# Movement of point

1. Aug 21, 2004

### TSN79

I have a structure that looks something like this; a steel-pole is vertical for 2m and then horizontal for 1m, like an uppside-down L-shape. At the end of the horizontal part, a force acts downwards (10kN). How do I go about to find both the horizontal and vertical movement of the point where the force acts?

2. Aug 21, 2004

### Clausius2

You have to use the Navier-Bresse equations. If you don't know them, say you don't and I will explain you the calculus.

The shape you describes is very simple, so you will not spend much effort in solving it. First of all, you must work out the flector's distribution of the structure. When you have this distribution, insert it in N-B equations. There are two theorems, first and second Mohr theorems that surely will help you much.

If you don't know what flector's distribution means (maybe this word does not exist in english), put it across next thread, please.

3. Aug 21, 2004

### enigma

Staff Emeritus
Is the structure free to rotate, or is it fixed somewhere and you're measuring deflection?

4. Aug 23, 2004

### TSN79

Is this the Navier-Bresse equation you mentioned? $$\delta=\frac{FL}{AE}$$ Because I have a pretty good idea that this is used in the solution, where F is a force, L is length, A is area, and E is a constant for steel. Delta is the movement I think. Flector's distribution I have not heard of, perhaps we call it something else in norwegian. By the way, the structure is fixed to the ground at the end of the vertical part. Explain please...?

5. Aug 23, 2004

### Tom Mattson

Staff Emeritus
In every problem of this type, your solution must contain the following 3 ingredients:

1. Equilibrium
You must write down the equilibrium equations, using Newton's second law.

2. Force-Deformation Relations
That's the relationship between F and &delta; that you just wrote down in your last post.

Now those two are easy. The tricky part is the third ingredient.

3. Compatibility
These deforming members are fighting over a fixed amount of space. If one bar pushes &epsilon; units to the right, the other bar must give by moving &epsilon; units to the left. You should let the displacement of each member be a vector such as ui+vj, where u and v are independent variables, and use right triangle trigonometry on the deformed member.

Last edited: Aug 23, 2004
6. Aug 24, 2004

### Clausius2

See this structure:

_______C
| B
|
|
|
A

A=point ground-clamped (ground fixed without posibility of rotation);
B=welding point between two girders;
C=force exerted point. A force F is exerted downwards in this point.
Lenghts: AB=L1; BC=L2;

All right, I'm going to solve this problem:

i) first of all we are going to calculate the bending moment distribution M(in spanish it is said "momento flector").

Force Reactions:
VA=vertical reaction in A (pointing upwards). HA=horizontal reaction (pointing rightwards) in A; MA=moment reaction in A (turning anticlockwise);

VA=F; HA=0; MA=F*L2; ok?

So that bending moments are M=MA in point A, M=MA in point B; and M=0 in point C. You should see bending moment is constant along AB, and linear along BC.

ii) Navier-Bresse equations:

Horizontal movement in C:

$$\overline{u_{c}}=\int{\frac{M}{EI}sds}$$ where E=Young modulus; I=section's inertia moment; s=doesn't matter.

You can employ 2nd Mohr theorem in order to solve this integral. Pay attention:
Take the bending distribution along AB. It's rectangular shaped isn't it?. Take the centroid of this distribution, namely G. It's trivial to see it's located at the middle point of AB. Proyect it over the girder AB. And then, join together points G and C with a straight line. The segment normal to this last line will be the tangent of the trayectory of point C due to ONLY AB bending moment distribution. You can draw a vector over this last line (it will point to right down side) to see spatially the path of point C. The horizontal component of this vector will be Uc. How is it calculated?. By handling the last equation:

$$\overline{v}=\sum(\frac{A_{i}}{EI}(d(G_{i}U)\overline{e_{x}}+d(G_{i}V)\overline{e_{y}}))$$

This equation is all what you need. The sum sweeps i=1,2 because of two bending moment distributions. A= area of each bending moment distribution (one is rectangular and the other one is triangular).
d(GV)=distance from each centroid calculated as stated before and V, a line which goes trough C point in vertical direction (y direction)
d(GU)=distance from each centroid and U, a line which goes trough C in horizontal direction (x direction).
e=unitary vector.
v=movement vector.

My solution is:

$$\overline{v}=\frac{F L_{1} L_{2}}{EI}(0.5L_{1}\overline{e_{x}}-L_{2}\overline{e_{y}})+\frac{F L_{2}^2}{3EI}(-L_{2}\overline{e_{y}})$$;

Anyway, you are solving an elastic body. If you don't have any knowledge about elastic theory or structural engineering, or you never have heard about N-B equations, then you are endangered fighting against this problem. I advice you to consult any structures book.