01-02ChapGere.0022

01-02ChapGere.0022 - IAMETER OF PIN BASED UPON SHEAR Double...

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Solution 1.8-7 Tube hoisted by a crane SECTION 1.8 Design for Axial Loads and Direct Shear 53 T T b 1 d b 2 T 5 tensile force in cable W 5 weight of steel tube d 5 diameter of pin b 1 5 inner dimension of tube 5 8.5in. b 2 5 outer dimension of tube 5 10.0in. L 5 length of tube 5 20ft t allow 5 8,700psi s b 5 13,000psi D IAMETER OF PIN BASED UPON SHEAR Double shear. 2 t allow A pin 5 W d 2 5 0.1382in. 2 d 1 5 0.372in. D IAMETER OF PIN BASED UPON BEARING ± b ( b 2 2 b 1 ) d 5 W (13,000 psi)(10.0 in. 2 8.5 in.) d 5 1,889 lb d 2 5 0.097 in. M INIMUM DIAMETER OF PIN Shear governs. d min 5 0.372 in. 2(8,700 psi) ¢ p d 2 4 5 1889 lb 5 1,889 lb ¢ 1 144 ft 2 in. (20 ft) 5 (490 lb / ft 3 )(27.75 in. 2 ) W 5 g s AL W EIGHT OF TUBE g s 5 weight density of steel 5 490lb/ft 3 5 27.75 in. 5 b 2 2 2 b 1 2 5 (10.0 in.) 2 2 (8.5 in.) 2 A 5 area of tube Solution 1.8-8 Tube hoisted by a crane T T b 1 d b 2 T 5 tensile force in cable W 5 weight of steel tube d 5 diameter of pin b 1 5 inner dimension of tube 5 210mm b 2 5 outer dimension of tube 5 250mm L 5 length of tube 5 6.0 m A 5 area of tube W 5 g s AL 5 (77.0 kN/m 3 )(18,400mm 2 )(6.0m) 5 8.501 kN D
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Unformatted text preview: IAMETER OF PIN BASED UPON SHEAR Double shear. 2 t allow A pin 5 W d 1 5 9.497mm D IAMETER OF PIN BASED UPON BEARING b ( b 2 2 b 1 ) d 5 W (90MPa)(40mm) d 5 8.501kN d 2 5 2.361mm M INIMUM DIAMETER OF PIN Shear governs. d min 5 9.50 mm 2(60 MPa) p 4 d 2 5 8.501 kN d 2 5 90.20 mm 2 A 5 b 2 2 2 b 1 2 5 18,400 mm 2 t allow 5 60 MPa s b 5 90 MPa W EIGHT OF TUBE g s 5 weight density of steel 5 77.0 kN/m 3 Problem 1.8-8 Solve the preceding problem if the length L of the tube is 6.0 m, the outer width is b 2 5 250 mm, the inner dimension is b 1 5 210 mm, the allow-able shear stress in the pin is 60 MPa, and the allowable bearing stress is 90 MPa. A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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