Chapter 7, Problem 6P

The figure shows a proposed design for the industrial roll shaft of Prob. 7–4. Hydrodynamic film bearings are to be used. All surfaces are machined except the journals, which are ground and polished. The material is 1035 HR steel. Perform a design assessment. Is the design satisfactory?

Problem 7–6

Bearing shoulder fillets 0.030 in, others $\frac{\text{1}}{\text{16}}$ in. Sled-runner keyway is $3\frac{\text{1}}{\text{2}}$ in long. Dimensions in inches.

To determine

Weather the design is satisfactory or not.

Answer to Problem 6P

The design is satisfactory for crown gear.

Explanation of Solution

Write the expression for force acting on roller.

    $F_{cy}=F\times l$                                                                                          (I)

Here, normal force is $F$ and length of roller is $l$.

Write the expression for force acting on C along z axis.

    $F_{cz}=\mu F_{cy}$                                                                                          (II)

Here, force on C in z axis is $F_{cz}$ and the friction coefficient is $\mu$.

Write the expression for torque.

    $T=F_{cz}\left(\frac{d}{2}\right)$                                                                                      (III)

Here, torque is $T$ and the roller diameter is $d$.

Write the expression for force at B along z axis.

    $F_{Bz}=\frac{T}{\frac{d_c}{2}}$                                                                                            (IV)

Here, force at B along z axis is $F_{Bz}$ and the pitch diameter is $d_c$.

Write the expression for force at B along y-axis

    $F_{By}=F_{Bz}\left(\tan 20°\right)$                                                                               (V)

Here, force at B along y-axis is $F_{By}$.

Write the expression for moment about O.

    $f_{cy}\left(5.75\right)-F_{Ay}\left(11.5\right)-F_{By}\left(14.25\right)=0$                                              (VI)

Write the expression for moment about A.

    $F_{oy}\left(11.5\right)-F_{By}\left(2.75\right)-F_{cy}\left(5.75\right)=0$                                               (VII)

Write the expression for moment at point C.

    $M_c=F_{oy}\left(5.75\right)$                                                                                (VIII)

Write the expression for moment at A.

    $M_A=F\times \left(2.75\right)$                                                                                (IX)

Write the expression for moment about O.

    $F_{cz}\left(5.75\right)-F_{Az}\left(11.5\right)+F_{Bz}\left(14.25\right)=0$                                              (X)

Write the expression for moment about A.

    $F_{oz}\left(11.5\right)+F_{Bz}\left(2.75\right)-F_{Cz}\left(5.75\right)=0$                                             (XI)

Write the expression for moment at point C along z-axis.

    $M_C=F_{oz}\left(5.75\right)$                                                                                 (XII)

Write the expression for moment at A along z-axis.

    $M_A=F_{Bz}\times 2.75$                                                                                (XIII)

Write the expression for total moment.

    $M=\sqrt{M_{xy}^2+M_{xz}^2}$                                                                              (XIV)

Here, bending moment in x-y plane is $M_{xy}$ and the moment in x-z plane is $M_{cxz}$.

Write the expression for total moment at D.

    $M_D=\sqrt{M_{Dxy}^2+M_{Dxz}^2}$                                                                            (XV)

Here, moment on x-y plane at D is $M_{Dxy}$ and the moment at D along xz plane is $M_{Dxz}$.

Write the moment equation in x-y plane.

    $M_{xy}=\left(-131.1\text{lbf}\right)x+15{\left(x-1.75\right)}^2-15{\left(x-9.75\right)}^2-\left(62.3\text{lbf}\right)\left(x-11.5\right)$ (XVI)

Integrate Equation (XVI).

$EI{y}^{\prime }=\left[\begin{array}{l}\left(\frac{-131.1\text{lbf}}{2}\right)x^2+5{\left(x-1.75\right)}^3-5{\left(x-9.75\right)}^3-\left(\frac{62.3\text{lbf}}{2}\right){\left(x-11.5\right)}^2\\ +C_1\end{array}\right]$ (XVII)

Integrate Equation (XVII).

    $EIy=\left[\begin{array}{l}\left(\frac{-131.1\text{lbf}}{6}\right)x^3+\frac{5}{4}{\left(x-1.75\right)}^4-\frac{5}{4}{\left(x-9.75\right)}^4\\ -\left(\frac{62.3\text{lbf}}{6}\right){\left(x-11.5\right)}^3+C_{1}x+C_2\end{array}\right]$ (XVIII)

Apply boundary Equation, substitute $0$ for $x$ and $0$ for $y$ in Equation (XVII).

