Chapter 5, Problem 39P

For the problem specified in the table, build upon the results of the original problem to determine the minimum factor of safety for yielding. Use both the maximum-shear-stress theory and the distortion-energy theory, and compare the results. The material is 1018 CD steel.

3–68* to 3–71*

A countershaft carrying two V-belt pulleys is shown in the figure. 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)    Determine the tensions in the belt on pulley B, assuming the shaft is running at a constant speed.

(b)    Find the magnitudes of the bearing reaction forces, assuming the bearings act as simple supports.

(c)    Draw shear-force and bending-moment diagrams for the shaft. If needed, make one set for the horizontal plane and another set for the vertical plane.

(d)    At the point of maximum bending moment, determine the bending stress and the torsional shear stress.

(e)    At the point of maximum bending moment, determine the principal stresses and the maximum shear stress.

Problem 3–68*

To determine

The factor of safety for yielding from maximum-shear-stress theory.

The factor of safety for yielding from distortion-energy theory.

Answer to Problem 39P

The factor of safety for yielding from maximum-shear-stress theory is $3.08$.

The factor of safety for yielding from distortion-energy theory is $3.15$.

Explanation of Solution

The Free body diagram of pulley $A$ is shown below.

Figure-(1)

The free body diagram of pulley $B$ is shown below.

Figure-(2)

The given assumption is that the belt tension on the loose side at $B$ is $15\%$ of the tension on the tight side.

Write the relationship between tension on the loose side with respect to tension on the tight side.

    $T_1=0.15T_2$                                                                                          (I)

Here, the tension on the loose side is $T_1$ and the tension on the tight side is $T_2$.

Write the equation to balance the tension on the counter shaft.

    $\begin{array}{c}{\displaystyle \sum T}=0\\ \left(T_{A1}-T_{A2}\right)\frac{d_A}{2}-\left(T_2-T_1\right)\frac{d_B}{2}=0\end{array}$                                                            (II)

Here, the tension on the tight side of pulley $A$ is $T_{A1}$, the tension on the loose side of pulley $A$ is $T_{A2}$, diameter of shaft $A$ is $d_A$ and the diameter of shaft $B$ is $d_B$.

Substitute $0.15T_2$ for $T_1$ in Equation (II).

Write the tension on the loose side.

    $\begin{array}{c}\left(T_{A1}-T_{A2}\right)\frac{d_A}{2}-\left(T_2-0.15T_2\right)\frac{d_B}{2}=0\\ T_2=\frac{1}{0.85}\left(\left(T_{A1}-T_{A2}\right)\frac{d_A}{d_B}\right)\end{array}$                        (III)

Write the magnitude of bearing reaction force at $C$.

    $\begin{array}{c}{\displaystyle \sum M_O}=0\\ \left[\begin{array}{l}-(T_{A1}+T_{A2})l_{OA}+(T_1+T_2)\left(l_{BA}+l_{OA}\right)\\ -R_C\left(l_{BA}+l_{OA}+l_{CB}\right)\end{array}\right]=0\\ R_C=\frac{\left[\begin{array}{l}-(T_{A1}+T_{A2})l_{OA}\\ +(T_1+T_2)\left(l_{BA}+l_{OA}\right)\end{array}\right]}{\left(l_{BA}+l_{OA}+l_{CB}\right)}\end{array}$              (IV)

Here, the magnitude of the bearing force at $C$ is $R_C$, distance between $O$ and $A$ is $l_{OA}$, distance between $A$ and $B$ is $l_{BA}$ and distance between $C$ and $B$ is $l_{CB}$.

Write the magnitude of bearing reaction force at $O$.

    $\begin{array}{c}{\displaystyle \sum F_O}=0\\ R_O+\left(T_{A1}+T_{A2}\right)-\left(T_2+T_1\right)+R_C=0\\ R_O=\left(T_2+T_1\right)-R_C-\left(T_{A1}+T_{A2}\right)\end{array}$         (V)

Here, the magnitude of bearing reaction force at $O$ is $R_O$.

