Question

A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in.2 The allowable stress for the hoops is σallos=11.4 ksi. Determine the maximum spacing s along the pipe so that the pipe can resist an internal gauge pressure of 4 psi.

Answer 1

Answer:

31.67 in

Explanation:

Given:

Diameter of the pipe, D = 3ft = 36 in

cross-sectional area of the steel = 0.2 in²

Note: Refer to the figure attached

From the free body diagram represented in the figure, we have

ΣFx = 0

or

pressure × projected area = 2 × Force in steel

Now, the projected area = spacing (s) × diameter of the wood pipe

force in steel = stress in steel (σ) × cross-sectional area of the steel

on substituting the values we get

4 ksi × (s × 36 in) = 2 × σ × 0.2 in²

also, allowable hoop stress = 11.4 ksi

thus,

σ = 11.4 ksi = 11.4 × 10³ psi

therefore, we have

4 psi × (s × 36 in) = 2 × 11.4 × 10³ psi × 0.2 in²

thus,

s = 31.67 in

hence the maximum spacing is 31.67 in

Answer 2

The maximum spacing s along the pipe so that the pipe can resist an internal gauge pressure of 4 psi 31.7 inches

What will be the maximum spacing s along the pipe so that the pipe can resist an internal gauge pressure of 4 psi?

It is given:

Diameter of the pipe, D = 3ft = 36 in

cross-sectional area of the steel = 0.2 in²

By balancing the forces

$\sum F_x=0$

$\rm Presuure \times Projected \ area=2\times Force\ in \ steel$

Now, the

$\rm projected \ area =spacing\times dia \ of\ the \ pipe$  

$\rm Force \ in \ steel = stress\ in \ steel \times cross \ section \ steel$

$4\times (s\times36)=2\times\sigma \times0.2$

thus,

$\sigma =11.4\times 10 ^3\ psi$

therefore, we have

$4\times (s\times36)=2\times10^3\times0.2$

Thus

$s=31.67\ inches$

Thus the maximum spacing s along the pipe so that the pipe can resist an internal gauge pressure of 4 psi 31.7 inches

To know more about the balancing of forces follow

https://brainly.com/question/517289