    $\begin{array}{l}EI\left(0\right)=\left[\begin{array}{l}\left(\frac{-131.1\text{lbf}}{6}\right)\times 0^3+\frac{5}{4}{\left(0-1.75\right)}^4-\frac{5}{4}{\left(x-9.75\right)}^4\\ -\left(\frac{62.3\text{lbf}}{6}\right){\left(0-11.5\right)}^3+C_1\times 0+C_2\end{array}\right]\\ C_2=0\end{array}$

Substitute $11.5$ for $x$ and $0$ for $y$ in Equation (XVII).

    $\begin{array}{l}EI\left(0\right)=\left[\begin{array}{l}\left(\frac{-131.1\text{lbf}}{6}\right){\left(11.5\text{in}\right)}^3+\frac{5}{4}{\left(11.5-1.75\right)}^4-\frac{5}{4}{\left(11.5-9.75\right)}^4\\ -\left(\frac{62.3\text{lbf}}{6}\right){\left(11.5\text{in}-11.5\text{in}\right)}^3+C_1\times 11.5+C_2\end{array}\right]\\ 11.5C_1=21646.6\text{lbf}\cdot {\text{in}}^{\text{3}}\\ C_1=\frac{21646.6}{11.5}\text{lbf}\cdot {\text{in}}^{\text{3}}\\ C_1=1908.4\text{lbf}\cdot {\text{in}}^{\text{3}}\end{array}$

Substitute $1908.4\text{lbf}\cdot {\text{in}}^{\text{3}}$ for $C_1$ and $0$ for $x$ in Equation (XVII).

    $\begin{array}{c}EI{y}^{\prime }=\left[\begin{array}{l}\left(\frac{-131.1\text{lbf}}{2}\right)0^2+5{\left(0-1.75\right)}^3-5{\left(0-9.75\right)}^3-\\ \left(\frac{62.3\text{lbf}}{2}\right){\left(0-11.5\right)}^2+1908.4\text{lbf}\cdot {\text{in}}^3\end{array}\right]\\ =0-4661.38+4661.38+1908.4\text{lbf}\cdot {\text{in}}^3\\ =1908.4\text{lbf}\cdot {\text{in}}^3\end{array}$

Substitute $11.5$ for $x$ and $1908.4\text{lbf}\cdot {\text{in}}^{\text{3}}$ for $C_1$ in Equation (XVII).

    $\begin{array}{c}EI{y}^{\prime }=\left[\begin{array}{l}\left(\frac{-131.1\text{lbf}}{2}\right){\left(11.5\right)}^2+5{\left(11.5-1.75\right)}^3-5{\left(11.5-9.75\right)}^3-\\ \left(\frac{62.3\text{lbf}}{2}\right){\left(11.5-11.5\right)}^2+1908.4\text{lbf}\cdot {\text{in}}^3\end{array}\right]\\ =-4061.5+1908.4\text{lbf}\cdot {\text{in}}^3\\ =-2153.1\text{lbf}\cdot {\text{in}}^3\end{array}$

Write the moment equation in x-z plane.

    $M_{xz}=\left[\begin{array}{l}\left(17.4\text{lbf}\cdot \text{in}\right)x-6{\left(x-1.75\right)}^2+6{\left(x-9.75\right)}^2\\ +\left(206.6\text{lbf}\cdot \text{in}\right)\left(x-11.5\right)\end{array}\right]$  (XVIII)

Integrate Equation (XVIII).

    $EI{z}^{\prime }=\left[\begin{array}{l}-\left(\frac{17.4\text{lbf}\cdot \text{in}}{2}\right)x^2-2{\left(x-1.75\right)}^3\\ +2{\left(x-9.75\right)}^3+\left(\frac{206.6\text{lbf}\cdot \text{in}}{2}\right){\left(x-11.5\right)}^2+C_3\end{array}\right]$  (XIX)

Integrate Equation (XIX).

    $EIz=\left[\begin{array}{l}-\left(\frac{17.4\text{lbf}\cdot \text{in}}{6}\right)x^3-2{\left(x-1.75\right)}^4\\ +\frac{1}{2}{\left(x-9.75\right)}^4+\left(\frac{206.6\text{lbf}\cdot \text{in}}{6}\right){\left(x-11.5\right)}^3+C_{3}x+C_4\end{array}\right]$    (XX)

Substitute $0$ for $x$ and $0$ for $z$ in Equation (XX).

    $\begin{array}{l}EI\left(0\right)=\left[\begin{array}{l}-\left(\frac{17.4\text{lbf}\cdot \text{in}}{6}\right)0^3-2{\left(0-1.75\right)}^4\\ +\frac{1}{2}{\left(0-9.75\right)}^4+\left(\frac{206.6\text{lbf}\cdot \text{in}}{6}\right){\left(0-11.5\right)}^3+C_3\left(0\right)+C_4\end{array}\right]\\ 0=C_4+0\\ C_4=0\end{array}$

Substitute $11.5$ for $x$ and $0$ for $z$ in Equation (XX).

