Answer:
Part A.
T = 65.248 N-m
Part B.
εx' = 0.00245
εy' = - 0.00245
Explanation:
Part A.
Given
R = 15 mm = 0.015 m
εx' = - 80*10⁻⁶
εy' = 80*10⁻⁶
Est = E = 200 GPa
υst = υ = 0.3
T = ?
In order to understand the question we can see the pic shown.
Pure shear.
εx = εy = 0
We apply the formula
εx' = εx*Cos²θ + εy*Sin²θ + γxy*Sin θ*Cos θ
If
θ = 45° we have
- 80*10⁻⁶ = 0 + 0 + γxy*Sin 45*Cos 45°
⇒ γxy = - 160*10⁻⁶
Also,
θ = 135° we have
80*10⁻⁶ = 0 + 0 + γxy*Sin 135*Cos 135°
⇒ γxy = - 160*10⁻⁶
Then we get G as follows:
G = E/(2*(1 + υ))
⇒ G = 200 GPa/(2*(1 + 0.3)) = 76.923 GPa = 76.923*10⁹Pa
we can use the equation
τ = G*γxy ⇒ τ = (76.923*10⁹Pa)(160*10⁻⁶) = 12.308*10⁶Pa
Finally, we can obtain T as follows
T = τ*Jz/R
where
Jz = π*R⁴/2 ⇒ Jz = π*(0.015 m)⁴/2 = 7.95*10⁻⁸m⁴
⇒ T = 12.308*10⁶Pa*7.95*10⁻⁸m⁴/0.015 m
⇒ T = 65.248 N-m
Part B.
The shaft has a radius of 15 mm and is made of L2 tool steel. Determine the strains in the x' and y' direction if a torque T = 2 kN-m is applied to the shaft.
Given:
R = 15 mm = 0.015 m
Jz = 7.95*10⁻⁸m⁴
E = 200 GPa
υ = 0.3
G = 76.923*10⁹Pa
T = 2 kN-m = 2000 N-m
We can apply the equation
T = τ*Jz/R ⇒ τ = T*R/Jz
⇒ τ = 2000 N-m*(0.015 m)/(7.95*10⁻⁸m⁴)
⇒ τ = 3.77*10⁸ Pa
We use the formula
τ = G*γxy ⇒ γxy = τ/G
⇒ γxy = 3.77*10⁸ Pa/76.923*10⁹Pa
⇒ γxy = 0.0049
Now, we apply the equations
εx' = εx*Cos²θ + εy*Sin²θ + γxy*Sin θ*Cos θ
If
θ = 45° we have
εx' = 0 + 0 + 0.0049*Sin 45°*Cos 45°
⇒ εx' = 0.00245
If
θ = 135° we have
εy' = 0 + 0 + 0.0049*Sin 135°*Cos 135°
⇒ εy' = - 0.00245
