Continuing from this example, let's learn how to model the shear and moment of a beam using equations and diagrams.

*The solution presented is in SI. The author will update the post soon to reflect English units.*

In this post, we explore how to construct the shear and moment equations of our beam example - an illustration of how to apply theory to a real-life example.

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Learn moreContinuing from this example, let's learn how to model the shear and moment of a beam using equations and diagrams.

*The solution presented is in SI. The author will update the post soon to reflect English units.*

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The following section shows how to use shear and moment equations and diagrams:

Let's demonstrate how to get the shear and moment equations between points \(A(0.0,0)\) and \(B(2.0,0)\)

We first need to place a section at a distance \(x\) from the origin between points \(A\) and \(B\). As a result, we will divide the beam into two sections - the left and the right. In addition, we'll exposed shear \(V_{AB}\) and moment \(M_{AB}\).

Let's consider the left section of the beam. All we have to do is to apply the shear and moment equilibrium equations.

Solving for shear \(V_{AB}\):

\(\left.V_{A B}=\sum F_{v_L}\right] \uparrow_{+}\)

Equation 1: \(V_{AB}=0\)

Equation 1 is the shear equation of segment \(AB\) Next, we solve for moment \(M_{AB}\):

\(\left.M_{A B}=\sum M_L\right] \circlearrowright_{+}\)

Equation 2: \(M_{AB}=27\)

Equation 2 is the moment equation of segment \(AB\).

These two equations are the shear and moment equations we need. With these functions, we can solve for the shear and moment between points \(A(0,0)\) and \(B(2,0)\).

If we wish to find the moment at \(x=1\), we plug in \(x=1\) at the equation to find it. In our example, since the moment is constant throughout \(A\) and \(B\), it is equal to \(27kN•m\)

After solving \(AB\), we proceed to other segments and repeat the process.

This section shows the complete solution of the shear and moment equations:

Let's place a cut section between points \(B(2.0,0)\) and \(C(6.5,0)\) and choose the left side.

Solving for its shear equation:

\(\left.V_{BC}=\sum F_{v_L}\right] \uparrow_{+}\)

\(V_{B C}=90-\frac{1}{2}(x-2)[(10)(x-2)]\)

Equation 3: \(V_{BC}=-5x^2+20x+70\)

Solving for its moment equation:

\(\left.M_{BC}=\sum M_L\right] \circlearrowright_{+}\)

\(M_{B C}=27+90(x-2)-\frac{1}{2}(x-2)[(10)(x-2)]\left[\frac{1}{3}(x-2)\right]\)

Equation 4: \(M_{BC}=-\frac{5}{3} x^3+10 x^2+70 x-139 \frac{2}{3}\)

Let's place a cut section between points \(C(6.5,0)\) and \(C(11.0,0)\) and choose the right side.

We chose the right side because the external loads are more straightforward to analyze than the left.

Solving for its shear equation:

\(\left.V_{CD}=\sum F_{v_R}\right] \downarrow_{+}\)

\(V_{C D}=90+(11-x)\left(36\right)-263.25\)

Equation 5: \(V_{CD}=-36x+222.75\)

Solving for its moment equation:

\(\left.M_{CD}=\sum M_R\right] \circlearrowleft_{+}\)

\(M_{C D}=-90[1.5+(11-x)]-(11-x)\left(36\right)\left[\frac{1}{2}(11-x)\right]+263.25(11-x)\)

Equation 6: \(M_{CD}=-18 x^2+222.75 x-407.25\)

Let's place a cut section between points \(D(11.0,0)\) and \(E(12.5,0)\) and choose the right side.

Solving for its shear equation:

\(\left.V_{DE}=\sum F_{v_R}\right] \downarrow_{+}\)

Equation 7: \(V_{DE}=90\)

Solving for its moment equation:

\(\left.M_{DE}=\sum M_R\right] \circlearrowleft_{+}\)

\(M_{DE}=-90(12.5-x)\)

Equation 8: \(M_{DE}=90x-1125\)

We have finally got all shear and moment equations for each segment. After analysis, it is always a good idea to summarise your results:

\(V_{AB}=0,\{0\leq{x}\lt {2}\}\)

\(V_{BC}=-5x^2+20x+70,\{2\leq{x}\lt {6.5}\}\)

\(V_{CD}=-36x+222.75,\{6.5\leq{x}\lt {11}\}\)

\(V_{DE}=90,\{11\leq{x}\lt {12.5}\}\)

\(M_{AB}=27,\{0\leq{x}\lt {2}\}\)

\(M_{BC}=-\frac{5}{3} x^3+10 x^2+70 x-139 \frac{2}{3},\{2\leq{x}\lt {6.5}\}\)

\(M_{CD}=-18 x^2+222.75 x-407.25,\{6.5\leq{x}\lt {11}\}\)

\(M_{DE}=90x-1125,\{11\leq{x}\lt {12.5}\}\)

We can represent these equations using a graph. If we plot shear \(V\) against position \(x\), we get the shear diagram of the structure.

Likewise, if we plot moment \(M\) against position \(x\), we get the moment diagram of the structure.

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Created On

June 5, 2023

Updated On

February 23, 2024

Contributors

Edgar Christian Dirige

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