Chapter 4, Problem 28P

To determine

The deflection of the shaft at point $A$.

The slope of the shaft at point $A$.

Answer to Problem 28P

The deflection of the shaft at point $A$ is $4.14\text{mm}$.

The slope of the shaft at point $A$ is $0.00704\text{rad}$.

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 in figure (1).

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

    $y_A=\frac{F_{1y}{b}_{1}x}{6EIl}\left(x^2+b_1^2-l^2\right)+\frac{F_{2y}{b}_{2}x}{6EIl}\left(x^2+b_2^2-l^2\right)$                                     (II)

Here, the force component at point $A$ along y-axis is $F_{1y}$, the location of point $A$ from right end is $b_1$ , the distance of point $A$ from left end is $x$, the location of point $B$ from right end is $b_2$  the total length of the beam is $l$, Young modulus of the material is $E$ and moment of inertia of the beam is $I$.

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

    $F_{1y}=F_1\sin 20°$                                                                                     (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.

    $F_{2y}=F_2\sin 25°$                                                                                     (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 using Table A-9.

    $z_A=\frac{F_{1z}{b}_{1}x}{6EIl}\left(x^2+b_1^2-l^2\right)+\frac{F_{2z}{b}_{2}x}{6EIl}\left(x^2+b_2^2-l^2\right)$                                          (V)

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.

    $F_{1z}=F_1\cos 20°$                                                                                          (VI)

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.

    $F_{2z}=-F_2\cos 25°$                                                                                       (VII)

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 $A$.

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

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

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

    $\begin{array}{c}{\left({\theta }_A\right)}_z=\left(\frac{dy}{dx}\right)\\ =\frac{d}{dx}\left\{\frac{F_{1y}{b}_{1}x}{6EIl}\left(x^2+b_1^2-l^2\right)+\frac{F_{2y}{b}_{2}x}{6EIl}\left(x^2+b_2^2-l^2\right)\right\}\\ =\frac{F_{1y}{b}_1}{6EIl}\left(3x^2+b_1^2-l^2\right)+\frac{F_{2y}{b}_2}{6EIl}\left(3x^2+b_2^2-l^2\right)\end{array}$                     (IX)

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

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

    $\begin{array}{c}{\left({\theta }_A\right)}_y=-\left(\frac{dz}{dx}\right)\\ =-\left\{\frac{d}{dx}\left[\frac{F_{1z}{b}_{1}x}{6EIl}\left(x^2+b_1^2-l^2\right)+\frac{F_{2z}{b}_{2}x}{6EIl}\left(x^2+b_2^2-l^2\right)\right]\right\}\\ =-\frac{F_{1z}{b}_1}{6EIl}\left(3x^2+b_1^2-l^2\right)-\frac{F_{2z}{b}_2}{6EIl}\left(3x^2+b_2^2-l^2\right)\end{array}$                    (X)

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

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

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

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

Conclusion:

Substitute $50\text{mm}$ for $d$ in the equation (I).

    $\begin{array}{c}I=\frac{\pi \times {\left(50\text{mm}\right)}^4}{64}\\ =\frac{19634954.08}{64}\\ =306.8\times {10}^3{\text{mm}}^{\text{4}}\end{array}$

Substitute $11\times {10}^3\text{N}$ for $F_1$ in the equation (III).

    $\begin{array}{c}{F}_{1y}=\left(11\times {10}^3\text{N}\right)\sin 20°\\ =\left(11\times {10}^3\text{N}\right)\left(0.3420\right)\\ =3762.22\text{N}\end{array}$

Substitute $22.8\times {10}^3\text{N}$ for $F_2$ in the equation (IV).

    $\begin{array}{c}{F}_{2y}=\left(22.8\times {10}^3\text{N}\right)\sin 25°\\ =\left(22.8\times {10}^3\text{N}\right)\left(0.4226\right)\\ =9635.7\text{N}\end{array}$

Substitute $11\times {10}^3\text{N}$ for $F_1$ in the equation (VI).

    $\begin{array}{c}{F}_{1z}=\left(11\times {10}^3\text{N}\right)\cos 20°\\ =\left(11\times {10}^3\text{N}\right)\left(0.9396\right)\\ =10336.62\text{N}\end{array}$

Substitute $22.8\times {10}^3\text{N}$ for $F_2$ in the equation (VII).

