Please answer with explanation.Question : These two shear pins have the same length and same cross-sectional area. Both are subject to the same external tension load. Which pin has the lower shear stress? O a. The solid pin on the left. O b. The hollow pin on the right. O c. Both pins have the same shear stress. These two shear pins have the same length and same cross-sectional area. Both are subject to the same external tension load. Which pin has the lower shearstress?O a. The solid pin on the left.O b. The hollow pin on the right.O c. Both pins have the same shear stress.

Answered: same external tension load. Which pin…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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19. The Knuckle Joint is made from SAE 1020 Steel having an allowable tensile strength of 15,000 psi and an allowable shear strength of 7500 psi. The axial load is 1 ton (recall, 1 ton = 2205 lb). Yoke Pin Rod -A The dimensions are as follows: D-0.652 in d-0.652 in c = 0.66 in b = 0.33 in m = 1.306 in e = 0.505 in Determine: a) Tensile stress (psi) at section A-A b) Factor of safety for a) c) Shearing stress (psi) in the pin d) Factor of safety for b) e) Bearing stress (psi) between the pin and the rod f) Factor of safety for e) g) Bearing stress (psi) between the pin and the yoke h) Factor of safety for g) i) Tensile stress (psi) across the hole in the rod j) Factor of safety for i) k) Tensile stress (psi) across the hole in the yoke 1) Factor of safety for k)
Problem 2 You will be determining the value for the external load P that will cause the diameter of BC to change to 5.8 mm. The original diameter of BC is 6 mm. Use the values given below unless otherwise stated by the questions. E = 200 GPa V = 0.38 300 mm 400 mm B 1 -200 mm-
Problem 17 A joint shown has the basic layout shown in the diagram below. (All dimensions on the diagram are in inches.) A single pin of diameter 0.5 inches (A = 0.2 in2) holds together the yoke and bar as shown. The material has an ultimate strength in tension of 80 ksi, and an ultimate strength in shear of 45 ksi. The load P has a value of 10000 lb. By examining the shear stress in the pin, the normal stress in the yoke, and the normal stress in the bar, find the overall factor of safety for the joint. P=10,000 lb YOKE -0.5 PIN BAR P=10,000 lb 1.5- 20 1.5 0.5 0.5

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