01-02ChapGere.0028 - P max Substitute d m into Eq. (1) or...

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Solution 1.8-14 Bar with a pin connection SECTION 1.8 Design for Axial Loads and Direct Shear 59 b 5 60mm t 5 10mm d 5 diameter of hole and pin s T 5 140MPa t S 5 80MPa s B 5 200MPa U NITS USED IN THE FOLLOWING CALCULATIONS : P is in kN s and t are in N/mm 2 (same as MPa) b , t , and d are in mm T ENSION IN THE BAR P T 5 s T (Net area) 5 s t ( t )( b 2 d ) 5 1.40 (60 2 d ) (Eq. 1) 5 (140 MPa)(10 mm)(60 mm 2 d ) ¢ 1 1000 S HEAR IN THE PIN 5 0.040 p d 2 5 0.12566 d 2 (Eq. 2) B EARING BETWEEN PIN AND BAR P B 5 s B td 5 2.0 d (Eq. 3) G RAPH OF E QS . (1), (2), AND (3) (a) P IN DIAMETER d m P T 5 P B or 1.40(60 2 d ) 5 2.0 d Solving, (b) L OAD
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Unformatted text preview: P max Substitute d m into Eq. (1) or Eq. (3): P max 5 49.4 kN d m 5 84.0 3.4 mm 5 24.7 mm 5 (200 MPa)(10 mm)( d ) 1 1000 5 2(80 MPa) p 4 ( d 2 ) 1 1000 P S 5 2 t S A pin 5 2 t S p d 2 4 d b t P P d d (mm) 100 75 50 25 10 20 30 40 P (kN) P T T ension Shear S Bearing B P max Eq.(3) Eq.(1) Eq.(2) d m 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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