01-02ChapGere.0014 - P allow if the bar is made of brass...

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SECTION 1.7 Allowable Stresses and Allowable Loads 45 Solution 1.7-11 Splice between two flat bars U LTIMATE LOAD BASED UPON TENSION IN THE BARS Cross-sectional area of bars: A 5 bt b 5 1.0in. t 5 0.4in. A 5 0.40in. 2 P 1 5 s ult A 5 (60ksi)(0.40in. 2 ) 5 24.0k U LTIMATE LOAD BASED UPON SHEAR IN THE RIVETS Double shear d 5 diameter of rivets d 5 5 / 8 in. A R 5 area of rivets A R 5 p d 2 4 5 p 4 ¢ 5 8 in. 2 5 0.3068 in. 2 U LTIMATE LOAD BASED UPON BEARING A b 5 bearing area 5 dt U LTIMATE LOAD Shear governs. P ult 5 15.34k A LLOWABLE LOAD P allow 5 P ult n 5 15.34 k 2.5 5 6.14 k P 3 5 s b A b 5 (80 ksi) ¢ 5 8 in. (0.4 in.) 5 20.0 k 5 15.34 k P 2 5 t ult (2 A R ) 5 2(25 ksi)(0.3068 in. 2 ) P P t Problem 1.7-12 A solid bar of circular cross section (diameter d ) has a hole of diameter d ±4 drilled laterally through the center of the bar (see figure). The allowable average tensile stress on the net cross section of the bar is s allow . (a) Obtain a formula for the allowable load P allow that the bar can carry in tension. (b) Calculate the value of
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Unformatted text preview: P allow if the bar is made of brass with diameter d 5 40 mm and s allow 5 80 MPa. ( Hint: Use the formulas of Case 15, Appendix D.) d d P P d 4 d 4 Solution 1.7-12 Bar with a hole C ROSS SECTION OF BAR (a) A LLOWABLE LOAD IN TENSION P allow 5 s allow A 5 0.5380 d 2 s allow (b) S UBSTITUTE NUMERICAL VALUES s allow 5 80MPa d 5 40mm P allow 5 68.9kN 5 d 2 2 arc cos 1 4 2 15 16 5 0.5380 d 2 A 5 2 d 2 2 B arc cos 1 4 2 d 8 d 8 15 ( d / 2) 2 R d d 4 From Case 15, Appendix D: b 5 B r 2 2 d 8 2 5 d B 15 64 5 d 8 15 r 5 d 2 a 5 d 8 A 5 2 r 2 a 2 ab r 2 5 arc cos 1 4 a 5 arc cos d / 8 r A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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