WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below member ABC is assumed to be a solid rigid member. Member CD is a cable. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the value of all the support forces acting on the structure.
C. Determine the force (tension ) in member CD.
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
PARTS A & B
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 = -D cos (37^o) + A_x = 0
Sum F_y = D sin (37^o) + A_y - 10,000 lbs - 8,000 lbs = 0
Sum T_A = (-10,000 lbs)(6.93 ft) - (8,000 lbs)(10.4 ft) + D sin (37^o)(5.61 ft) + D cos (37^o)(9.61 ft) = 0
Solving for the unknowns:
D = 13,800 lbs; A_x = 11,100 lbs; A_y = 9,710 lbs

PART C - Now find internal force in member DC.
In this problem member DC is a single axial member connected to the wall at point D. Therefore, the force in member DC is equal and opposite the force exerted on DC by the wall. From parts A and B, the force on DC due to the wall is 13,800 lbs. Therefore, force in DC is also 13,800 lbs (in tension since DC is a cable).