Solve Static Equilibrium Qs: 17.55kN & 17.45kN, 15.63kN & 16.37kN

[kiteboy]

Hi

I have some questions to go through for static equilibrium...I have the answers but I have no idea how to do these really - I am guessing most of them without really understanding

I can't remember doing physics in shcool at all and since its been the best part of 20 years since school some help would be REALLY appreciated

Ok ill put two sample questions with the answers that were also given

A horizontal beam AB is simply supported at each end. The beam carries a point load of 12kN at a point 0.8 m from A, a point load of 8kN at a point 1.9m from A and a point load of 15kN at a point 3m from A. Calculate the reactions exerted by the supports on the beam.
The answers are : 17.55kB at A and 17.45kN at B

Another one

A horizontal beam AB is 6m long and simply supported at each end. The beam carries a uniformly distributed load of 3kN/m from each end A to the mid point of the beam, a point load of 9kN at a point 2.1m from A, and a point load of 14kN at a point 4.7m from A. Calculate the reactions exerted by the supports on the beam.
(answers are 15.63kN at A and 16.37kN at B)


If anybody can find the time to really explain how these are generally worked it out would be great

Ive already ordered books from amazon to hopefully help me out as this is beyond me at the moment - even google is failing me!

Thanks

 

[jhae2.718]

For static equilibrium, two conditions must be met:
$$\Sigma \vec{F} = 0$$

$$\Sigma \vec{\tau} = 0$$

 

[kiteboy]

For static equilibrium, two conditions must be met:
$$\Sigma \vec{F} = 0$$

$$\Sigma \vec{\tau} = 0$$

I think that means the sum of the forces in the x and y direction must be 0
I think the sum of the moments must also be 0

I kinda get that but not really when it comes to the examples if you know what i mean

 

[jhae2.718]

To show you the basic process, I'll walk you through a simpler example, given by this image:

Lets say we have two forces, $$F_1$$ and $$F_2$$ that are evenly spaced along a beam of length L. (For this, we assume that the mass of the beam is negligible. Sometimes the beam is given a mass; it acts at the center of mass.) There are simple reaction forces acting at A and B. That is, there is just a force in the vertical direction at each point.

The beam is in static equilibrium, so the following equations hold:
$$\Sigma \vec{F}=0$$

$$\Sigma \vec{M} = 0$$

We have two unknowns, A and B. Recall that the definition of the magnitude of a moment is $$M=Fd$$, where F is the magnitude of the force and d is the length of the moment arm, which is the distance between the line of action through the force and the line through the point which the moment is about.

(More generally, $$\vec{M}=\vec{r} \times \vec{F}$$, the cross product of the position vector and the force vector.)

If we take the net moment about one of the points, the reaction force at that point will produce no moment. This will give us an expression we can solve for the other reaction force.

I'll find $$\Sigma M_A$$, the net moment at A:
$$\Sigma M_A = F_1\frac{L}{3}+F_2\frac{2L}{3}-BL$$
This has to equal zero for it to be in static equilibrium, so:
$$0 = F_1\frac{L}{3}+F_2\frac{2L}{3}-BL$$
$$BL= F_1\frac{L}{3}+F_2\frac{2L}{3}$$
$$B= \frac{1}{3}\left( F_1 + 2F_2\right)$$

Now that we know B, we can use ∑F=0 to find A. There are no forces in the x direction, so we only need to look at the y direction. This gives:
$$\Sigma F_y = A+B-F_1-F_2$$
Letting ∑F=0 and solving for A gives:
$$A= F_1+F_2-B$$
$$A= F_1+F_2-\frac{1}{3}\left( F_1 + F_2\right)$$
$$A=\frac{1}{3}\left( 2F_1 + F_2\right)$$

Generally, you want to do the following:
1. Draw a diagram showing the forces.
2. Apply ∑M=0 through the point that eliminates the most unknown forces and solve for the remaining unknowns.
3. Apply ∑F=0 to solve for the remaining forces.

For your second problem, you have a part of the beam where a uniform load is applied. Calling the linear load density $$\lambda$$, the force it applies is given by $$F=\lambda d$$, where d is the length along which the load is applied. The load can then be treated as a point mass halfway between the points where the load is applied.