    $\begin{array}{c}{F}_{2z}=-\left(22.8\times {10}^3\text{N}\right)\cos 25°\\ =-\left(22.8\times {10}^3\text{N}\right)\left(0.9063\right)\\ =-20663.82\text{N}\end{array}$

Substitute $3762.22\text{N}$ for $F_{1y}$ , $400\text{mm}$ for $x$ , $650\text{mm}$ for $b_1$ , $1050\text{mm}$ for $l$ , $300\text{mm}$ for $b_2$ , $207\times {10}^3\text{MPa}$ for $E$ , $306.8\times {10}^3{\text{mm}}^{\text{4}}$ for $I$ and $9635.7\text{N}$ for $F_{2y}$ in the equation (II).

    $\begin{array}{c}{y}_A=\left\{\begin{array}{l}\frac{\left(3762.22\text{N}\right)\left(650\text{mm}\right)\left(400\text{mm}\right)}{6\left(207\times {10}^3\text{MPa}\right)\left(306.8\times {10}^3{\text{mm}}^{\text{4}}\right)\left(1050\text{mm}\right)}\times \left(\begin{array}{l}{\left(400\text{mm}\right)}^2\\ +{\left(650\text{mm}\right)}^2\\ -{\left(1050\text{mm}\right)}^2\end{array}\right)\\ +\frac{\left(9635.7\right)\left(300\text{mm}\right)\left(400\text{mm}\right)}{6\left(207\times {10}^3\text{MPa}\right)\left(306.8\times {10}^3{\text{mm}}^{\text{4}}\right)\left(1050\text{mm}\right)}\times \\ \left({\left(400\text{mm}\right)}^2+{\left(300\text{mm}\right)}^2-{\left(1050\text{mm}\right)}^2\right)\end{array}\right\}\\ =\left\{\left[\begin{array}{l}\left(2.44\times {10}^{-6}\right)\left(-520000\right)+\\ \left(2.89\times {10}^{-6}\right)\left(-852500\right)\end{array}\right]\left(\text{N}/\text{MPa}\cdot \text{mm}\right)\left(\frac{\text{1}\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\right\}\\ =\left(-1.2688\right)\text{mm}+\left(-2.4637\right)\text{mm}\\ =-3.735\text{mm}\end{array}$

Substitute $10336.62\text{N}$ for $F_{1z}$, $400\text{mm}$ for $x$, $650\text{mm}$ for $b_1$, $1050\text{mm}$ for $l$, $300\text{mm}$ for $b_2$, $207\times {10}^3\text{MPa}$ for $E$, $306.8\times {10}^3{\text{mm}}^{\text{4}}$ for $I$ and $-20663.82\text{N}$ for $F_{2z}$ in the equation (V).

    $\begin{array}{c}{z}_A=\left\{\begin{array}{l}\frac{\left(10336.62\text{N}\right)\left(650\text{mm}\right)\left(400\text{mm}\right)}{6\left(207\times {10}^3\text{MPa}\right)\left(306.8\times {10}^3{\text{mm}}^{\text{4}}\right)\left(1050\text{mm}\right)}\times \\ \left({\left(400\text{mm}\right)}^2+{\left(650\text{mm}\right)}^2-{\left(1050\text{mm}\right)}^2\right)\\ +\frac{\left(-20663.82\right)\left(300\text{mm}\right)\left(400\text{mm}\right)}{6\left(207\times {10}^3\text{MPa}\right)\left(306.8\times {10}^3{\text{mm}}^{\text{4}}\right)\left(1050\text{mm}\right)}\times \\ \left(\left(400\right){\text{mm}}^2+{\left(300\text{mm}\right)}^2-{\left(1050\text{mm}\right)}^2\right)\end{array}\right\}\\ =\left\{\left[\begin{array}{l}\left(6.7171\times {10}^{-6}\right)\left(-520000\right)+\\ \left(-6.1976\times {10}^{-6}\right)\left(-852500\right)\end{array}\right]\left(\text{N}/\text{MPa}\cdot \text{mm}\right)\left(\frac{\text{1}\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\right\}\\ =\left(-3.4928\right)\text{mm}+\left(5.2834\right)\text{mm}\\ =1.791\text{mm}\end{array}$

Substitute $-3.735\text{mm}$ for $y_A$ and $1.791\text{mm}$ for $z_A$ in the equation (VI).

    $\begin{array}{c}\delta =\sqrt{{\left(-3.735\text{mm}\right)}^2+{\left(1.791\text{mm}\right)}^2}\\ =\sqrt{13.95{\text{mm}}^2+3.2076{\text{mm}}^2}\\ =4.14\text{mm}\end{array}$

Thus, the net displacement of the shaft at point $A$ is $4.14\text{mm}$.

