PS5-08-solutions - CE166 Fall 2008 Problem Set 5 Solutions...

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CE166 Fall 2008 Problem Set 5 Solutions Page 1 of 8 Professor Arpad Horvath Department of Civil and Environmental Engineering University of California, Berkeley Question 1 – Part A – Wall Form Design Design Load: P = 150 + 43400/110 + 2800(7)/110 = 723 lb/ft 2 Stud Spacing: w = 1 ft × 723 lb/ft 2 = 723 lb/ft = w Bending: in 0 . 10 723 10 604 . 0 95 . 10 3 Shear: in 71 . 14 4 3 2 723 10 477 . 0 20 3 Deflection: in 1 . 15 723 10 257 . 0 13 . 2 3 6 Bending governs: use 8" stud spacing for modularity This requires 19 studs/side × 8ft/stud × 2 sides = 304LF of 2×4 studs Double Wale Spacing: w = (8"/12") × 723 lb/ft 2 = 482 lb/ft Bending: in 7 . 32 482 063 . 3 1400 95 . 10 Shear:   in 1 . 33 5 . 3 2 482 25 . 5 180 3 . 13 Deflection: in 6 . 55 482 359 . 5 10 6 . 1 13 . 2 3 6 Bending governs: use 32" wale spacing for modularity This requires (4×2) wales/side × 12ft/wale × 2 sides= 192 LF studs + wales = 304LF + 192LF = 496LF < 500 LF 2×4 available Okay Tie Spacing: w = (32"/12") × 723 lb/ft 2 = 1,927 lb/ft Bending: in 1 . 23 1927 063 . 3 2 1400 95 . 10 Shear:   in 0 . 20 5 . 3 2 1927 25 . 5 2 180 3 . 13 Deflection: in 1 . 44 1927 359 . 5 2 10 6 . 1 13 . 2 3 6 Shear governs: use 18" tie spacing for modularity
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CE166 Fall 2008 Problem Set 5 Solutions Page 2 of 8 Professor Arpad Horvath Department of Civil and Environmental Engineering University of California, Berkeley Check Tie Load: tie lb tie lb ft lb P / 000 , 2 / 891 , 2 723 " 12 " 18 " 12 " 32 2 so try 12” tie spacing: tie lb tie lb ft lb ft P / 000 , 2 / 927 , 1 723 1 " 12 " 32 2 Okay Tie load governs, use 12” tie spacing for modularity Check Bearing (Wales on Studs): A = 2 × 1.5" × 1.5" = 4.5 in 2 lb ft lb P 1285 482 " 12 " 32 2 psi
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This note was uploaded on 02/27/2011 for the course CE 166 taught by Professor Horvath during the Spring '08 term at University of California, Berkeley.

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PS5-08-solutions - CE166 Fall 2008 Problem Set 5 Solutions...

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