Chapter 4, Problem 27P

To determine

The deflection of the shaft at point $A$.

The slope of the shaft at point $A$.

Answer to Problem 27P

The deflection of the shaft at point $A$ is $0.142\text{in}$.

The slope of the shaft at point $A$ is $0.00115\text{rad}$.

Explanation of Solution

Write the expression for the moment of inertia of the shaft.

    $I=\frac{\pi d^4}{64}$                                                                 (I)

Here, the diameter of the shaft is $d$ and the moment of inertia is $I$.

Draw the free body diagram of the beam.

Figure (1)

The free body diagram of the beam in the direction of y-axis is shown figure (1).

Write the deflection equation along y-axis for beam 6 and beam 10 using Table A-9.

    $y_A=\frac{F_{1y}{b}_{1}x}{6EIl}\left(x^2+b_1^2-l^2\right)+\frac{F_{2y}{a}_{2}x}{6EIl}\left(l^2-x^2\right)$                                               (II)

Here, the force component at point A along y-axis is $F_{1y}$, the location of point A from the point C is $b_1$, the distance of point A from left end is $x$, the total length of the beam between point O point C is $l$ , Young modulus of the material is $E$ , moment of inertia of the beam is $I$, the force component at point B along y-axis is $F_{2y}$ and the location of point B from the point C is $a_2$.

Write the force component at point $A$ along y-axis.

    $F_{1y}=-F_1\cos 20°$                                                                                 (III)

Here, the force component at $A$ along y-axis is $F_{1y}$ and the force at point $A$ is $F_1$.

Write the force component at point $B$ along y-axis.

    $F_{2y}=F_2\sin 20°$                                                                                  (IV)

Here, the force component at $B$ along y-axis is $F_{2y}$ and the force at point $B$ is $F_2$.

The free body diagram of the beam in the direction of z-axis is shown below.

Figure (2)

Write the deflection equation along z-axis for beam 6 and beam 10 using Table A-9.

    $z_A=\frac{F_{1z}{b}_{1}x}{6EIl}\left(x^2+b_1^2-l^2\right)+\frac{F_{2z}{a}_{2}x}{6EIl}\left(l^2-x^2\right)$                                           (IV)

Here, the force component at point $A$ along z-axis is $F_{1z}$ and the force component at point $B$ along z-axis is $F_{2z}$.

Write the force component at point $A$ along z-axis.

    $F_{1z}=F_1\sin 20°$                                                                                   (V)

Here, the force component at $A$ along z-axis is $F_{1z}$ and the force at point $A$ is $F_1$.

Write the force component at point $B$ along z-axis.

    $F_{2z}=-F_2\cos 20°$                                                                                 (VI)

Here, the force component at $B$ along z-axis is $F_{2z}$ and the force at point $B$ is $F_2$.

Write the expression for net displacement at point $A$.

    $\delta =\sqrt{y_A^2+z_A^2}$                                                                                       (VII)

Here, the net deflection at point $A$ is $\delta$.

Write the slope equation at along z-axis at point $A$.

    $\begin{array}{c}{\left({\theta }_A\right)}_z=\left(\frac{dy}{dx}\right)\\ =\frac{d}{dx}\left\{\frac{F_{1y}{b}_{1}x}{6EIl}\left(x^2+b_1^2-l^2\right)+\frac{F_{2y}{a}_{2}x}{6EIl}\left(l^2-x^2\right)\right\}\\ =\frac{F_{1y}{b}_1}{6EIl}\left(3x^2+b_1^2-l^2\right)+\frac{F_{2y}{a}_2}{6EIl}\left(l^2-3x^2\right)\end{array}$                            (VIII)

Here, the slope along z-axis at point is ${\left({\theta }_A\right)}_z$.

Write the slope equation at along y-axis at point $A$.

    $\begin{array}{c}{\left({\theta }_A\right)}_y=-\left(\frac{dz}{dx}\right)\\ =-\frac{d}{dx}\left\{\frac{F_{1z}{b}_{1}x}{6EIl}\left(x^2+b_1^2-l^2\right)+\frac{F_{2z}{a}_{2}x}{6EIl}\left(l^2-x^2\right)\right\}\\ =-\frac{F_{1z}{b}_1}{6EIl}\left(3x^2+b_1^2-l^2\right)-\frac{F_{2z}{a}_2}{6EIl}\left(l^2-3x^2\right)\end{array}$                           (IX)

Here, the slope along z-axis at point is ${\left({\theta }_A\right)}_y$.

Write the expression for the net slope at point $A$.

    ${\Theta }_A=\sqrt{{\left({\theta }_A\right)}_z^2+{\left({\theta }_A\right)}_y^2}$                                                                                (X)

Here, the net slope is ${\Theta }_A$.

Conclusion:

Substitute $1.25\text{in}$ for $d$ in the equation (I).

