Chapter 4, Problem 4.5.14P

The beam AB shown in the figure supports a uniform load of intensity 3000 N/m acting over half the length of the beam. The beam rests on a foundation that produces a uniformly distributed load over the entire length.

1. Draw the shear-force and bending-moment diagrams for this beam.

Repeat part (a) for the distributed load variation shown in Fig. b.

  

  

To determine

The shear force and bending moment diagram for the given beam.

Answer to Problem 4.5.14P

Maximum shear force V_max= y13 $1200{\rm N}$

Maximum bending moment M_max= $960{\rm N}-m$

Explanation of Solution

Given information: The given beam and parameters are shown in the figure below:

  

For calculating the maximum shear force (V) and bending moment (M) of the given figure, we need to find the amount of force acting upwards on the entire span length.

  ${\displaystyle \sum f_q}=0$

  $\begin{array}{l}q\left(1.6+2\times 0.8\right)-3000\times 1.6=0\\ 3.2q=4800\\ q=\frac{4800}{3.2}\\ q=1500{\rm N}/m\end{array}$

Shear Force Diagram:

To find the shear force of the given figure, we divide the above figure in a number of sections.

1. Firstly taking a section from length 0 to 0.8 m.

  $\begin{array}{l}{V}_1=qx\\ V_1=1500x\end{array}$

The value of shear force when x= 0 at point A is

  $\begin{array}{l}{V}_1=qx\\ V_1=1500\times 0\\ V_1=0\end{array}$

The value of shear force when x= 0.8 at point A is

  $\begin{array}{l}{V}_1=qx\\ V_1=1500\times 0.8\\ V_1=1200{\rm N}\end{array}$

2. Again taking a section from length 0.8 m to mid-span.

  $\begin{array}{l}{V}_2=1500x-3000\left(x-0.8\right)\\ V_2=1500x-3000x+2400\\ V_2=2400-1500x\end{array}$

The value of shear force when x= 0.8 is,

  $\begin{array}{l}{V}_2=2400-1500\times 0.8\\ V_2=2400-1200\\ V_2=1200{\rm N}\end{array}$

The value of shear force at mid- span when x= 1.6 is,

  $\begin{array}{l}{V}_2=2400-1500\times 1.6\\ V_2=2400-2400\\ V_2=0{\rm N}\end{array}$

3. For the next half of the beam, the shear values can be obtained from the concept of symmetry and the obtained values is shown below in the given shear force diagram.

Bending Moment Diagram:

To find the bending moments of the given figure, we divide the above figure in a number of sections.

1. Firstly taking a section from length 0 to 0.8 m.

  $\begin{array}{l}{M}_1=qx\left(\frac{x}{2}\right)\\ M_1=1500x\left(\frac{x}{2}\right)\\ M_1=750x^2\end{array}$

As the equation is of second order of degree, the curve obtained is a parabola.

The value of Moment when x= 0 at point A is

  $\begin{array}{l}{M}_1=750x^2\\ M_1=750\times 0\\ M_1=0\end{array}$

The value of moment when x= 0.8 at point A is

  $\begin{array}{l}{M}_1=750x^2\\ M_1=750\times {\left(0.8\right)}^2\\ M_1=480{\rm N}-m\end{array}$

2. Again taking a section from length 0.8 m to mid-span.

  $\begin{array}{l}{M}_2=1500x\left(\frac{x}{2}\right)-3000\left(x-0.8\right)\left(\frac{x-0.8}{2}\right)\\ M_2=750x^2-1500{\left(x-0.8\right)}^2\\ \end{array}$

The value of bending moment when x= 0.8 is,

  $\begin{array}{l}{M}_2=750\times {\left(0.8\right)}^2-1500{\left(0.8-0.8\right)}^2\\ M_2=480{\rm N}-m\\ \end{array}$

The value of bending moment at mid- span when x= 1.6 is,

  $\begin{array}{l}{M}_2=750\times {\left(1.6\right)}^2-1500{\left(1.6-0.8\right)}^2\\ M_2=1920-1500\times {\left(0.8\right)}^2\\ M_2=1920-960\\ M_2=960{\rm N}-m\\ \end{array}$

3. For the next half of the beam, the bending moment values can be obtained from the concept of symmetry and the obtained values is shown below in the given bending moment diagram.

On the basis of above calculation the shear force and bending moment diagram for the given beam is as follows:

  

To determine

The shear force and bending moment diagram for the given beam.

Answer to Problem 4.5.14P

Maximum shear force V_max= $4.5{\rm K}{\rm N}$

Maximum bending moment M_max= $-12.67{\rm K}{\rm N}$

Explanation of Solution

Given information: The given beam and parameters are shown in the figure below:

  

For calculating the maximum shear force (V) and bending moment (M) of the given figure, we need to find the amount of force acting upwards on the entire span length.

