A nylon [E = 360 ksi; ν = 0.4] rod (1) having a diameter of d1 = 2.50 in. is placed inside a steel [E = 29,000 ksi; ν = 0.29] tube (2) as shown in Figure P3.7. The inside diameter of the steel tube is d2 = 2.52 in. An external load P is applied to the nylon rod, compressing it. At what value of P

Question

A nylon [E = 360 ksi; ν = 0.4] rod (1) having a diameter of [latex]d_{1}[/latex] = 2.50 in. is placed inside a steel [E = 29,000 ksi; ν = 0.29] tube (2) as shown in Figure P3.7. The inside diameter of the steel tube is [latex]d_{2}[/latex] = 2.52 in. An external load P is applied to the nylon rod, compressing it. At what value of P will the space between the nylon rod and the steel tube be closed?

Accepted Answer

To close the space between the nylon rod and the steel tube, the lateral strain of the nylon rod must be:

[latex]\varepsilon_{ lat }=\frac{2.52 in .-2.50 in .}{2.50 in .}=0.0080 in . / in .[/latex]

From the given Poisson’s ratio, the longitudinal strain in the nylon rod must be

[latex]\varepsilon_{ ext {long }}=-\frac{\varepsilon_{ ext {lat }}}{v}=-\frac{0.0080 ext { in./in. }}{0.4}=-0.0200 ext { in./in. }[/latex]

Use Hooke’s Law to calculate the corresponding normal stress:

[latex]\sigma=E \varepsilon_{ ext {long }}=(360 ksi )(-0.0200 ext { in./in. })=-7.2000 ksi[/latex]

The cross-sectional area of the nylon rod is

[latex]A=\frac{\pi}{4}(2.50 in .)^{2}=4.9087 in .^{2}[/latex]

Therefore, the force required to make the nylon rod touch the inner wall of the tube is

[latex]P=\sigma A=(-7.2000 ksi )\left(4.9087 in .^{2} ight)=-35.3429 kips =35.3 kips ( C )[/latex]

Question 3.7: A nylon [E = 360 ksi; ν = 0.4] rod (1) having a diameter of ...

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