Chapter 4, Problem 57P

To determine

The required pair of structural steel channels.

Answer to Problem 57P

The required pair of structural steel channels is $5\text{in}\times \text{6}\text{.7}\text{lbf}/\text{ft}$.

Explanation of Solution

The Figure (1) shows the free body diagram of the beam.

Figure (1)

Here, the reaction force at point $O$ is $R_O$ and at the point $C$ is $R_C$.

Write the net moment at point O.

    $R_C\times l-\left(F_1\times a\right)-\left(F_2\times b\right)=0$                                                    (I)

Here, the reaction force at point C is $R_C$, the length of the beam is $l$, the reaction force at point A is $F_1$, the distance between the point Aand point O is $a$, the reaction force at point B is $F_2$, the distance between the point Band point O is $b$.

Write the expression for the net force on the beam.

    $R_O+R_C-F_1-F_2=0$                                                            (II)

Here, the net reaction force at point O is $R_O$.

Refer to Table 3-1 “Singularity (Macaulay) functions”.

Write the expression for load intensity acting on the beam.

    $q=R_O{\langle x\rangle }^{-1}-F_1{\langle x-a\rangle }^{-1}-F_2{\langle x-b\rangle }^{-1}+R_C{\langle x-l\rangle }^{-1}$   (III)

Here, the load intensity on the beam is $q$, the distance from the left end is $x$.

Write the expression for the moment.

    $M={\displaystyle \int {\displaystyle \int qdx}}$                                                                     (IV)

Here, the net bending moment is $M$.

Substitute the value from Equation (III) to Equation (IV).

    $\begin{array}{c}M={\displaystyle \int {\displaystyle \int R_O{\langle x\rangle }^{-1}-F_1{\langle x-a\rangle }^{-1}-F_2{\langle x-b\rangle }^{-1}+R_C{\langle x-l\rangle }^{-1}}}\\ =R_O{\langle x\rangle }^1-F_1{\langle x-a\rangle }^1-F_2{\langle x-b\rangle }^1+R_C{\langle x-l\rangle }^1\end{array}$    (V)

Write the equation for the deflection across the beam.

    $EI\frac{d^{2}y}{d{x}^2}=M$      (VI)

Here, the modulus of elasticity is $E$, the area moment of inertia is $I$ and the deflection is $y$.

Substitute the value from Equation (V) to Equation (VI).

    $EI\frac{d^{2}y}{d{x}^2}=R_O{\langle x\rangle }^1-F_1{\langle x-a\rangle }^1-F_2{\langle x-b\rangle }^1+R_C{\langle x-l\rangle }^1$             (VII)

Integrate Equation (VII) with respect to $x$.

    $EI\frac{dy}{dx}=\frac{R_O}{2}{\langle x\rangle }^2-\frac{F_1}{2}{\langle x-a\rangle }^2-\frac{F_2}{2}{\langle x-b\rangle }^2+\frac{R_C}{2}{\langle x-l\rangle }^2+C_1$   (VIII)

Integrate the Equation (IX) with respect to $x$.

    $EIy=\frac{R_O}{6}{\langle x\rangle }^3-\frac{F_1}{6}{\langle x-a\rangle }^3-\frac{F_2}{6}{\langle x-b\rangle }^3+\frac{R_C}{6}{\langle x-l\rangle }^3+C_{1}x+C_2$ (IX)

Write the expression for the bending stress in the beam.

    ${\sigma }_{\max }=\frac{M_{\max }c}{I_{1-1}}$                                                                                (X)

Here, the bending stress is $\sigma$, the deflection caused by two channels is $I_{1-1}$ and the distance between the edge and neutral axis is $c$.

Write the expression for deflection at the midspan using similar triangles.

    $y_{mid}=\frac{I}{I_{1-1}}\left(-\frac{1}{2}\right)$                                                                    (XI)

Here, the deflection at the mid-span is $y_{mid}$.

Write the expression for the area moment of inertia.

    $I=2I_{1-1}$                                                                                       (XII)

Here, the moment of inertia of single channel is $I_{1\text{-}1}$.

Conclusion:

Refer to Table A-5 “Physical Constants of Materials”; obtain the properties of modulus of elasticity for carbon steel as, $30\times {10}^6\text{psi}$

Substitute $450\text{lbf}$ for $F_1$, $300\text{lbf}$ for $F_2$, $20\text{ft}$ for $l$, $6\text{ft}$ for $a$ and $10\text{ft}$ for $b$ in Equation (I).