    $\begin{array}{l}EIz=\left[\begin{array}{l}-\left(\frac{17.4\text{lbf}\cdot \text{in}}{6}\right){\left(11.5\right)}^3-2{\left(11.5-1.75\right)}^4\\ +\frac{1}{2}{\left(11.5-9.75\right)}^4+\left(\frac{206.6\text{lbf}\cdot \text{in}}{6}\right){\left(11.5-11.5\right)}^3+C_3\left(11.5\right)+C_4\end{array}\right]\\ 11.5C_3=103.2125\\ C_3=\frac{103.2125}{11.5}\\ C_3=8.975\text{lbf}\cdot {\text{in}}^3\end{array}$

Substitute $0$ for $x$ and $8.975\text{lbf}\cdot {\text{in}}^3$ for $C_3$ in Equation (XIX).

    $\begin{array}{c}EI{z}^{\prime }=\left[\begin{array}{l}-\left(\frac{17.4\text{lbf}\cdot \text{in}}{2}\right)0^2-2{\left(0-1.75\right)}^3\\ +2{\left(0-9.75\right)}^3+\left(\frac{206.6\text{lbf}\cdot \text{in}}{2}\right){\left(0-11.5\right)}^2+8.975\text{lbf}\cdot {\text{in}}^3\end{array}\right]\\ =0-2{\left(-1.75\right)}^3+2{\left(-9.75\right)}^3+\left(\frac{206.6\text{lbf}\cdot \text{in}}{2}\right){\left(11.5\right)}^2+8.975\text{lbf}\cdot {\text{in}}^3\\ =8.975\text{lbf}\cdot {\text{in}}^3\end{array}$

Substitute $11.5$ for $x$ and $8.975\text{lbf}\cdot {\text{in}}^3$ for $C_3$ in equation (XIX).

    $\begin{array}{c}EI{z}^{\prime }=\left[\begin{array}{l}-\left(\frac{17.4\text{lbf}\cdot \text{in}}{2}\right){\left(11.5\right)}^2-2{\left(11.5-1.75\right)}^3\\ +2{\left(11.5-9.75\right)}^3+\left(\frac{206.6\text{lbf}\cdot \text{in}}{2}\right){\left(11.5-11.5\right)}^2+8.975\text{lbf}\cdot {\text{in}}^3\end{array}\right]\\ =\left(-8.7\text{lbf}\cdot \text{in}\right){\left(11.5\right)}^2-2{\left(9.75\right)}^3+2{\left(1.75\right)}^3+0+8.975\text{lbf}\cdot {\text{in}}^3\\ =-683\text{lbf}\cdot {\text{in}}^3\end{array}$

Write the expression for flexural rigidity.

    $EI\theta =\sqrt{{\left(EI{y}^{\prime }\right)}^2+{\left(EI{z}^{\prime }\right)}^2}$                                                                    (XX)

Here, flexural rigidity is $EI$ and the angle is $\theta$.

Write the expression for deflection at A.

    ${\dot{y}}_A=\frac{-2153.1}{E{I}_1}$                                                                                        (XXI)

Here, speed is ${\dot{y}}_A$,

Write the expression for relative deflection.

    ${\dot{y}}_{\frac{B}{A}}=\frac{Fx\left(x-2l\right)}{2E{I}_2}$                                                                                (XXII)

Here, force is $F$, length is $l$ and the young’s modulus is $E$.

Write the expression for deflection at B.

    ${\dot{y}}_B={\dot{y}}_A+{\dot{y}}_{\frac{B}{A}}$                                                                                      (XXIII)

Write the expression for deflection at z.

    ${\dot{z}}_A=\frac{-683.5}{E{I}_1}$                                                                                        (XXIV)

Here, deflection at A is ${\dot{z}}_A$.

Write the expression for relative deflection.

    ${\dot{z}}_{\frac{B}{A}}=\frac{Fx\left(x-2l\right)}{2E{I}_2}$                                                                                 (XXV)

Here, force is $F$, length is $l$ and the young’s modulus is $E$.

Write the expression for deflection at B.

    ${\dot{z}}_B={\dot{z}}_A+{\dot{z}}_{\frac{B}{A}}$                                                                                        (XXVI)

Here, deflection at B is $\dot{z}$.

Write the expression for gear mesh.

    ${\theta }_B=\sqrt{{\left({\dot{y}}_B\right)}^2+{\left({\dot{z}}_B\right)}^2}$                                                                           (XXVII)

Here, gear mesh at B is ${\theta }_B$.

Write the expression for concentration factor for bending.

