Chapter 3, Problem 48P

3–48 and 3–49 The beam shown is loaded in the xy and xz planes.

1. (a) Find the y- and z-components of the reactions at the supports.
2. (b) Plot the shear-force and bending-moment diagrams for the xy and xz planes. Label the diagrams properly and provide the values at key points.
3. (c) Determine the net shear-force and bending-moment at the key points of part (b).
4. (d) Determine the maximum tensile bending stress. For Prob. 3–48, use the cross section given in Prob. 3–34, part (a). For Prob. 3–49, use the cross section given in Prob. 3–39, part (b).

To determine

The $y$-components of the reactions at the supports of the beam.

The $z$-components of the reactions at the supports of the beam.

Answer to Problem 48P

The $y$-components of the reactions at the supports $O$ and of the beam are $0.649\text{kN}$ and $1.949\text{kN}$ respectively.

The $z$-components of the reactions at the supports $O$ and $C$ of the beam are $1.625\text{kN}$ and $1.375\text{kN}$ respectively.

Explanation of Solution

The following figure shows the forces acting on the beam along x-z plane.

Figure-(1)

Write the equilibrium equation of force applied on the beam in $z$-direction.

    $\begin{array}{l}{\displaystyle \sum F_z}=0\\ R_{1z}-\left(1.5\text{kN}\right)-\left(2\left(\sin 30°\right)\text{kN}/\text{m}\right)\left(1.5\text{m}\right)+R_{2z}=0\\ R_{1z}+R_{2z}-3\text{kN}=0\\ R_{1z}+R_{2z}=3\text{kN}\end{array}$                                     (I)

Write the bending moment at the support $O$ in $x\text{-}z$ plane.

    $\begin{array}{l}{\displaystyle \sum M_O}=0\\ \left[\begin{array}{l}\left(1.5\text{kN}\right)\left(0.5\text{m}\right)+\left(2\left(\sin 30°\right)\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(0.5\text{m}+1\text{m}+\frac{1.5\text{m}}{2}\right)\\ -R_{2z}\left(3\text{m}\right)\end{array}\right]=0\\ \left(0.75\text{kN}\cdot \text{m}\right)+\left(1.5\text{kN}\right)\left(2.25\text{m}\right)-R_{2z}\left(3\text{m}\right)=0\end{array}$ (II)

The following figure shows the forces acting on the beam along x-y plane.

Figure-(2)

Write the equilibrium equation of force applied on the beam in $y$-direction.

    $\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_{1y}-\left(2\cos \left(30°\right)\text{kN}/\text{m}\right)\left(1.5\text{m}\right)+R_{2y}=0\\ R_{1y}+R_{2y}-2.598\text{kN}=0\\ R_{1y}+R_{2y}=2.598\text{kN}\end{array}$                                                 (III)

Write the bending moment at the support $O$ in $x\text{-}y$ plane.

    $\begin{array}{l}{\displaystyle \sum M_O}=0\\ -\left(2\cos \left(30°\right)\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(0.5\text{m}+1\text{m}+\frac{1.5\text{m}}{2}\right)+R_{2y}\left(3\text{m}\right)=0\\ -\left(2.598\text{kN}\right)\left(2.25\text{m}\right)+R_{2y}\left(3\text{m}\right)=0\end{array}$          (IV)

Conclusion:

From Equation (II), $R_{2z}=1.375\text{kN}$.

Substitute $1.375\text{kN}$ for $R_{2z}$ in Equation (I).

    $\begin{array}{l}{R}_{1z}+1.375\text{kN}=3\text{kN}\\ R_{1z}=\left(3\text{kN}\right)-\left(1.375\text{kN}\right)\\ R_{1z}=1.625\text{kN}\end{array}$

Thus, the $z$-components of the reactions at the supports $O$ and $C$ of the beam are $1.625\text{kN}$ and $1.375\text{kN}$ respectively.

Simplify Equation (IV).

    $\begin{array}{l}-\left(2.598\text{kN}\right)\left(2.25\text{m}\right)+R_{2y}\left(3\text{m}\right)=0\\ R_{2y}=\frac{5.8455\text{kN}\cdot \text{m}}{3}\\ R_{2y}\approx 1.949\text{kN}\end{array}$

Substitute $1.949\text{kN}$ for $R_{2y}$ in Equation (III).

