WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown on the right, 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 .75 in².
The structure is initially unstressed and then experiences a temperature increase of 40 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 member CD.
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.

In this problem we first examine 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 in 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. The forces in the brass member are equal from symmetry of the structure.

The horizontal member BDF moves to a middle position, as shown in diagram 2.
We now write the equilibrium equations:
Sum F_x = 0
Sum F_y = F_br - F_st + F_br = 0
Sum T_D = -F_br (6 ft) + F_br (6 ft) = 0
The torque equation simply gives us: F_br = F_br
and the sum of y forces gives us: F_st = 2 F_br

We do not have enough equation to solve the problem. We obtain an additional independent equation for the formation relationship.

PART B: DEFORMATION
d _br = d_st from the geometry of the structure…expanding using:
d _total = [a DTL ⁺ FL / EA ], we 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

From our static equilibrium equation in part I, we have : F_st = 2 F_br
Substituting, we obtain:
0.64x10^-5 (2 F_br ) + 1.28x10^-5 (F_br ) = 0.0307
solving: F_br = 1,200 lbs ; F_st = 2,400 lbs
Stress in steel = F/A = 2,400 lbs/.75 in² = 3,200 lbs/in²

PART C: MOVEMENT
Movement of point D. Point D is connected to steel member CD and so moves the amount CD deforms or
Mov. D = [(12x10^-6 / ^oC)(40^oC)(96 in) + (2,400 lbs)(96 in) / (30x10⁶ lbs/in² )(.75 in²)]
Mov. D = 0.05632 in