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Lady_Fox [76]
3 years ago
13

A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r

pm and transmits 40 hp through a gear. The hub upon which the gear is seated is 1.5 in wide. You have in stock 1137 OQT 1300 carbon steel that you are considering using as a key to seat the gear hub. What dimensions will the key need to be? Is this material appropriate for the application? If not, select a more appropriate material.

Engineering
2 answers:
MrMuchimi3 years ago
8 0

Answer:

The key dimensions; (4.54mm× 6.75mm×4.5mm)

Explanation

Check attachment

velikii [3]3 years ago
4 0
Answer:










Explanation:



4140-40 I’d pick wood




I hope this helps! :)
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Which tool is used to pull the tapered shaft on a tie-rod end from its mating steering component?
kaheart [24]

Answer:

A Pitman arm puller

Explanation:

this heavy-duty puller is made specially for removing the pressed-on pitman arm from the sector shaft. Tie-rod end puller: This tool is used to pull the tapered shaft on a tie-rod end from its mating steering component.

5 0
3 years ago
In order to be a Mechanical Engineer, you need to:
djyliett [7]

Answer:

3

Explanation:

it is compulsory to have a bachelor's degree

4 0
3 years ago
Read 2 more answers
An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is
devlian [24]

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= 2\pi\sqrt{\frac{L}{g}}

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= 2\pi\sqrt{\frac{L}{9.81}}

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = \frac{\textup{AE}}{\textup{L}}

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = \frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = \sqrt{\frac{K}{m}}

or

mass, m = \sqrt{\frac{K}{\omega_n^2}}

or

mass, m = \sqrt{\frac{20.66\times10^8}{0.95^2}}

mass, m = 22.90 × 10⁸ kg

4 0
3 years ago
Jack has been concerned about the rapidly changing green regulations in his state and his ability as a mechanical engineer to ke
uysha [10]

Answer:

Option A, B and D

Explanation:

Jack can easily convince boss if he focus around two major aspects of the company

a) Revenue enhancement - Jack must outline the benefits of his research that can be used to improvise customer offerings and  hence can be further used to devise more energy-efficient options to customer

b) Reduction in mistakes - Issues such as poor implementation can be avoided with better approach and understanding.

Hence, option A, B and D are correct

3 0
3 years ago
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
kramer

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

7 0
3 years ago
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