    $\begin{array}{l}{R}_{1y}+1.949\text{kN}=2.598\text{kN}\\ R_{1y}=\left(2.598\text{kN}\right)-\left(1.949\text{kN}\right)\\ R_{1y}=0.649\text{kN}\end{array}$

Thus, the $y$-components of the reactions at the supports $O$ and of the beam are $0.649\text{kN}$ and $1.949\text{kN}$ respectively.

To determine

The shear force and bending moment diagram of the beam.

Explanation of Solution

Write the expression for the moment at point A.

    $M_{Az}=-R_{2z}\left(2.5\text{m}\right)+\left(2\sin 30°\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(\frac{1.5\text{m}}{2}+1\text{m}\right)$        (V)

Write the expression for the moment at point B.

    $M_{Bz}=-R_{2z}\left(1.5\text{m}\right)+\left(2\sin 30°\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(\frac{1.5\text{m}}{2}\right)$                 (VI)

Write the equation for the shear force.

    $V_z=-2\sin 30°x+R_{2z}$                                                                     (VII)

Write the expression for the maximum bending moment.

    $M_{z\max }=-R_{2z}\left(x\right)+\left(2\sin 30°\text{kN}/\text{m}\right)\left(x\right)\left(\frac{x}{2}\right)$                                 (VIII)

Write the expression for the bending moment at B.

    $M_{By}=R_{2y}\left(1.5\text{m}\right)-\left(2\cos 30°\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(\frac{1.5\text{m}}{2}\right)$                    (IX)

Write the equation for the shear force.

    $V_y=2\cos 30°x-R_{2y}$                                                                          (X)

Write the expression for the maximum bending moment.

    $M_{y\max }=R_{2y}\left(x\right)-\left(2\cos 30°\text{kN}/\text{m}\right)\left(x\right)\left(\frac{x}{2}\right)$                                     (XI)

Conclusion:

Substitute $1.375\text{kN}$ for $R_{2z}$ in Equation (V).

    $\begin{array}{c}{M}_{Az}=-\left(1.375\text{kN}\right)\left(2.5\text{m}\right)+\left(2\sin 30°\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(\frac{1.5\text{m}}{2}+1\text{m}\right)\\ =-\left(1.375\text{kN}\right)\left(2.5\text{m}\right)+\left(1\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(1.75\text{m}\right)\\ =-0.8125\text{kN}\cdot \text{m}\end{array}$

Substitute $1.375\text{kN}$ for $R_{2z}$ in Equation (VI).

    $\begin{array}{c}{M}_{Bz}=-\left(1.375\text{kN}\right)\left(1.5\text{m}\right)+\left(2\sin 30°\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(\frac{1.5\text{m}}{2}\right)\\ =-\left(1.375\text{kN}\right)\left(1.5\text{m}\right)+\left(1\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(0.75\text{m}\right)\\ =-0.9375\text{kN}\cdot \text{m}\end{array}$

The maximum bending moment occur at $V_z=0\text{kN}$.

Substitute $0\text{kN}$ for $V_z$, and $1.375\text{kN}$ for $R_{2z}$ in Equation (VII).

    $\begin{array}{l}0\text{kN}=-2\sin 30°x+1.375\text{kN}\\ 0\text{kN}=x\text{kN}/\text{m}+1.375\text{kN}\\ x=1.375\text{m}\end{array}$

Substitute $1.375\text{m}$ for $x$, and $1.375\text{kN}$ for $R_{2z}$ in Equation (VIII).

    $\begin{array}{c}{M}_{z\max }=-\left(1.375\text{kN}\right)\left(1.375\text{m}\right)+\left(2\sin 30°\text{kN}/\text{m}\right)\left(1.375\text{m}\right)\left(\frac{1.375\text{m}}{2}\right)\\ =-\left(1.375\text{kN}\right)\left(1.375\text{m}\right)+\left(1\text{kN}/\text{m}\right)\left(1.375\text{m}\right)\left(\frac{1.375\text{m}}{2}\right)\\ =-0.9453\text{kN}\cdot \text{m}\end{array}$

The following figure shows the shear force and bending moment diagram of the beam along x-z plane.

Figure-(3)

Substitute $1.949\text{kN}$ for $R_{2y}$ in Equation (IX).

