Given: Consider frame ABC. Length AB is 10 ft, BC is 20 ft, and angle ABC is 60°. The cross-sectional area of AB is 10 in² and of BC is 2 in². Young's modulus is 30x10⁶ psi. Load P = 20,000 lb.

Req'd: Determine the downward deflection of Joint B.

Sol'n: From the FBD of joint B, vertical and horizontal equilibrium give:

Solving for the internal forces:

F_BC = 1.15 P     and      F_AB = –0.577 P

The internal energy of a system is:

U = (2.71x10^–6 ) P²     lb-in. (P in pounds)

The work done by the applied force P equals the internal energy stored:

Solving for d:

d = 2(2.71x10^–6) P

For P = 20,000 lb:

d = 0.108 in.