    $k_f=1+q\left(k_t-1\right)$                                                                                 (XXVIII)

Here, concentration factor for bending is $k_f$, notch sensitivity is $q$ and the concentration factor due to bending stress is $k_t$.

Write the expression for concentration factor torsion.

    $K_{fs}=1+q_s\left(k_{fs}-1\right)$                                                                              (XXIX)

Here, concentration factor for torsion is $K_{fs}$.

Write the expression for endurance limit.

    $S_e{}^{\prime }=0.5S_{ut}$                                                                                          (XXX)

Here, endurance limit is $S_e{}^{\prime }$.

Write the expression for surface factor.

    $k_a=a{S}_{ut}^b$                                                                                              (XXXI)

Here, surface factor is $k_a$, minimum tensile strength is $S_{ut}$ and the exponent is $b$.

Write the expression for size factor.

    $k_b={\left(\frac{d}{0.3}\right)}^{-.107}$                                                                                    (XXXII)

Here, size factor is $k_b$.

Write the expression for endurance limit at critical path.

    $S_e=k_a{k}_b{k}_c{S}_e{}^{\prime }$                                                                                     (XXXIII)

Here, surface modification factor is $k_a$ and reliability factor is $k_e$.

Conclusion:

The following figure shows roller forces.

Figure-(1)

Substitute $30\text{lbf}/\text{in}$ for $F$ and $8\text{in}$ for $l$ in Equation (I).

    $\begin{array}{c}{F}_{cy}=30\text{lbf}/\text{in}\times 8\text{in}\\ =240\text{lbf}\end{array}$

Substitute $240\text{lbf}$ for $F_{cy}$ and $0.4$ for $\mu$ in Equation (II).

    $\begin{array}{c}{F}_{cz}=240\text{lbf}\times \text{0}\text{.4}\\ =\text{96}\text{lbf}\end{array}$

Substitute $\text{96}\text{lbf}$ for $F_{cz}$ and $4\text{in}$ for $d$ in Equation (III).

    $\begin{array}{c}{F}_{Bz}=\text{96}\text{lbf}\times \frac{4\text{in}}{2}\\ =192\text{lbf}\cdot \text{in}\end{array}$

Substitute $192\text{lbf}\cdot \text{in}$ for $F_{Bz}$ an d $3\text{in}$ for $d_c$ in Equation (IV).

    $\begin{array}{c}{F}_{Bz}=\frac{192\text{lbf}\cdot \text{in}}{\frac{3\text{in}}{2}}\\ =\frac{192\text{lbf}\cdot \text{in}}{1.5\text{in}}\\ =128\text{lbf}\end{array}$

Substitute $128\text{lbf}$ for $F_{Bz}$ in Equation (V).

    $\begin{array}{c}{F}_{By}=128\text{lbf}\left(\tan 20°\right)\\ =128\text{lbf}\left(0.36397\right)\\ =46.588\text{lbf}\\ \approx 46.6\text{lbf}\end{array}$

The figure below shows the free body diagram and bending moment diagram.

Figure-(2)

Substitute $240\text{lbf}$ for $F_{cy}$ and $46.6\text{lbf}$ for $F_{By}$ in Equation (VI).

    $\begin{array}{l}\left(240\text{lbf}\right)\left(5.75\text{in}\right)-F_{Ay}\left(11.5\text{in}\right)-\left(46.6\text{lbf}\right)\left(14.25\text{in}\right)=0\\ F_{Ay}=\frac{\left(240\text{lbf}\right)\left(5.75\text{in}\right)-\left(46.6\text{lbf}\right)\left(14.25\text{in}\right)}{11.5\text{in}}\\ F_{Ay}=62.3\text{lbf}\end{array}$

Substitute $46.6\text{lbf}$ for $F_{By}$ and $240\text{lbf}$ for $F_{cy}$ in Equation (VII).

    $\begin{array}{l}{F}_{oy}\left(11.5\text{in}\right)-\left(46.6\text{lbf}\right)\left(2.75\text{in}\right)-\left(240\text{lbf}\right)\left(5.75\text{in}\right)=0\\ F_{oy}=\frac{\left(46.6\text{lbf}\right)\left(2.75\text{in}\right)+\left(240\text{lbf}\right)\left(5.75\text{in}\right)}{11.5\text{in}}\\ F_{oy}=131.1\text{lbf}\end{array}$

Substitute $131.1\text{lbf}$ for $F_{oy}$ in Equation (VIII).

    $\begin{array}{c}{M}_c=\left(131.1\text{lbf}\right)\left(5.75\text{in}\right)\\ =753.8\text{lbf}\cdot \text{in}\end{array}$

The figure below shows the bending moment diagram for x-y plane.

