Q1
SOLUTIONS
Q2
𝑃 = 𝑇𝜔
where 𝑃 is the power, 𝑇 is the torque and 𝜔 is the angular velocity (in rads/s).
=> 𝑇 =
𝑃
60 × 103
=
= 95.49 𝑁. 𝑚
𝜔 (6000 × 2𝜋)/60
Taking the equation for torsion:
𝐽 𝑇
=
𝑟 𝜏
The maximum shear stress occurs when 𝑟 = 𝑅.
Applying the Tresca yield criteria:
𝜏𝑚𝑎𝑥 =
𝜎𝑦
= 100 𝑀𝑃𝑎
2
For a solid circular shaft:
𝐽=
𝜋𝑅 4
2
Therefore:
3 2𝑇
𝑅 = √ = 0.0085 𝑚
𝜋𝜏
So, mass per unit length:
𝑚 = 𝜋𝑅 2 𝜌 = 1.83 𝑘𝑔/𝑚
For a thin-walled circular shaft:
𝐽 = 2𝜋𝑅 3 𝑡
𝑅
for the case when 𝑡 = 10
𝐽=
𝜋𝑅 4
5
Therefore:
3 5𝑇
𝑅 = √ = 0.0115 𝑚
𝜋𝜏
And mass per unit length:
𝑚 = 2𝜋𝑅𝑡𝜌 =
2𝜋𝑅 2 𝜌
= 0.36 𝑘𝑔/𝑚
10
The thin-walled section is much lighter. Hollow circular shafts are much more efficient in torsion than solid
circular shafts. Just as in beams, where we wish to have material as far as possible from the neutral axis, in
shafts, we wish to keep it as far as possible away from the axis about which the torque is applied.
Q3
Cross-sectional area of cylindrical tube:
𝐷 = 0.025 𝑚, 𝑡 = 0.0005 𝑚
∴ 𝐴 = 3.85 × 10−5 𝑚2
Axial stress in the tube due to applied load:
𝜎=
𝐹
19.5 𝑘𝑁
=
= 507 𝑀𝑃𝑎
𝐴 3.85 × 10−5 𝑚2
To determine the torque required to cause yield, we need to select a value for the shear stress at yield. In this
case, we will use von Mises yield criterion to determine the shear stress at yield. For pure shear, the von
Mises yield criterion states:
𝜏=
𝜎𝑦
√3
=
507 𝑀𝑃𝑎
√3
= 292.7 𝑀𝑃𝑎
Check the radius-to-thickness ratio => thin-walled shaft assumption applies.
Torque in thin-walled walled shaft:
𝑇=
𝜏𝐽
= 𝜏2𝜋𝑟 2 𝑡 = 143.6 𝑁𝑚
𝑟
So yes, there will be just enough capacity to cause yield under torsional loading.
To determine whether the von Mises or Tresca yield criterion fits the data better, we need to determine the
von Mises and Tresca stresses for the three loading cases.
First, we get the equation for the principal stresses:
𝜎𝐼,𝐼𝐼 = (
𝜎𝑥 + 𝜎𝑦
𝜎𝑥 − 𝜎𝑦 2
𝐹
𝐹 2
𝑇 2
2
) ± √(
) + (𝜏𝑥𝑦 ) = 𝜎𝐼,𝐼𝐼 =
± √( ) + (
)
2
2
2𝐴
2𝐴
2𝜋𝑟 2 𝑡
Then we make a table to calculate the principal stresses and subsequently von Mises and Tresca stresses (all
in MPa):
F
5
10
15
T
140
120
90
𝝈𝑰
357
407
462
𝝈𝑰𝑰𝑰
-227
-147
-73
von Mises σVM
509
497
503
Tresca σT
585
554
535
where 𝜎𝑉𝑀 = √𝜎𝐼 2 − 𝜎𝐼 𝜎𝐼𝐼𝐼 + 𝜎𝐼𝐼𝐼 2 and 𝜎𝑇 = 𝜎𝐼 − 𝜎𝐼𝐼𝐼 were 𝜎𝑇 is the “Tresca Stress” that can be compared
𝜎 −𝜎
𝜎
directly with σY since the criteria requires 𝐼 𝐼𝐼𝐼 ≤ 𝑌 and the “twos” can be cancelled to give: 𝜎𝐼 − 𝜎𝐼𝐼𝐼 ≤ 𝜎𝑌
2
2
From the table above it can be seen that von Mises yield criterion is closer to the actual yield stress, so the
von Mises yield criterion is a better fit.
Q4
Axial stress in the shaft due to applied load:
𝜎=
𝐹
125
=𝜋
= 63.66 𝑀𝑃𝑎
2
𝐴
×
0.05
4
The shear stress in the shaft due to applied torque:
𝐽 𝑇
=
𝑟 𝜏
𝜏=
𝑇𝑟 2.4 × 0.025
=
= 97.78 𝑀𝑃𝑎
𝜋 × 0.0254
𝐽
2
Determine the principal stresses:
𝜎𝑥 + 𝜎𝑦
𝜎𝑥 − 𝜎𝑦 2
0 + 63.66
0 − 63.66 2
2
𝜎𝐼,𝐼𝐼 = (
) ± √(
) + (𝜏𝑥𝑦 ) = (
) ± √(
) + (97.78)2
2
2
2
2
𝜎𝐼 = 135 𝑀𝑃𝑎
𝜎𝐼𝐼 = −71 𝑀𝑃𝑎
Check von Mises yield criterion:
𝜎𝑉𝑀 = √𝜎𝐼 2 − 𝜎𝐼 𝜎𝐼𝐼 + 𝜎𝐼𝐼 2 = √1352 − (135 × −71) + 712 = 181.2 𝑀𝑃𝑎
Yield stress is greater than von Mises stress, therefore shaft will not yield.
Determine the principal strains using Hooke’s Law:
1
𝜀𝐼 = 𝐸 [𝜎𝐼 − 𝜐(𝜎𝐼𝐼 + 𝜎𝐼𝐼𝐼 )] = 7.4 × 10−4
1
𝜀𝐼𝐼 = 𝐸 [𝜎𝐼𝐼 − 𝜐(𝜎𝐼𝐼𝐼 + 𝜎𝐼 )] = −5.3 × 10−4
1
𝜀𝐼𝐼𝐼 = 𝐸 [𝜎𝐼𝐼𝐼 − 𝜐(𝜎𝐼 + 𝜎𝐼𝐼 )] = −9.1 × 10−5