WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown on right member ABCD is pinned to the wall at point A, and supported by a brass member, BE, and by a steel member, CF. Both BE and CF have cross sectional areas of .5 in².
The structure is initially unstressed and then experiences a temperature increase of 50 degrees Celsius. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress that develops in brass member BE.
C. Determine the resulting movement of point D .

E_st = 30 x 10 ⁶ psi; E_br = 15 x 10 ⁶ psi; E_al = 10 x 10 ⁶ psi
a_st = 12 x 10^-6 /^oC; a_br = 20 x 10^-6 /^oC; a_al = 23 x 10^-6 /^oC

Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
PART A: 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.

First determine the direction of support forces by examining what thermal deformation are trying to occur and how the structure will respond.

As shown in Diagram 2, the brass member (BE) would like to expand more than the steel member (CF) due to thermal effects. (Since the thermal coefficient of expansion of brass is larger than the thermal coefficient of expansion of steel.) Since we assume member ABCD will not bend, it will rotate about point A to a middle position as shown in Diagram 3.

That is, as the brass member expands, the steel member want to expand less and pulls back on the brass member putting it into compression, and stopping it from expanding as much as it would like to. From the steel member's point of view, it expands to a point (thermal expansion) and wants to stop, but the brass member continues to expand, pulling on the steel member causing it to expand more than it would like (putting the steel member in tension).

Thus the brass member is in compression and the steel member is in tension, and the free body diagram may now be drawn as shown in Diagram 4.

Apply equilibrium conditions: Sum F_x = A_x = 0
Sum F_y = A_y - F_BR + F_ST = 0 Sum T_A = F_ST (10 ft) - F_BR (6 ft) = 0

PART B: DEFORMATION
Too many unknowns; we now find a relation ship between the deformations to develop an additional equation. From the geometry of the problem, we have:
+ d_{BR/6ft =}+ d_ST/10ft or d_{BR total} = .6 d _{ST total}
The total deformation depends on the thermal deformation and the mechanical deformation and can be expressed as:
d _total = (a DTL ⁺ FL/EA);
substituting this expression into our deformation relationship gives us:
(a DTL - FL/EA)_BR = .6(a DTL + FL/EA)_ST

Substituting in values, we have:
[ (20x10^-6/^oC) (+50^oC) (72 in) - F_BR (72 in) / (15x10⁶ lbs/in² ) (.5 in² )] = .6 [ (12x10^-6/^oC) (+50^oC) (72 in) + F_ST (72 in) / (30x10⁶ lbs/in² ) (.5 in² )] OR 0.072 in - 9.6x10^-6 F_BR = 0.026 in + 2.88x10^-6 F_ST OR 2.88x10^-6 F_ST + 9.6x10^-6 F_BR = 0.046

From our statics torque equation we have: F_ST (10 ft) - F_BR (6 ft) = 0 OR F_ST = .6F_BR

We now substitute into our deformation expression
2.88x10^-6 (.6 F_BR) + 9.6x10^-6 F_BR = 0.046
Solving for F_BR = 4,060 lbs F_ST = 2,440 lbs
Then the stress in brass member BE is s = F/A = 4,060 lbs/ .5 in² = 8,120 lbs/in²

PART C: MOVEMENT
Finally, point D moves in proportion to the movement of point C (which is equal to the deformation of member CF), and we can write:
Mov. D / 12 ft = d_CF / 10 ft
Mov. D / 12 ft = [ (12x10^-6/^oC) ( +50^oC) (72 in) + (2,440 lbs) (72 in) / (30x10⁶ lbs/in²) (.5 in²) ] / 10 ft
or Mov. D = (12 ft) [ 0.0549 in / 10 ft ] = 0.0659 in