5-47 A round shaft supports a transverse load of F = 15 000...

5-47 A round shaft supports a transverse load of $F = 15\ 000$ lbf and carries a torque of $T = 7$ kip \cdot in, as shown in the figure. The shaft does not rotate. The shaft is machined from AISI 4140 steel, quenched and tempered at 400°F. Document the location of the critical stress element and how you determined that element to be critical. Do not assume transverse shear stress is negligible without proving it. Then, (a) determine the minimum factor of safety based on yielding according to the maximum- shear-stress theory. (b) determine the minimum factor of safety based on yielding according to the distortion-energy theory.

Transcript

00:03 Here it is given that here it is given that it is given that the force that is equal to 15 ,000 lbf and the t is equal to 7 kip depit dot in from the sap that is machined at massin at aisi 4140 steel so from the standard table from the standard table for the steel properties still properties properties the yield strength of the aisi 4140 steel quansed u n c h d quansed and temper temper at 400 degree fahrenheit will be equal to my way that will be equal to the sigma y will be equal to 219 .7 k s i so on calculating on calculating calculating calculating calculating the support reaction the support reaction for why for the the way direction we'll get the summation of f y is equal to zero so here you can write you can write r of a plus r of d is equal to 15 ,000 lbf and bf so now the summation of n a will be equal to 0 so here we can write r d into 3 is equal to 15 ,000 into 2 so the so the value of r d will be equal to 10 ,000 lbf so here you can calculate the value of r a which will be equal to 5 ,000 lbf so for the share force you can calculate for the shear force for the shear force you can calculate the vd is equal to minus rd which will be equal to minus of 10 ,000 lbf and bf now the vc will be equal to minus of rd plus f which will be equal to 5 ,000 lbf now the bd will be equal to minus of rd plus f which will be also equal to 5 ,000 lbf and the value of b a will be equal to 5 ,000 lbf so now for the bending moment you can write for the bending moment for the bending moment we can write the md is equal to 0 the mc is equal to r d into 1 r d into 1 which will be equal to 10 kip dot in now the mb will be equal to r d into 2 minus f into 1 so the value will be 5 kip dot in so now the value of m a will be equal to 0 the torsural shear stress will be so here the torsural shear stress will be so here the torsural shear stress shear stress will be the tau b will be equal to 16 t upon pi of d b to the power cube so here on putting the value will get 16 into 7 upon pi into 1 .1 to the power key so on calculating you will get this value it will be 26 .784 k so now for the tau c we can calculate 16 t upon pi dc to the power cube.

05:27 So on putting the value we will get 16 into 7 upon pi into 1 .3 to the power cube.

05:36 So the value will be 16 .22 k s i...

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