# Elasticity / Diving Board problem

1. Dec 8, 2008

### joshncsu

1. The problem statement, all variables and given/known data

The board's left end is fastened to a fixed support by a connection that is free to pivot. The board is supported to its middle on another fixed support point. A diver of mass 60 kg stands at the end of the board. (ignore the mass of the board for this problem)

The board extends 2.2 m from the pivot.
The diver is 60 kg.

I) Suppose the board deflects by 3 cm when the 60 kg diver stands on it. Now he jumps up and comes down on the board; at the lowest point of his motion, the deflection of the board is 6 cm. What is the net force on the diver?

II)A new board is installed that is twice as thick as the old board but is made of the same material and is the same length and width. Approximately how much does this board deflect when the 60 kg diver stands on it?

2. Relevant equations

$$\vec{F}_{sp} = k \Delta x$$

$$\Delta L = \frac{FL}{AY}$$

$$\Delta x = \frac{\Vert \vec{F} \Vert}{k}$$

thus,

$$k = \frac{AY}{L}$$

3. The attempt at a solution

I solved part I.

$$k = \frac{\vec{F}_{sp}}{\Delta x}$$

$$k = \frac{60 \cdot 9.8}{.03} = 19600 N/m$$

$$\vec{F}_{sp} = k \Delta x = 19600 \cdot .06 = 1176 N$$

I'm having a problem with part II. I figure since the area is doubled that I would use Young's modulus (Y) to solve this. But the following formula needs and area (A) and I don't know what to put for it.

$$k = \frac{AY}{L} = 1176 = \underbrace{\frac{AY}{2.2}}_\text{two unknowns}$$

2. Dec 8, 2008

### PhanthomJay

You are using the wrong formula for k. The formula you are using applies to members under tensile or compressive axial loads. You have, rather, a bending load deflection, which is a function of I, the area moment of inertia of the board, not A. If you double the width, I increases by??

3. Dec 8, 2008

### joshncsu

I'm having trouble finding the correlation between I (inertia) and k (spring constant).

The formulas that I have thus far (in the textbook) for either are:

$$\vec{F}_{sp} = k \Delta x \hspace{1in} I = \sum mr^2$$

Note: Momentum or energy has not been covered yet.

4. Dec 8, 2008

### PhanthomJay

You seem, however, to have chanced upon the spring constant, k, for a beam under axial load, which is k=AY/L. For a cantilever beam in bending, k is 3YI/L^3. I don't think you are expected to know that, but you should know the formula for calculating I for a rectangular shape, which is bt^3/12. So if the thickness doubles, and all else stays the same, I must increse by ??, and the deflection must be reduced by a factor of ????

5. Dec 8, 2008

### joshncsu

Thanks for your help but I'm still confused. I've searched my book and I don't see that formula. I would try it anyway but I don't know what b and t are. I did find this:

Plane or slab, about edge (M = mass, a = length from axis):

$$I = 1/3 Ma^2$$

I know that a = 2.2, but the original mass is unknown or ignored. Even by going with that I get:

$$I = 1.613 \cdot M$$

So I can only conclude that if the mass is doubled then I is doubled. Though that doesn't seem right. And if it was I still don't know what to do with that data and apply it to the deflection of the board (delta x).

Sorry if I'm making this problem more difficult than it really is or if I'm missing something obvious.

6. Dec 9, 2008

### PhanthomJay

You are confusing mass moment of inertia with area moment of inertia (also called the second moment of the area). It is the area moment of inertia that you need to be concerned with in this problem. Unless I've missed something, I'm not sure how you can solve for the deflection of the board when it's thickness is doubled, without having knowledge of the area moment of inertia of a rectangular cross section, which, as I noted earlier, is bt^3/12, where b is the width of the board, and t is its thickness. What this formula tells you, is that if the thickness is doubled, that is, multiplied by 2, and b stays the same, then the moment of inertia must increase by (2)^3 , or by a factor of 8. In other words, the board is going to be 8 times stiffer if you double its width. Accordingly, since the beam deflection is inversely proportional to the Moment of Inertia, when the Moment of inertia increases by 8, the deflection under the same load must decrease by 8. Thus, if it deflected 3 cm originally, when you double the width, its deflection must now be only ????. You don't have to know 'a' to solve this problem.