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01-02ChapGere.0028

# 01-02ChapGere.0028 - P max Substitute d m into Eq(1 or Eq(3...

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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 60 mm t 10 mm d diameter of hole and pin T 140 MPa S 80 MPa B 200 MPa U NITS USED IN THE FOLLOWING CALCULATIONS : P is in kN and are in N/mm 2 (same as MPa) b , t , and d are in mm T ENSION IN THE BAR P T T (Net area) t ( t )( b d ) 1.40 (60 d ) (Eq. 1) (140 MPa)(10 mm)(60 mm d ) ¢ 1 1000 S HEAR IN THE PIN 0.040 d 2 0.12566 d 2 (Eq. 2) B EARING BETWEEN PIN AND BAR P B B td 2.0 d (Eq. 3) G RAPH OF E QS . (1), (2), AND (3) (a) P IN DIAMETER d m P T P B or 1.40(60 d ) 2.0 d Solving,
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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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