313 Maximum Torque That Can be Applied to a Hollow Steel Shaft | Strength of Materials Review at MATHalino

Problem 313
Determine the maximum torque that can be applied to a hollow circular steel shaft of 100-mm outside diameter and an 80-mm inside diameter without exceeding a shearing stress of 60 MPa or a twist of 0.5 deg/m. Use G = 83 GPa.

Solution 313

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Based on maximum allowable shearing stress:

$\tau_{max} = \dfrac{16TD}{\pi(D^4 - d^4)}$

$60 = \dfrac{16T(100)}{\pi (100^4 - 80^4)}$

$T = 6\,955 486.14 \, \text{N}\cdot\text{mm}$

$T = 6\,955.5 \, \text{N}\cdot\text{m}$

Based on maximum allowable angle of twist:

$\theta = \dfrac{TL}{JG}$

$0.5^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{T(1000)}{\frac{1}{32}\pi (100^4 - 80^4)(83\,000)}$

$T = 4\,198\,282.97 \, \text{N}\cdot\text{mm}$

$T = 4\,198.28 \, \text{N}\cdot\text{m}$

Use the smaller torque, T = 4 198.28 N·m. answer

Tags:

Torque

Angle of Twist

Torsional Shearing Stress

Steel Shaft

Shearing Stress

Hollow Shaft

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Based on maximum allowable

Permalink Submitted by fatigue01 on April 25, 2021 - 3:56pm.

Based on maximum allowable angle of twist:

where did you get the value of L?
im still confuse about the value of L being 1000

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The maximum allowable twist

Permalink Submitted by Jhun Vert on April 26, 2021 - 10:13am.

The maximum allowable twist is 0.5 deg/meter, that is, 0.5 degree for every L = 1 meter length.

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if u see here 0.5 degree per

Permalink Submitted by zilam (guest) on November 13, 2022 - 11:54pm.

if u see here 0.5 degree per meter given,but we have to calculate mm unit thats why here 1000 multiplied

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how is MPa got canceled by mm

Permalink Submitted by fatigue01 on April 26, 2021 - 9:34pm.

how is MPa got canceled by mm?

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Note: 1 MPa = 1 MN/m2 = 1 N

Permalink Submitted by Jhun Vert on April 26, 2021 - 11:58pm.

Note: 1 MPa = 1 MN/m² = 1 N/mm².

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