Rod deflection

1. Nov 27, 2003

alexbib

Here's a lil question:

If you have a rod of which an end is clamped to a table and you apply a force somewhere on the rod, what will the deflection be (in terms of difference in the height of the other end of the rod? For sure it will be proportional to the force and to some power of the lenght at which it is applied. What would be the complete formula, solved for deflection?

2. Nov 28, 2003

Tyro

For symmetrical beams with $$I_{xy} = 0$$, the curvature, $$v^{''}$$ is given by:

$$v^{''} = -{\frac{M_y}{EI_{xx}}}$$

The deflection is thus:
$$v(L) = -\int_{0}^{L}\int_{0}^{L}{\frac{M_y}{EI_{xx}}}\cdot{dz}\cdot{dz}$$

where L is the length of the rod, z is the axial length along the rod, F is the applied force, $$I_{xx}$$ is the moment of inertia about the horizontal axis, E is the elastic modulus of the material and $$M_y$$ for a point load at the end is $$M_y = {F}\cdot{(L - z)}$$.

Last edited: Nov 28, 2003
3. Nov 28, 2003

alexbib

Thx a lot for answering. $$I_{yy}$$ is the moment of inertia along the horizontal axis you say? Then what is the $$I_{xx}$$found in the formula? The moment of inertia along another axis?

What do you mean by "For symmetrical beams with$$I_{xy} = 0$$"?

Also, I'm not sure I fully understand what z is...
Sorry for being a newb lol.

Finally how is the curvature expressed. Is it a scalar quantity? What is its unit?

Thanks a lot,

Alex

4. Nov 28, 2003

Tyro

Oops, I made a typo. $$I_{yy}$$ should be $$I_{xx}$$. Sorry for the confusion caused (has now been edited). If you wanted to calculate $$u^{''}$$ which is the horizontal deflection due to an asymmetric section and/or horizontal load, then $$I_{yy}$$ is the moment of inertia about the vertical Y-axis.

Non-symmetrical section beams (e.g. L-section, as opposed to I or T-section) will have non-zero $$I_{xy}$$.

z is a variable, L is the total length (fixed). So z varies from 0 to L.

For the moment, don't worry about curvature. Use the integral equation to get the deflections you want. But since you asked, curvature has units m-1. It is a scalar quantity.

5. Nov 29, 2003

alexbib

Ok, thanks. So $$M_y$$ would be the torque, attaining it's maximum value when z=0, when the force is applied derectly at the free end?

6. Nov 29, 2003

Tyro

Actually, M was an abbreviation for "moment". But torque will do

Yup, the moment is maximum at the root of the rod where it is supported and goes to zero the closer you get to where you apply the load - the moment arm gets shorter as you go closer to the applied force.

BTW, you solve the integration constants based on the initial conditions. E.g. the slope at a rigid, unhinged support @ z=0 is 0, initial tip deflection @ z=L is 0, etc.

7. Nov 29, 2003

alexbib

Alright, thanks for the help!