    $\begin{array}{c}I=\frac{\pi \times {\left(1.25\text{in}\right)}^4}{64}\\ =\frac{7.6699}{64}{\text{in}}^{\text{4}}\\ =0.1198{\text{in}}^{\text{4}}\end{array}$

Substitute $300\text{lbf}$ for $F_1$ in the equation (III).

    $\begin{array}{c}{F}_{1y}=-\left(300\text{lbf}\right)\cos 20°\\ =-\left(300\text{lbf}\right)\left(0.9396\right)\\ =-281.91\text{lbf}\end{array}$

Substitute $750\text{lbf}$ for $F_2$ in the equation (IV).

    $\begin{array}{c}{F}_{2y}=\left(750\text{lbf}\right)\sin 20°\\ =\left(750\text{lbf}\right)\left(0.3420\right)\\ =256.52\text{lbf}\end{array}$

Substitute $300\text{lbf}$ for $F_1$ in the equation (V).

    $\begin{array}{c}{F}_{1z}=\left(300\text{lbf}\right)\sin 20°\\ =\left(300\text{lbf}\right)\left(0.3420\right)\\ =102.61\text{lbf}\end{array}$

Substitute $750\text{lbf}$ for $F_2$ in the equation (VI).

    $\begin{array}{c}{F}_{2z}=-\left(750\text{lbf}\right)\cos 20°\\ =-\left(750\text{lbf}\right)\left(0.9396\right)\\ =-704.77\text{lbf}\end{array}$

Substitute $-281.91\text{lbf}$ for $F_{1y}$, $16\text{in}$ for $x$, $14\text{in}$ for $b_1$ , $30\text{in}$ for $l$, $30\times {10}^6\text{psi}$ for $E$, $0.1198{\text{in}}^{\text{4}}$ for $I$ and $256.52\text{lbf}$ for $F_{2y}$ in the equation (II).

    $\begin{array}{c}{y}_A=\left\{\begin{array}{l}\frac{\left(-281.91\text{lbf}\right)\left(14\text{in}\right)\left(16\text{in}\right)}{6\left(30\times {10}^6\text{psi}\right)\left(0.1198{\text{in}}^{\text{4}}\right)\left(30\text{in}\right)}\left({\left(16\text{in}\right)}^2+{\left(14\text{in}\right)}^2-{\left(30\text{in}\right)}^2\right)+\\ \frac{\left(256.52\text{lbf}\right)\left(9\text{in}\right)\left(16\text{in}\right)}{6\left(30\times {10}^6\text{psi}\right)\left(0.1198{\text{in}}^{\text{4}}\right)\left(30\text{in}\right)}\left({\left(30\text{in}\right)}^2-{\left(16\text{in}\right)}^2\right)\end{array}\right\}\\ =\left\{\left(-9.7613\times {10}^{-5}\right)\left(-448\right)+\left(5.709\times {10}^{-5}\right)\left(644\right)\right\}\left(\text{lbf}/\text{psi}\cdot {\text{in}}^2\right)\left(\frac{\text{1}\text{psi}}{\text{1}\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =\left[\left(0.0437\right)+\left(0.0367\right)\right]\text{in}\\ =0.0805\text{in}\end{array}$

Substitute $102.61\text{lbf}$ for $F_{1z}$, $16\text{in}$ for $x$, $14\text{in}$ for $b_1$ , $30\text{in}$ for $l$, $30\times {10}^6\text{psi}$ for $E$, $0.1198{\text{in}}^{\text{4}}$ for $I$ and $-704.77\text{lbf}$ for $F_{2z}$ in the equation (IV).

    $\begin{array}{c}{z}_A=\left\{\begin{array}{l}\frac{\left(102.61\text{lbf}\right)\left(14\text{in}\right)\left(16\text{in}\right)}{6\left(30\times {10}^6\text{psi}\right)\left(0.1198{\text{in}}^{\text{4}}\right)\left(30\text{in}\right)}\left({\left(16\text{in}\right)}^2+{\left(14\text{in}\right)}^2-{\left(30\text{in}\right)}^2\right)+\\ \frac{\left(-704.77\text{lbf}\right)\left(9\text{in}\right)\left(16\text{in}\right)}{6\left(30\times {10}^6\text{psi}\right)\left(0.1198{\text{in}}^{\text{4}}\right)\left(30\text{in}\right)}\left({\left(30\text{in}\right)}^2-{\left(16\text{in}\right)}^2\right)\end{array}\right\}\\ =\left\{\left(3.5529\times {10}^{-5}\right)\left(-448\right)+\left(-1.5687\times {10}^{-4}\right)\left(644\right)\right\}\left(\text{lbf}/\text{psi}\cdot {\text{in}}^2\right)\left(\frac{\text{1}\text{psi}}{\text{1}\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =\left[\left(-0.0159\right)+\left(-0.101\right)\right]\text{in}\\ =-0.1169\text{in}\end{array}$

Substitute $0.0805\text{in}$ for $y_A$ and $-0.1169\text{in}$ for $z_A$ in the equation (VII).

