Assignment___5__2.1_A_-_H - A = P f A = 50 kips 20 ksi A =...

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CIVE 335 Due Date 9/11/08 Assignment # 5 Chapter 2 Problems 2.1 A - H NAME (Last, First) Problem 2.1 A. Find the cross-sectional area of the bar in tension P = 40 kips f = 12 ksi f = P/ A A = P / f A = 40 kips / 12 ksi A = B. Find the maximum axial load placed on a short timbe A = 9.5 in. square f = 1,100 psi f = P/ A A = 9.5 in. x 9.5 in = P = P = 99,275 lb C. Find the diameter of a bolt subjected to a force P = 9,000 lbs f = 15 ksi f = P/ A A = P / f A = = 4 A / 3.14 = = 0.764 in. D = 0.874 in D. Find the length of the side of a square footing P = 240 kips f = 8,000 psf f = P/ A A = P / f A = 240,000 lb / 8,000 lb / sq. ft. A = A = L = 5.48 ft E. Find the total shearing force across a joint D = 1.25 in f = 15 ksi f = P/ A P = P = 18.4 kips F. Find the total load of a hollow column f = 9 ksi = 5 in - 0.75 in = 4.25 in P = 196 kips G. Find the minimum x-sectional area of steel bar in tensio P = 50 kips f = 20 ksi f = P/ A
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Unformatted text preview: A = P / f A = 50 kips / 20 ksi A = H. Find the nominal size square timber used P = 115 kips f = 1,000 psi f = P/ A A = P / f A = 115,000 lbs / 1,000 psi A = o.k. 12" x 12" 3.33 in 2 P = f x A 90.25 in 2 1,100 psi x 90.25 in 2 9,000 lb / 15,000 lb/in 2 A = 0.60 in 2 A = ∏ x D 2 / 4 D 2 (4) (0.60 in 2 ) / 3.14 30 ft 2 A = L 2 30 ft 2 P = f x A 15 ksi x (3.14 x 1.25 2 / 4) D o = 10 in = 2 x R o Thickness shell = 0.75 in A od = 3.14 x D od 2 / 4 A od = 3.14 x (10 in) 2 / 4 = 78.5 in 2 A id = 3.14 x R id 2 R id = R od- 0.75 in A id = 3.14 x (4.25 in) 2 = 56.7 in 2 A act = A od- A id A act = 78.5 in 2- 56.7 in 2 = 21.8 in 2 f = P / A act P = f x A act P = 9 ksi x 21.8 in 2 2.50 in 2 115 in 2 According to table A.8, the nominal size square timber is 12 x 12 = 144 in 2 Therefore: 144 in 2 > 115 in 2...
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This note was uploaded on 03/27/2012 for the course CIVE 335 taught by Professor Desormeaux during the Spring '12 term at University of Louisiana at Lafayette.

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