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2 years ago

7

A gear motor can develop 2 hp when it turns at 450rpm. If the motor turns a solid shaft with a diameter of 1 in., determine the

maximum shear stress developed in the shaft. (30 pts)

1 answer:

kramer2 years ago

7 0

Answer:

Maximum shear stress is;

τ_max = 1427.12 psi

Explanation:

We are given;

Power = 2 HP = 2 × 746 Watts = 1492 W

Angular speed;ω = 450 rev/min = 450 × 2π/60 rad/s = 47.124 rad/s

Diameter;d = 1 in

We know that; power = shear stress × angular speed

So,

P = τω

τ = P/ω

τ = 1492/47.124

τ = 31.66 N.m

Converting this to lb.in, we have;

τ = 280.2146 lb.in

Maximum shear stress is given by the formula;

τ_max = (τ•d/2)/J

J is polar moment of inertia given by the formula; J = πd⁴/32

So,

τ_max = (τ•d/2)/(πd⁴/32)

This reduces to;

τ_max = (16τ)/(πd³)

Plugging in values;

τ_max = (16 × 280.2146)/((π×1³)

τ_max = 1427.12 psi

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<u>AMMONIA </u>

state(1)

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b)

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Answer:

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Explanation:

1) Notation

$\sigma_c$ = tensile stress = 200 Mpa

$K$ = plane strain fracture toughness= 55 Mpa $\sqrt{m}$

$\lambda$ = length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

$\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}$ (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for $\lambda$

Multiplying both sides of equation (1) by $Y\sqrt{\pi\lambda}$

$\sigma_c Y\sqrt{\pi\lambda}=K$ (2)

Sequaring both sides of equation (2):

$(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2$

$\sigma^2_c Y^2 \pi\lambda=K^2$ (3)

Dividing both sides by $\sigma^2_c Y^2 \pi$ we got:

$\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2$ (4)

Replacing the values into equation (4) we got:

$\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m$

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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