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
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Question “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”
Answer
Beam:
A beam is a structural member for which loads are applied perpendicularly to its longitudinal axis.
Shear force
The 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 the bending moment or the internal moment is called the bending moment.
Sign conventions Shear force and bending moment
These are the conventions that apply to the shear force as well as the bending moment.
The following beam is an example of the AB in which the loads H, F, and QR act as shown in this figure.
Take into account the forces on the left side in the 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.
\begin{array}{l}\\\sum {{F_y}} = 0\\\\{R_a} - P - F - V = 0\\\\V = {R_a} - P - F\\\end{array}Calculate the bending moment by applying moment equilibrium to section a/a.
\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}Below is the body diagram for the beam.
Use equations of equilibrium to determine the beam.
\sum {{F_y} = 0} {A_y} + {B_y} = \frac{1}{2}{w_0}LConsider the following: 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}Taken from equation (1)
{A_y} + {B_y} = \frac{1}{2}{w_0}LSubstitute $math_tag_6
\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}Use similar triangles to calculate the height H at a distance x from your left end.
\begin{array}{l}\\\frac{{{w_0}}}{L} = \frac{h}{x}\\\\h = \frac{{{w_0}x}}{L}\\\end{array}Calculate the shear force inside the beam of the span $math_tag_9
\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 shear force inside the beam for span $math_tag_11
\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 that acts on the beam of the span 0 \le x \le \frac{L}{3} \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 that acts on the beam of the span $math_tag_11
$math_tag_16
Part B: The internal shear force acting upon the beam for 0 \le x \le \frac{L}{3}
The beam’s internal shear force for the span \frac{L}{3} \le x \le LPart B
The internal moment that acts on the beam of the span 0 \le x \le \frac{L}{3}Part B
The internal moment that acts on the beam of the span \frac{L}{3} \le x \le L \left( {\frac{{{w_0}L}}{4}} \right)\left( {x - \frac{L}{3}} \right) - \frac{{{w_0}{x^3}}}{{6L}}
Conclusion
Above is the solution for “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“. We hope that you find a good answer and gain the knowledge about this topic of Mechanics-Of-Materials.