Community Answer

【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: \( \mathrm{{M}{\left({x}\right)}={F}\times{\left(\frac{{L}}{{2}}-{X}\right)}} \)where F is the transverse load, L is the length of the beam, and x is the distance along the beam. In this case, \( \mathrm{{L}={100}{m}{m}} \) and x is measured from the left bearing, so we have:\( \mathrm{{M}{\left({x}\right)}={26}\times{\left({50}-{X}\right)}} \)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: \( \mathrm{\sigma_{{x}}={M}\times\frac{{c}}{{I}}} \)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: \( \mathrm{{c}=\frac{{d}}{{2}}} \) \( \mathrm{{I}=\frac{{\pi\times{d}^{{4}}}}{{64}}} \)In this case,\( \mathrm{{d}={25}{m}{m}} \) , so we have:\( \mathrm{{c}={12.5}{m}{m}} \)\( \mathrm{{I}=\frac{{\pi\times{25}^{{4}}}}{{64}}={122{,}934}{m}{m}^{{4}}} \) Therefore, the normal stress on the top surface of the shaft as a function of x is:\( \mathrm{\sigma_{{x}}{\left({x}\right)}={M}{\left({x}\right)}\frac{{c}}{{I}}=\frac{{{26}{\left({50}-{x}\right)}{12.5}}}{{\frac{{\pi{25}^{{4}}}}{{64}}}}={131.7}\times{\left({50}-{x}\right)}} \) ExplanationPlease refer the solution in this step.Explanation:Please refer to solution in this step.Step2/4(b) The maximum normal stress occurs at the location where \( \mathrm{\sigma_{{x}}} \)is a maximum. To find this, we can take the derivative of \( \mathrm{\sigma_{{x}}} \)with respect to x and set it equal to zero:\( \mathrm{\frac{{{d}\sigma_{{x}}}}{{{\left.{d}{x}\right.}}}=-{131.7}={0}} \)This gives \( \mathrm{{x}={50}{m}{m}} \) as the location of the maximum stress. To find the maximum stress, we can substitute \( \mathrm{{x}={50}{m}{m}} \) into the expression for \( \mathrm{\sigma_{{x}}} \): \( \mathrm{\sigma_{{x}}{\left(\max\right)}={131.7}\times{\left({50}-{50}\right)}={0}{M}{P}{a}} \)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:\( \mathrm{\sigma_{{x}}={2}\times\sigma_{{x}}{\left(\max\right)}={0}{M}{P}{a}} \)This is below the yield stress of \( \mathrm{{A}{I}{S}{I}{1},{030}} \ ... See the full answer