Chapter 4, Problem 4.5.29P

A beam of length L is designed to support a uniform load of intensity q (see figure). If the supports of the beam are placed at the ends, creating a simple beam, the maximum bending moment in the beam is qL²/8. However, if the supports of the beam are moved symmetrically toward the middle of the beam (as shown), the maximum bending moment is reduced.

1. Determine the distance a between the supports so that the maximum bending moment in the beam has the smallest possible numerical value. Draw the shear-force and bending-moment diagrams for this condition.
2. Repeat part (a) if the uniform load is replaced with a triangularly distributed load with peak intensity q₀= q at mid-span (see Fig. b).

  

To determine

The distance a between the support by having the small numerical value of maximum bending moment and to draw the diagrams of shear force and bending moment.

Answer to Problem 4.5.29P

a=0.5858L

Explanation of Solution

Given:

The given figure

The length of the beam L supports the load which is uniform having the intensity as q. The bending moment that is maximum is given as $q{L}^2/8$

Concept Used: $$

Resultant force,

$R_A=R_B=\frac{qL}{2}$

Calculation:

As the forces are symmetry,

  $R_A=R_B$

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_A+R_B-qL=0\\ R_A=R_B=\frac{qL}{2}\end{array}$

At a distance from x to C, the section between CA is considered,

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ -V-qx=0\Rightarrow V=-qx\\ {\displaystyle \sum M_x}=0\\ M+q(x)\left(\frac{x}{2}\right)=0\Rightarrow M=-q\frac{x^2}{2}\end{array}$

When x=0,

  $V_C=0,M_C=0,$

When $x=\frac{L-a}{2}$ ,

  $\begin{array}{l}{V}_A=-\frac{q}{2}(L-a)\to (1)\\ M_A=-\frac{q}{2}{\left(\frac{L-a}{2}\right)}^2\Rightarrow M_A=-\frac{q}{8}{\left(L-a\right)}^2\to (2)\end{array}$

At a distance from x to C, the section between AB is considered,

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ \frac{qL}{2}-V-qx=0\Rightarrow V=-qx+\frac{qL}{2}\\ {\displaystyle \sum M_x}=0\\ M+q(x)\left(\frac{x}{2}\right)-\frac{qL}{2}\left(x-\frac{L-a}{2}\right)=0\Rightarrow M=-q\frac{x^2}{2}+\frac{qL}{2}\left(x-\frac{L-a}{2}\right)\end{array}$

When $x=\frac{L-a}{2}$ ,

  $\begin{array}{l}{V}_A=-q\left(\frac{L-a}{2}\right)+\frac{qL}{2}\Rightarrow V_A=\frac{qa}{2}\to (3)\\ M_A=-\frac{q}{2}{\left(\frac{L-a}{2}\right)}^2-\frac{qL}{2}(0)\Rightarrow M_A=-\frac{q}{8}{\left(L-a\right)}^2\to (4)\end{array}$

When, $x=L-\frac{L-a}{2}$ ,

  $=\frac{L+a}{2}$

  $\begin{array}{l}{V}_B=-q\left(\frac{L+a}{2}\right)+\frac{qL}{2}\Rightarrow V_B=-\frac{qa}{2}\to (5)\\ M_B=-\frac{q}{2}{\left(\frac{L+a}{2}\right)}^2+\frac{qL}{2}\left(\frac{L+a}{2}-\frac{L-a}{2}\right)\Rightarrow M_B=-\frac{q}{8}{\left(L+a\right)}^2+\frac{qL}{2}(a)\\ M_B=-\frac{q}{8}\left(L^2+a^2+2aL-4aL\right)\\ M_B=-\frac{q}{8}\left(L^2+a^2-2aL\right)\Rightarrow M_B=-\frac{q}{8}{\left(L-a\right)}^2\to (6)\end{array}$

The diagram of the shear force and the bending moment at the midpoint is symmetry,

At $x=\frac{L}{2}$ the mid-section has the maximum bending moment,

  $M_{\max }=-\frac{q}{2}{\left(\frac{L}{2}\right)}^2+\frac{qL}{2}\left(\frac{L}{2}-\frac{L-a}{2}\right)$

