# Determine the reaction forces at the supports

• MNWO
In summary, the conversation involves a student seeking help with a static equilibrium problem involving calculating the reactions at support points on a beam. The student presents their calculations and asks for feedback on their approach. The expert points out mistakes in the calculation and gives suggestions on how to correctly solve the problem.

#### MNWO

Hello, so I've been given this question and I get a different anwser then my friends. Can you check if what I wrote is right? If its wrong then can you tell me what have I done wrong? Sorry for the quality of the picture but my Paint skills are not so great :)
I was measuring to the right support
RA is the left support, RB is the right support. Here is my calculations:
Sum of anticlockwise movement = Sum of clockwise movement
forces to the right support= 2.4 + 5 + 3.5 = 10.9kN
force x distance to the support= (3.5kN x 2m) + ( 5kN x 5.5m) + (2.4kN x 10.5m)=59.7kN
59.7kN=sum of clockwise movement
1.8kN + RB12 = 59.7kN
59.7-1.8=57.9kN
57.9/12=RB
RB=4.825
so RA= 59.7kN- 4.825
RA=54.875 kN

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Hello MNWO. You should use the formatting template when you create a thread in the homework sections (in fact it's a requirement).

As it stands there's no clear statement of what the problem is: You haven't stated exactly what it is you hope to calculate. That statement should be given in the Problem Statement section of the template. You also need to define what the forces are that are indicated on your drawing; one of them is given as 1 kN/m. How are we to interpret that? As a distributed load? If so, how is it distributed?

Assuming that the beam is at static equilibrium, why don't you just set the moment equal to zero and sum the moments across the beam?

For instance, taking the moments about RA

∑Mra = 0 = -1.5m(2.4kN) + ... you do the rest.

Where are you getting 3.5kN x 2m?

The 3.5kN is not a distributive force on your diagram, it's a concentrated force that is 10m from RA.

You cannot solve this problem by just summing the forces in the Y direction - you have two unknowns - you need two equations. Namely a moment equation and a sum of the forces equation, or two moment equations.

With your solution the beam would be accelerating in the Y-direction - making it into a dynamics problem.

Edit: by clockwise and anti-clockwise movements I'm assuming you mean moments in which case you're adding them up wrong. Moments depend on TWO tings, the force and the distance from the axis of rotation. Remember that torque is the cross product of the distance and the force T = R X F . You can't sum up the forces by themselves first. Remember your order of operations from math.

gneill said:
Hello MNWO. You should use the formatting template when you create a thread in the homework sections (in fact it's a requirement).

As it stands there's no clear statement of what the problem is: You haven't stated exactly what it is you hope to calculate. That statement should be given in the Problem Statement section of the template. You also need to define what the forces are that are indicated on your drawing; one of them is given as 1 kN/m. How are we to interpret that? As a distributed load? If so, how is it distributed?

I think the 1kN/m is distributed over the 5m that it is located under because the two arrows on either side don't have any loads on them. That's how I interpreted it. I'm pretty sure we're looking at a basic static equilibrium problem.

MNWO said:
Hello, so I've been given this question and I get a different anwser then my friends. Can you check if what I wrote is right? If its wrong then can you tell me what have I done wrong? Sorry for the quality of the picture but my Paint skills are not so great :)
View attachment 76408I was measuring to the right support
RA is the left support, RB is the right support. Here is my calculations:
Sum of anticlockwise movement = Sum of clockwise movement
forces to the right support= 2.4 + 5 + 3.5 = 10.9kN
force x distance to the support= (3.5kN x 2m) + ( 5kN x 5.5m) + (2.4kN x 10.5m)=59.7kN
59.7kN=sum of clockwise movement

You mean 'moment' here, rather than 'movement'. The units of moment are in kN-m and not kN.

You've taken the support at B as the reference in writing the moment equation above.
1.8kN + RB12 = 59.7kN
You can't use the reaction at B in this equation, because the moment @ RB = 0, since it is the refererence
59.7-1.8=57.9kN
57.9/12=RB
RB=4.825
so RA= 59.7kN- 4.825
RA=54.875 kN

You are subtracting forces from moments, which you can't do, because forces and moments have different units. You must add up all the known forces applied to the beam and use them to solve for the unknown reaction.

## What is the concept of "reaction forces at the supports"?

The reaction forces at the supports refer to the forces that are exerted on a structure at its points of support, such as the ground or a wall. These forces are equal in magnitude and opposite in direction to the forces applied to the structure, allowing it to remain in equilibrium.

## Why is it important to determine the reaction forces at the supports?

Knowing the reaction forces at the supports is crucial for understanding the stability and structural integrity of a system. It helps engineers and designers ensure that a structure can withstand the applied forces and prevent any potential failures.

## How are reaction forces at the supports calculated?

The reaction forces at the supports are calculated using the principles of equilibrium, which state that the sum of all forces in the horizontal and vertical directions must equal zero. By analyzing the external and internal forces acting on a structure, the reaction forces at the supports can be determined.

## What factors can affect the reaction forces at the supports?

The reaction forces at the supports can be affected by various factors, such as the type of support (fixed, pinned, or roller), the geometry and material properties of the structure, and the magnitude and direction of the applied loads. Changes in any of these factors can alter the reaction forces and may need to be considered in the design process.

## What are the limitations of determining the reaction forces at the supports?

While calculating the reaction forces at the supports can provide valuable information about a structure, it is important to note that it is an idealized representation and may not fully reflect the real-world conditions. Factors such as material deformations, joint stiffness, and external disturbances can affect the actual reaction forces experienced by a structure.