STATICS / STRENGTH OF MATERIALS - Example

In the structure shown on right horizontal member BCF is supported by vertical brass members, AB, and EF, and by 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 decrease 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 steel member CD.
C. Determine the resulting movement of point E .
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.
STEP 2: Break any forces into x and y components

STEP 3: Apply equilibrium conditions:
Sum F_x = 0
Sum F_y = D_y - A_y - F_y =0
Sum T_c = A_y (6 ft) - F_y (6 ft) = 0

From torque equation we get A_y = F_y; If we put this into the force equation we get: D_y - 2A_y = 0 or A_y = D_y / 2

To solve for forces A_y, D_y and F_y, we need a third equation - which we obtain from the relationship between the deformations.

PART B: DEFORMATION
Since the forces in members AB and FE are equal, since they are made of the same material - brass, and since they are the same length and area, they will deform the same amount (from d = FL / EA).

Both the bottom brass member (AB and FE) and the top steel member (CD) try to contract (because the change in temperature is negative).

However clearly both top and bottom members can not contract, one will "win," contracting while forcing the other member(s) to expand. We will assume the steel member "wins" actually shrinking a certain amount. The brass members will elongate the same amount. The steel contracts, as shown in diagram 2.

We write the deformation relationship as: - d (steel) = + d (brass)
(negative indicates that the deformation of the steel is a contraction)

The deformation of the members depend on both the thermal deformation and the mechanical deformation due to the forces that develop in the members.
This may be written as follows:
d = [aDTL ⁺ (FL/EA) ]_material
("+" sign if member is in tension, "-" sign if member is in compression)
where:
a = thermal coefficient of expansion
DT = change in temperature
L = length of member
F = force in member
E = Young's modulus for material
A = cross sectional area of member

We now substitute the expression into our deformation relationship so:
- d (steel) = + d (brass) becomes
-[aDTL + (FL/EA) ]_steel = [aDTL + (FL/EA) ]_brass

We now substitute in values and get the expression:
-[(12x10^-6/^oC) (-50^oC) (72 in) + D_y (72 in) / ( 30x10⁶ lbs/in²) (.75 in²)] =
+[(20x10^-6/^oC) (-50^oC) (96 in) + A_y (96 in) / ( 15x10⁶ lbs/in²) (.5 in²)]

Combining terms, the equation becomes:
-[-0.0432 + (3.2x10^-6 ) D_y ] = [-0.096 + ( 12.8x10^-6 ) A_y ] OR
3.2x10^-6 D_y + 12.8x10^-6 A_y = 0.139

This is our third equation, we can now substitute our relationship from statics: A_y =D_y / 2, into the above expression and get:
3.2x10^-6 D_y + 12.8x10^-6 (D_y / 2) = 0.139
Solving for
D_y = 14,500 lbs (force in steel member); A_y = 7,250 lbs (force in brass member(s))
Then to find stress in member CD
s = D_y / A = 14,500 lbs / 0.75 in² = 19,300 lbs/in²

PART C: MOVEMENT
Point E moves since it is attached to member FE, and its movement is equal to the deformation of member FE.
d_FE = [ (20x10^-6/^oC) (-50^oC) (96 in) + (7,250 lbs) (96 in) / (15x10⁶ lbs/in²) (.5 in²)] = -0.0032 in

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