Write the shear force at $O$.

    $S{F}_O=-R_O$                                                                     (VI)

Here, the shear force at $O$ is $S{F}_O$.

Write the shear force at $A$.

    $S{F}_A=-R_O-\left(T_{A1}+T_{A2}\right)$                                                 (VII)

Here, the shear force at $A$ is $S{F}_A$.

Write the shear force at $B$.

    $S{F}_B=S{F}_A+\left(T_1+T_2\right)$                                                     (VIII)

Here, the shear force at $B$ is $S{F}_B$.3

Write the shear force at $C$.

    $S{F}_C=S{F}_B$                                                                        (IX)

Here, the shear force at $C$ is $S{F}_C$.

We know that, the moment at the supports of the simply supported beam is zero.

Write the moment at $O$ and $C$.

    $M_O=M_C=0$                                                                 (X)

Here, the moment at $O$ is $M_O$ and the moment at $C$ is $M_C$.

Write the moment at $A$.

    $M_A=-R_O\times l_{OA}$                                                                (XI)

Here, the moment at $A$ is $M_A$.

Write the moment at $B$.

    $M_B=-R_C\times l_{CB}$                                                               (XII)

Here, the moment at $B$ is $M_B$.

It is clear from the shear force diagram and bending moment diagram of the shaft that the point of maximum bending moment is at $x=10\text{in}$, and the bending moment at that point is $2932.5\text{lbf}\cdot \text{in}$.

Write the torque acting at pulley $A$.

    $T=\left(T_{A1}-T_{A2}\right)\frac{d_A}{2}$                                                       (XIII)

Here, the torque acting on pulley $A$ is $T$.

Write the bending stress.

    $\sigma =\frac{32M_A}{\pi d^3}$                                                                (XIV)

Here, the bending stress is $\sigma$ and diameter of shaft is $d$.

Write the shear stress.

    $\tau =\frac{16T}{\pi d^3}$                                                                    (XV)

Here, the shear stress is $\tau$.

Write the maximum principal stress.

    ${\sigma }_1=\frac{\sigma }{2}+\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$                                                                   (XVI)

Here, the maximum principal stress is ${\sigma }_1$.

Calculate the minimum principal stress.

    ${\sigma }_2=\frac{\sigma }{2}-\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$                                                                    (XVII)

Here, the minimum principal stress is ${\sigma }_2$.

Write the maximum shear stress.

    ${\tau }_{\max }=\sqrt{{\left(\frac{\sigma }{2}\right)}^2+{\tau }^2}$                                                                        (XVIII)

Here, maximum shear stress is ${\tau }_{\max }$.

Calculate the factor of safety from maximum-shear-stress theory.

    $n=\frac{S_y}{{\sigma }_1-{\sigma }_2}$                                                                                           (XIX)

Here, the maximum yield stress for $1018$ CD steel is $S_y$.

Calculate the factor of safety from distortion-energy theory.

    $n=\frac{S_y}{\sigma \text{'}}$                                                                                                   (XX)

Here, the Von Mises stress is $\sigma \text{'}$.

Write the expression for von Mises stress.

    $\sigma \text{'}={\left({\sigma }_1^2-{\sigma }_1.{\sigma }_2+{\sigma }_2^2\right)}^{\frac{1}{2}}$

Substitute $\left[{\left({\sigma }_1^2-{\sigma }_1.{\sigma }_2+{\sigma }_2^2\right)}^{\frac{1}{2}}\right]$ for $\sigma \text{'}$ in Equation (XX).

    $n=\frac{S_y}{\left[{\left({\sigma }_1^2-{\sigma }_1.{\sigma }_2+{\sigma }_2^2\right)}^{\frac{1}{2}}\right]}$                                                                      (XXI)

Conclusion:

Substitute $500\text{lbf}$ for $T_{A1}$, $75\text{lbf}$ for $T_{A2}$, $8\text{in}$ for $d_A$ and $10\text{in}$ for $d_B$  in Equation (III).

