WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below member ABCD is assumed to be a solid rigid member. It is pinned to the floor at point A, and is supported by cable CE. Cable CE is made of steel and has a diameter of 1 inch. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress in cable CE.
Determine the movement of point D due to the applied loads.
Est = 30 x 10 6 psi; Ebr = 15 x 10 6 psi; Eal = 10 x 10 6 psi
ast = 12 x 10-6 /oC; abr = 20 x 10-6 /oC; aal = 23 x 10-6 /oC
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
Part A. External support reaction - 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 not already in x and y direction into their x and y components.
STEP 3: Apply the equilibrium conditions.
Sum F_x = A_x - E cos (30^o) = 0
Sum F_y = A_y + E sin (30^o) - 10,000 lbs - 12,000 lbs = 0
Sum T_A = -E sin (30^o)(4 ft) + E cos (30^o)(16 ft) - (10,000 lbs)(4.8 ft)- (12,000 lbs)(9.6 ft)= 0
Solving for the unknowns:
E = 13,790 lbs; A_x = 11940 lbs; A_y = 15,100 lbs
Part B. Stress_CE = F/A = 13,790 lbs/ .785 in² = 17,600 psi
Part C. Def = Deformation
STEP 1: Point D moves due to the deformation of cable EC. So we first determine the deformation of EC.
Def_EC = (FL / EA)_EC = (13,790 lbs)(12.9 ft)(12 in/ft) / (30*10⁶ lbs/in²)(3.14 * .5 in²) = .0912 in
STEP 2: Point D moves in proportion to how much point C moves. And point C moves the amount cable EC stretches. So we can write:
Mov. C / 12 ft = Mov. D / 16 ft
.0912 in / 12 ft = Mov. D / 16 ft and so Mov. D = .122 in