What is the support reaction at E for the 9m length beam?

[MMCS]

Calculate the support reactions at B and E for the 9m length beam shown below.
I have the answers at 4.71KN and 1.79KN but it is old work that i am revising and i can't remember how i got to them. One of the supports not being at the end is throwing me off. If anyone could show a rough working out it would be appreciated.

 

[PhanthomJay]

You should refresh your memory on the 3 equilibrium equations
$\Sigma F_x = 0, \Sigma F_y = 0,$ and $\Sigma M_{any-point} = 0$.

There is nothing going on in the x direction, so you just need the last 2 equations. The reaction forces must add up to the applied loads; but the key is summing moments about a point = 0 (hint: choose one of the reaction points as your point to sum moments about. Watch plus/minus signs).

Please show an attempt at a solution for further assistance.

 

[MMCS]

Hi thanks for your reply

I have (-1500 x 2) + (-3000 x 4) + (-2000 x 9) + 7B = 0
-33000 + 7B + E= 0
B = 33000/7
B = 4714N

I don't know where to go next to calculate E?

 

[PhanthomJay]

Hi thanks for your reply

I have (-1500 x 2) + (-3000 x 4) + (-2000 x 9) + 7B = 0
-33000 + 7B + E= 0
B = 33000/7
B = 4714N

I don't know where to go next to calculate E?

Yes, good. For E, the sum of forces in the vertical y direction must be zero. That will give you E. You can also find E by summing moments about B. That serves as a good check on your work.

 

[asteriatic]

The support reaction at E for the 9m length beam can be calculated using the principles of statics. Since the beam is in equilibrium, the sum of all the vertical forces acting on it must be equal to zero.

To calculate the support reaction at E, we need to consider the forces acting on the beam at that point. These include the weight of the beam itself, any applied loads, and the reactions at the other support (B).

Assuming that the beam is uniform and there are no applied loads, we can calculate the weight of the beam using its length and density. Let's say the beam has a density of 7850 kg/m^3 and a cross-sectional area of 1m^2, then its weight would be 9m x 7850 kg/m^3 x 1m^2 = 70,650 kg.

Now, let's consider the reaction at support B. Since the beam is in equilibrium, the reaction at B must be equal and opposite to the weight of the beam. This means that the reaction at B would be 70,650 kg x 9.8 m/s^2 = 693,570 N or approximately 693.6 kN.

To calculate the reaction at support E, we can use the principle of moments. This states that the sum of all the moments (torques) acting on the beam about any point must be equal to zero.

Since we are interested in the reaction at E, let's consider the moments about that point. The weight of the beam will create a clockwise moment (negative) and the reaction at B will create a counterclockwise moment (positive).

Using the equation for moments (M = F x d), we can set up the following equation:

-70,650 kg x 9.8 m/s^2 x 9m + RE x 9m = 0

Solving for RE, we get RE = 70,650 kg x 9.8 m/s^2 = 692,670 N or approximately 692.7 kN.

So, the support reaction at E for the 9m length beam is approximately 692.7 kN.

It is important to note that this calculation assumes that the beam is in a static state and there are no applied loads. If there are any applied loads, the support reactions at B and E would change accordingly. It is