University:
Michigan State University
Course:
ERG 100 | Introduction to Engineering Design
Academic year:

2021
Views:

416

Pages:

1
Author:

Nesbyenvkk

PROBLEM 1.51 In the steel structure shown, a 6-mm-diameter pin is used at C and 10-mm-diameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3.0 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes. SOLUTION Use free body ABC. ΣMC = 0 : 0.280 P − 0.120 FBD = 0 P= 3 FBD 7 (1) ΣMB = 0 : 0.160 P − 0.120 C = 0 P= 3 C 4 (2) Tension on net section of link BD. § 400 × 106 · −3 −3 3 A net = ¨ ¸¸ (6 × 10 )(18 − 10)(10 ) = 6.40 × 10 N ¨ F. S. 3 © ¹ Shear in pins at B and D. τ π 2 § 150 × 106 · § π · −3 2 3 FBD = τ Apin = U d =¨ ¸¸ ¨ ¸ (10 × 10 ) = 3.9270 × 10 N ¨ F. S. 4 3 © ¹© 4 ¹ FBD = σ A net = σU Smaller value of FBD is 3.9270 × 10 3 N. From (1) § 3· P = ¨ ¸ (3.9270 × 103 ) = 1.683 × 103 N © 7¹ Shear in pin at C. C = 2τ Apin = 2 From (2) §3· P = ¨ ¸ (2.8274 × 10 3) = 2.12 × 10 3 N ©4¹ Smaller value of P is allowable value. §150 × 106 d 2 = (2) ¨ ¨ F. S. 4 3 © τU π P = 1.683 × 103 N ·§ π · −3 2 3 ¸¸¨ ¸ (6 × 10 ) = 2.8274 × 10 N ¹© 4 ¹ P = 1.683 kN W