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2 years ago

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The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smooth bearings that allow it to rot

ate freely. The tubes have an outer diameter of 30 mm and an inner diameter of 20 mm.The solid section has a diameter of 40 mm.
If the gears, fixed to its ends, are subjected to 85?N?m torques, determine the angle of twist of gear A relative to gear D.

1 answer:

timurjin [86]2 years ago

6 0

Missing Details in Question

I'll assume the missing details to be the attached file

Answer:

0.879°

Explanation:

Given

Do = External Diameter of tube = 30mm = 0.03m

D1 = Internal Diameter of tube = 20mm = 0.02m

D = Diameter of shaft = 40mm

= 0.04m

The shear modulus of A-36 steel is as follows!;

G = 75GPa

Calculating polar moment of inertia of segments AB and CD;

This is given by π(Do⁴ - D1⁴)/32

= π(0.03⁴ - 0.02⁴)/32

= 22/7 * (0.03⁴ - 0.02⁴)/32

= 6.3839E−8m⁴

Calculating polar moment of inertia of segments BC

This is given by π(D⁴)/32

= π(0.04⁴)/32

= 22/7 * (0.04⁴⁴)/32

= 2.5143E-7m⁴

Lastly, the angle of twist of gear A relative to gear D is calculated by

ϕ = ϕAB + ϕBC + ϕCD

ϕ = [TL/GJ]ab + [TL/GJ]bc + [TL/GJ]cd

ϕ = T/G( [L/J]ab + [L/J]bc + [L/J]cd

ϕ = 85/(75*10^8) (0.4/(6.3839E−8) + 0.4/(6.3839E−3) + 0.25/(2.5143E-7)

Solving the above

ϕ = 0.01534 rad -- Convert to degrees

ϕ = 0?879°

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Read 2 more answers

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(b) Change in specific entropy=0.0044 $\frac{KJ}{Kg-K}$

Explanation:

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$h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}$

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Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

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$s_f= 0.57\frac{KJ}{Kg-K} ,s_g= 8.25\frac{KJ}{Kg-K}$

So the enthalpy of steam at the exit of turbine

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$h_2=2134.57\frac{KJ}{Kg}$

$s_2=s_f+x(s_g-s_f)\frac{KJ}{Kg-K}$

$s_2=0.57+0.83(8.25-0.57)\frac{KJ}{Kg-K}$

$s_2=6.6944\frac{KJ}{Kg-K}$

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(b) Change in specific entropy

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Change in specific entropy =0.0044 $\frac{KJ}{Kg-K}$

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