  ${\displaystyle \sum f_q}=0$

  $\begin{array}{l}q\left(1.6+2\times 0.8\right)-\frac{1}{2}\times 1.6\times 3000=0\\ 3.2q=2400\\ q=\frac{2400}{3.2}\\ q=750{\rm N}/m\end{array}$

Shear Force Diagram:

To find the shear force of the given figure we divide the above figure in a number of sections.

1. Firstly taking a section from length 0 to 0.8 m.

  $\begin{array}{l}{V}_1=qx\\ V_1=750x\end{array}$

The value of shear force when x= 0 at point A is

  $\begin{array}{l}{V}_1=qx\\ V_1=750\times 0\\ V_1=0\end{array}$

The value of shear force when x= 0.8 at point A is

  $\begin{array}{l}{V}_1=qx\\ V_1=750\times 0.8\\ V_1=600{\rm N}\end{array}$

2. Again taking a section from length 0.8 m to mid-span.
For obtaining the required shear force value, consider a section X-X for the given uniformly varying load.

From the triangle similarity,

  $\begin{array}{l}\frac{q_x}{x-0.8}=\frac{q_0}{1.6}\\ q_x=\frac{3000}{1.6}\left(x-0.8\right)\\ q_x=1875\left(x-0.8\right)\end{array}$

  $\begin{array}{l}{V}_2=qx-\frac{1}{2}\times \left(x-0.8\right)\times 1875\left(x-0.8\right)\\ V_2=qx-\frac{1}{2}\times {\left(x-0.8\right)}^2\times 1875\\ \end{array}$

The value of shear force when x= 0.8 is,

  $\begin{array}{l}{V}_2=750\times 0.8-\frac{1}{2}\times {\left(0.8-0.8\right)}^2\times 1875\\ V_2=600\end{array}$

The value of shear force at mid- span when x= 2.4 is,

  $\begin{array}{l}{V}_2=750\times 2.4-\frac{1}{2}\times {\left(2.4-0.8\right)}^2\times 1875\\ V_2=1800-2400\\ V_2=-600{\rm N}\end{array}$

3. For the next half of the beam, the shear values can be obtained from the concept of symmetry and the obtained values is shown below in the given shear force diagram.

Bending Moment Diagram:

To find the bending moments of the given figure we divide the above figure in a number of sections.

1. Firstly taking a section from length 0 to 0.8 m.

  $\begin{array}{l}{M}_1=qx\left(\frac{x}{2}\right)\\ M_1=750x\left(\frac{x}{2}\right)\\ M_1=375x^2\end{array}$

As the equation is of second order of degree, the curve obtained is a parabola.

The value of Moment when x= 0 at point A is

  $\begin{array}{l}{M}_1=375x^2\\ M_1=375\times 0\\ M_1=0\end{array}$

The value of moment when x= 0.8 at point A is

  $\begin{array}{l}{M}_1=375x^2\\ M_1=375\times {\left(0.8\right)}^2\\ M_1=240{\rm N}-m\end{array}$

2. For obtaining the required shear force value, consider a section X-X for the given uniformly varying load. From the triangle similarity,

  $\begin{array}{l}\frac{q_x}{x-0.8}=\frac{q_0}{1.6}\\ q_x=\frac{3000}{1.6}\left(x-0.8\right)\\ q_x=1875\left(x-0.8\right)\end{array}$

  $\begin{array}{l}{M}_2=750x\left(\frac{x}{2}\right)-\frac{1}{2}\left(x-0.8\right)\times 1875\left(x-0.8\right)\left(\frac{x-0.8}{3}\right)\\ M_2=375x^2-\frac{1875}{6}{\left(x-0.8\right)}^3\\ \end{array}$

The value of bending moment when x= 0.8 is,

  $\begin{array}{l}{M}_2=375{\left(0.8\right)}^2-\frac{1875}{6}{\left(0.8-0.8\right)}^3\\ M_2=240{\rm N}-m\end{array}$

The value of bending moment at mid- span when x= 1.6 is,

  $\begin{array}{l}{M}_2=375{\left(1.6\right)}^2-\frac{1875}{6}{\left(1.6-0.8\right)}^3\\ M_2=960-250\\ M_2=710{\rm N}-m\end{array}$

3. For the next half of the beam, the bending moment values can be obtained from the concept of symmetry and the obtained values is shown below in the given bending moment diagram.

On the basis of above calculation the shear force and bending moment diagram for the given beam is as follows: