WORLD FOR CIVIL

Topic 3.84: Exam - Solution Problem 1

The following values may be needed for the problem below
Linear coefficient of Expansion:
_Steel = 6.5 x 10^-6/^oF; _Aluminum = 13 x 10^-6/^oF ; _Brass = 11 x 10^-6/^oF; _concrete = 11 x 10^-6/^oF
Young's Modulus:
E_Steel = 30 x 10 ⁶ lb/in²; E_Aluminum = 10 x 10 ⁶ lb/in²; E_Brass = 15 x 10 ⁶ lb/in²; E_Concrete = 5 x 10 ⁶ lb/in²

1. In the structure shown 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.

Solution:
PART I: STATICS
Our first step is to draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. In order to do this accurately, we 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.) Although 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
We have too many unknowns at this point, so we now examine the deformations in the problem.

PART II: DEFORMATION
We now find a relationship between the deformations to develop an additional equation. From the geometry of the problem, we have:
+ _BR / 6 ft = + _ST / 10 ft or _BR = .6 _ST
The total deformation depends on the thermal deformation and the mechanical deformation and can be expressed as: _total = ( ⁺ FL/EA); then substituting this expression into our deformation relationship gives us:
( - FL/EA)_BR = .6( + 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 static 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, and 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 III: 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: Movement of D / 12 ft = d_CF / 10 ft.
Movement of D / 12 ft = [ (12x10^-6/^oC) ( +50^oC) (72 in) + (2,440 lbs) (72 in) / (30x10⁶ lbs/in²) (.5 in²) ] / 10 ft or Movement of D = (12 ft) [ 0.0549 in / 10 ft ] = 0.0659 in