Question

A steel tube having an outer diameter of 2.5 in. is used to transmit 9 hp...

A steel tube having an outer diameter of 2.5 in. is used to transmit 9 hp when turning at 27 rev/min. 1 hp = 550 ft?lb/s. Determine the inner diameter d of the tube if the allowable shear stress is ?allow = 10ksi . it is a hollow tube.

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Answer #1
Concepts and reason

Angular velocity:

The rate at which the angular position of a turning object changes with respect to time is referred as angular velocity. It is denoted by and its unit israd/s
.

Power:

Power is defined as the rate of doing work per unit time. It is denoted by Р
and its unit is Watts.

Torque:

Torque is referred as the turning force which makes the object to turn about an axis. It is denoted by T
and its unit isN m
.

Allowable shear stress:

Allowable shear stress is the ratio between the yield stress and the safety factor. It is also referred as the quantity of shear stress that the material can withstand without failure.

Fundamentals

The formula for calculating the angular velocity(a)
is as follows:

2πΝ
ω=
60

Here, rotating speed is N
in rev/min
.

The formula for calculating the power is as follows:

P Ta

Here, power isР
and torque is T
.

The polar moment of inertia is expressed as follows:

ла
J
2

Here, diameter of the cylinder is .

Calculate the angular velocity.

2πΝ
ω=
60

Here, turning speed of the steel tube isN
.

Substitute 27 rev/min
forN
.

27 (27 rev/min)
60
=2.82 rad/s

Calculate the torque (т)
produced by the steel tube.

P Ta

Substitute 9 hp
forР
and 2.82 rad/s
for.

9 hp T(2.82 rad/s)
9 hpx 550 t-lb/s
hp
=T(2.82 rad/s)
4950 ft lb/s
T=
2.82 rad/s
T 1755.31 lb ft

Calculate the polar moment of inertia.

Т
т
J

Here, allowable shear stress is, outer radius of the tube is and polar moment of inertia isJ
.

Substitute10 ksi
for, 1755.31 lb ft
forT
and2.5 in
2
for.

1755.31 lb ft
10 ksi
2.5 in
J
2
in
ft
:
1755.31 lb ftx12
10x10 psi7
2.5 in
J
2
J=2.633 in.4

Calculate the inner diameter of the tube using the formula of the polar moment of inertia.

d
J =
2
d
2
2

Here, outer diameter of the tube is and inner diameter of the tube isd
.

Substitute 2.633 in.4
forJ
and 2.5 in
for.

2.5 in
d
2.633 in.4
2
2
2
d 1.87 in

Ans:

The inner diameter of the tube (d)
is1.87 in
.

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