Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The largest offset that can be used is [tex]h = 0.455 \ in[/tex]
Explanation:
From the question we are told that
The diameter of the metal tube is [tex]d_m = 0.75 \ in[/tex]
The thickness of the wall is [tex]D = 0.08 \ in[/tex]
Generally the inner diameter is mathematically evaluated as
[tex]d_i = d_m -2D[/tex]
[tex]= 0.75 - 2(0.08)[/tex]
[tex]= 0.59 \ in[/tex]
Generally the tube's cross-sectional area can be evaluated as
[tex]a = \frac{\pi}{4} (d_m^2 - d_i^2)[/tex]
[tex]= \frac{\pi}{4} (0.75^2 - 0.59^2)[/tex]
[tex]= 0.1684 \ in^2[/tex]
Generally the maximum stress of the metal is mathematically evaluated as
[tex]\sigma = \frac{P}{A}[/tex]
[tex]\sigma = \frac{P}{ 0.1684}[/tex]
The diagram showing when the stress is been applied is shown on the second uploaded image
Since the internal forces in the cross section are the same with the force P and the bending couple M then
[tex]M = P * h[/tex]
Where h is the offset
The maximum stress becomes
[tex]\sigma_n = \frac{P}{A} + \frac{M r_m }{I}[/tex]
Where [tex]r_m[/tex] is the radius of the outer diameter which is evaluated as
[tex]r_m = \frac{0.75}{2}[/tex]
[tex]r_m = 0.375 \ in[/tex]
and I is the moment of inertia which is evaluated as
[tex]I = \frac{\pi}{64} (d_m^4 - d_i^4 )[/tex]
[tex]= \frac{\pi}{64}(0.75^4 - 0.59^4)[/tex]
[tex]= 0.009583 \ in^4[/tex]
So the maximum stress becomes
[tex]\sigma' = \frac{P}{0.1684} + \frac{Phr}{0.009583}[/tex]
Now the question made us to understand that the maximum stress when the offset was introduced must not exceed the 4 times the original stress
So
[tex]\sigma ' = 4 \sigma[/tex]
=> [tex]\frac{P}{0.1684} + \frac{Phr_m }{0.009583} = 4 [\frac{P}{0.1684} ][/tex]
The P would cancel out
[tex]\frac{1}{0.1684} + \frac{h(0.375)}{0.009583} = \frac{4}{0.1684}[/tex]
[tex]5.94 + 39.13h = 23.753[/tex]
[tex]39.13h = 17. 813[/tex]
[tex]h = 0.455 \ in[/tex]