Substitute $3762.22\text{N}$ for $F_{1y}$, $400\text{mm}$ for $x$, $650\text{mm}$ for $b_1$ , $1050\text{mm}$ for $l$, $300\text{mm}$ for $b_2$, $207\times {10}^3\text{MPa}$ for $E$, $306.8\times {10}^3{\text{mm}}^{\text{4}}$ for $I$ and $9635.7\text{N}$ for $F_{2y}$ in the equation (IX).

    $\begin{array}{c}{\left({\theta }_A\right)}_z=\left\{\begin{array}{l}\frac{\left(3762.22\text{N}\right)\left(650\text{mm}\right)}{6\left(207\times {10}^3\text{MPa}\right)\left(306.8\times {10}^3{\text{mm}}^{\text{4}}\right)\left(1050\text{mm}\right)}\times \\ \left(3\times {\left(400\text{mm}\right)}^2+{\left(650\text{mm}\right)}^2-{\left(1050\text{mm}\right)}^2\right)\\ +\frac{\left(9635.7\text{N}\right)\left(300\text{mm}\right)}{6\left(207\times {10}^3\text{MPa}\right)\left(306.8\times {10}^3{\text{mm}}^{\text{4}}\right)\left(1050\text{mm}\right)}\times \\ \left(3\times {\left(400\text{mm}\right)}^2+{\left(300\text{mm}\right)}^2-{\left(1050\text{mm}\right)}^2\right)\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\left(6.1121\times {10}^{-9}\right)\left(-200000\right)\right]\text{N}/\text{MPa}\cdot {\text{mm}}^2\\ +\left[\left(7.2250\times {10}^{-9}\right)\left(-532500\right)\right]\text{N}/\text{MPa}\cdot {\text{mm}}^2\end{array}\right\}\\ =\left(\left(-0.00122\right)+\left(-0.00384\right)\right)\text{N}/\text{MPa}\cdot {\text{mm}}^2\left(\frac{1\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =-0.00507\text{rad}\end{array}$

Substitute $10336.62\text{N}$ for $F_{1z}$ , $400\text{mm}$ for $x$ , $650\text{mm}$ for $b_1$ , $1050\text{mm}$ for $l$ , $300\text{mm}$ for $b_2$ , $207\times {10}^3\text{MPa}$ for $E$ , $306.8\times {10}^3{\text{mm}}^{\text{4}}$ for $I$ and $-20663.82\text{N}$ for $F_{2z}$ in the equation (X).

    $\begin{array}{c}{\left({\theta }_A\right)}_y=\left\{\begin{array}{l}\frac{\left(10336.62\text{N}\right)\left(650\text{mm}\right)}{6\left(207\times {10}^3\text{MPa}\right)\left(306.8\times {10}^3{\text{mm}}^{\text{4}}\right)\left(1050\text{mm}\right)}\times \\ \left(3\times {\left(400\text{mm}\right)}^2+{\left(650\text{mm}\right)}^2-{\left(1050\text{mm}\right)}^2\right)\\ +\frac{\left(-20663.82\text{N}\right)\left(300\text{mm}\right)}{6\left(207\times {10}^3\text{MPa}\right)\left(306.8\times {10}^3{\text{mm}}^{\text{4}}\right)\left(1050\text{mm}\right)}\times \\ \left(3\times {\left(400\text{mm}\right)}^2+{\left(300\text{mm}\right)}^2-{\left(1050\text{mm}\right)}^2\right)\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\left(1.6792\times {10}^{-8}\right)\left(-200000\right)\right]\text{N}/\text{MPa}\cdot {\text{mm}}^2\\ +\left[\left(-1.5494\times {10}^{-8}\right)\left(-532500\right)\right]\text{N}/\text{MPa}\cdot {\text{mm}}^2\end{array}\right\}\\ =\left(\left(-0.00335\right)+\left(0.00825\right)\right)\text{N}/\text{MPa}\cdot {\text{mm}}^2\left(\frac{1\text{MPa}}{\text{1}\text{N}/{\text{mm}}^{\text{2}}}\right)\\ =-0.00489\text{rad}\end{array}$

Substitute $-0.00507\text{rad}$ for ${\left({\theta }_A\right)}_z$ and $-0.00489\text{rad}$ for ${\left({\theta }_A\right)}_y$ in the equation (XI).

    $\begin{array}{c}{\Theta }_A=\sqrt{{\left(-0.00507\text{rad}\right)}^2+{\left(-0.00489\text{rad}\right)}^2}\\ =\sqrt{0.0000257{\text{rad}}^2+0.0000239\text{rad}}\\ =0.00704\text{rad}\end{array}$

Thus, the net slope of the shaft at point $A$ is $0.00704\text{rad}$.