    $\begin{array}{c}{T}_2=\frac{1}{0.85}\left(\left(500\text{lbf}-75\text{lbf}\right)\times \frac{8\text{in}}{10\text{in}}\right)\\ =\frac{1}{0.85}\left(\left(425\text{lbf}\right)\times 0.8\right)\\ =400\text{lbf}\end{array}$

Substitute $400\text{lbf}$ for $T_2$ in Equation (I)

    $\begin{array}{c}{T}_1=0.15\times 400\text{lbf}\\ =60\text{lbf}\end{array}$

Substitute $500\text{lbf}$ for $T_{A1}$, $75\text{lbf}$ for $T_{A2}$, $400\text{lbf}$ for $T_2$, $60\text{lbf}$ for $T_1$, $10\text{in}$ for $l_{OA}$, $18\text{in}$ for $l_{BA}$ and $12\text{in}$ for $l_{CB}$ in Equation (IV).

    $\begin{array}{c}{R}_C=\frac{\left[\left(-(500\text{lbf}+75\text{lbf})10\text{in}\right)+\left((400\text{lbf}+60\text{lbf})\left(18\text{in}+10\text{in}\right)\right)\right]}{\left(18\text{in}+10\text{in}+12\text{in}\right)}\\ =\frac{\left[\left(-(575\text{lbf})10\text{in}\right)+\left((460\text{lbf})\left(28\text{in}\right)\right)\right]}{\left(40\text{in}\right)}\\ =178.25\text{lbf}\end{array}$

Substitute $500\text{lbf}$ for $T_{A1}$, $75\text{lbf}$ for $T_{A2}$, $400\text{lbf}$ for $T_2$, $60\text{lbf}$ for $T_1$ and $178.25\text{lbf}$ for $R_C$ in Equation (V).

    $\begin{array}{c}{R}_O=\left(400\text{lbf}+60\text{lbf}\right)-178.25\text{lbf}-\left(500\text{lbf}+75\text{lbf}\right)\\ =-293.25\text{lbf}\end{array}$

Substitute $-293.25\text{lbf}$ for $R_O$ in Equation (VI).

    $S{F}_O=293.25\text{lbf}$

Substitute $500\text{lbf}$ for $T_{A1}$, $75\text{lbf}$ for $T_{A2}$ and $-293.25\text{lbf}$ for $R_O$ in Equation (VII).

    $\begin{array}{c}S{F}_A=-\left(-293.25\text{lbf}\right)-\left(500\text{lbf}+75\text{lbf}\right)\\ =-281.75\text{lbf}\end{array}$

Substitute $400\text{lbf}$ for $T_2$, $60\text{lbf}$ for $T_1$ and $-281.75\text{lbf}$ for $S{F}_A$ in Equation VIII).

    $\begin{array}{c}S{F}_B=-281.75\text{lbf}+\left(400\text{lbf}+60\text{lbf}\right)\\ =178.25\text{lbf}\end{array}$

Substitute $178.25\text{lbf}$ for $S{F}_B$ in Equation (IX).

    $S{F}_C=178.25\text{lbf}$

Substitute $-293.25\text{lbf}$ for $R_O$ and $10\text{in}$ for $l_{OA}$ in Equation (XI).

    $\begin{array}{c}{M}_A=-\left(-293.25\text{lbf}\right)\times 10\text{in}\\ =293.25\text{lbf}\cdot \text{in}\end{array}$

Substitute $178.25\text{lbf}$ for $R_C$ and $12\text{in}$ for $l_{CB}$ in Equation (XII)

    $\begin{array}{c}{M}_B=-178.25\text{lbf}\times 12\text{in}\\ =-2139\text{lbf}\cdot \text{in}\end{array}$

The shear force diagram and bending moment diagram for the shaft is as follows.

Figure-(3)

Substitute $500\text{lbf}$ for $T_{A1}$, $75\text{lbf}$ for $T_{A2}$ and $8\text{in}$ for $d_A$ in Equation (XIII).

