4c-ex. Energy Methods Examples
Ex. 4c.1
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Example 4c.1

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:

Frame ABC

FBCsin(60°) = P     and      FAB + FBCcos(60°) = 0

Solving for the internal forces:

FBC = 1.15 P     and      FAB = –0.577 P

FBD of Joint B

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.

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Updated: 05/24/2009 DJD