mechanical term since both members are in compression. We now substitute in numeric values given in our problem and get:
[(20x10^-6 / ^oC)(40^oC)(96 in) - F_br (96 in) / (15x10⁶ lbs./in² )(.5 in²)] = -1.75[(23x10^-6 / ^oC)(40^oC)(120 in) - F_al (120 in) / (10x10⁶ lbs./in² )(.75 in²)].
(Notice that we need not be concerned that the temperature is in Celsius rather than Fahrenheit, since the linear coefficient of expansion is also per ^o Celsius, the units cancel.) Then after combining terms we have:
(0.0768 - 1.28x10^-5 F_br ) = (-0.193 + 2.8x10^-5 F_al ) or 1.28x10^-5 F_br + 2.8x10^-5 F_al = 0.27
This is our additional equation. Now from our torque equation from static equilibrium in Part I we have F_al = 1.75 F_br We now substitute this into our deformation equation above obtaining:
1.28x10^-5 F_br + 2.8x10^-5 (1.75 F_br ) = 0.27 or
6.18 x 10-5 (in./lb.) F_br = .27 in.
solving:
F_br = 4,370 lbs.; F_al = 7,650 lbs. The stress in brass = F/A = 4,370 lbs./.5 in² = 8,740 lbs./in² The stress in aluminum = F/A = 7,650 lbs./.75 in² = 10,200 lbs./in²

PART III: Deformation of Brass Member
The deformation of the brass member may now be found from the general expression for the deformation: [- FL / EA ]_br or putting in values

_Brass= [(20x10^-6 / ^oC)(40^oC)(96 in) -(4,370 lbs.)(96 in) / (15x10⁶ lbs./in² )(.5 in²)] = 0.210 in. (Since this deformation is positive, it means brass member CD does expand as we assumed. If the value were negative it would have meant that the brass member was actually compressed by the aluminum member, which then would have expanded. However the values found for the forces and the stress would have been correct in either case.)