    $\begin{array}{c}T=\left(500\text{lbf}-75\text{lbf}\right)\frac{8\text{in}}{2}\\ =\left(425\text{lbf}\right)\frac{8\text{in}}{2}\\ =1700\text{lbf}\cdot \text{in}\end{array}$

Substitute $2932.5\text{lbf}\cdot \text{in}$ for $M_A$ and $1\frac{1}{4}\text{in}$ for $d$ in Equation (XIV).

    $\begin{array}{c}\sigma =\frac{32\times 2932.5\text{lbf}}{\pi {\left(1\frac{1}{4}\text{in}\right)}^3}\\ =\left(15293.5\text{psi}\right)\left(\frac{1\text{kpsi}}{1000\text{psi}}\right)\\ =15.2935\text{kpsi}\\ \approx 15.3\text{kpsi}\end{array}$

Substitute $1700\text{lbf}\cdot \text{in}$ for $T$ and $1\frac{1}{4}\text{in}$ for $d$ in Equation (XV).

    $\begin{array}{c}\tau =\frac{16\times 1700\text{lbf}}{\pi {\left(1\frac{1}{4}\text{in}\right)}^3}\\ =\left(4432.9\text{psi}\right)\left(\frac{1\text{kpsi}}{1000\text{psi}}\right)\\ =4.4329\text{kpsi}\\ \approx 4.43\text{kpsi}\end{array}$

Substitute $15.3\text{kpsi}$ for $\sigma$ and $4.43\text{kpsi}$ for $\tau$ in Equation (XVI).

    $\begin{array}{c}{\sigma }_1=\frac{15.3\text{kpsi}}{2}+\sqrt{{\left(\frac{15.3\text{kpsi}}{2}\right)}^2+{\left(4.43\text{kpsi}\right)}^2}\\ =7.65\text{kpsi}+8.84\text{kpsi}\\ =16.49\text{kpsi}\\ \approx 16.5\text{kpsi}\end{array}$

Substitute $15.3\text{kpsi}$ for $\sigma$ and $4.43\text{kpsi}$ for $\tau$ in Equation (XVII).

    $\begin{array}{c}{\sigma }_2=\frac{15.3\text{kpsi}}{2}-\sqrt{{\left(\frac{15.3\text{kpsi}}{2}\right)}^2+{\left(4.43\text{kpsi}\right)}^2}\\ =7.65\text{kpsi}-8.84\text{kpsi}\\ =-1.19\text{kpsi}\end{array}$

Substitute $15.3\text{kpsi}$ for $\sigma$ and $4.43\text{kpsi}$ for $\tau$ in Equation (XVIII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{15.3\text{kpsi}}{2}\right)}^2+{\left(4.43\text{kpsi}\right)}^2}\\ =8.84\text{kpsi}\end{array}$

Refer to the Table A-20 “Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels” to obtain the yield strength of $1018$ CD steel as $54\text{kpsi}$.

Substitute $16.5\text{kpsi}$ for ${\sigma }_1$, $54\text{kpsi}$ for $S_y$ and $-1.19\text{kpsi}$ for ${\sigma }_2$ in the Equation (XIX).

    $\begin{array}{c}n=\frac{54\text{kpsi}}{16.5\text{kpsi}-\left(-1.19\text{kpsi}\right)}\\ =3.052\\ \approx 3.05\end{array}$

Thus, the factor of safety for yielding from maximum-shear-stress theory is $3.05$.

Substitute $16.5\text{kpsi}$ for ${\sigma }_1$, $54\text{kpsi}$ for $S_y$ and $-1.19\text{kpsi}$ for ${\sigma }_2$ in Equation (XXI).

    $\begin{array}{c}n=\frac{54\text{kpsi}}{{\left({\left(16.5\text{kpsi}\right)}^2-\left(16.5\text{kpsi}\right)\left(-1.19\text{kpsi}\right)+{\left(-1.19\text{kpsi}\right)}^2\right)}^{\frac{1}{2}}}\\ =3.153\\ \approx 3.15\end{array}$

Thus, the factor of safety for yielding from distortion-energy theory is $3.15$.