    $\begin{array}{l}\left[\begin{array}{l}\left\{R_C\times \left(20\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\right\}-\left\{\left(450\text{lbf}\right)\times \left(6\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\right\}\\ -\left\{\left(300\text{lbf}\right)\times \left(10\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\right\}\end{array}\right]=0\\ R_C=\frac{68400\text{lbf}\cdot \text{in}}{240\text{in}}\\ R_C=285\text{lbf}\end{array}$

Substitute $450\text{lbf}$ for $F_1$, $300\text{lbf}$ for $F_2$ and $285\text{lbf}$ for $R_C$ in Equation (I).

    $\begin{array}{l}{R}_O+285\text{lbf}-450\text{lbf}-300\text{lbf}=0\\ R_O=465\text{lbf}\end{array}$

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

    $\begin{array}{c}EI\times 0=\frac{R_O}{6}{\langle 0\rangle }^3-0-0+0+C_1\times 0+C_2\\ C_2=0\end{array}$

Substitute $465\text{lbf}$ for $R_O$, $10\text{ft}$ for $x$,$450\text{lbf}$ for $F_1$, $300\text{lbf}$ for $F_2$, $285\text{lbf}$ for $R_C$, $6\text{ft}$ for $a$ and $10\text{ft}$ for $b$ in Equation (V).

    $\begin{array}{c}{M}_{\max }=\left[\begin{array}{l}\left(465\text{lbf}\right){\langle \left(10\text{ft}\right)\rangle }^1-\left(450\text{lbf}\right){\langle \left(10\text{ft}\right)-\left(6\text{ft}\right)\rangle }^1\\ -\left(300\text{lbf}\right){\langle \left(10\text{ft}\right)-\left(10\text{ft}\right)\rangle }^1+0\end{array}\right]\\ =\left[\begin{array}{l}\left(465\text{lbf}\right){\langle \left(10\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\rangle }^1-\left(450\text{lbf}\right){\langle \begin{array}{l}\left(10\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\\ -\left(6\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\end{array}\rangle }^1\\ -\left(300\text{lbf}\right){\langle \left(10\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)-\left(10\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\rangle }^1+0\end{array}\right]\\ =55800\text{lbf}\cdot \text{in}-21600\text{lbf}\cdot \text{in}\\ =34.2\times {10}^3\text{lbf}\cdot \text{in}\end{array}$

Substitute $240\text{in}$ for $x$, $0$ for $C_2$,$450\text{lbf}$ for $F_1$, $300\text{lbf}$ for $F_2$, $285\text{lbf}$ for $R_C$, $6\text{ft}$ for $a$, $10\text{ft}$ for $b$, $20\text{ft}$ for $l$ and 0 for $y$ in Equation (IX).

    $\begin{array}{l}EI\times 0=\left[\begin{array}{l}\frac{465\text{lbf}}{6}{\langle \left(20\text{ft}\right)\rangle }^3-\frac{450\text{lbf}}{6}{\langle \left(20\text{ft}\right)-\left(6\text{ft}\right)\rangle }^3\\ -\frac{300\text{lbf}}{6}{\langle \left(20\text{ft}\right)-\left(10\text{ft}\right)\rangle }^3\\ +\frac{R_C}{6}{\langle \left(20\text{ft}\right)-\left(20\text{ft}\right)\rangle }^3+C_1\times \left(20\text{ft}\right)+0\end{array}\right]\\ 0=\left[\begin{array}{l}\frac{465\text{lbf}}{6}{\langle \left(20\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\rangle }^3-\frac{450\text{lbf}}{6}{\langle \begin{array}{l}\left(20\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\\ -\left(6\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\end{array}\rangle }^3\\ -\frac{300\text{lbf}}{6}{\langle \left(20\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)-\left(10\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\rangle }^3\\ +\frac{R_C}{6}{\langle \left(20\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)-\left(20\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\rangle }^3\\ +C_1\times \left(20\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)+0\end{array}\right]\\ C_1=\frac{-77.5\text{lbf}\times {\left(240\text{in}\right)}^3+75\text{lbf}\times {\left(168\text{in}\right)}^3+50\text{lbf}\times {\left(120\text{in}\right)}^3}{240}\\ C_1=-2.622\times {10}^6\text{lbf}\cdot {\text{in}}^{\text{2}}\end{array}$

Substitute $0$ for $C_2$ and $-2.622\times {10}^6\text{lbf}\cdot {\text{in}}^{\text{2}}$ for $C_1$ in Equation (IX).

