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I want to calculate how much a solid bar will flex when force is applied to it. The set up looks like this:

Sketch

The rod (in green) rests on two stationary points, and the force is applied in the center between the two points. The rod has a rectangular cross section.

What I want to know is the length that the middle of the rod will be moved in the direction of the force. The force can be assumed to be small enough not to deform the rod.

What properties of the material in the rod do I need to know to calculate this? And how do I do the actual calculation?

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    $\begingroup$ A general bar's deformation is governed by the Euler-Bernoulli equation, $(EI u''(x))'' \sim f(x)$, under certain assumptions. $\endgroup$
    – JamalS
    Jan 24, 2015 at 19:04
  • $\begingroup$ I don't understand how to use that. Could you describe what all the variables are? $\endgroup$
    – Findus
    Jan 24, 2015 at 19:47
  • $\begingroup$ Wait a second, you want to know "how much a solid bar will flex when force is applied to it", yet "The force can be assumed to be small enough not to deform the rod" ? That doesn't seem consistent. $\endgroup$
    – paisanco
    Jan 24, 2015 at 20:00
  • $\begingroup$ Okay, maybe I am using the wrong terms here (English is not my native language). What I mean is that force is applied, which makes the rod bend, but it is not permanently deformed. I.e. when the force is no longer applied, the rod returns to straight. $\endgroup$
    – Findus
    Jan 24, 2015 at 20:46
  • $\begingroup$ OK, you meant deformations are small enough to be elastic, not plastic. $\endgroup$
    – paisanco
    Jan 24, 2015 at 21:06

2 Answers 2

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It is all explained here if you search simply supported beams.

You will find the equation $$w = \frac{F \ell^3}{48 E I}$$

Here $\ell$ is the distance between the supports, $F$ is the force applied, $E$ is the elastic modulus of the material and $I$ is area moment of the section. Rectangular sections have $I=\frac{1}{12} b h^3$ where $b$ is width and $h$ is height. The caveat here is the use of consistent units. You cannot mix metric with inches with feet.

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  • $\begingroup$ Thanks! Exactly what I needed, and also a very helpful link! I think the reason I did not fins this on google myself is that I did not know the correct terms in english for this. $\endgroup$
    – Findus
    Jan 25, 2015 at 12:12
  • $\begingroup$ The two common types of supports are called: simply supported beam and cantelever beam. Those are the correct terms. $\endgroup$
    – John Alexiou
    Jan 25, 2015 at 16:26
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You want to know the vertical deflection of the rod, which will pull the ends of the rod inward toward the red arrow.

This is a center point loaded simple beam.

The shear diagram plots as two horizontal rectangles ( up 0.5 on the left, down 1.0 in the middle, up 0.5 on the right.

The bending moment diagram plots like the low pitched roof of a house ( 2 back to back triangles ).

The next integration is a quadratic equation ( K Y^2 ) The next integration is what you want, a cubic equation. ( K/3 Y^3 ) The distance that was originally horizontal now lies along the curved surface of the cubic equation.

You can model this using two aluminum yard sticks ( or meter sticks ) in parallel over two fulcrums ( like engineers scales ). One is horizontal and unloaded, the other is center point loaded, and deflects more and more with increasing load.

Because the rod is a rectangle, the top half gets shorter ( inside the curve ), while the outside ( bottom ) gets longer, but the average length does not change, it just lies along the curved line of the cubic equation. ( the neutral axis of the rectangular rod bends along the "Minus" K/3 Y^3 line ). ( deflects downward )

Note if you change to a uniformly distributed load, the curve would be a fourth order equation ( minus K/12 Y^4 equation ).

Coins, or steel washers equally spaced along the aluminum yard ( meter ) stick would make a good distributed load.

Remember the center point must move down vertically, so the distance the rod moves inward from both ends is symmetrical.

Using a one kilogram mass at a distance of 0.5 meters from both ends should work if it does not bend the meter stick in half.

The deflection would also be dependent upon the height and width of the rectangle, and the strength of the material selected. You would need to look up the Modulus of elasticity of aluminum, or of steel, or wood depending upon what material is used in the rod.

Mike Clark Golden, Colorado, USA

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