    $\begin{array}{c}\delta =\sqrt{{\left(0.0805\text{in}\right)}^2+{\left(-0.1169\text{in}\right)}^2}\\ =\sqrt{0.00648{\text{in}}^2+0.0136{\text{in}}^2}\\ =0.142\text{in}\end{array}$

Thus, the net displacement of the shaft at point $A$ is $0.142\text{in}$.

Substitute $-281.91\text{lbf}$ for $F_{1y}$ , $16\text{in}$ for $x$ , $14\text{in}$ for $b_1$ , $30\text{in}$ for $l$ , $30\times {10}^6\text{psi}$ for $E$ , $0.1198{\text{in}}^{\text{4}}$ for $I$ and $256.52\text{lbf}$ for $F_{2y}$ in the equation (VIII).

    $\begin{array}{c}{\left({\theta }_A\right)}_z=\left\{\begin{array}{l}\frac{\left(-281.91\text{lbf}\right)\left(14\text{in}\right)}{6\left(30\times {10}^6\text{psi}\right)\left(0.1198{\text{in}}^{\text{4}}\right)\left(30\text{in}\right)}\left(3{\left(16\text{in}\right)}^2+{\left(14\text{in}\right)}^2-{\left(30\text{in}\right)}^2\right)\\ +\frac{\left(256.52\text{lbf}\right)\left(9\text{in}\right)}{6\left(30\times {10}^6\text{psi}\right)\left(0.1198{\text{in}}^{\text{4}}\right)\left(30\text{in}\right)}\left({\left(30\text{in}\right)}^2-3{\left(16\text{in}\right)}^2\right)\end{array}\right\}\\ =\left\{\left(-6.1008\times {10}^{-6}\right)\left(64\right)+\left(3.5687\times {10}^{-6}\right)\left(132\right)\right\}\left(\text{lbf}/\text{psi}\cdot {\text{in}}^2\right)\left(\frac{\text{1}\text{psi}}{\text{1}\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =\left[\left(-0.000390\right)+\left(0.000471\right)\right]\text{in}\\ =8.06\times {10}^{-5}\text{rad}\end{array}$

Substitute $102.61\text{lbf}$ for $F_{1z}$ , $16\text{in}$ for $x$ , $14\text{in}$ for $b_1$ , $30\text{in}$ for $l$ , $30\times {10}^6\text{psi}$ for $E$ , $0.1198{\text{in}}^{\text{4}}$ for $I$ and $-704.77\text{lbf}$ for $F_{2z}$ in the equation (IX).

    $\begin{array}{c}{\left({\theta }_A\right)}_y=\left\{\begin{array}{l}\frac{\left(102.61\text{lbf}\right)\left(14\text{in}\right)}{6\left(30\times {10}^6\text{psi}\right)\left(0.1198{\text{in}}^{\text{4}}\right)\left(30\text{in}\right)}\left(3{\left(16\text{in}\right)}^2+{\left(14\text{in}\right)}^2-{\left(30\text{in}\right)}^2\right)\\ +\frac{\left(-704.77\text{lbf}\right)\left(9\text{in}\right)}{6\left(30\times {10}^6\text{psi}\right)\left(0.1198{\text{in}}^{\text{4}}\right)\left(30\text{in}\right)}\left({\left(30\text{in}\right)}^2-3{\left(16\text{in}\right)}^2\right)\end{array}\right\}\\ =\left\{\left(2.2205\times {10}^{-6}\right)\left(64\right)+\left(-9.8048\times {10}^{-6}\right)\left(132\right)\right\}\left(\text{lbf}/\text{psi}\cdot {\text{in}}^2\right)\left(\frac{\text{1}\text{psi}}{\text{1}\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =\left[\left(0.000142\right)+\left(0.00129\right)\right]\text{in}\\ =0.00115\text{rad}\end{array}$

Substitute $8.06\times {10}^{-5}\text{rad}$ for ${\left({\theta }_A\right)}_z$ and $0.00115\text{rad}$ for ${\left({\theta }_A\right)}_y$ in the equation (X).

    $\begin{array}{c}{\Theta }_A=\sqrt{{\left(8.06\times {10}^{-5}\text{rad}\right)}^2+{\left(0.00115\text{rad}\right)}^2}\\ =\sqrt{6.49\times {10}^{-9}{\text{rad}}^2+1.3225\times {10}^{-6}{\text{rad}}^2}\\ =0.00115\text{rad}\end{array}$

Thus, the net slope of the shaft at point $A$ is $0.00115\text{rad}$.