STATICS / STRENGTH OF MATERIALS - Example

In the structure shown on right horizontal member BCD is supported by vertical brass member, AB, a steel member DE, and is pinned to the floor at point C. Both AB and DE have cross sectional areas 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 member DE.
C. Determine the resulting movement of point B .
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 members would shrink if they were free to do so due to the temperature decrease. The brass member, AB, would shrink much more than the steel member since it's coefficient of linear expansion is nearly twice that of steel. (see diagram 2)

As the brass shrinks, it causes the horizontal member BCD to rotate about point C. As it does so, point D moves downward - which is all right to a certain point, as the steel is also shrinking. At a certain pint the steel is done shrinking due to the temperature decrease and "wants" to stop; however the brass "wants" to shrink more and continues to pull upward on point B causing more rotation of member BCD about point C, and causing point D to move downward an additional amount, putting steel member DE into compression. The steel member DE pushes back on point D, stopping the brass from shrinking any more and putting the brass into tension. This thought analysis process is how the directions of the support reactions direction in diagram 1 were arrives at.

The horizontal member BCD ends up in an intermediate position as shown in diagram 2. We now write the equilibrium equations:
Sum F_x = C_x = 0
Sum F_y = F_br - C_y + F_st = 0
Sum T_C = -F_br(4 ft) + F_st(8 ft) = 0

There are not enough independent equations to solve for our three unknowns, so we need another independent equation - which we will obtain from the deformation relationship.

PART B: DEFORMATION
From the geometry of the problem we see that
- d _brass / 4 ft = - d _steel / 8 ft
(negative signs since deformation and contractions - members get shorter.)
2d_brass = d_steel

We now expand our expression for our deformations using:
d_total = [a DTL ⁺ FL / EA]_material ;
so, we obtain
2[a DTL + FL / EA]_brass = [a DTL - FL / EA]_steel
and putting values of materials and the structure into this equation we obtain:
2[(20x10^-6 / ^oC)(60^oC)(96 in) + F_br (96 in) / (15x10⁶ lbs/in² )(.5 in²)] =
[(12x10^-6 / ^oC)(60^oC)(48 in) + F_st (48 in) / (30x10⁶ lbs/in² )(.5 in²)]
or (-0.2304 + 2.56x10^-5 F_br ) = (-0.0346 - 0.32x10^-5 F_st )
or 2.56x10^-5 F_br + 0.32x10^-5 F_st = 0.196

from our static equilibrium torque equilibrium we had:
-F_br(4) + F_st(8) = 0 or F_br = 2F_st , we now substitute this into our deformation equation.
and obtain:
2.56x10^-5 (2 F_st ) + 0.32x10^-5 F_st = 0.196
solving:
F_st = 3,600 lbs ; F_br = 7,200 lbs
and stress in steel = F/A = 3,600 lbs / .5 in² = 7,200 lbs/in²

PART C: MOVEMENT
Point B is attached to member AB and so movement of point B is equal to the deformation of brass member AB.
Mov. B = [(20x10^-6 / ^oC)(60^oC)(96 in) +(7,200 lbs)(96 in) / (15x10⁶ lbs/in² )(.5 in²)]
Mov. B = -0.023 in