A circular solid shaft of span L = 5 m is fixed at one end and free at the other end. A twisting moment T = 100 kN - m is applied at the free end. The torsional rigidity Gj is 5000 kNm2 /red. Fllowing statements are made for this shaft.I. The maximum rotation is 0.01 radII. The torsional strain energy is 1 kN-m With reference to the above statements,which of the following applies?

Haw the torsional stress can be calculated?Question 4Select one:a.By twisting moment and modulus of elasticityb.By torque and polar moment of resistancec.By torque and area of cross-section

The bending effect of force is Question 2Select one:a.Torsionb.Strainc.Torqued.Moment

For a shaft subject to torsional stress, which statement is correct?Group of answer choicesAt the center of the shaft, torsional stress is maximum.Torsional stress does not depend on the applied torque.Torsional stress gradually decreases from the top surface towards the center of the shaft.Torsional stress does not depend on the radius of the shaft.

The maximum torsional stresses will be created:Question 3Select one:a.On the external surface of the shaftb.In the centroid of cross-sectionc.At the supports

To solve for the maximum torsional stress and the angle of twist, we need to use the following formulas: 1. **Maximum Torsional Stress:** \[ \tau_{\text{max}} = \frac{T \cdot c}{J} \] where: - \( T \) is the applied torque. - \( c \) is the outer radius of the section. - \( J \) is the polar moment of inertia. 2. **Angle of Twist:** \[ \theta = \frac{T \cdot L}{G \cdot J} \] where: - \( T \) is the applied torque. - \( L \) is the length of the section. - \( G \) is the shear modulus. - \( J \) is the polar moment of inertia. ### Step 1: Calculate the Polar Moment of Inertia (J) For a thin-walled closed section, the polar moment of inertia \( J \) can be approximated by: \[ J \approx 2 \cdot A_m \cdot t^2\] where: - \( A_m \) is the mean area enclosed by the midline of the section. - \( t \) is the thickness of the wall. The mean area \( A_m \) can be calculated as: \[ A_m = \text{Perimeter} \times \text{Mean thickness} \] The perimeter of the midline is: \[ \text{Perimeter} = 2 \times (45 - 12.5) + \pi \times (15 - 12.5) \] \[ \text{Perimeter} = 2 \times 32.5 + \pi \times 2.5\] \[ \text{Perimeter} = 65 + 7.85 \approx 72.85 \, \text{mm} \] The mean thickness is: \[ t = 12.5 \, \text{mm} \] So, \[ A_m = 72.85 \times 12.5 \approx 910.625 \, \text{mm}^2\] Now, the polar moment of inertia \( J \) is: \[ J \approx 2 \times 910.625 \times (12.5)^2\] \[ J \approx 2 \times 910.625 \times 156.25\] \[ J \approx 284570.3125 \, \text{mm}^4\] ### Step 2: Calculate the Maximum Torsional StressThe outer radius \( c \) is: \[ c = 15 \, \text{mm} \] The applied torque \( T \) is: \[ T = 500 \, \text{Nm} = 500 \times 10^3 \, \text{Nmm} \] So, the maximum torsional stress \( \tau_{\text{max}} \) is: \[ \tau_{\text{max}} = \frac{500 \times 10^3 \times 15}{284570.3125} \] \[ \tau_{\text{max}} \approx 26.37 \, \text{N/mm}^2\] ### Step 3: Calculate the Angle of TwistThe length \( L \) is: \[ L = 2 \, \text{m} = 2000 \, \text{mm} \] The shear modulus \( G \) is: \[ G = 26 \, \text{GPa} = 26 \times 10^3 \, \text{N/mm}^2\] So, the angle of twist \( \theta \) is: \[ \theta = \frac{500 \times 10^3 \times 2000}{26 \times 10^3 \times 284570.3125} \] \[ \theta \approx 0.134 \, \text{radians} \] ### Summary: 1. The maximum torsional stress developed is approximately \( 26.37 \, \text{N/mm}^2 \). 2. The angle of twist for a 2 m length is approximately \( 0.134 \, \text{radians} \).