Question

A countershaft carrying two V-belt pulleys is shown in figure 1 (dimensions in inches). Pulley A receives power from a motor through a belt with the belt tensions shown. The power is transmitted through the shaft and delivered to the belt on pulley B. Assume the belt tension on the loose side at B is
15 percent of the tension on the tight side.
a) At the point of maximum bending moment, determine the principal stresses and the maximum shear stress.
b) Determine the minimum yield strength that should be considered to obtain a minimum factor of safety of 3 based on the distortion-energy theory.

Answer 1

The torque acting on the shaft, \(T=(300-50)\left(\frac{8}{2}\right)\)

\(T=250 \times 4\)

\(T=1000 \mathrm{lbf}\)

Given that at the pulley B tension on loose side is \(15 \%\) of the tension on the tight side \(T_{2}=0.15 T_{1}\)

We also know that \(T=\left(T_{1}-T_{2}\right)\left(\frac{6}{2}\right)\)

$$ 1000=\left(T_{1}-0.15 T_{1}\right)(3) $$

\(0.85 T_{1}=\frac{1000}{3}\)

\(T_{1}=392.16 \mathrm{lbf}\)

Therefore, the tension on the right side is \(T_{1}=392.16\) lbf

We have, \(T_{2}=0.15 T_{1}\)

\(T_{2}=0.15(392.16)\)

\(T_{2}=58.82\) tbf

Therefore, the tension on the loose side is \(T_{2}=58.82\) lbf

The free-body diagram of the shaft in the \(x\) -y-axis is shown below.

Taking moments about O \(\sum M_{O}=0\)

\(\left(R_{C}\right)_{z}(22)-(392.16+58.82)(16)=0\)

\(\left(R_{C}\right)_{z}=\frac{7215.68}{22}\)

\(\left(R_{C}\right)_{z}=327.98 \mathrm{lbf}\)

\(\sum F_{z}=0\)

\(-\left(R_{0}\right)_{z}-\left(R_{C}\right)_{z}+392.16+58.82=0\)

\(\left.R_{0}\right)_{z}+327.98-392.16-58.82=0\)

\(\left(R_{O}\right)_{z}=123 \mathrm{lbf}\)

Along \(x-y\) plane:

Shear force cal culations:

Shear force at \(A, V_{A}=350-222.73=127.27\) lbf Shear force at B, \(V_{B}=127.27\) lbf Shear force at \(\mathrm{C}, V_{c}=127.27-127.27=0\)

ding moment cal culations:

ending moment at \(O, M_{0}=0\) Sending moment at A, \(M_{A}=-(222.73)(8)=-1781.84\) lbf-in Bending moment at \(\mathrm{B}\), \(M_{B}=(127.27 \times 8)-1781.84=-763.68\) lbf-in

Bending moment at C, \(M_{C}=(127.27 \times 6)-763.68=0\)

SFD and BM diagram along

Along \(x-z\) plane:

Shear force cal culations:

Shear force at \(\mathrm{O}, V_{O}=123 \mathrm{lbf}\)

Shear force at \(\mathrm{A}, V_{A}=123 \mathrm{lbf}\)

Shear force at \(\mathrm{B}, V_{B}=123-392.16-58.82=-327.98 \mathrm{lbf}\)

Shear force at \(\mathrm{C}, V_{C}=-327.98+327.98=0\)

Bending moment cal culations:

Bending moment at \(\mathrm{O}, M_{Q}=0\) Bending moment at \(A, M_{A}=(123)(8)=9841 \mathrm{bf}-\mathrm{in}\)

Bending moment at \(\mathrm{B}, M_{B}=(123 \times 8)+984=19681 \mathrm{bf}-\mathrm{in}\)

Bending moment at \(\mathrm{C}, M_{C}=-(327.98 \times 6)+1968=0\)

SFD and BM diagram al ong \(X-Z\) plane