The following section shows how to relate load, shear, and moment. It makes it possible to create shear and moment diagrams without formulating equations:

For this post, we'll see how to relate shear and moment:

Set-Up References

Before relating load, shear, and moment, we must set up references. Naturally, we do this just before we analyze the structure's reactions. It's vital to identify all external loads and points of interest. Below are the references we have established in this example:

\(A(0.0m, 0.0m), A(0.0ft, 0.0ft)\). The \(27kN•m (19.91kip•ft)\) clockwise moment.

\(B(2.0m, 0.0m), B(6.56ft, 0.0ft)\). The roller support and the start of varying linear load \(0\).

\(C(6.5m, 0.0m), C(21.33ft, 0.0ft)\). The end of the varying linear load of \(45kN/m (3.08kip/ft)\) ↓ and the start of the uniform distributed load of \(36kN/m (2.47kip/ft)\) ↓,

\(D(11.0m, 0.0m), D(36.09ft, 0.0ft)\). The end of the uniform distributed load \(36kN/m (2.47kip/ft)\) ↓ and the hinge support, and

\(E(12.5m, 0.0m), E(41.01ft, 0.0ft)\). The \(90kN (20.23kip)\) downwardconcentrated load.

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To draw the moment diagram, we must have its shear diagram. We have covered how to obtain it in the last part.

Shear Diagram

Moment Diagram (Shear-Moment)

Let's draw the moment diagram based on the beam's shear. There are many strategies to create it; however, no matter what approach we use, we need to know two things: (1) the moment value and (2) its slope at each critical beam point per point and segment.

Typically, the way we solve it is to analyze each part successively - from the first point to its adjacent segment to the next end and so on. Each component has different objectives we need to accomplish, which we will see later.

Let's see this process in action:

Point A

We first start by examining point \(A\) by itself. Here, we are looking at the value of the moment at the said point. Another thing we need to see here is to check if the shear will be affected by discontinuities (moments).

In this example, the external load acting on it is a 27kN•m clockwise (+) moment; hence, discontinuity exists. In this case, the moment becomes:

\(M_{A_2}=M_{A_1}+27\)

\(M_{A_2}=0+27\)

\(M_{A_2}=27kN•m\)

When we analyze the next segment after, we use \(M_{A_2}=27kN•m\) instead of \(M_{A_1}=0\).

Segment AB

After examining point \(A\) and its moment, we move on to the adjacent segment \(AB\). When analyzing segments, we need to find two things: (1) the moment and (2) its slope at its endpoints \(A\) and \(B\).

Endpoint A

Moment Value

From examining point \(A\), we saw that \(M_A=27kN•m\). In this segment, it is still equal to \(27kN•m\)

Slope

In terms of the slope, we use the moment slope equation:

\(V_A=\frac{dM}{dx}=M_{A'}\)

\(M_{A}'=0\)

The shear at point \(A\) is zero; hence, we have a zero slope. It means that the tangent line plot must be a horizontal line.

Endpoint B

Moment Value

Next, we consider the moment at point \(B\). To solve it, we will use the change in moment expression to find the value:

\(M_B-M_A=\Delta{M}=\int_{x_1}^{x_2}V\times{dx}\)

\(M_B-27=\int_{0}^{2}0\times{dx}\)

\(M_B=27kN•m\)

There is no shear in segment \(AB\), so \(M_B\) is still zero.

Slope

In terms of the slope, we use the moment slope equation:

\(V_B=\frac{dM}{dx}=M_{B'}\)

\(M_{B}'=0\)

A zero slope means the tangent line plot must be horizontal.

Point B

Next, we examine point \(B\) by itself. From the previous segment, the moment \(M_{B_1}\) is \(27kN•m\); Since there is no discontinuity at this point, the moment at \(B\) is still \(27kN•m\).

Segment BC

After examining point \(B\) and its moment, we move on to the adjacent segment \(BC\). Again, we need to find two things: (1) the moment and (2) its slope at its endpoints \(B\) and \(C\).

Endpoint B

Moment Value

From examining point \(B\), we saw that \(M_B\) is \(27kN•m\). Likewise, this segment's moment \(M_B\) is \(27kN•m\).

Slope

In terms of the slope, we use the moment slope equation:

\(V_B=\frac{dM}{dx}=M_{B'}\)

\(M_{B}'=90\)

A positive slope means the tangent line plot must go upward (bottom-left to top-right).