    $\begin{array}{c}{M}_{By}=\left(1.949\text{kN}\right)\left(1.5\text{m}\right)-\left(2\cos 30°\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(\frac{1.5\text{m}}{2}\right)\\ =\left(1.949\text{kN}\right)\left(1.5\text{m}\right)-\left(1.732\text{kN}/\text{m}\right)\left(1.5\text{m}\right)\left(0.75\text{m}\right)\\ =0.975\text{kN}\cdot \text{m}\end{array}$

The maximum bending moment occur at $V_y=0\text{kN}$.

Substitute $0\text{kN}$ for $V_y$, and $1.949\text{kN}$ for $R_{2y}$ in Equation (X).

    $\begin{array}{l}0\text{kN}=2\cos 30°x-1.949\text{kN}\\ 0\text{kN}=1.732x\text{kN}/\text{m}-1.949\text{kN}\\ x=1.125\text{m}\end{array}$

Substitute $1.125\text{m}$ for $x$, and $1.949\text{kN}$ for $R_{2y}$ in Equation (XI).

    $\begin{array}{c}{M}_{y\max }=1.949\text{kN}\left(1.125\text{m}\right)-\left(2\cos 30°\text{kN}/\text{m}\right)\left(1.125\text{m}\right)\left(\frac{1.125\text{m}}{2}\right)\\ =1.949\text{kN}\left(1.125\text{m}\right)-\left(1.732\text{kN}/\text{m}\right)\left(1.125\text{m}\right)\left(0.5625\text{m}\right)\\ =1.096\text{kN}\cdot \text{m}\end{array}$

The following figure shows the shear force and bending moment diagram of the beam along x-y plane.

Figure-(4)

To determine

The net bending moment at $x=3\text{m}$.

The net shear force at $x=3\text{m}$.

The net bending moment at $x=1.5\text{m}$.

The net shear force at $x=1.5\text{m}$.

The net bending moment at $x=0.5\text{m}$.

The net shear force at $x=0.5\text{m}$.

The net bending moment at $x=0\text{m}$.

The net shear force at $x=0\text{m}$.

Answer to Problem 48P

The net bending moment at $x=3\text{m}$ is $0\text{kN}\cdot \text{m}$.

The net shear force at $x=3\text{m}$ is $2.385\text{kN}$.

The net bending moment at $x=1.5\text{m}$ is $1.352\text{kN}\cdot \text{m}$.

The net shear force at $x=1.5\text{m}$ is $0.661\text{kN}$.

The net bending moment at $x=0.5\text{m}$ is $0.8749\text{kN}\cdot \text{m}$.

The net shear force at $x=0.5\text{m}$ is $1.749\text{kN}$.

The net bending moment at $x=0\text{m}$ is $0\text{kN}\cdot \text{m}$.

The net shear force at $x=0\text{m}$ is $1.749\text{kN}$.

Explanation of Solution

Write the expression for the net shear force on the beam.

    $V=\sqrt{V_z^2+V_y^2}$                                                                        (XII)

Write the expression for the net bending moment on the beam.

    $M=\sqrt{M_z^2+M_y^2}$                                                                  (XIII)

Conclusion:

The following table shows the magnitudes of bending moment and shear force along x-y and x-z plane from the shear force and bending moment diagram.

$x\left(\text{m}\right)$	$V_z\left(\text{kN}\right)$	$V_y\left(\text{kN}\right)$	$M_y\left(\text{kN}\cdot \text{m}\right)$	$M_z\left(\text{kN}\cdot \text{m}\right)$
$0$	$-1.625$	$0.6491$	$0$	$0$
$0.5$	$-1.625$	$0.6491$	$0.8125$	$0.3246$
$1.5$	$-0.125$	$0.6491$	$0.9375$	$0.9737$
$3$	$1.375$	$-1.949$	$0$	$0$

At $x=0\text{m}$:

Substitute $-1.625\text{kN}$ for $V_z$, and $0.6491\text{kN}$ for $V_y$ in Equation (XII).

    $\begin{array}{c}{V}_0=\sqrt{{\left(-1.625\text{kN}\right)}^2+{\left(0.6491\text{kN}\right)}^2}\\ =\sqrt{2.640625+0.42133}\text{kN}\\ =1.749\text{kN}\end{array}$

Thus, the net shear force at $x=0\text{m}$ is $1.749\text{kN}$.

