Required Information The cantilever bar in the figure is...

Required Information The cantilever bar in the figure is made from a ductile material and is statically loaded with $F_y = 250.0$ lbf, $F_x = 300$ lbf, and $F_z = 0$ lbf. The length of the rod DC denoted by $L_1$ is given to be 13.00 in. The length of the rod AB denoted by $L_2$ is given to be 9.00 in. Determine the magnitudes for all stresses acting on the critical stress element. (Transverse shear may be neglected if you can justify this decision.) The bending stress is determined to be ____ kpsi. The shear stress is determined to be ____ kpsi. (Round the final answer to three decimal places.)

Transcript

00:01 So here in this question, we are considering a cantilever which is shown here in this figure so this is a cantilever which we are considering from here here is the joint we can say that this is a pipe another pipe from here so this is a cantilever which we are given here here we we are given the value of f of x that is equals to 300 lbf we are given the value of f of y that is equals to 250 lbf we are given the value of f of c that is equals to 0.

00:31 So this is given so here we have to analyze the stress situation from here.

00:36 So here we are considering the critical stress element from here.

00:43 So from here we can say that the critical stress element from here is given as t that is equals to f of y which is multiplied by c of t that from here is equals to 250 which is multiplied by 2 so simplifying the term we get the value of t that is equals to 3000 lbf inch.

01:03 So this is the value of t now we have to calculate the torsional shear stress acting on the cross -sectional that is represented by t of a that is equals to 16 of t divided by pi which is multiplied by d raised to the power 3 that is equals to 16 multiplied by 3000 divided by pi multiplied by 1 to its whole raised to the power 3.

01:23 So the value of t of a from here is equals to 15 .278 of ksi so this is the value of t of a from here.

01:30 Now, we have to calculate the shear bending shear which is acting on the cross -sectional so this from here is given as m that is equals to f of y which is multiplied by a of c that is equals to 250 which is multiplied by the 9 plus 2 so simplifying the term we get the value of m that is equals to 2750 lbf inch so this is the value of m from here now the bending stress is given as sigma that is equals to 32 of m 32 of m that is divided by pi multiplied by d to its whole raised to the power 3 that is equals to 32 multiplied by 2750 that is divided by pi multiplied by 1 to its whole raised to the power 3 so the value of sigma f from here is equals to 28 .011 ksi so this is the value of the sigma e now transverse shear stress is given as tau that is equals to 4 of v that is divided by 3 of a plugging into the value that is equals to 4 of v divided by 3 multiplied by pi that is divided by 4 multiplied by d to its whole squared.

02:33 Plugging into the value we get the value of tau that is equals to 424 .41 ksi...

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