Question

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Answer 1

Concepts and reason

Beam:

A beam is a structural member, for which loads are applied perpendicular to their longitudinal axis.

Shear force:

Shear force or internal shear at any section of a beam is the algebraic sum of the lateral forces acting on either side of the section.

Bending moment:

The bending moment or internal moment about any section of a beam is the algebraic sum of the moments of all the forces about the section on either side of the section.

Sign conventions for shear force and bending moment:

The following are the conventions for the shear force and the bending moment.

Fundamentals

Consider the following beam AB on which the loads P , F and Q are acting as shown in the following figure.

Consider the forces to the left side of the section a-a and the shear force and bending moment are calculated as shown below:

Shear force calculations:

Apply vertical force equilibrium and calculate the shear force.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{R_a} - P - F - V = 0\\\\V = {R_a} - P - F\\\end{array}$

Apply moment equilibrium about section a-a and calculate the bending moment.

$\begin{array}{l}\\\sum {{M_{a - a}}} = 0\\\\M - \left( {F \times {x_1}} \right) - \left( {P \times {x_2}} \right) + \left( {{R_a} \times {x_3}} \right) = 0\\\\M = F{x_1} + P{x_2} - {R_a}{x_3}\\\end{array}$

Draw the free body diagram of the beam as shown below.

Apply equations of equilibrium for the beam.

$\sum {{F_y} = 0}$

${A_y} + {B_y} = \frac{1}{2}{w_0}L$ …… (1)

Take the moment about point B.

$\begin{array}{l}\\\sum {{M_B} = 0} \\\\{A_y} \times \frac{{2L}}{3} - \left( {\frac{1}{2} \times L \times {w_0} \times \frac{L}{3}} \right) = 0\\\\{A_y} = \frac{{{w_0}L}}{4}\\\end{array}$

From equation (1),

${A_y} + {B_y} = \frac{1}{2}{w_0}L$

Substitute $\frac{{{w_0}L}}{4}$ for ${A_y}$ .

$\begin{array}{l}\\\frac{{{w_0}L}}{4} + {B_y} = \frac{1}{2}{w_0}L\\\\{B_y} = \frac{{{w_0}L}}{4}\\\end{array}$

Calculate the height h at a distance x from the left end by using similar triangles.

$\begin{array}{l}\\\frac{{{w_0}}}{L} = \frac{h}{x}\\\\h = \frac{{{w_0}x}}{L}\\\end{array}$

Calculate the internal shear force acting on the beam for the span $0 \le x \le \frac{L}{3}$ as follows:

$\begin{array}{c}\\V = - \frac{1}{2}\left( x \right)\left( {\frac{{{w_0}x}}{L}} \right)\\\\ = - \frac{1}{2}\left( {\frac{{{w_0}{x^2}}}{L}} \right)\\\end{array}$

Calculate the internal shear force in the beam for the span $\frac{L}{3} \le x \le L$ as follows:

$\begin{array}{c}\\{V_x} = {A_y} - \left( {\frac{1}{2} \times x \times h} \right)\\\\{V_x} = \left( {\frac{{{w_0}L}}{4}} \right) - \left( {\frac{1}{2} \times x \times \left( {\frac{{{w_0}x}}{L}} \right)} \right)\\\\ = \left( {\frac{{{w_0}L}}{4}} \right) - \left( {\frac{{{w_0}{x^2}}}{{2L}}} \right)\\\end{array}$

Calculate the internal moment acting on the beam for the span $0 \le x \le \frac{L}{3}$ as follows:

$\begin{array}{c}\\M = - \frac{1}{2}\left( x \right)\left( {\frac{{{w_0}x}}{L}} \right)\left( {\frac{x}{3}} \right)\\\\ = - \frac{1}{6}\left( {\frac{{{w_0}{x^3}}}{L}} \right)\\\end{array}$

Calculate the internal moment acting on the beam for the span $\frac{L}{3} \le x \le L$ as follows:

$\begin{array}{c}\\{M_x} = {A_y} \times \left( {x - \frac{L}{3}} \right) - \left( {\frac{1}{2} \times x \times h} \right)\left( {\frac{x}{3}} \right)\\\\ = \left( {\frac{{{w_0}L}}{4}} \right)\left( {x - \frac{L}{3}} \right) - \left( {\frac{1}{2} \times x \times \left( {\frac{{{w_0}x}}{L}} \right)} \right)\left( {\frac{x}{3}} \right)\\\\ = \left( {\frac{{{w_0}L}}{4}} \right)\left( {x - \frac{L}{3}} \right) - \frac{{{w_0}{x^3}}}{{6L}}\\\end{array}$

Ans: Part A

The internal shear force acting on the beam for the span $0 \le x \le \frac{L}{3}$ is $- \frac{1}{2}\left( {\frac{{{w_0}{x^2}}}{L}} \right)$ .

Part B

The internal shear force in the beam for the span $\frac{L}{3} \le x \le L$ is $\left( {\frac{{{w_0}L}}{4}} \right) - \left( {\frac{{{w_0}{x^2}}}{{2L}}} \right)$ .

Part C

The internal moment acting on the beam for the span $0 \le x \le \frac{L}{3}$ is $- \frac{1}{6}\left( {\frac{{{w_0}{x^3}}}{L}} \right)$ .

Part D

The internal moment acting on the beam for the span $\frac{L}{3} \le x \le L$ is

$\left( {\frac{{{w_0}L}}{4}} \right)\left( {x - \frac{L}{3}} \right) - \frac{{{w_0}{x^3}}}{{6L}}$ .