To determine the required diameter of the shaft, we need to analyze the torsion using two conditions: the shear stress and the angle of twist.
Step 1: Calculate the Diameter using Shear Stress
The shear stress \(\tau\) in a solid circular shaft is given by the formula:
\[ \tau = \frac{T \cdot r}{J} \]
Where:
- \(\tau\) is the shear stress
- \(T\) is the torque (50 N·m)
- \(r\) is the radius of the shaft
- \(J\) is the polar moment of inertia for a solid circular shaft, which is given by:
\[ J = \frac{\pi d^4}{32} \]
Where \(d\) is the diameter of the shaft. The radius \(r\) is \(\frac{d}{2}\).
Rearranging the first equation for diameter \(d\):
\[ \tau = \frac{T \cdot \frac{d}{2}}{\frac{\pi d^4}{32}} \]
This simplifies to:
\[ \tau = \frac{32 T}{\pi d^3} \]
Now, substituting the maximum shear stress of 72 MPa (or \(72 \times 10^6\) Pa) and the torque (50 N·m) into the equation:
\[ 72 \times 10^6 = \frac{32 \cdot 50}{\pi d^3} \]
Now, solving for \(d^3\):
\[ d^3 = \frac{32 \cdot 50}{\pi \cdot 72 \times 10^6} \]
Calculating the right side:
\[ d^3 = \frac{1600}{\pi \cdot 72 \times 10^6} \]
\[ = \frac{1600}{226.19467 \times 10^6} \]
\[ = \frac{1600}{226194670} = 7.0726 \times 10^{-6} , m^3 \]
Taking the cube root to find \(d\):
\[ d = (7.0726 \times 10^{-6})^{1/3} \approx 0.00195 , m \approx 1.95 , cm \]
Step 2: Calculate the Diameter using Angle of Twist
The angle of twist \(\theta\) can be calculated using the formula:
\[ \theta = \frac{T \cdot L}{J \cdot G} \]
Where:
- \(L\) is the length of the shaft (700 mm = 0.7 m)
- \(G\) is the modulus of rigidity (79 GPa = \(79 \times 10^9\) Pa)
Rearranging to solve for \(J\):
\[ J = \frac{T \cdot L}{G \cdot \theta} \]
Substituting values for \(\theta\) (in radians) - since \(2°\) must be converted to radians:
\[ \theta = 2° = \frac{2 \cdot \pi}{180} \approx 0.0349 , radians \]
And substituting the values into the equation:
\[ J = \frac{50 \cdot 0.7}{79 \times 10^9 \cdot 0.0349} \]
Calculating the right side:
\[ J \approx \frac{35}{2.758 \times 10^6} \approx 1.270 \times 10^{-5} , m^4 \]
Setting this equal to the formula for \(J\):
\[ \frac{\pi d^4}{32} = 1.270 \times 10^{-5} \]
Rearranging to solve for \(d^4\):
\[ d^4 = \frac{32 \cdot 1.270 \times 10^{-5}}{\pi} \]
Calculating:
\[ d^4 \approx \frac{4.064 \times 10^{-5}}{\pi} \approx \frac{4.064 \times 10^{-5}}{3.1416} \approx 1.295 \times 10^{-5} , m^4 \]
Now taking the fourth root:
\[ d \approx (1.295 \times 10^{-5})^{1/4} \]
Calculating this gives:
\[ d \approx 0.01063 , m \approx 1.063 , cm \]
Step 3: Final Comparisons and Decisions
We have calculated two diameters:
- From shear stress: \(d \approx 1.95 , cm\)
- From angle of twist: \(d \approx 1.063 , cm\)
Since we should satisfy both conditions, the larger diameter will be used. Therefore, we need to size the shaft to actually transmit the required torque without exceeding the maximum shear stress or the angle of twist.
Thus, the required diameter of the shaft is approximately 1.95 cm.
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