ARC322 Homework 4 2008 SOLUTION

ARC322 Homework 4 2008 SOLUTION - 4. Assume that the beam...

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ARC322 Homework 4 2008 1. A WF has a maximum moment of 124,500 ft-lb. Select a WF for this condition. W16x40 Check: S = M/ f = (124,500 ft-lb x 12 in/ft)/24,000 lb/in 2 = 62.25 in 3 W16x40 S = 64.7 in 3 … OK (Note … a W14x43 will also work with an S = 62.7 in 3 , but the W16x40 is more economical AND much more available from suppliers … note its bold type in the chart). 2. What is the unbraced length without a moment reduction for the beam in #1? 7.4 ft 3. What is the maximum moment that a W24x207 beam can carry? 1,050,000 ft-lb Check: S = M/ f rearranging M = S f = 531 in 3 x 24,000 lb/in 2 = M = 12,744,000 in-lb/12in/ft = 1,062,000 ft-lb … why is it not 1,050,000 ft-lb??? Because we used f = 24,000 pb/in 2 which is a ROUNDED value … plus, AISC adds a little for a factor of safety in its AISD Tables.
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Unformatted text preview: 4. Assume that the beam in #3 is a simply supported beam and that its load is 3500 lb/ft. What is the length of the beam? Here, we need to stop and think a little w = 3500 lb/ft we dont know W because we dont know the length. But, we do know the maximum moment is 1,050,000 ft-lb and we know f for A36 steel is 24,000 lb/in 2 so M = w L 2 /8 rearranging L 2 = 8M/ w = (8 x 1,050,000 ft-lb)/3500 lb/ft = 2400 ft L 2 = 2400 ft L = (2400 ft) 1/2 using exponent since MS-Word doesnt do square root symbols easily L = 48.99 ft you could probably round to 49 ft if you want. 5. What is the bending stress on the beam? f = M/S = (1,050,000 ft-lb x 12 in/ft)/531 in 3 = 23,728.81 lb/in 2 < 24,000 lb/in 2 OK...
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This note was uploaded on 03/30/2008 for the course ARC 322 taught by Professor Mcgavin during the Winter '08 term at Cal Poly Pomona.

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