WORLD FOR CIVIL

Topic 3.84: Exam - Solution Problem 2

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


2. In the structure shown the L-shaped member BCD is supported by Steel rod AB and Aluminum member DE, and pinned at point C, as shown. Member DE has a cross sectional area of 1 in² and member AB have a cross sectional area of .5 in². The structure is initially unstressed and then experiences a temperature decrease of 60 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 steel rod AB.
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 deformations are trying to occur and how the structure will respond.
As shown in Diagram 2, the steel member (AB) and the aluminum member (DE) would both like to contract, however both can not win. We will assume that the aluminum "wins" and does contract, but not as much as it would like to - due to the steel resisting, putting the aluminum in tension. The aluminum is, of course, also pulling on the steel, putting the steel into tension - and in our assumption, forcing the steel to actually expand. The result is that the bar moves to position as shown in Diagram 3, with the external forces, as shown, acting on the structure.

Diagram 3 is now our free body diagram of the structure with all forces in either the x or y-direction, and we now apply static equilibrium conditions:
Sum F_x = -F_st + C_x = 0;
Sum F_y = C_y - F_al = 0;
Sum T_C = F_st (4 ft) - F_al (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:
+ _st / 4 ft = - _al / 6 ft or 1.5_st = - _al
(Where the + sign indicates the steel member expands, and the - sign indicates the aluminum contract) The total deformation depends on the thermal deformation and the mechanical deformation and can be expressed as: _total = ( ⁺ FL/EA);
substituting this expression into our deformation relationship gives us:
1.5 ( + FL/EA)_st = - ( + FL/EA)_al
(Where the + sign in front of each the mechanical deformation terms is used because both member are in tension.) Next, substituting in values, we have:
1.5 [(12x10^-6/^oC) (-60^oC) (24 in) + F_st (24 in) / (30x10⁶ lbs/in² ) (.5 in² )] = - [(23x10^-6/^oC) (-60^oC) (24 in) + F_al (24 in) / (10x10⁶ lbs/in² ) (1 in² )] , or
-.0259 + 2.4x10^-6 F_st = 0.0331 - 2.4x10^-6 F_al, or
2.4x10^-6 F_st + 2.4x10^-6 F_al = 0.059
From our static torque equation we have: F_st (4 ft) - F_al (6 ft) = 0 OR F_st = 1.5 F_al
We now substitute into our deformation expression
2.4x10^-6 (1.5 F_al ) + 2.4x10^-6 F_al = 0.059, or 6 x 10^-6 (in/lb.) F_al = .059 in
Solving for F_al = 9,830 lbs F_st = 14,750 lbs
Then the stress in steel member AB is s = F/A = 14,750 lbs/ .5 in² = 29,500 lbs/in²

PART III: MOVEMENT
Finally, point D movement is equal to the deformation of member DE, and we can write: Movement. D = [(23x10^-6/^oC) (-60^oC) (24 in) + 9830 lb (24 in) / (10x10⁶ lbs/in² ) (1 in² )] Movement. D = [-.0331 in + .0236 in ] = - .0095 in. The negative sign indicates we made a correct assumption, and the aluminum member actually does contract a small amount.