Problem 17A joint shown has the basic layout shown in the diagram below. (All dimensions on the diagramare in inches.) A single pin of diameter 0.5 inches (A = 0.2 in2) holds together the yoke and baras shown. The material has an ultimate strength in tension of 80 ksi, and an ultimate strength inshear of 45 ksi. The load P has a value of 10000 lb. By examining the shear stress in the pin, thenormal stress in the yoke, and the normal stress in the bar, find the overall factor of safety forthe joint.P=10,000 lbYOKE-0.5PINBARP=10,000 lb1.5-201.50.50.5

Answered: A joint shown has the basic layout…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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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

Expert Solution

Step 1

The point at which a material begins to break is a safety factor. The material is not yet in danger of breaking, but it sags to the point where cracks may be seen. It is also referred to as the beginning of a crack's propagation. The point at which a material completely fractures under stress is known as the ultimate stress. This is the maximum amount of stress that a material can tolerate under typical circumstances.

Given data:

$\begin{array}{rcl} the diameter of the pin {(d)}_{p} & = & 0.5 in \\ the area of the pin {(A)}_{p} & = & 0.2 {in}^{2} \\ the ultimate strength of the material {(S)}_{ut} & = & 80 ksi \\ the ultimate shear strength (ζ) & = & 45 ksi \\ the applied load (P) & = & 10000 lb \end{array}$

Find out:

The overall factor of safety (F).

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