  $M_{\max }=-\frac{q}{2}{L}^2+\frac{qL}{2}\left(a\right)\Rightarrow =-\frac{q}{2}(L^2-2aL)\to (7)$

As the maximum bending moment of the beam has the least numerical value,

  $\left|M_{\max }\right|=-\left|M_{\max }\right|$

  $\begin{array}{l}\frac{q}{8}\left(L^2-2aL\right)=-\frac{q}{8}\left(L-a\right)2\\ L^2-2aL=-L^2-a^2+2aL\\ a^2-4aL+2L^2=0\\ a=\frac{-(4L)\pm \sqrt{{(-4L)}^2}-4(1)(2L^2)}{2(1)}\\ a=\frac{(4L)\pm 2L\sqrt{2}}{2(1)}\Rightarrow \left(\left(2\right)\pm \sqrt{2}\right)L,\\ But\\ a<L\\ a=(2-\sqrt{2})L\Rightarrow 0.5858L\end{array}$

With the equation (7),

  $M_{\max }=-\frac{q}{8}(L^2-2(0.5858L)L)\to (7)$

  $=-\frac{q}{8}{L}^2(1-1.1716)\Rightarrow 0.02145q{L}^2$

From equation 1,

  $\begin{array}{l}{V}_A=-\frac{q}{2}(L-0.5858L)\Rightarrow -0.2071qL\\ \end{array}$

From equation 2,

  $M_A=-\frac{q}{8}{(L-0.5858L)}^2\Rightarrow -0.0214q{L}^2$

From equation 3,

  $V_A=\frac{q}{2}(0.5858L)\Rightarrow 0.2929qL$

From equation 4,

  $M_A=-\frac{q}{8}{(L-0.5858L)}^2\Rightarrow -0.021q{L}^2$

From equation (5),

  $V_A=-\frac{q}{2}(0.5858L)\Rightarrow -0.2929qL$

From equation 6,

  $M_B=-\frac{q}{8}{\left(L-a\right)}^2\Rightarrow -0.0214q{L}^2$

The bending moment is 0 from the given diagram, so we have,

  $\begin{array}{l}0=-\frac{q}{2}{(}^b+\frac{qL}{2}\left(b-\frac{L-a}{2}\right)\Rightarrow -(b^2)+L\left(b-\frac{L-0.5858L}{2}\right)\\ 0=-(b^2)+L\left(b-0.2071L\right)\\ (b^2)-bL+0.2071L^2=0\end{array}$

  $\begin{array}{l}b=\frac{-(-L)\pm \sqrt{{(-L)}^2-4(1)(0.2071L^2)}}{2}=\frac{1\pm 0.41425}{2}\\ b=0.2929L\end{array}$

Conclusion:

Thus the distance a between the support is determined as a=0.5858L.

To determine

The distance ‘a’ between the support by having the small numerical value of maximum bending moment and to draw the diagrams of shear force and bending moment.

Answer to Problem 4.5.29P

  $b=0.3333L$

Explanation of Solution

Given:

The given figure

$$

The length of the beam L supports the load which is uniform having the intensity as $q_0=q$ at the mid- point.

Concept Used: $$

Hence the above figure is in symmetry so, the calculations is done by considering this.

Resultant force,

  $R_A=R_B=\frac{1}{2}\times \frac{q_{0}L}{2}=\frac{q_{0}L}{4}$

Calculation: Now the maximum moment is occurs at the centre of the beam then take,

  $\Sigma {\rm M}=0$

  $\begin{array}{l}{R}_B*\frac{a}{2}=\frac{1}{2}(\frac{L}{2})q_0\times \frac{L}{6}\\ \frac{q_{0}L}{4}\times \frac{a}{2}=\frac{q_0{L}^2}{24}\\ a=\frac{L}{3}=0.3333L\end{array}$

Conclusion:

Thus the distance a between the support is determined as $a=0.3333L$ .

The shear force and bending moment diagram is as follows: