Chapter 3, Problem 76P

(a)

To determine

The free body diagram of the shaft.

The reactions at $A$ and $B$.

Answer to Problem 76P

The free body-diagram of the shaft is shown below.

The reactions at $A$ and $B$ are:

• The reaction at $A$ in $y-$ direction is $639.43\text{lbf}$.
• The reaction at $A$ in $z-$ direction is $1513.7\text{lbf}$.
• The reaction at $B$ in $y-$ direction is $276.6\text{lbf}$.
• The reaction at $B$ in $z-$ direction is $705.7\text{lbf}$.

Explanation of Solution

The figure below shows the arrangement of the shafts.

Figure (1)

The free body diagram of the arrangement of shafts is shown below.

Figure (2)

Write the expression of moment at $B$ in $z\text{-}$ direction.

    $\begin{array}{l}{\displaystyle \sum {\left(M_B\right)}_z}=0\\ \left[\begin{array}{l}\left(l_{AB}\right)A_y+\left(l_{PE}\times {\left(F_x\right)}_E\right)\\ -\left(l_{BP}\times {\left(F_y\right)}_E\right)\end{array}\right]=0\\ A_y=\frac{-\left(d_E\times {\left(F_x\right)}_E\right)+\left(\left(l_{AP}+l_{AB}\right){\left(F_y\right)}_E\right)}{\left(l_{AB}\right)}\end{array}$                                    (I)

Here, the reaction at $A$ in $y$ direction is $A_y$, the length of the shaft $AB$ is $l_{AB}$, the length of the shaft $AP$ is $l_{AP}$, the component of force at $E$ in $x\text{-}$ direction is ${\left(F_x\right)}_E$, the component of force at $E$ in $y\text{-}$ direction is ${\left(F_y\right)}_E$ and the diameter of Bevel gear at $E$ is $d_E$.

Write the expression of moment at $A$ in $z\text{-}$ direction.

    $\begin{array}{l}{\displaystyle \sum {\left(M_A\right)}_z}=0\\ \left[\begin{array}{l}\left(l_{AB}\right)B_y-\left(d_E\times {\left(F_x\right)}_E\right)\\ +\left(l_{AP}\times {\left(F_y\right)}_E\right)\end{array}\right]=0\\ B_y=\frac{-\left(d_E\times {\left(F_x\right)}_E\right)+\left(l_{AP}\times {\left(F_y\right)}_E\right)}{\left(l_{AB}\right)}\end{array}$                                       (II)

Here, the reaction at $B$ in $\text{y-}$ direction is $B_y$.

Write the expression of moment at $B$ in $\text{y-}$ direction.

    $\begin{array}{l}{\left({\displaystyle \sum M_B}\right)}_y=0\\ A_z\left(l_{AB}\right)-\left(l_{AB}+l_{AP}\right){\left(F_z\right)}_E=0\\ A_z=\frac{\left(l_{AB}+l_{AP}\right){\left(F_z\right)}_E}{\left(l_{AB}\right)}\end{array}$                                              (III)

Here, the reaction at $A$ in $z\text{-}$ direction is $A_z$.

Write the expression of moment at $A$ in $\text{y-}$ direction.

    $\begin{array}{l}{\left({\displaystyle \sum M_A}\right)}_y=0\\ \left(l_{AB}\right)B_z-\left(l_{AP}\times {\left(F_z\right)}_E\right)=0\\ B_z=\frac{\left(l_{AP}\times {\left(F_z\right)}_E\right)}{\left(l_{AB}\right)}\end{array}$                                                   (IV)

Here, the reaction at $B$ in $z\text{-}$ direction is $B_z$.

Conclusion:

Substitute $92.8\text{lbf}$ for ${\left(F_x\right)}_E$, $1.3\text{in}$ for $d_E$, $362.8\text{lbf}$ for ${\left(F_y\right)}_E$, $2.62\text{in}$ for $l_{AP}$ and $3\text{in}$ for $l_{AB}$ in Equation (I).

    $\begin{array}{c}{A}_y=\frac{-\left(\left(1.3\text{in}\right)\left(92.8\text{lbf}\right)\right)+\left(\left(3\text{in}+\text{2}\text{.62}\text{in}\right)\left(362.8\text{lbf}\right)\right)}{\left(3\text{in}\right)}\\ =\frac{1918.296\text{lbf}\cdot \text{in}}{3\text{in}}\\ =639.43\text{lbf}\end{array}$

Thus, the reaction at $A$ in $y-$ direction is $639.43\text{lbf}$.

Substitute $92.8\text{lbf}$ for ${\left(F_x\right)}_E$, $1.3\text{in}$ for $d_E$, $362.8\text{lbf}$ for ${\left(F_y\right)}_E$, $2.62\text{in}$ for $l_{AP}$ and $3\text{in}$ for $l_{AB}$ in Equation (II).

    $\begin{array}{c}{B}_y=\frac{-\left(\left(1.3\text{in}\right)\left(92.8\text{lbf}\right)\right)+\left(\left(2.62\text{in}\right)\left(362.8\text{lbf}\right)\right)}{3\text{in}}\\ =\frac{829.89\text{lbf}\cdot \text{in}}{3\text{in}}\\ \approx 276.6\text{lbf}\end{array}$

Thus, the reaction at $B$ in $y-$ direction is $276.6\text{lbf}$.

Substitute $808\text{lbf}$ for ${\left(F_z\right)}_E$, $3\text{in}$ for $l_{AB}$ and $2.62\text{in}$ for $l_{AP}$ in Equation (III).

    $\begin{array}{c}{A}_z=\frac{\left(\left(3\text{in}+2.62\text{in}\right)\left(808\text{lbf}\right)\right)}{\left(3\text{in}\right)}\\ =\frac{4540.96\text{lbf}}{\left(3\text{in}\right)}\\ =1513.653\text{lbf}\\ \approx 1513.7\text{lbf}\end{array}$

Thus, the reaction at $A$ in $z\text{-}$ direction is $1513.7\text{lbf}$.

Substitute $808\text{lbf}$ for ${\left(F_z\right)}_E$, $3\text{in}$ for $l_{AB}$ and $2.62\text{in}$ for $l_{AP}$ in Equation (IV).

    $\begin{array}{c}{B}_z=\frac{\left(\left(2.62\text{in}\right)\left(808\text{lbf}\right)\right)}{\left(3\text{in}\right)}\\ =\frac{2116.96\text{lbf}\cdot \text{in}}{3\text{in}}\\ =705.65\text{lbf}\\ \approx 705.7\text{lbf}\end{array}$

Thus, the reaction at $B$ in $z\text{-}$ direction is $705.7\text{lbf}$.

(b)

To determine

The shear force and bending moment diagrams.

Answer to Problem 76P

The following figure shows the shear force and bending moment diagram for the forces in x-y plane.

The following figure shows the shear force and bending moment diagram for the forces in x-z plane.

Explanation of Solution

The calculations for shear force diagram in $y-$ direction on shaft $BA$.

Write the expression of Shear force at $B$ in $y-$ direction.

    $S{F}_{By}=-B_y$                                                                                                 (V)

Here, the shear at $B$ in $y-$ direction is $S{F}_{By}$.

Write the expression of Shear force at $A$ in $y-$ direction.

    $S{F}_{Ay}=S{F}_{By}+\left(A_y\right)$                                                                                   (VI)

Here, the shear force at $A$ in $y-$ direction is $S{F}_{Ay}$.

Write the expression of Shear force at $P$ in $y-$ direction.

    $S{F}_{Py}=S{F}_{Ay}-{\left(F_y\right)}_E$                                                                                (VII)

Here, the shear force at $A$ in $y-$ direction is $S{F}_{Py}$.

The calculations for bending moment diagram in $x-$ direction on shaft $BA$.

It is know that the bending moment at the supports of the simply supported beam is zero.

Write the bending moment at $B$ and $P$ in $x-$ direction.

    $\begin{array}{l}{M}_{Px}=0\\ M_{Bx}=0\end{array}$

Here, the bending moment at $B$ in $x-$ direction is $M_{Bx}$ and the bending moment at $P$ in $x-$ direction is $M_{Px}$.

Write the expression of bending moment at $A$ in $y-$ direction due to shear force at $B$.

    $M_{Ay}=B_y\times l_{AB}$                                                                          (VIII)

Here, the bending moment at $A$ in $y-$ direction due to shear force at $B$ is $M_{Ay}$.

Write the expression of bending moment at $P$ in $y-$ direction due to shear force at $P$.

    $M_{Py}=B_y\left(l_{PA}+l_{AB}\right)-A_y\times l_{AP}$                                                     (IX)

Here, the bending moment at $A$ in $y-$ direction due to shear force at $P$ is $M_{Py}$.

The calculations for shear force diagram in $z\text{-}$ direction on shaft $BA$.

Write the expression of Shear force at $B$ in $z\text{-}$ direction.

    $S{F}_{Bz}=-B_z$                                                                                       (X)

Here, the shear at $B$ in $z\text{-}$ direction is $S{F}_{Bz}$.

Write the expression of Shear force at $A$ in $z\text{-}$ direction.

    $S{F}_{Az}=A_z-S{F}_{Bz}$                                                                           (XI)

Here, the shear force at $A$ in $z\text{-}$ direction is $S{F}_{Az}$.

Write the expression of Shear force at $P$ in $z\text{-}$ direction.

    $S{F}_{Pz}=S{F}_{Az}-{\left(F_z\right)}_E$                                                                    (XII)

Here, the shear force at $P$ in $z\text{-}$ direction is $S{F}_{Pz}$.

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

Write the bending moment at $B$ in $x\text{-}$ direction.

    $M_{Bx}=0$

Here, the bending moment at $B$ in $x\text{-}$ direction is $M_{Bx}$.

Write the bending moment at $P$ in $x\text{-}$ direction.

    $M_{Px}=0$

Here, the bending moment at $P$ in $x\text{-}$ direction is $M_{Px}$.

Write the expression of bending moment at $A$ in $\text{z-}$ direction due to shear force at $B$.

    $M_{Az}=\left(B_z\times l_{AB}\right)$                                                                            (XIII)

Here, the bending moment at $A$ in $x\text{-}$ direction due to shear force at $B$ is $M_{Az}$.

Conclusion:

Substitute $276.6\text{lbf}$ for $B_y$ in Equation (V).

    $S{F}_{By}=-176.6\text{lbf}$

Substitute $-176.6\text{lbf}$ for $S{F}_{By}$ and $639.4\text{lbf}$ for $A_y$ in Equation (VI).

    $\begin{array}{c}S{F}_{Ay}=-276.6\text{lbf}+639.4\text{lbf}\\ =362.8\text{lbf}\end{array}$

Substitute $362.8\text{lbf}$ for $S{F}_{Ay}$ and $362.8\text{lbf}$ for ${\left(F_y\right)}_E$ in Equation (VII).

    $\begin{array}{c}S{F}_{py}=362.8\text{lbf}-362.8\text{lbf}\\ =0\text{lbf}\end{array}$

Substitute $276.6\text{lbf}$ for $B_y$ and $3\text{in}$ for $l_{BA}$ in Equation (VIII).

    $\begin{array}{c}{M}_{Ay}=\left(176.6\text{lbf}\right)\left(3\text{in}\right)\\ =829.8\text{lbf}\cdot \text{in}\\ \approx 829.8\text{lbf}\cdot \text{in}\end{array}$

Substitute $639.4\text{lbf}$ for $A_y$, $3\text{in}$ for $l_{AB}$  $276.6\text{lbf}$ for $B_y$ and $2.62\text{in}$ for $l_{AP}$ in Equation (IX).

    $\begin{array}{c}{M}_{Py}=276.6\text{lbf}\left(3\text{in}+2.62\text{in}\right)-\left(639.4\text{lbf}\right)\left(2.62\text{in}\right)\\ =1554.492\text{lbf}\cdot \text{in}-1675.228\text{lbf}\cdot \text{in}\\ =-120.7\text{lbf}\cdot \text{in}\end{array}$

The figure-(3) shows the shear force and bending moment diagram for the forces in x-y plane.

Figure (3)

Substitute $705.7\text{lbf}$ for $B_z$ in Equation (X).

    $S{F}_{Bz}=-705.7\text{lbf}$

Substitute $705.7\text{lbf}$ for $B_z$ and $1513.7\text{lbf}$ for $A_z$ in Equation (XI).

    $\begin{array}{c}S{F}_{Az}=1513.7\text{lbf}-705.7\text{lbf}\\ =808\text{lbf}\end{array}$

Substitute $808\text{lbf}$ for $S{F}_{Az}$ and $808\text{lbf}$ for ${\left(F_z\right)}_E$ in Equation (XII).

    $\begin{array}{c}S{F}_{Pz}=808\text{lbf}-808\text{lbf}\\ =0\text{lbf}\end{array}$

Substitute $705.7\text{lbf}$ for $B_z$ and $3\text{in}$ for $l_{AB}$ in Equation (XIII).

    $\begin{array}{c}{M}_{Az}=\left(705.7\text{lbf}\right)\left(3\text{in}\right)\\ =2117.1\text{lbf}\cdot \text{in}\end{array}$

The figure-(4) shows the shear force and bending moment diagram for the forces in x-z plane.

Figure (4)

(c)

To determine

The torsional shear stress for critical stress element.

The bending stress for critical stress element.

The axial stress for critical stress element.

Answer to Problem 76P

The torsional shear stress for critical stress element is $7850.12\text{psi}$.

The bending stress for critical stress element is $-33990\text{psi}$.

The axial stress for critical stress element is $-153\text{psi}$.

Explanation of Solution

Write the expression of maximum torque acting on the shaft $BC$.

    $T={\left(F_z\right)}_E\times d_E$                                                                          (XIV)

Here, the maximum torque acting on the shaft $BC$ is $T$.

Write the expression of maximum bending moment acting on the shaft $BC$.

    $M=\sqrt{{\left(M_{Ay}\right)}^2+{\left(M_{Az}\right)}^2}$                                                            (XV)

Here, the maximum bending moment acting on the shaft $BC$ is $M$.

Write the expression of torsional shear stress for critical stress element.

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

Here, the torsional shear stress for critical stress element is $\tau$ and diameter of the shaft is $d$.

Write the expression of bending stress for critical stress element.

    ${\sigma }_b=\pm \frac{32M}{\pi d^3}$                                                                               (XVII)

Here, the bending stress for critical stress element is ${\sigma }_b$.

Write the expression of axial stress for critical stress element.

    ${\sigma }_a=-\frac{4F}{\pi d^2}$                                                                                    (XVIII)

Here, the axial stress for critical stress element is ${\sigma }_a$.

Conclusion:

Substitute $808\text{lbf}$ for ${\left(F_z\right)}_E$ and $1.3\text{in}$ for $d_E$ in Equation (XIV).

    $\begin{array}{c}T=\left(808\text{lbf}\right)\left(1.3\text{in}\right)\\ =1050.4\text{lbf}\cdot \text{in}\end{array}$

Substitute $829.8\text{lbf}\cdot \text{in}$ for $M_{Ay}$ and $2117\text{lbf}\cdot \text{in}$ for $M_{Az}$ in Equation (XV).

    $\begin{array}{c}M=\sqrt{{\left(829.8\text{lbf}\cdot \text{in}\right)}^2+{\left(2117\text{lbf}\cdot \text{in}\right)}^2}\\ =\sqrt{5170257.04{\text{lbf}}^2\cdot {\text{in}}^2}\\ =2273.81\text{lbf}\cdot \text{in}\\ \approx 2274\text{lbf}\cdot \text{in}\end{array}$

Substitute $1050.4\text{lbf}\cdot \text{in}$ for $T$ and $0.88\text{in}$ for $d$ in Equation (XVI).

    $\begin{array}{c}\tau =\frac{16\times \left(1050.4\text{lbf}\cdot \text{in}\right)}{\pi {\left(0.88\text{in}\right)}^3}\\ =7850.12\text{lbf}/{\text{in}}^{\text{2}}\left(\frac{\text{1}\text{psi}}{\text{1}\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =7850.12\text{psi}\end{array}$

Thus, the torsional shear stress for critical stress element is $7850.12\text{psi}$.