Substitute $0\text{kN}\cdot \text{m}$ for $M_z$, and $0\text{kN}\cdot \text{m}$ for $M_y$ in Equation (XIII).

    $\begin{array}{c}{M}_0=\sqrt{{\left(0\text{kN}\cdot \text{m}\right)}^2+{\left(0\text{kN}\cdot \text{m}\right)}^2}\\ =0\text{kN}\cdot \text{m}\end{array}$

Thus, the net bending moment at $x=0\text{m}$ is $0\text{kN}\cdot \text{m}$.

At $x=0.5\text{m}$:

Substitute $-1.625\text{kN}$ for $V_z$, and $0.6491\text{kN}$ for $V_y$ in Equation (XII).

    $\begin{array}{c}{V}_{0.5}=\sqrt{{\left(-1.625\text{kN}\right)}^2+{\left(0.6491\text{kN}\right)}^2}\\ =\sqrt{2.640625+0.42133}\text{kN}\\ =1.749\text{kN}\end{array}$

Thus, the net shear force at $x=0.5\text{m}$ is $1.749\text{kN}$.

Substitute $0.8125\text{kN}\cdot \text{m}$ for $M_z$, and $0.3246\text{kN}\cdot \text{m}$ for $M_y$ in Equation (XIII).

    $\begin{array}{c}{M}_{0.5}=\sqrt{{\left(0.8125\text{kN}\cdot \text{m}\right)}^2+{\left(0.3246\text{kN}\cdot \text{m}\right)}^2}\\ =\sqrt{\left(0.66015+0.10536\right){\text{kN}}^2\cdot {\text{m}}^2}\\ =0.8749\text{kN}\cdot \text{m}\end{array}$

Thus, the net bending moment at $x=0.5\text{m}$ is $0.8749\text{kN}\cdot \text{m}$.

At $x=1.5\text{m}$:

Substitute $-0.125\text{kN}$ for $V_z$, and $0.6491\text{kN}$ for $V_y$ in Equation (XII).

    $\begin{array}{c}{V}_{1.5}=\sqrt{{\left(-0.125\text{kN}\right)}^2+{\left(0.6491\text{kN}\right)}^2}\\ =\sqrt{0.015625+0.42133}\text{kN}\\ =0.661\text{kN}\end{array}$

Thus, the net shear force at $x=1.5\text{m}$ is $0.661\text{kN}$.

Substitute $0.9737\text{kN}\cdot \text{m}$ for $M_z$, and $0.9375\text{kN}\cdot \text{m}$ for $M_y$ in Equation (XIII).

    $\begin{array}{c}{M}_{1.5}=\sqrt{{\left(0.9737\text{kN}\cdot \text{m}\right)}^2+{\left(0.9375\text{kN}\cdot \text{m}\right)}^2}\\ =\sqrt{1.82699}\text{kN}\cdot \text{m}\\ =1.352\text{kN}\cdot \text{m}\end{array}$

Thus, the net bending moment at $x=1.5\text{m}$ is $1.352\text{kN}\cdot \text{m}$.

At $x=3\text{m}$:

Substitute $1.375\text{kN}$ for $V_z$, and $-1.949\text{kN}$ for $V_y$ in Equation (XII).

    $\begin{array}{c}{V}_3=\sqrt{{\left(1.375\text{kN}\right)}^2+{\left(-1.949\text{kN}\right)}^2}\\ =\sqrt{1.890625+3.798601}\text{kN}\\ =2.385\text{kN}\end{array}$

Thus, the net shear force at $x=3\text{m}$ is $2.385\text{kN}$.

Substitute $0\text{kN}\cdot \text{m}$ for $M_z$, and $0\text{kN}\cdot \text{m}$ for $M_y$ in Equation (XIII).

    $\begin{array}{c}{M}_3=\sqrt{{\left(0\text{kN}\cdot \text{m}\right)}^2+{\left(0\text{kN}\cdot \text{m}\right)}^2}\\ =0\text{kN}\cdot \text{m}\end{array}$

Thus, the net bending moment at $x=3\text{m}$ is $0\text{kN}\cdot \text{m}$.

To determine

The maximum tensile bending stress on the beam.

Answer to Problem 48P

The maximum tensile bending stress on the beam is $73.52\text{MPa}$.

Explanation of Solution

The following diagram shows the cross-section of the beam.

Figure-(5)

Write the expression for the second moment of area of the section about z-axis.