Figure-(3)

Substitute $46.6\text{lbf}$ for $F$ in Equation (IX).

    $\begin{array}{c}{M}_A=46.6\text{lbf}\times \left(2.75\right)\\ =128.15\text{lbf}\cdot \text{in}\end{array}$

The figure below shows free body diagram and the bending moment diagram in x-z plane.

Figure-(4)

Substitute $\text{96}\text{lbf}$ for $F_{cz}$ and $128\text{lbf}$ for $F_{Bz}$ in Equation (X).

    $\begin{array}{l}\left(\text{96}\text{lbf}\right)\left(5.75\text{in}\right)-F_{Az}\left(11.5\text{in}\right)+\left(128\text{lbf}\right)\left(14.25\text{in}\right)=0\\ F_{Az}=\frac{\left(\text{96}\text{lbf}\right)\left(5.75\text{in}\right)+\left(128\text{lbf}\right)\left(14.25\text{in}\right)}{11.5\text{in}}\\ F_{Az}=206.6\text{lbf}\end{array}$

Substitute $\text{96}\text{lbf}$ for $F_{cz}$ and $128\text{lbf}$ for $F_{Bz}$ in Equation in Equation (XI).

    $\begin{array}{l}{F}_{oz}\left(11.5\text{in}\right)+128\text{lbf}\times \left(2.75\text{in}\right)-\text{96}\text{lbf}\times \left(5.75\text{in}\right)=0\\ F_{oz}=\frac{\text{96}\text{lbf}\times \left(5.75\text{in}\right)-128\text{lbf}\times \left(2.75\text{in}\right)}{11.5\text{in}}\\ F_{oz}=17.4\text{lbf}\end{array}$

Substitute $17.4\text{lbf}$ for $F_{oz}$ in Equation (XII).

    $\begin{array}{c}{M}_C=17.4\text{lbf}\times \left(5.75\text{in}\right)\\ =100.05\text{lbf}\cdot \text{in}\end{array}$

Substitute $128\text{lbf}$ for $F_{Bz}$ in Equation (XIII).

    $\begin{array}{c}{M}_A=128\text{lbf}\times 2.75\text{in}\\ =-352\text{lbf}\cdot \text{in}\end{array}$

Substitute $100\text{lbf}\cdot \text{in}$ for $M_{Cxy}$ and $-754\text{lbf}\cdot \text{in}$ for $M_{cxz}$ in Equation (XIV).

    $\begin{array}{c}{M}_C=\sqrt{{\left(100\text{lbf}\cdot \text{in}\right)}^2+{\left(-754\text{lbf}\cdot \text{in}\right)}^2}\\ =\sqrt{578516\text{lbf}\cdot {\text{in}}^2}\\ =761\text{lbf}\cdot \text{in}\end{array}$

Substitute $-128\text{lbf}\cdot \text{in}$ for $M_{Dxy}$  and $-352\text{lbf}\cdot \text{in}$ for $M_{Dxz}$ in Equation (XV).

    $\begin{array}{c}{M}_D=\sqrt{{\left(-128\text{lbf}\cdot \text{in}\right)}^2+{\left(-352\text{lbf}\cdot \text{in}\right)}^2}\\ =\sqrt{140288\text{lbf}\cdot {\text{in}}^2}\\ =375\text{lbf}\cdot \text{in}\end{array}$

The figure below shows the bending moment diagram.

Figure-(5)

The figure below shows the forces in x-z plane.

Figure-(6)

Substitute $6.25$ for $x$ in equation (XVIII).

    $\begin{array}{c}{M}_{xz}=\left[\begin{array}{l}\left(17.4\text{lbf}\cdot \text{in}\right)x-6{\left(6.25-1.75\right)}^2+6{\left(6.25-9.75\right)}^2\\ +\left(206.6\text{lbf}\cdot \text{in}\right)\left(6.25-11.5\right)\end{array}\right]\\ =\left(17.4\text{lbf}\cdot \text{in}\right)x-6{\left(4.5\right)}^2+6{\left(-3.5\right)}^2+\left(206.6\text{lbf}\cdot \text{in}\right)\left(-5.25\right)\\ =-1023\text{lbf}\cdot \text{in}\end{array}$

The figure below shows the bending moment diagram.

Figure-(7)

Substitute $1908.4\text{lbf}\cdot {\text{in}}^3$ for $EI{y}^{\prime }$, $8.975\text{lbf}\cdot {\text{in}}^3$ for $EI{z}^{\prime }$ , $30\times {10}^6\text{psi}$ for $E$ and $0.324{\text{in}}^{\text{4}}$ for $I$ in Equation (XX).

