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

2021
Views:

226

Pages:

1
Author:

Accinkiquicpqw

PROBLEM 1.52 Solve Prob. 1.51, assuming that the structure has been redesigned to use 12-mm-diameter pins at B and D and no other change has been made. PROBLEM 1.51 In the steel structure shown, a 6-mm-diameter pin is used at C and 10-mmdiameter 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. ΣM C = 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. F BD = σ A net = σU F .S . § 400 × 106 A net = ¨¨ 3 © · −3 −3 3 ¸¸ (6 × 10 )(18 − 12)(10 ) = 4.80 × 10 N ¹ Shear in pins at B and D. FBD = τ A pin = τU π F. S. 4 § 150 × 10 6 ·§ π · 3 −3 2 d2 = ¨ ¸¸¨ ¸ (12 × 10 ) = 5.6549 × 10 N ¨ 3 © ¹© 4 ¹ Smaller value of FBD is 4.80 × 10 3 N. From (1), § 3· P = ¨ ¸ (4.80 × 103 ) = 2.06 × 103 N ©7¹ Shear in pin at C. C = 2 τ A pin = 2 From (2), τU π F .S . 4 § 150 × 10 6 · § π · 3 −3 2 d 2 = (2) ¨ ¸¸ ¨ ¸ (6 × 10 ) = 2.8274 × 10 N ¨ 3 © ¹© 4 ¹ § 3· P = ¨ ¸ (2.8274 × 103 ) = 2.12 × 103 N © 4¹ Smaller value of P is the allowable value. P = 2.06 × 103 N P = 2.06 kN W