What are Load and Shear Relationships?

We continue our analysis of our beam example by constructing the shear diagram. This time, we shall relate the external loads applied to the beam to draw it.

WeTheStudy lets you connect ideas

Learn more

Continuing from this example, let's learn how to model the shear and moment of a beam by relating its external loads, shear, and moment.

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

This lesson is a two-part article. This post is the first in the series.

For the second part, we can visit this link.

Main Solution

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 loads and shear:

Set-Up References

Before we start relating, 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:

$A (0.0 m, 0.0 m), A (0.0 f t, 0.0 f t)$ . The $27 k N • m (19.91 k i p • f t)$ clockwise moment.
$B (2.0 m, 0.0 m), B (6.56 f t, 0.0 f t)$ . The roller support and the start of varying linear load $0$ .
$C (6.5 m, 0.0 m), C (21.33 f t, 0.0 f t)$ . The end of the varying linear load of $45 k N / m (3.08 k i p / f t)$ ↓ and the start of the uniform distributed load of $36 k N / m (2.47 k i p / f t)$ ↓,
$D (11.0 m, 0.0 m), D (36.09 f t, 0.0 f t)$ . The end of the uniform distributed load $36 k N / m (2.47 k i p / f t)$ ↓ and the hinge support, and
$E (12.5 m, 0.0 m), E (41.01 f t, 0.0 f t)$ . The $90 k N (20.23 k i p)$ downward concentrated load.

Want to access the remaining content?

You're a Member!

Click to expand on exclusive content

Want to access the remaining content?

Become a Member

When you sign-up and subscribe to WeTheStudy, you’ll get the following benefits:

No ads! (yey!)

Complete access to all articles

Ability to track your progress in the tree

SIGN-UP

Complete Your Checkout

When you complete your account, here are the following benefits:

No ads! (yey!)

Complete access to all articles

Ability to track your progress in the tree

PROCEED CHECKOUT

Shear Diagram (Load-Shear)

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

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 shear at the said point. Another thing we need to see here is to check if the shear will be affected by discontinuities (concentrated loads).

In this example, the external load acting on it is a moment; hence, it doesn't affect the shear. In this case, the shear is zero:

$V_{A} = 0$

Segment AB

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

Endpoint A

Shear Value

From examining point $A$ , we saw that $V_{A} = 0$ . In this segment, it is still equal to zero!

Slope

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

$w_{A} = \frac{d V}{d x} = V_{A}^{'}$

$V_{A}^{'} = 0$

There is no concentrated (or varying load) at point $A$ ; hence, we have a zero slope. It means that the tangent line plot must be a horizontal line.

Endpoint B

Shear Value

Next, we consider the shear at point $B$ . To solve it, we will use the change in shear expression to find the value:

$V_{B} - V_{A} = Δ V_{B A} = \int_{x_{1}}^{x_{2}} w \times d x$

$V_{B} - 0 = \int_{0}^{2} 0 \times d x$

$V_{B} = 0$

There is no $w$ loads in segment $A B$ ; that's why $V_{B}$ is still zero.

Slope

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

$w_{B} = \frac{d V}{d x} = V_{B}^{'}$

$V_{B}^{'} = 0$

A zero slope will mean the tangent line plot must be a horizontal line.

Point B

Next, we examine point $B$ by itself. From the previous segment, the shear $V_{B_{1}}$ is zero; however, point $B$ is the location of the concentrated roller with a value of $90 k N$ . There is a discontinuity at this point. The shear at $B$ becomes:

$V_{B_{2}} = 0 + 90$

$V_{B_{2}} = 90 k N$

When we analyze the next segment after, we use $V_{B_{2}} = 90 k N$ instead of $V_{B_{1}} = 0$ .

Segment BC

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

Endpoint B

Shear Value

From examining point $B$ , we saw that $V_{B}$ became $90 k N$ . Likewise, this segment's shear $V_{B}$ is $90 k N$ .

Slope

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

$w_{B} = \frac{d V}{d x} = V_{B}^{'}$

$V_{B}^{'} = 0$

A zero slope will mean the tangent line plot must be a horizontal line.

Endpoint C

Shear Value

Next, we consider the shear at point $C$ . To solve it, we will use the change in shear expression to find the value:

$V_{C} - V_{B} = Δ V_{C B} = \int_{x_{1}}^{x_{2}} w \times d x$

$V_{C} - 90 = \frac{1}{2} (6.5 - 2) (- 45)$

$V_{C} = - 11.25 k N$

The right side of the equation is the area of the uniform triangular $w$ load in segment $B C$ .

Slope

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

$w_{C} = \frac{d V}{d x} = V_{C}^{'}$

$V_{C}^{'} = - 45$

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

Point of Zero Shear

Going off-track for a bit, let's analyze the shear at endpoints $B$ and $C$ :

$V_{B} = 90 k N$

$V_{C} = - 11.25 k N$

We can see that the shear at $B$ was positive. When it reached point $C$ , it became negative. There is a point between $B$ and $C$ in which the shear became zero. It is at this point wherein the moment is at its maximum.

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 shear from the origin is at:

$x = 6.2426$

Point C

Next, we examine point $C$ by itself. From the previous segment, the shear $V_{C}$ is $- 11.25 k N$ ; There is no discontinuity at this point; hence the shear is still the same.

Segment CD

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

Endpoint C

Shear Value

From examining point $C$ , we saw that $V_{C}$ is $- 11.25 k N$ . Likewise, this segment's shear $V_{C}$ is $- 11.25 k N$ .

Slope

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

$w_{C} = \frac{d V}{d x} = V_{C}^{'}$

$V_{C}^{'} = - 36$

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

Endpoint D

Shear Value

Next, we consider the shear at point $D$ . To solve it, we will use the change in shear expression to find the value:

$V_{D} - V_{C} = Δ V_{D C} = \int_{x_{1}}^{x_{2}} w \times d x$

$V_{D} - (- 11.25) = (11 - 6.5) (- 36)$

$V_{D} = - 173.25 k N$

The right side of the equation is the area of the uniform rectangular $w$ load in segment $C D$ .

Slope

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

$w_{D} = \frac{d V}{d x} = V_{D}^{'}$

$V_{D}^{'} = - 36$

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

Point D

Next, we examine point $D$ by itself. From the previous segment, the shear $V_{D_{1}}$ is $- 173.25 k N$ ; however, point $D$ is the location of the hinged support with a value of $263.25 k N$ . There is a discontinuity at this point. The shear at $D$ becomes:

$V_{D_{2}} = - 173.25 + 263.25$

$V_{D_{2}} = 90 k N$

When we analyze the next segment after, we use $V_{D_{2}} = 90 k N$ instead of $V_{D_{1}} = - 173.25$ .

Segment DE

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

Endpoint D

Shear Value

From examining point $D$ , we saw that $V_{D}$ is $90 k N$ . Likewise, this segment's shear $V_{D}$ is $90 k N$ .

Slope

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

$w_{D} = \frac{d V}{d x} = V_{D}^{'}$

$V_{D}^{'} = 0$

A zero slope will mean the tangent line plot must be a horizontal line.

Endpoint E

Shear Value

Next, we consider the shear at point $E$ . To solve it, we will use the change in shear expression to find the value:

$V_{E} - V_{D} = Δ V_{E D} = \int_{x_{1}}^{x_{2}} w \times d x$

$V_{E} - 90 = \int_{11}^{12.5} 0 \times d x$

$V_{E} = 90 k N$

There is no $w$ loads in segment $D E$ ; that's why $V_{E}$ is still zero.

Slope

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

$w_{E} = \frac{d V}{d x} = V_{E}^{'}$

$V_{E}^{'} = 0$

A zero slope will mean the tangent line plot must be a horizontal line.

Point E

Finally, we examine point $E$ by itself. From the previous segment, the shear $V_{E_{1}}$ is $90 k N$ ; however, point $E$ is the location of a concentrated load with a value of $- 90 k N$ . There is a discontinuity at this point. The shear at $E$ becomes:

$V_{E_{2}} = 90 + (- 90)$

$V_{E_{2}} = 0 k N$

We have reached the end of the beam. One way to check if our solution is correct is to evaluate if the shear at the end is zero (to comply with equilibrium)

Drawing the Shear Diagram

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

The shear values and slopes per segment are the following:

Segment AB

$V_{A} = 0$ , $V_{A}^{'} = 0$

$V_{B} = 0$ , $V_{B}^{'} = 0$

Segment BC

$V_{B} = 90 k N$ , $V_{B}^{'} = 0$

$V_{C} = - 11.25 k N$ , $V_{C}^{'} = - 45$

Segment CD

$V_{C} = - 11.25 k N$ , $V_{C}^{'} = - 36$

$V_{D} = - 173.25 k N$ , $V_{D}^{'} = - 36$

Segment DE

$V_{D} = 90 k N$ , $V_{D}^{'} = 0$

$V_{E} = 90 k N$ , $V_{E}^{'} = 0$

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

Shear Diagram

‍

How do we know if our diagram is correct? The shear must close back to zero when we analyze the final point. In this case, it closed $V_{E} = 0 k N$ when we investigated point $E$ .

No ads?

Tree Navigation

See how this article relates to other ideas

Children

No children (yet)

Siblings

How to Form Shear and Moment Equations?

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.

What is the Point of Zero Shear?

The point of zero shear is the position along the beam in which the shear is zero. Let's explore different ways to calculate this position using functions or graphs.

What are Shear and Moment Relationships?

We continue our analysis of our beam example by constructing the moment diagram. This time, we shall relate the shear values to draw the moment graph.

What is the Point of Zero Moment?

The point of zero moment is the position along the beam in which the moment is zero. Let's explore different ways to calculate this position using functions or graphs.

Parent

Analyzing the Internal Force and Stresses of a Beam

In this series of posts, we'll explore how to solve our beam example's internal force and stresses using various methods and techniques - another essential requirement for analysis.

Connections

What is the First Derivative?

The first derivative tells us the steepness of a mathematical function at a given point. In other words, it is the slope of the math function at said point.

How to Solve the Area Between Curves Using Integration?

We can determine the area between curves using definite integrals. In this post, let's explore how it expands the approximate solution of solving area.

Created On

June 5, 2023

Updated On

February 23, 2024

Contributors

Edgar Christian Dirige

Founder

References

WeTheStudy original content

Revision

1.00

Got some questions? Something wrong? Contact us

This account is currently deactivated due to a
failed payment. Update your payment method now