    $\begin{array}{l}\left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)\theta =\sqrt{{\left(1908.4\text{lbf}\cdot {\text{in}}^3\right)}^2+{\left(8.975\text{lbf}\cdot {\text{in}}^3\right)}^2}\\ \left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)\theta =\sqrt{3642071.11\text{lbf}\cdot {\text{in}}^3}\\ \theta =\frac{1908.4\text{lbf}\cdot {\text{in}}^3}{\left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)}\\ \theta =0.001\text{rad}\end{array}$

Substitute $-2153.1\text{lbf}\cdot {\text{in}}^3$ for $EI{y}^{\prime }$, $-683\text{lbf}\cdot {\text{in}}^3$ for $EI{z}^{\prime }$, $30\times {10}^6\text{psi}$ for $E$ and $0.324{\text{in}}^{\text{4}}$ for $I$ in Equation (XX).

    $\begin{array}{l}\left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)\theta =\sqrt{{\left(-2153.1\text{lbf}\cdot {\text{in}}^3\right)}^2+{\left(-683\text{lbf}\cdot {\text{in}}^3\right)}^2}\\ \left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)\theta =2259\text{lbf}\cdot {\text{in}}^3\\ \theta =\frac{2259\text{lbf}\cdot {\text{in}}^3}{\left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)}\\ \theta =0.000628\text{rad}\end{array}$

Calculate factor of safety.

    $\begin{array}{c}n=\frac{0.001\text{rad}}{0.000628\text{rad}}\\ =1.59\end{array}$

The figure below shows forces in x-y plane.

Figure-(8)

Substitute $46.6\text{lbf}$ for $F$ $2.75$ for $x$ and $2.75$ for $L$ in Equation (XXIII).

    $\begin{array}{c}{\dot{y}}_{\frac{B}{A}}=\frac{\left(46.6\text{lbf}\right)\left(2.75\right)\left(2.75-2\left(2.75\right)\right)}{2E{I}_2}\\ =\frac{-176.2}{2E{I}_2}\end{array}$

Substitute $\frac{-176.2}{2E{I}_2}$ for ${\dot{y}}_{\frac{B}{A}}$ $\frac{-2153.1}{E{I}_1}$ for ${\dot{y}}_A$, $30\times {10}^6\text{psi}$ for $E$ and $0.324{\text{in}}^{\text{4}}$ for $I$ in Equation (XXIII).

    $\begin{array}{c}{\dot{y}}_B=\frac{-2153.1}{\left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)}+\frac{-176.2}{2\left(30\times {10}^6\text{psi}\right)\frac{\pi }{64}{\left(0.875\right)}^4}\\ =\frac{-2153.1}{\left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)}+\frac{-176.2}{2\left(30\times {10}^6\text{psi}\right)\left(0.028774{\text{in}}^4\right)}\\ =-0.000803\text{rad}\end{array}$

The figure below shows the forces on the system.

Figure-(9)

Substitute $128\text{lbf}$ for $F$, $2.75$ for $x$ and $2.75$ for $L$ in Equation (XXV).

    $\begin{array}{c}{\dot{z}}_{\frac{B}{A}}=\frac{\left(128\text{lbf}\right)\left(2.75\right)\left(2.75-2\left(2.75\right)\right)}{2E{I}_2}\\ =\frac{-484}{2E{I}_2}\end{array}$

Substitute $30\times {10}^6\text{psi}$ for $E$ and $0.324{\text{in}}^{\text{4}}$ for $I$, $\frac{-484}{2E{I}_2}$ for ${\dot{z}}_{\frac{B}{A}}$, and $\frac{-683.5}{E{I}_1}$ for ${\dot{z}}_A$ in Equation (XXVI).

    $\begin{array}{c}{\dot{z}}_B=\frac{-683.5}{\left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)}+\frac{-484}{2\left(30\times {10}^6\text{psi}\right)\frac{\pi }{64}{\left(0.875\right)}^4}\\ =\frac{-683.5}{\left(30\times {10}^6\text{psi}\right)\left(0.324{\text{in}}^{\text{4}}\right)}+\frac{-484}{2\left(30\times {10}^6\text{psi}\right)\frac{\pi }{64}{\left(0.028774{\text{in}}^4\right)}^4}\\ =-0.000751\text{rad}\end{array}$

Substitute $-0.000803\text{rad}$ for ${\dot{y}}_B$ and $-0.000751\text{rad}$ for ${\dot{z}}_B$ in Equation (XXVII).

    $\begin{array}{c}{\theta }_B=\sqrt{{\left(-0.000803\text{rad}\right)}^2+{\left(-0.000751\text{rad}\right)}^2}\\ =\sqrt{\left(1.21\times {10}^{-6}{\text{rad}}^2\right)}\\ =0.0011\text{rad}\end{array}$

Refer to table A-20, “Deterministic ASTM minimum tensile and yield strength for some hot rolled and cold drawn steels” to obtain tensile strength as $72\text{kpsi}$ and yield strength as $39.5\text{kpsi}$.