Point of Maximum Moment

Moment Value

In addition to finding the moment at endpoints, we need to compute the moment at the location of zero shear \(B'\). From our previous discussion, this point is at \(x=6.24\) from the origin. It is at this point where the maximum moment occurs. To solve it, we still use the change in moment expression to find the value:

The right side of the equation is the area of the positive plot of the shear region at segment \(BC\). In other words, it is the area of the shear region \(BB'\).

Finding The Area

This section is a refresher on how to get the right side of the expression above. To get the area of the shear region, we use principles from Geometry or Integral Calculus. In the solution above, we used the former approach.

We can think of the area as the difference between the rectangular area \(A_a\) and the spandrel area \(A_b\), \(A=A_a-A_b\).

The area of the rectangle \(A_a\) is \(bh\) where \(b\) and \(h\) are the base and height, respectively.

Variable \(b\) or the distance from \(B\) to \(B'\) is \(6.24-2\)

Variable \(h\) is \(90\)

The area of a spandrel \(A_b\) is \(\frac{1}{n+1}bh\) where \(n\), \(b\), and \(h\) are the degree of the shear function, base, and height respectively. We base the variable \(h\) on the tangent line of the origin.

The degree \(n\) is equal to 2

Variable \(b\) or the distance from \(B\) to \(B'\) is \(6.24-2\)

Variable \(h\) is \(90\) (based from the tangent line at \(B\))

The degree \(n\) of the shear function at segment \(BC\) is two (2) since we have a quadratic function. How do we know if it is such? From the moment and shear slope equations, we can deduce that the degree of an algebraic function decrements from moment to shear to load. The same is true in reverse - the degree increases from load to shear to moment.

In our example, the load in segment \(BC\) is linear, meaning it has a degree of one (linear function). From this, the degree of the shear graph must be two (quadratic function). If we continue even further, the degree of the moment graph will be three (cubic function).

We substitute these values to get the right side of the equation:

The right side of the equation is the area of the negative plot of the shear region at segment \(BC\). In other words, it is the area of the shear region \(B'C\).

Finding the Area

This section is a refresher on how to get the right side of the expression above. To get the area of the shear region, we use principles from Geometry or Integral Calculus. In the solution above, we used the former approach.

We can think of the area as the sum of the triangular area \(A_a\) and the spandrel area \(A_b\), \(A=A_a+A_b\).

The area of the triangle \(A_a\) is \(\frac{1}{2}bh\) where \(b\) and \(h\) are the base and height respectively.

Variable \(b\) or the distance from \(B'\) to \(C\) is \(6.5-6.24\)

The variable \(h\) would require additional steps. To solve it, we must use ratio and proportion with the shear slope at \(B'\).

To get the shear slope at \(B'\), we need to know that load at \(B'\)

\(\frac{45}{4.5}=\frac{w_{B'}}{6.42-2}\)

\(w_{B'}=42.4\)

From the shear slope equation, the slope of the shear at point \(B'\) is:

\(w_{B'}=\frac{dV}{dx}=V_{B'}\)

\(V_{B'}=42.4\)

We can now use ratio and proportion to solve for \(h\):

\(\frac{h}{6.5-6.24}=\frac{42.4}{1}\)

\(h=10.91\)

Since it is in the negative region, \(h=-10.91\).

The area of a spandrel \(A_b\) is \(\frac{1}{n+1}bh\) where \(n\), \(b\), and \(h\) are the degree of the shear function, base, and height respectively. We base the variable \(h\) on the tangent line of the origin.

The degree \(n\) is equal to 2

Variable \(b\) or the distance from \(B\) to \(B'\) is \(6.5-6.24\)

The variable \(h\) would require some additional steps. Remember, we must base \(h\) in the spandrel from the tangent line at \(B'\). We have the shear at point \(C\), which is \(-11.25\) and \(h\) from the triangle, which is \(10.91\). From these values, it is easy to solve it:

\(h=11.25-10.91\)

\(h=0.34\)

Since it is in the negative region, \(h=-0.34\).

We substitute these values to get the right side of the equation:

In terms of the slope at said point, we use the moment slope equation:

\(V_C=\frac{dM}{dx}=M_{C'}\)

\(M_{C}'=-11.25\)

A negative slope means the tangent line plot must go downward (top-left to bottom-right).

Point C

Next, we examine point \(C\) by itself. From the previous segment, the moment \(M_{C}\) is \(280.12kN•m\); There is no discontinuity at this point; hence the moment is still the same.

Segment CD

After examining point \(C\) and its moment, we move on to the adjacent segment \(CD\). Again, we need to find two things: (1) the moment and (2) its slope at its endpoints \(C\) and \(D\).

Endpoint C

Moment Value

From examining point \(C\), we saw that \(M_C\) is \(280.12kN•m\). Likewise, this segment's moment \(M_C\) is \(280.12kN•m\).

Slope

In terms of the slope, we use the moment slope equation:

\(V_C=\frac{dM}{dx}=M_{C'}\)

\(M_{C}'=-11.25\)

A negative slope means the tangent line plot must go downward (top-left to bottom-right).

Endpoint D

Moment Value

Next, we consider the moment at point \(D\). To solve it, we will use the change in moment expression to find the value:

The right side of the equation is the area of the shear region in segment \(CD\). The value is the sum of the rectangular and triangular areas.

Slope

In terms of the slope at said point, we use the moment slope equation:

\(V_D=\frac{dM}{dx}=M_{D'}\)

\(M_{D}'=-173.25kN•m\)

A negative slope means the tangent line plot must go downward (top-left to bottom-right).

Point of Zero Moment

Going off-track for a bit, let's analyze the moment at endpoints \(C\) and \(D\):

\(M_C=280.12kN•m\)

\(M_D=-135kN•m\)

We can see that the moment at \(C\) was positive. When it reached point \(D\), it became negative. This incident means a point exists between \(C\) and \(D\) in which the moment becomes zero. At this point, the inflection point in the deflected shape will occur.

There are many ways to find this point. We can see the solution in this separate post. At the end of the article, the location of zero moment from the origin is at:

\(x=10.15\)

Point D

Next, we examine point \(D\) by itself. From the previous segment, the moment \(M_{D}\) is \(-135kN•m\); There is no discontinuity at this point; hence the moment is still the same.

Segment DE

After examining point \(D\) and its moment, we move on to the adjacent segment \(DE\). Again, we need to find two things: (1) the moment and (2) its slope at its endpoints \(D\) and \(E\).

Endpoint D

Moment Value

From examining point \(D\), we saw that \(M_D\) is \(-135kN•m\). Likewise, this segment's moment \(M_D\) is \(-135kN•m\).

Slope

In terms of the slope, we use the moment slope equation:

\(V_D=\frac{dM}{dx}=M_{D'}\)

\(M_{D}'=90\)

A positive slope means the tangent line plot must go upward (bottom-left to top-right).

Endpoint E

Moment Value

Next, we consider the moment at point \(E\). To solve it, we will use the change in moment expression to find the value:

\(M_E-M_D=\Delta{M}=\int_{x_1}^{x_2}V\times{dx}\)

\(M_E-(-135)=(12-11.5)(90)\)

\(M_E=0kN•m\)

The right side of the equation is the area of the shear diagram at segment \(DE\)

Slope

In terms of the slope at said point, we use the moment slope equation:

\(V_E=\frac{dM}{dx}=M_{E'}\)

\(M_{E}'=90\)

A positive slope means the tangent line plot must go upward (bottom-left to top-right).

Point E

Finally, we examine point \(E\) by itself. From the previous segment, the moment \(M_{E}\) is \(0kN•m\); There is no discontinuity at this point; hence the moment is still the same.

Drawing the Moment Diagram

Now that we know the moment value and slope for each critical point per point and segment, we can finally construct the moment diagram per segment.

The moment values and slopes per segment are the following:

Segment AB

\(M_A=27kN•m\), \(M_{A}'=0\)

\(M_B=27kN•m\), \(M_{B}'=0\)

Segment BC

\(M_B=27kN•m\), \(M_{B}'=90\)

\(M_B'=281.56kN•m\), \(M_{B'}'=0\)

\(M_C=280.12kN•m\), \(M_{C}'=-11.25\)

Segment CD

\(M_C=280.12kN•m\), \(M_{C}'=-11.25\)

\(M_D=-135kN•m\), \(M_{D}'=-173.25\)

Segment DE

\(M_D=-135kN•m\), \(M_{D}'=90\)

\(M_E=0kN•m\), \(M_{E}'=90\)

These values are the information we need to draw the moment diagram. Plotting the moment values on a graph and drawing a tangent line along these points, we have the following chart:

Moment Diagram

How do we know if our diagram is correct? The moment must close back to zero when we analyze the final point. In this case, it closed \(M_E=0kN•m\) when we investigated point \(E\).