# How to Correctly Calculate Bending Moments in Loaded Beams?

• David J
In summary: This is unnecessary and can be avoided by using the slope of the SF curve to find the change in BM. For example, the BM at x = 3 is equal to the BM at x = 2 + the area under the shear curve between x = 2 m and x = 3 m, thussince BM = -10 kN-m at x = 2 m,then the BM at x = 3 m would be BM = -10 kN-m + [(1/2)*(16.7 + 11.7)] = -10 kN-m + 14.20 kN-m = 4.2 kN-M and so on.Therefore, the change in BM at
David J
Gold Member

## Homework Statement

I need to find vertical reactions, sketch the shear force and bending moment diagrams, calculate bending moments, etc

## The Attempt at a Solution

This is attached, I think my answers are correct, I just need help in identifying any errors, etc

#### Attachments

79.3 KB · Views: 2,312
David J said:

## Homework Statement

I need to find vertical reactions, sketch the shear force and bending moment diagrams, calculate bending moments, etc

## The Attempt at a Solution

This is attached, I think my answers are correct, I just need help in identifying any errors, etc
Solving for the reactions R1 and R2 appears to be correct.

I see a problem when you start to construct your shear force diagram.

You haven't shown the shear force correctly at the location of R1 at x = 2 m, the point load 10 kN at x = 3m, and the point load 20 kN at x = 10m. For some reason, you show the correct jump in the shear force curve where R2 is located, which is why these other errors are so puzzling.

Did I forget to include the udl in the first 2 meters? The first 2 meters starts in the positive direction when it should maybe start heading in the negative direction due to the udl

David J said:
Did I forget to include the udl in the first 2 meters? The first 2 meters starts in the positive direction when it should maybe start heading in the negative direction due to the udl
The problem is not with the UDL. What happens to the shear force curve at the point where a concentrated load is located?

As quoted in my lesson :-
For a series of concentrated vertical forces on a beam, the shear force diagram consists of a series of vertical steps and horizontal lines. For a beam subjected to concentrated vertical forces and a udl, the shear force diagram will again have vertical steps but with lines of constant slope joining the steps. I guess the curve is going to react in the opposite direction to the force

David J said:
As quoted in my lesson :-
For a series of concentrated vertical forces on a beam, the shear force diagram consists of a series of vertical steps and horizontal lines. For a beam subjected to concentrated vertical forces and a udl, the shear force diagram will again have vertical steps but with lines of constant slope joining the steps. I guess the curve is going to react in the opposite direction to the force

As I said in a previous post above:
1. You show the correct jump in the shear force curve where R2 is located
2. You have not shown any jump in the shear force curve at the location of R1 at x = 2 m, the point load 10 kN at x = 3m, or at the point load 20 kN at x = 10m.

There should be jumps or steps in the shear force curve at any location where there is a concentrated force or reaction. It's not clear to me why you showed the step in shear force at R2, but neglected to show a step in shear force at any of the other locations where a concentrated force or reaction was located.

I have spent most of the day going back through the lessons and looking at the examples. I think the attached is correct (or close) now. I messed the first section of the SFD up

#### Attachments

• Question 1b.pdf
28.2 KB · Views: 194
David J said:
I have spent most of the day going back through the lessons and looking at the examples. I think the attached is correct (or close) now. I messed the first section of the SFD up
This shear force curve looks much better.

Now, calculate the bending moment curve from it.

I am hoping this is close to correct

#### Attachments

• Question 1c.pdf
28.5 KB · Views: 182
David J said:
I am hoping this is close to correct
The bending moments seem to be OK, but it looks like you keep doing a lot of unnecessary calculation as you move along the length of the beam.

For example, the BM at x = 3 is equal to the BM at x = 2 + the area under the shear curve between x = 2 m and x = 3 m, thus

since BM = -10 kN-m at x = 2 m,

then the BM at x = 3 m would be BM = -10 kN-m + [(1/2)*(16.7 + 11.7)] = -10 kN-m + 14.20 kN-m = 4.2 kN-M and so on.

Since the SF curve is composed of straight lines, you are essentially calculating the area under a series of trapezoids to find the change in BM as you move along the length of the beam.

## 1. What is a loaded beam?

A loaded beam is a structural element that is designed to support a load or weight across a span. It can be made of various materials such as wood, steel, or concrete and is commonly used in construction and engineering projects.

## 2. How are loaded beams different from other types of beams?

Loaded beams are specifically designed to withstand a load or weight, whereas other types of beams may have different purposes such as providing support or framing. Loaded beams are also designed to distribute the load evenly along the length of the beam.

## 3. What factors affect the strength of a loaded beam?

The strength of a loaded beam is affected by several factors including the material used, the shape and size of the beam, the placement of the load, and any additional external forces such as wind or vibrations. The design and construction of the beam also play a crucial role in its strength.

## 4. How do engineers determine the appropriate size and material for a loaded beam?

Engineers use mathematical calculations and computer simulations to determine the appropriate size and material for a loaded beam. They take into account the type and amount of load, the span of the beam, and other factors to ensure that the beam can safely support the intended weight without failure.

## 5. What are some common applications of loaded beams?

Loaded beams have a wide range of applications in various industries, including construction, architecture, and manufacturing. They are commonly used in buildings, bridges, and other structures where weight or load needs to be distributed across a span. Loaded beams are also used in machines and equipment to support heavy components or materials.

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