Substitute $10.75$ for $x$ in Equation (XVII).

    $\begin{array}{c}{M}_{xy}=\left[\begin{array}{l}\left(-131.1\text{lbf}\right)\left(10.75\right)+15{\left(10.75-1.75\right)}^2-\\ 15{\left(10.75-9.75\right)}^2-\left(62.3\text{lbf}\right)\left(10.75-11.5\right)\end{array}\right]\\ =\left(-131.1\text{lbf}\right)\left(10.75\right)+15\left(9\right)-15\left(1\right)-\left(62.3\text{lbf}\right)\left(10.75-11.5\right)\\ =-209.3\text{lbf}\cdot \text{in}\end{array}$

Substitute $10.75$ for $x$ in Equation (XVIII).

    $\begin{array}{c}{M}_{xz}=\left[\begin{array}{l}\left(17.4\text{lbf}\cdot \text{in}\right)\left(10.75\right)-6{\left(10.75-1.75\right)}^2+6{\left(10.75-9.75\right)}^2\\ +\left(206.6\text{lbf}\cdot \text{in}\right)\left(10.75-11.5\right)\end{array}\right]\\ =\left(17.4\text{lbf}\cdot \text{in}\right)\left(10.75\right)-6{\left(9\right)}^2+6{\left(1\right)}^2+\left(206.6\text{lbf}\cdot \text{in}\right)\left(10.75-11.5\right)\\ =-293\text{lbf}\cdot \text{in}\end{array}$

Substitute $-209.3\text{lbf}\cdot \text{in}$ for $M_{xy}$ and $-293\text{lbf}\cdot \text{in}$ for $M_{xz}$ in Equation (XIV).

    $\begin{array}{c}M=\sqrt{{\left(-209.3\text{lbf}\cdot \text{in}\right)}^2+{\left(-293\text{lbf}\cdot \text{in}\right)}^2}\\ =\sqrt{129655.49\text{lbf}\cdot \text{in}}\\ =360\text{lbf}\cdot \text{in}\end{array}$

Refer to table 7-1, “First iteration estimated for stress concentration factors” to obtain concentration factor for bending stress as $2.3$ and for torsional stress as $1.8$.

Refer to figure6-20, to obtain notch sensitivity as $0.65$.

Substitute $0.65$ for $q$, $2.3$ for $k_t$ in Equation (XXVIII).

    $\begin{array}{c}{k}_f=1+0.65\left(2.3-1\right)\\ =1+0.845\\ \approx 1.85\end{array}$

Substitute $0.7$ for $q_s$ and $1.8$ for $k_{ts}$ in Equation (XXIX).

    $\begin{array}{c}{K}_{fs}=1+0.7\left(1.8-1\right)\\ =1+0.56\\ =1.56\end{array}$

Substitute $72\text{kpsi}$ for $S_{ut}$ in Equation (XXX).

    $\begin{array}{c}{S}_e{}^{\prime }=0.5\times 72\text{kpsi}\\ =\text{36}\text{kpsi}\end{array}$

Refer to table 6-2, “Parameters for main surface condition factor” to obtain $a$ as $2.7$ and $b$ as $-0.265$.

Substitute $2.7$ for $a$, $-0.265$ for $b$  and $72\text{kpsi}$ for $S_{ut}$ in Equation (XXXI).

    $\begin{array}{c}{k}_a=2.7{\left(72\text{kpsi}\right)}^{-0.265}\\ =2.7\left(0.321963\right)\\ =0.869\end{array}$

Substitute $1\text{in}$ for $d$ in Equation (XXXII).

    $\begin{array}{c}{k}_b={\left(\frac{1\text{in}}{0.3}\right)}^{-.107}\\ ={\left(3.334\right)}^{-0.107}\\ =0.879\end{array}$

Substitute $0.869$ for $k_a$, $0.879$ for $k_b$, and $36\text{kpsi}$ for $S_e{}^{\prime }$ in Equation (XXXIII).

    $\begin{array}{c}{S}_E=0.869\times 0.879\times 36\text{kpsi}\\ \text{=0}\text{.763851}\times 36\text{kpsi}\\ =27.49\text{kpsi}\end{array}$

Substitute $1E$ for $d$, $27.49\text{kpsi}$ for $S_e$, $1.85$ for $k_f$, $39500\text{psi}$ for $S_{ut}$, $1.56$ for $k_{fs}$, $192\text{lbf}\cdot \text{in}$ for $T_m$ in Equation (XXXIV).

