Question 5. (8 points total) The rotating circular driveshaft in figure 3 is simply supported by bearing reaction forces \( R_{1} \) and \( R_{2} \). The shaft has a shoulder fillet with radius \( r=2 \mathrm{~mm} \) that reduces the diameter from \( D=15 \mathrm{~mm} \) to \( d=10 \mathrm{~mm} \). A transverse load \( F=300 \mathrm{~N} \) is applied as shown. As well as a steady torque to turn the shaft of \( T=20 \mathrm{~N} \). The shaft is made steel with material properties \( S_{u t}=424 \mathrm{MPa} \) and \( S_{y}=193 \mathrm{MPa} \). Assume that the fully corrected endurance limit is \( S_{c}=200 \mathrm{MPa} \). The For long, thin components undergoing bending, the normal stress from bending is typically much greater than the transverse shear stress from 2 bending. In this question we will assume the transverse shear stress is negligible and the normal stress from bending dominates. Let us assume that the maximum stresses occur at the shoulder fillet rather than at the point where the force \( F \) is applied. In a real design problem we would have to calculate both and compare them. (a) (4 points) Find the mean and alternating values for normal stress due to bending and the shear stress from the applied torque at the shoulder fllet. Assume that as the shaft rotates, the normal stress from bending at the surface alternates between positive and negative values of the same magnitude. Hint: If a steady (constant) torque is applied, what will the alternating component of torque be? (b) (4 points) Find the alternating and mean von Mises stress (from distortion energy theory) and use this to calculate the fatigue safety factor of the driveshaft using both the DEGoodman, DE-Morrow and DE-Gerber criteria. Figure 3: Driveshaft with a shoulder fillet.

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Transcribed Image Text: 5. (8 points total) The rotating circular driveshaft in figure 3 is simply supported by bearing reaction forces \( R_{1} \) and \( R_{2} \). The shaft has a shoulder fillet with radius \( r=2 \mathrm{~mm} \) that reduces the diameter from \( D=15 \mathrm{~mm} \) to \( d=10 \mathrm{~mm} \). A transverse load \( F=300 \mathrm{~N} \) is applied as shown. As well as a steady torque to turn the shaft of \( T=20 \mathrm{~N} \). The shaft is made steel with material properties \( S_{u t}=424 \mathrm{MPa} \) and \( S_{y}=193 \mathrm{MPa} \). Assume that the fully corrected endurance limit is \( S_{c}=200 \mathrm{MPa} \). The For long, thin components undergoing bending, the normal stress from bending is typically much greater than the transverse shear stress from 2 bending. In this question we will assume the transverse shear stress is negligible and the normal stress from bending dominates. Let us assume that the maximum stresses occur at the shoulder fillet rather than at the point where the force \( F \) is applied. In a real design problem we would have to calculate both and compare them. (a) (4 points) Find the mean and alternating values for normal stress due to bending and the shear stress from the applied torque at the shoulder fllet. Assume that as the shaft rotates, the normal stress from bending at the surface alternates between positive and negative values of the same magnitude. Hint: If a steady (constant) torque is applied, what will the alternating component of torque be? (b) (4 points) Find the alternating and mean von Mises stress (from distortion energy theory) and use this to calculate the fatigue safety factor of the driveshaft using both the DEGoodman, DE-Morrow and DE-Gerber criteria. Figure 3: Driveshaft with a shoulder fillet.

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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/4(a) To find the internal bending moment in the shaft as a function of x, we can use the formula for the bending moment in a simply supported beam with a point load at the center:  where F is the transverse load, L is the length of the beam, and x is the distance along the beam. In this case,  and x is measured from the left bearing, so we have:  To find the normal stress on the top surface of the shaft as a function of a, we can use the formula for the normal stress due to bending:  where c is the distance from the neutral axis to the top surface of the shaft, and I is the second moment of area. For a circular shaft with diameter d, we have:   In this case, , so we have:   Therefore, the normal stress on the top surface of the shaft as a function of x is: 137.7 * 50-xExplanationPlease refer to solution in this step.Explanation:Please refer to solution in this step.Step2/4b) The maximum normal stress occurs at the location where  is a maximum. To find this, we can take the derivative of  with respect to x and set it equal to zero:  This gives  as the location of the maximum stress. To find the maximum stress, we can substitute  into the expression for   However, this is the stress on the neutral axis, which is not the location of maximum stress. To find the maximum stress, we need to add the normal stress due to bending on the bottom surface of the shaft, which is equal in magnitude to the stress on the top surface but opposite in sign. Therefore, the maximum normal stress is:  This is below the yield stress of  cold-drawn steel, which is about . Therefore, the safety factor for static failure is: SF = yield stress/maximum stress  This means that th ... See the full answer