How to Form Shear and Moment Equations?

Sample Solution

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

Segment AB

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

Place the Cutting Plane Between Segment Points

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 \(M_{AB}\).

Apply Shear and Moment Equations

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.

Full Solution

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

Segment BC

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

Segment CD

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

Segment DE

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

Summarise Your Results

Shear and Moment Equations

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

Shear Equations

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

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

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

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

Moment Equations

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

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

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

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

Shear and Moment Diagrams

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

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

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