  $I_z=I_a-2I_b$                                                                       (XIV)

Here, the second moment area of the area $a$ is $I_a$ and the second moment of area of the area $b$ is $I_b$.

Write the moment of inertia of the rectangular section.

    $I=\frac{B{H}^3}{12}$                                                                               (XV)

Here total height of the section is $H$, and the width is $H$.

Write the expression for the moment of inertia about y-axis.

    $I_y=2\left(\frac{h{B}^3}{12}\right)+\frac{h_2{b}_2^3}{12}$                                                             (XVI)

Here, the height of the flange is $h$, height of the web is $h_2$, and the width is $b_2$.

Write the expression for maximum tensile bending stress.

    ${\sigma }_{\max }=\frac{-M_z{y}_A}{I_z}+\frac{M_y{z}_A}{I_y}$                                                       (XVII)

Conclusion:

Substitute $40\text{mm}$ for $B$ and $75\text{mm}$ for $H$ in Equation (XV).

    $\begin{array}{c}{I}_a=\frac{\left(40\text{mm}\right){\left(75\text{mm}\right)}^3}{12}\\ =1406250{\text{mm}}^4\end{array}$

Substitute $34\text{mm}$ for $B$ and $25\text{mm}$ for $H$ in Equation (XV).

    $\begin{array}{c}{I}_b=\frac{\left(34\text{mm}\right){\left(25\text{mm}\right)}^3}{12}\\ =44270.83{\text{mm}}^4\end{array}$

Substitute $1406250{\text{mm}}^4$ for $I_a$ and $44270.83{\text{mm}}^4$ for $I_b$ in Equation (XIV).

    $\begin{array}{c}{I}_z=1406250{\text{mm}}^4-2\times 44270.83{\text{mm}}^4\\ =\left(1317708.34{\text{mm}}^4\right)\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)\\ \approx 1.32\times {10}^{-6}{\text{m}}^4\end{array}$

Substitute $40\text{mm}$ for $B$, $25\text{mm}$ for $h$, $25\text{mm}$ for $h_2$, and $6\text{mm}$ for $b_2$ in Equation (XVI).

    $\begin{array}{c}{I}_y=2\left(\frac{\left(25\text{mm}\right){\left(40\text{mm}\right)}^3}{12}\right)+\frac{\left(25\text{mm}\right){\left(6\text{mm}\right)}^3}{12}\\ =2\left(133333.33{\text{mm}}^4\right)+450{\text{mm}}^4\\ =\left(267116.66{\text{mm}}^4\right)\left(\frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)\\ \approx 2.67\times {10}^{-7}{\text{m}}^4\end{array}$

Substitute $1.075\text{kN}\cdot \text{m}$ for $M_z$, $-0.9408\text{kN}\cdot \text{m}$ for $M_y$, $-37.5\text{mm}$ for $y_A$, $-20\text{mm}$ for $z_A$, $1.32\times {10}^{-6}{\text{m}}^4$ for $I_z$, and $2.67\times {10}^{-7}{\text{m}}^4$ for $I_y$ in Equation (XVII).

    $\begin{array}{c}{\sigma }_{\max }=\frac{-\left(1.075\text{kN}\cdot \text{m}\right)\left(-37.5\text{mm}\right)}{1.32\times {10}^{-6}{\text{m}}^4}+\frac{\left(-0.9408\text{kN}\cdot \text{m}\right)\left(-20\text{mm}\right)}{2.67\times {10}^{-7}{\text{m}}^4}\\ =\left[\begin{array}{l}\frac{-\left(1.075\text{kN}\cdot \text{m}\right)\left(-37.5\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{1.32\times {10}^{-6}{\text{m}}^4}\\ +\frac{\left(-0.9408\text{kN}\cdot \text{m}\right)\left(-20\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{2.67\times {10}^{-7}{\text{m}}^4}\end{array}\right]\\ =\left[3.0539\times {10}^3\text{kN}/{\text{m}}^{\text{2}}+7.0471\times {10}^4\text{kN}/{\text{m}}^{\text{2}}\right]\left(\frac{1\text{MPa}}{{10}^3\text{kN}/{\text{m}}^{\text{2}}}\right)\\ =73.52\text{MPa}\end{array}$

Thus, the maximum tensile bending stress on the beam is $73.52\text{MPa}$.