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Alchen [17]
3 years ago
12

A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the

tube if the stress is 120 MPa?
Engineering
2 answers:
Olegator [25]3 years ago
7 0

Answer:

119.35 mm

Explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe = \frac{\pi}{4}(D^2-d^2)

or

Area of hollow pipe = \frac{\pi}{4}(D^2-100^2)

thus,

400 × 10³ N = 120 × \frac{\pi}{4}(D^2-100^2)

or

D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]

or

D = 119.35 mm

IgorC [24]3 years ago
3 0

Answer:

D =119.35 mm

Explanation:

given data:

inside diameter = 100 mm

load = 400 kN

stress = 120MPa

we know that load is given as

P = \sigma A  

where:

P=400kN = 400000N

\sigma = 120MPa

A =(\frac{1}{4} \pi D^2 - \frac{1}{4}\pi (100^2)

A=\frac{1}{4} \pi (D^2 - 10000)

putting all value in the above equation to get the required diameter value

400 =  120*\frac{1}{4} \pi (D^2 - 10000)

solving for

D =119.35 mm

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