    $\begin{array}{c}EIy=\left[\begin{array}{l}\frac{R_O}{6}{\langle x\rangle }^3-\frac{F_1}{6}{\langle x-a\rangle }^3-\frac{F_2}{6}{\langle x-b\rangle }^3+\frac{R_C}{6}{\langle x-l\rangle }^3\\ -\left(2.622\times {10}^6\text{lbf}\cdot {\text{in}}^2\right)x\end{array}\right]\\ y=\frac{1}{EI}\left[\begin{array}{l}\frac{R_O}{6}{\langle x\rangle }^3-\frac{F_1}{6}{\langle x-a\rangle }^3-\frac{F_2}{6}{\langle x-b\rangle }^3\\ +\frac{R_C}{6}{\langle x-l\rangle }^3-\left(2.622\times {10}^6\text{lbf}\cdot {\text{in}}^2\right)x\end{array}\right]\end{array}$   (XIII)

Substitute $465\text{lbf}$ for $R_O$, $120\text{in}$ for $x$,$450\text{lbf}$ for $F_1$, $300\text{lbf}$ for $F_2$, $285\text{lbf}$ for $R_C$, $6\text{ft}$ for $a$, $10\text{ft}$ for $b$, $20\text{ft}$ for $l$, $30\times {10}^6\text{psi}$ for $E$, $-2.622\times {10}^6\text{lbf}\cdot {\text{in}}^{\text{2}}$ for $C_1$ and $-0.5\text{in}$ for $y$ in Equation (XIII).

    $\begin{array}{l}-0.5=\frac{1}{\left(30\times {10}^6\right)I}\left[\begin{array}{l}\frac{465\text{lbf}}{6}{\langle 120\text{in}\rangle }^3-\frac{450\text{lbf}}{6}{\langle 120\text{in}-\left(6\text{ft}\right)\rangle }^3\\ -\frac{300\text{lbf}}{6}{\langle 120\text{in}-\left(10\text{ft}\right)\rangle }^3\\ +0-\left(2.622\times {10}^6\text{lbf}\cdot {\text{in}}^{\text{2}}\right)\times 120\text{in}\end{array}\right]\\ -0.5=\frac{1}{\left(30\times {10}^6\right)I}\left[\begin{array}{l}\frac{465\text{lbf}}{6}{\langle 120\text{in}\rangle }^3-\frac{450\text{lbf}}{6}{\langle 120\text{in}-\left(6\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\rangle }^3\\ \text{'}-\frac{300\text{lbf}}{6}{\langle 120\text{in}-\left(10\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\rangle }^3\\ +0-\left(2.622\times {10}^6\text{lbf}\cdot {\text{in}}^{\text{2}}\right)\times 120\text{in}\end{array}\right]\\ I=12.60{\text{in}}^{\text{4}}\end{array}$

The obtain value of second area of moment is $12.60{\text{in}}^{\text{4}}$. Select and check the results for a value of second area of moment that is greater than the obtain value for the deflection to maintain at the same level of $0.5\text{in}$.

Refer to Table A-7 “Properties of StructuralSteel Channels”. Select two steel channels of $5\text{in}\times \text{6}\text{.7}\text{lbf}/\text{ft}$.

    $I_{1-1}=7.49{\text{in}}^{\text{4}}$

Substitute $7.49{\text{in}}^{\text{4}}$ for $I_{1-1}$ in Equation (XII).

    $\begin{array}{c}I=2\times 7.49{\text{in}}^4\\ =14.98{\text{in}}^{\text{4}}\end{array}$

Substitute $34.2\times {10}^3\text{lbf}\cdot \text{in}$ for $M_{\max }$, $2.5\text{in}$ for $c$ and $14.98{\text{in}}^{\text{4}}$ for $I_{1-1}$ in Equation (X).

    $\begin{array}{c}{\sigma }_{\max }=\frac{34.2\times {10}^3\text{lbf}\cdot \text{in}\times 2.5\text{in}}{14.98{\text{in}}^4}\\ =\left(5710\text{psi}\right)\left(\frac{1\text{kpsi}}{1000\text{psi}}\right)\\ =5.7\text{kpsi}\end{array}$

Substitute $14.98{\text{in}}^{\text{4}}$ for $I_{1-1}$, $12.60{\text{in}}^{\text{4}}$ for $I$ in Equation (XI).

    $\begin{array}{c}{y}_{mid}=\frac{12.60{\text{in}}^4}{14.98{\text{in}}^4}\left(-\frac{1}{2}\right)\\ =-0.421\end{array}$

Since the deflection at the midspan and the maximum stress in the beam is lying within the limits, so the selected pair of channels are significant.

Thus, the pair of structural steel channels is $5\text{in}\times \text{6}\text{.7}\text{lbf}/\text{ft}$.