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

Problem 3. The torsion member has an elliptical cross section shown in the figure withmajor and minor dimensions of 80 mm and 40 mm respectively. The yield stress of thetorsion member is 350 MPa.a. Neglecting stress concentrations, mark the location(s) of maximum shear stresson the cross section shown in Fig. P2 with an X.b. Calculate the maximum torque that can be applied to the torsion member basedon a factor of safety of SF = 2.5 using the maximum shear stress criterion of failure.c. The company that makes this torsion member decides to start making a circularcross section member that utilizes the same amount of material per length as theoriginal elliptical cross section design. Assuming the same safety factor SF = 2.5using the maximum shear stress criterion of failure, what is the maximum torquethat can be applied to the circular cross section torsion member (hint: Circle A=πr2,Ellipse A=πbh).d. Calculate the percentage increase in the maximum torque resulting from the crosssection change.Elliptical Cross Section Member

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

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 \( 0.134 \, \text{radians} \).

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.

A shaft has an outer diameter of 137 mm and an inner diameter of 80 mm. The shaft is subjected to three different torques as shown in the figure.  Determine the torsional shear stress at the inner surface of the shaft in MPa.