    $\begin{array}{l}1\text{in}={\left[\frac{16\times n}{\pi }\left(\begin{array}{l}\frac{1}{27500\text{psi}}{\left({\left(4\left(1.85\right)\left(360\text{lbf}\cdot \text{in}\right)\right)}^2\right)}^{\frac{1}{2}}\\ +\frac{1}{39500\text{psi}}{\left({\left(3\left(1.56\right)\left(192\text{lbf}\cdot \text{in}\right)\right)}^2\right)}^{\frac{1}{2}}\end{array}\right)\right]}^{\frac{1}{3}}\\ 1\text{in}={\left[\frac{16\times n}{\pi }\left(\begin{array}{l}\frac{1}{27500\text{psi}}\left(\left(4\left(1.85\right)\left(360\text{lbf}\cdot \text{in}\right)\right)\right)\\ +\frac{1}{39500\text{psi}}\left(\left(3\left(1.56\right)\left(192\text{lbf}\cdot \text{in}\right)\right)\right)\end{array}\right)\right]}^{\frac{1}{3}}\\ n=3.91\end{array}$

Refer to table 7-1, “First iteration estimated for stress concentration factors” to obtain concentration factor for bending stress as $2.14$ and for torsional stress as $3$.

Refer to figure6-20, to obtain notch sensitivity as $0.75$.

Substitute $0.75$ for $q$, $2.14$ for $k_t$ in Equation (XVI).

    $\begin{array}{c}{k}_f=1+0.75\left(2.14-1\right)\\ =1+0.855\\ \approx 1.9\end{array}$

Substitute $0.8$ for $q_s$ and $3$ for $k_{ts}$ in Equation (XVII).

    $\begin{array}{c}{K}_{fs}=1+0.8\left(3-1\right)\\ =1+1.6\\ =2.6\end{array}$

Substitute $72\text{kpsi}$ for $S_{ut}$ in equation (XVIII).

    $\begin{array}{c}{S}_e{}^{\prime }=0.5\times 72\text{kpsi}\\ =\text{36}\text{kpsi}\end{array}$

Refer to table 6-2, “Parameters for main surface condition factor” to obtain $a$ as $2.7$ and $b$ as $-0.265$.

Substitute $2.7$ for $a$, $-0.265$ for $b$  and $68\text{kpsi}$ for $S_{ut}$ in Equation (XXXI).

    $\begin{array}{c}{k}_a=2.7{\left(72\text{kpsi}\right)}^{-0.265}\\ =2.7\left(0.32196\right)\\ =0.869\end{array}$

Substitute $1\text{in}$ for $d$ in Equation (XXXII).

    $\begin{array}{c}{k}_b={\left(\frac{1\text{in}}{0.3}\right)}^{-.107}\\ ={\left(3.334\right)}^{-0.107}\\ =0.879\end{array}$

Substitute $0.869$ for $k_a$, $0.879$ for $k_b$, and $36\text{kpsi}$ for $S_e{}^{\prime }$ in Equation (XXXIII).

    $\begin{array}{c}{S}_E=0.869\times 0.879\times 36\text{kpsi}\\ \text{=0}\text{.763851}\times 36\text{kpsi}\\ =27.49\text{kpsi}\end{array}$

Substitute $1E$ for $d$, $27.49\text{kpsi}$ for $S_e$, $1.85$ for $k_f$, $1089.48\text{lbf}\cdot \text{in}$ for $M$, $39500\text{psi}$ for $S_{ut}$, $1.56$ for $k_{fs}$, $192\text{lbf}\cdot \text{in}$ for $T_m$ in Equation (XXXIV).

    $\begin{array}{l}1\text{in}={\left[\frac{16\times n}{\pi }\left(\begin{array}{l}\frac{1}{27500\text{psi}}{\left({\left(4\left(1.9\right)\left(1089.48\text{lbf}\cdot \text{in}\right)\right)}^2\right)}^{\frac{1}{2}}\\ +\frac{1}{39500\text{psi}}{\left({\left(3\left(2.6\right)\left(192\text{lbf}\cdot \text{in}\right)\right)}^2\right)}^{\frac{1}{2}}\end{array}\right)\right]}^{\frac{1}{3}}\\ 1\text{in}={\left[\frac{16\times n}{\pi }\left(\begin{array}{l}\frac{1}{27500\text{psi}}\left(\left(4\left(1.9\right)\left(1089.48\text{lbf}\cdot \text{in}\right)\right)\right)\\ +\frac{1}{39500\text{psi}}\left(\left(3\left(2.6\right)\left(192\text{lbf}\cdot \text{in}\right)\right)\right)\end{array}\right)\right]}^{\frac{1}{3}}\\ n=1.29\end{array}$

The factor of safety is acceptable for crowned gear.

Thus, crowned gear is more reliable.