WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below member BCDFG is assumed to be a solid rigid member. It is supported by a two cables, AB, and DE. Cable AB is brass, and cable DE is steel. Both cables have a cross sectional area of .5 square inches. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress in cable AB.
C. Determine the movement of point F due to the applied loads.
E_st = 30 x 10⁶ psi
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_y = A_y + E_y - 20,000 lbs - 10,000 lbs = 0
Sum T_B = (-20,000 lbs)(4 ft) + E_y(10 ft) - (10,000 lbs)(12ft) = 0

Solving for the unknowns: E_y = 20,000 lbs; A_y = 10,000 lbs
Part B. Stress_AB = F/A = 10,000 lbs/.5 in² = 20,000 psi
Part C. Def = Deformation
Movement of point F. Point F moves since both member AB and ED deform and member BCDFG moves downward according to these deformations.
STEP 1: Calculate deformation of members AB and ED.
Def_AB = (FL / AE)_AB = (10,000 lbs)(120 in) / (15*10⁶ lbs/in²)(.5 in²) = .16 in
Def_ED = (FL / AE)_ED = (20,000 lbs)(120 in) / (30*10⁶ lbs/in²)(.5 in²) = .16 in
STEP 2: Movement of point F. In this problem since both AE and ED deform the same amount (.16 in), then cross member BCDFG simply moves downward that amount, as does point F. So, movement of F = .16 in