Chapter 14, Problem 12P

To determine

Using excel, plot the graph shows the deflection of a beam and also calculate the maximum deflection of the beam.

Answer to Problem 12P

Using excel, a table and graph is created to shows the deflection of a beam and the maximum deflection of the beam is $\begin{array}{c}-0.03942\end{array}\text{m}$.

Explanation of Solution

Given data:

The length of the cantilever beam is $5\text{m}$.

The modulus of elasticity $\left(E\right)$ is $200\text{GPa}$.

That is,

$\begin{array}{c}E=200\text{GPa}\\ =200\times {10}^9\text{Pa}\end{array}$

The second moment of area $\left(I\right)$ is $99.1\times {10}^6{\text{mm}}^{\text{4}}$.

That is,

$\begin{array}{c}I=99.1\times {10}^6{\text{mm}}^{\text{4}}\\ =99.1\times {10}^6\times {\left({10}^{-3}\right)}^4{\text{m}}^{\text{4}}\left[\therefore 1\text{m}={10}^{-3}\right]\\ =99.1\times {10}^6\times {10}^{-12}{\text{m}}^{\text{4}}\end{array}$

The distributed load $\left(w\right)$ is $10000\frac{\text{N}}{\text{m}}$,

Formula used:

Formula to calculate the deflection of the cantilever of the beam is,

$y=\frac{-w{x}^2}{24EI}\left(x^2-4Lx+6L^2\right)$ (1)

Here,

$y$ is the deflection at a given x location in m,

$w$ is the distributed load in $\frac{\text{N}}{\text{m}}$,

$x$ is the distance from the support as shown in m,

$E$ is the modulus of elasticity in $\frac{\text{N}}{{\text{m}}^{\text{2}}}$,

$I$ is the second moment of area ${\text{m}}^4$, and

$L$ is the length of the beam in m.

Calculation:

Consider the distance (x) from the support as shown with the range from $0\text{m}$ to $5\text{m}$ in increment of $0.2\text{m}$.

Refer to the Figure 1:

Column A shows the distance $\left(x\right)$ from the support as shown in m with range from $0\text{m}$ to $5\text{m}$. That is, A2 cell representing the value of $0$ and extend the cell with increment of $0.2$ along in column wise up to 5.

Column B shows the deflection $\left(y\right)$ at a given x location in m. That is, in B2 cell the formula used to find the deflection $\left(y\right)$ as “$=\frac{-w{x}^2}{24EI}\left(x^2-4Lx+6L^2\right)$” is written as “=(-(10000*A2^2)/(24*(200*10^9)*(99.1*10^6*10^-12)))*((A2^2)-(4*5*A2)+(6*5^2))”.

Here, A2 cell represent the distance $\left(x\right)$ from the support as shown in m, $w$ is $10000$, $E$ is $200\times {10}^9$ , $I$ is $99.1\times {10}^6\times {10}^{-12}$ and $L$ is $5$. Likewise, the formula is extended from B2 to B27 in column wise.

Table 1 shows the distance $\left(x\right)$ from the support as shown in m and the deflection $\left(y\right)$ at a given x location in m with the range from $0\text{m}$ to $5\text{m}$ in increment of $0.2\text{m}$.

Refer to the Figure 14.16 in the textbook.

Draw the graph for the distance $\left(x\right)$ from the support as shown in m and the deflection $\left(y\right)$ at a given x location in m values in excel by selecting the respective of two columns. Next, pick the “Insert tab” and select “scatter” then choose “Scatter with Smooth Line and Markers” in it. Thus, the graph is displayed as shown in figure 2. Change the axis title as “Deflection of a cantilever beam”, X axis title as “Distance from the support, x (m)” and Y-axis tile as “Deflection at given x (m) location, y (m)”.

For X-axis, scale change is done by clicking the Layout on toolbar and picks the “Axes”. Choose the primary horizontal axis and select the “More Primary Horizontal Axis Options”. A dialog box of Format axis shows the axis option in that click the fixed and type the minimum value as 0, maximum value as 5, and major unit as 1 and minor unit as 0.2 and click the close option.

Likewise for Y-axis, scale change is done by clicking the Layout on toolbar and picks the “Axes”. Choose the primary vertical axis and select the “More Primary Vertical Axis Options”. A dialog box of Format axis shows the axis option in that click the fixed and type the minimum value as $-0.045$, maximum value as 0, and major unit as $0.005$ and minor unit as $0.005$ and click the close option.

Figure 2 shows the curve of the deflection of a cantilever beam.

For the maximum deflection of the beam, the distance $\left(x\right)$ from the support as shown in m is equal to the length $\left(L\right)$ of the beam.

That is,

$x=L=5\text{m}$

Given,

$E=200\times {10}^9\text{Pa}$ (2)

Substitute the unit $\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for Pa in equation (2),

$E=200\times {10}^9\frac{\text{N}}{{\text{m}}^{\text{2}}}$

Substitute $5\text{m}$ for x, $5\text{m}$ for L, $10000\frac{\text{N}}{\text{m}}$ for w, $200\times {10}^9\frac{\text{N}}{{\text{m}}^{\text{2}}}$ for $E$, and $99.1\times {10}^6\times {10}^{-12}{\text{m}}^{\text{4}}$ for I in equation (1) to find y.

$\begin{array}{c}y=\frac{-\left(10000\frac{\text{N}}{\text{m}}\right){\left(5\text{m}\right)}^2}{\left(24\left(200\times {10}^9\frac{\text{N}}{{\text{m}}^{\text{2}}}\right)\left(99.1\times {10}^6\times {10}^{-12}{\text{m}}^{\text{4}}\right)\right)}\left({\left(5\text{m}\right)}^2-\left(4\left(5\text{m}\right)\left(5\text{m}\right)\right)+6{\left(5\text{m}\right)}^2\right)\\ =\begin{array}{c}-0.03942\end{array}\text{m}\end{array}$

Thus, the maximum deflection of the beam is $\begin{array}{c}-0.03942\end{array}\text{m}$.

Conclusion:

Hence, a table and graph is created using excel to shows the deflection of a beam and the maximum deflection of the beam is $\begin{array}{c}-0.03942\end{array}\text{m}$ has been explained.