Question

An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. Knowing that the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The largest offset that can be used is
`h = 0.455 \ in`

Step-by-step explanation:

From the question we are told that

The diameter of the metal tube is
`d_m = 0.75 \ in`

The thickness of the wall is
`D = 0.08 \ in`

Generally the inner diameter is mathematically evaluated as

`d_i = d_m -2D`

`= 0.75 - 2(0.08)`

`= 0.59 \ in`

Generally the tube's cross-sectional area can be evaluated as

`a = (\pi)/(4) (d_m^2 - d_i^2)`

`= (\pi)/(4) (0.75^2 - 0.59^2)`

`= 0.1684 \ in^2`

Generally the maximum stress of the metal is mathematically evaluated as

`\sigma = (P)/(A)`

`\sigma = (P)/( 0.1684)`

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

`M = P * h`

Where h is the offset

The maximum stress becomes

`\sigma_n = (P)/(A) + (M r_m )/(I)`

Where
`r_m` is the radius of the outer diameter which is evaluated as

`r_m = (0.75)/(2)`

`r_m = 0.375 \ in`

and I is the moment of inertia which is evaluated as

`I = (\pi)/(64) (d_m^4 - d_i^4 )`

`= (\pi)/(64)(0.75^4 - 0.59^4)`

`= 0.009583 \ in^4`

So the maximum stress becomes

`\sigma' = (P)/(0.1684) + (Phr)/(0.009583)`

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

`\sigma ' = 4 \sigma`

=>
`(P)/(0.1684) + (Phr_m )/(0.009583) = 4 [(P)/(0.1684) ]`

The P would cancel out

`(1)/(0.1684) + (h(0.375))/(0.009583) = (4)/(0.1684)`

`5.94 + 39.13h = 23.753`

`39.13h = 17. 813`

`h = 0.455 \ in`