Substitute $2274\text{lbf}\cdot \text{in}$ for $M$ and $0.88\text{in}$ for $d$ in Equation (XVII).

    $\begin{array}{c}{\sigma }_b=\pm \frac{32\times \left(2274\text{lbf}\cdot \text{in}\right)}{\pi \times {\left(0.88\text{in}\right)}^3}\\ =\pm 33989.32\text{lbf}/{\text{in}}^{\text{2}}\left(\frac{\text{1}\text{psi}}{\text{1}\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ \approx \pm 33990\text{psi}\end{array}$

Thus, the bending stress for critical stress element is $-33990\text{psi}$.

Substitute $92.8\text{lbf}$ for $F$ and $0.88\text{in}$ for $d$ in Equation (XVIII).

    $\begin{array}{c}{\sigma }_a=-\frac{4\times \left(92.8\text{lbf}\right)}{\pi {\left(0.88\text{in}\right)}^2}\\ =-152.578\text{lbf}/{\text{in}}^{\text{2}}\left(\frac{\text{1}\text{psi}}{\text{1}\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ \approx -153\text{psi}\end{array}$

Thus, the axial stress for critical stress element is $-153\text{psi}$.

(d)

To determine

The principal stresses for critical stress element.

The maximum shear stress for critical stress element.

Answer to Problem 76P

The maximum and minimum principal stresses for critical stress element are $1718.40\text{psi}$ and $-35861.4\text{psi}$ respectively.

The maximum shear stress for critical stress element is $18789.9041\text{psi}$.

Explanation of Solution

Write the expression of maximum bending stress on the critical stress element.

    ${\sigma }_{\max }=-{\sigma }_b+{\sigma }_a$                                                                             (XIX)

Here, the maximum bending stress on the critical stress element is ${\sigma }_{\max }$.

Write the expression of principal stresses on the critical stress element.

    ${\sigma }_1,{\sigma }_2=\frac{{\sigma }_{\max }}{2}\pm \sqrt{{\left(\frac{{\sigma }_{\max }}{2}\right)}^2+{\tau }^2}$                                                        (XX)

Here, the maximum and minimum principal stresses on the critical stress element are ${\sigma }_1$ and ${\sigma }_2$ respectively.

Write the expression of maximum shear stress on the critical stress element.

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

Here, the maximum shear stress on the critical stress element is ${\tau }_{\max }$.

Conclusion:

Substitute $-33990\text{psi}$ for ${\sigma }_b$ and $-153\text{psi}$ for ${\sigma }_a$ in Equation (XIX).

    $\begin{array}{c}{\sigma }_{\max }=\left(-33990\text{psi}\right)+\left(-153\text{psi}\right)\\ =-34143\text{psi}\end{array}$

Substitute $34143\text{psi}$ for ${\sigma }_{\max }$ and $7850.12\text{psi}$ for $\tau$ in Equation (XX).

    $\begin{array}{c}{\sigma }_1,{\sigma }_2=\frac{\left(-34143\text{psi}\right)}{2}\pm \sqrt{{\left(\frac{34143\text{psi}}{2}\right)}^2+{\left(7850.12\text{psi}\right)}^2}\\ {\sigma }_1,{\sigma }_2=-\left(17071.5\text{psi}\right)\pm \left(18789.90\text{psi}\right)\\ {\sigma }_1=1718.40\text{psi}\\ {\sigma }_2=-35861.4\text{psi}\end{array}$

Thus, the maximum and minimum principal stresses for critical stress element are $1718.40\text{psi}$ and $-35861.4\text{psi}$ respectively.

Substitute $-34143\text{psi}$ for ${\sigma }_{\max }$ and $7850.12\text{psi}$ for $\tau$ in Equation (XXI).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{\left(-34143\text{psi}\right)}{2}\right)}^2+{\left(7850.12\text{psi}\right)}^2}\\ =\sqrt{353060496.3{\text{psi}}^2}\\ =18789.9041\text{psi}\end{array}$

Thus, the maximum shear stress for critical stress element is $18789.9041\text{psi}$.