ARC322 Homework 1B SOLUTION

ARC322 Homework 1B SOLUTION - you came up with the correct...

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ARC322 Homework #1B Since A = B, R 1 = R 2 = P/2 = 12,500 LB/2 = 6,250 LB W = w L = (1200 LB/ft)(28 ft) = 33,600 LB W would then be placed at the center of the span, so R 1 = R 2 = W/2 = 33,600 LB/2 = 16,800 LB R 1 = PB/L = (7750 LB)(6 ft)/21 ft = 2214.29 LB R 2 = PA/L = (7750 LB)(15 ft)/21 ft = 5535.71 LB Note: The Reactions can be Rounded to whole pounds.
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ARC322 Homework #1 Page 2 R 1 x 27 ft = (1000 LB x 21 ft) + (11,000 LB x 12 ft) R 1 = 153,000 ft-lb/27 ft R 1 = 5,666.67 LB R 2 x 27 ft = (1000 LB x 6 ft) + (11,000 LB x15 ft) R 2 = (171,000 ft-lb)/27 ft R 2 = 6,333.33 LB There is another way to solve this . .. using R 1 = Pb/L and summing … R 1P1 = (P1)(B+C)/L = (1000 LB x 21 ft)/27 ft = 777.78 LB R 1P2 = (P2)(C)/L = (11,000 LB x 12 ft)/27 ft = 4,888.89 LB R 1 = R 1P1 + R 1P2 = 777.78 LB + 4,888.89 LB = 5,666,67 LB You can solve for R 2 in the same manner. You should also sum the Vertical Forces to make sure that
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Unformatted text preview: you came up with the correct reactions … R 1 + R 2 = P1 + P2 Reactions = 5,666.67 LB + 6,333.33 LB = 12,000 LB Loads = 1,000 LB + 11,000 LB = 12,000 LB 12,000 LB = 12,000 LB therefore OK ARC322 Homework #1 Page 3 W = wL = 180 LB/ft x 20 ft = 3,600 LB R 2 – 30 ft = (1175 LB x 10 ft) + (3600 LB x 20 ft) R 2 = 83,750 ft-lb/30 ft R 2 = 2,791.67 LB R 1 – 30 ft = (1175 lb x 20 ft) + (3600 LB x 10 ft) R 1 = 59,500 LB/30 ft R 1 = 1,983.33 LB Summing the Vertical Forces: R 1 + R 2 = P + W Reactions = 1,983.33 LB + 2,791.67 LB = 4,775 LB Loads = 1,175 LB + 3,600 LB = 4,775 LB 4,775 LB = 4,775 LB therefore OK Note: When your Engineers do the check, they will sum the forces as those pushing UP = positive and those pushing DOWN = negative … this then gives a sum of the vertical forces as 0 (ZERO). You are welcome to do it this way also....
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ARC322 Homework 1B SOLUTION - you came up with the correct...

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