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Consider a beam shown in the figure below. (Figure 1) Express the internal shear in the beam as a function of x for 0≤x≤L/3, and for L/3≤x≤L, and the internal moment in the beam as a function of x for 0≤x≤L/3 and L/3≤x≤L

The beam is an horizontal structure employed for load-bearing in the construction of a complicated structure. Here are some kinds of beams:

- Overhanging beam
- Simply supported beam
- Cantilever beam

Concepts and ReasonBeam:A beam is a structural member for which loads are applied perpendicularly to its longitudinal axis.

Shear forceThe algebraic sum of any lateral forces on either side of a section of a beam’s shear force or internal herar is called the shear force.

Bending moment:The algebraic sum of all moments around a section of beam that is either bending or internal is called the bending moment.

Sign conventionsShear force and bending momentThese are the conventions that apply to the shear force as well as the bending moment.

FundamentalsThe following beam is an example of the AB in which the loads

H,F, andQRact as shown in this figure.Take into account the forces on the left side of section

a-a. The shear force, bending moment and shear force are calculated as follows:Calculation of shear force:

Calculate the shear force by applying vertical force equilibrium.

[katex]\begin{array}{l}\\\sum {{F_y}} = 0\\\\{R_a} – P – F – V = 0\\\\V = {R_a} – P – F\\\end{array}[/katex]

Calculate the bending moment by applying moment equilibrium to section

a/a.[katex]\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}[/katex]

Below is the body diagram for the beam.

Use equations of equilibrium to determine the beam.

[katex]\sum {{F_y} = 0}[/katex]

[katex]{A_y} + {B_y} = \frac{1}{2}{w_0}L[/katex]

Consider the following:

B[katex]\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}[/katex]

Taken from equation (1)

[katex]{A_y} + {B_y} = \frac{1}{2}{w_0}L[/katex]

Substitute $equaTag6

[katex]\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}[/katex]

Use similar triangles to calculate the height

Hat a distancexfrom your left end.[katex]\begin{array}{l}\\\frac{{{w_0}}}{L} = \frac{h}{x}\\\\h = \frac{{{w_0}x}}{L}\\\end{array}[/katex]

Calculate the shear force inside the beam that acts on the span [katex]0 \le x \le \frac{L}{3}[/katex]

[katex]\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}[/katex]

Calculate the shear force inside the beam for $equaTag11

[katex]\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}[/katex]

Calculate the moment internal acting on the beam of the span [katex]0 \le x \le \frac{L}{3}[/katex]

[katex]\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}[/katex]

Calculate the moment internal acting on the beam of the span [katex]\frac{L}{3} \le x \le L[/katex]

[katex]\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}[/katex]Ans Part A

The internal shear force acting upon the beam for [katex]0 \le x \le \frac{L}{3}[/katex]Part A

The beam’s internal shear force for the span [katex]\frac{L}{3} \le x \le L[/katex]Part B

The internal moment that acts on the beam of the span [katex]0 \le x \le \frac{L}{3}[/katex]Part C

The internal moment that acts on the beam of the span [katex]\frac{L}{3} \le x \le L[/katex]

[katex]\left( {\frac{{{w_0}L}}{4}} \right)\left( {x – \frac{L}{3}} \right) – \frac{{{w_0}{x^3}}}{{6L}}[/katex]