# Shear Force and Bending Moment Diagram

• jasonbot
In summary, the conversation discusses finding the Shear Force and Bending Moment diagrams for a problem involving a beam with lengths of 1.2m each and a peak force of 4kN/m. The speaker also mentions calculating the support reactions as Ay=4.8kN (up) and Ma= 5.76kN.m (anticlockwise), and using similar triangles to solve for Vx and Mx. The conversation concludes with a mention of possible errors and a request for advice on using triangle forces correctly.
jasonbot

## Homework Statement

I have to find the Shear Force and Bending Moment diagrams for this problem. The lengths in the picture are 1.2m each and the peak force is 4kN/m and I calculated the support reactions to be:

Ay=4,8kN (up)
Ma= 5.76kN.m (anticlockwise)

## Homework Equations

These are what I'm looking for but failed to do...

## The Attempt at a Solution

I first section the beam 0<x<4

Efy=0

+4.8 -.5x*3.33x-Vx=0

I got the 3.33x by using similar triangles; if triangle 1 is (1,2;4) then sectioned triangle should be (x;y) so therefore 4/1.2=y/x x=3.33xthis gave me Vx=-1.67x^3+4.8 which is wrong.

Then EMo=0

+Mx +.5x*3.33x*(x/3) -4.8x +5.76=0

which gets me Mx=-0.56x^3 +4.8x +5.76The correct answers for this question are:

Vx=4.8 -(x^3/6)
Mx=4.8x -(x^3/18) -5.76I have a feeling I'm not using the triangle force correctly. Please advise.

#### Attachments

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jasonbot, welcome to PF!

jasonbot said:

## Homework Statement

I have to find the Shear Force and Bending Moment diagrams for this problem. The lengths in the picture are 1.2m each
Is it 12 m or 1.2 m?
and the peak force is 4kN/m and I calculated the support reactions to be:

Ay=4,8kN (up)
Ma= 5.76kN.m (anticlockwise)
yes, looks good
I first section the beam 0<x<4
0<x<12 m
Efy=0

+4.8 -.5x*3.33x-Vx=0

I got the 3.33x by using similar triangles; if triangle 1 is (1,2;4) then sectioned triangle should be (x;y) so therefore 4/1.2=y/x x=3.33x

this gave me Vx=-1.67x^3+4.8 which is wrong.
if you correct the 1.2 m to 12m , then Vx = -.167x^2 +4.8
Then EMo=0

+Mx +.5x*3.33x*(x/3) -4.8x +5.76=0

which gets me Mx=-0.56x^3 +4.8x +5.76
something is afoul between 1.2 m and 12 m, the decimal point is messed.., but you are still not quite right, don't let the plus and minus signs wear you down...
`
The correct answers for this question are:

Vx=4.8 -(x^2/6)
typo error
Mx=4.8x -(x^3/18) -5.76
taking clockwise moments as plus, I don't know about that decimal point, however

I have a feeling I'm not using the triangle force correctly. Please advise.
Looks like you did well, but check out your plus and minus signs, they'll bite you every time...

Last edited:

I would first like to commend you on your attempt at solving this problem. It is clear that you have a good understanding of the basic principles involved in finding the Shear Force and Bending Moment diagrams for a beam.

However, I would suggest that you take a step back and review your approach. It seems that you are using the method of sections to find the Shear Force and Bending Moment at a particular point on the beam. While this method can be effective in some cases, it may not be the best approach for this problem.

Instead, I would recommend using the equations for Shear Force and Bending Moment at a point on a beam in terms of the applied loads and reactions. These equations can be derived from the equilibrium equations and are much simpler to use in this case. You can then use these equations to find the Shear Force and Bending Moment at any point along the beam.

In addition, it is important to carefully consider the signs of the forces and moments. In your attempt, you have not taken into account the direction of the forces and moments, which has resulted in incorrect solutions.

I hope this helps and good luck with your future calculations!

## 1. What is a Shear Force Diagram?

A Shear Force Diagram (SFD) is a graphical representation of the shear force at different points along a beam. It shows the magnitude and direction of the vertical forces acting on a beam, caused by external loads and reactions.

## 2. What is a Bending Moment Diagram?

A Bending Moment Diagram (BMD) is a graphical representation of the bending moment at different points along a beam. It shows the magnitude and direction of the internal moments caused by external loads and reactions. BMDs are important in determining the strength and stability of a beam.

## 3. How are Shear Force and Bending Moment related?

Shear Force and Bending Moment are closely related as they both represent the internal forces and moments on a beam. The Shear Force at a point on a beam is equal to the slope of the Bending Moment Diagram at that point. This means that changes in the Shear Force will result in changes in the Bending Moment, and vice versa.

## 4. How is a Shear Force and Bending Moment Diagram calculated?

Shear Force and Bending Moment Diagrams are calculated using the equations of equilibrium and the concept of free body diagrams. The external loads and reactions on a beam are analyzed, and then the Shear Force and Bending Moment are calculated at different points along the beam using these equations.

## 5. Why are Shear Force and Bending Moment Diagrams important?

Shear Force and Bending Moment Diagrams are important in structural engineering as they help in analyzing and designing beams and other structural elements. They provide valuable information about the internal forces and moments that a beam experiences, which is crucial in determining its strength and stability. These diagrams also help in identifying critical points and sections of a beam that may need additional reinforcement.

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