# Deflection of Cantilever Beam Arrangement

1. Mar 14, 2005

### phiska

If two cantilever beams are fixed at Left Hand Side (LHS) and joined to eachother by means of a prop at RHS, how do i calculate the downwards vertical deflection of the bottom beam?

The solution i need to prove is (5WL^3)/(48EI)-top deflection.

However, i always seem to get a 7 where there should be a 5!!

Any clues as to where i am going wrong?

The beams are of length L, the prop length a.
There is a point force on the bottom beam of magnitude W, downwards, at L/2 m.

I already have calculated the downwards vertical deflection of the top beam to be (d of top)= (PL^3)/(3EI)
EI=constant

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2. Mar 14, 2005

### Speed

Can we see the full question?

3. Mar 14, 2005

### brewnog

A quick guess would be that you've got your constraints wrong on the little link which joins the two beams. Are those supposed to be pin jointed or built in?

4. Mar 15, 2005

### phiska

It simply states in the question that the distance, a, between the beams is maintained by a prop of diameter, d.

5. Mar 15, 2005

### FredGarvin

The deflection you stated for the top beam is for a concentrated load at the end of the cantilevered beam, not the mid span.

$$\delta_m = \frac{Pa^2}{6EI}(3L-a)$$

Where:
$$\delta_m$$ = Max deflection
P = Concentrated load at any point of application
a = Distance from cantilever end to load P
L = Total length of beam
E = Young's modulus
I = Area moment of inertia

$$\delta_m = \frac{W(\frac{L}{2})^2}{6EI}(3L-\frac{L}{2})$$

$$\delta_m = \frac{WL^2}{24EI}(\frac{6L}{2}-\frac{L}{2})$$

$$\delta_m = \frac{6WL^3}{48EI}-\frac{WL^3}{48EI}$$

$$\delta_m = \frac{5WL^3}{48EI}$$

Last edited: Mar 15, 2005
6. Mar 15, 2005

### phiska

why is it that for the maximum deflection it is (3L-L/2) rather than just (L-L/2)?

I want to understand what is going on aswell as getting the correct answer.

7. Mar 15, 2005

### FredGarvin

Admittedly, that is simply the equation I remember for a cantilever beam with a load anywhere along it's span. I will have to dig up it's derivation later. Give me some time and I'll find it (unless someone else can do it first).

8. Mar 16, 2005

### FredGarvin

It took me a bit, but I got it (and a nice refresher in beams!):

For the following I use x as the distance from the free end of the beam to the point of load application and P as the load value.

Using the general curvature deflection equation

$$\frac{d^2y}{dx^2}=\frac{M(x)}{EI}$$

$$EI\frac{d^2y}{dx^2}=-Px$$

$$EI\frac{dy}{dx}=-\frac{1}{2}Px^2 + C_1$$
Using the B.C.'s at the cantilever (pt. B) x=L and dy/dx=0 we get:

$$C_1 =\frac{1}{2}PL^2$$ and that leads to:

$$EI\frac{dy}{dx}=-\frac{1}{2}Px^2 +\frac{1}{2}PL^2$$

$$EI Y =-\frac{1}{6}Px^3 +\frac{1}{2}PL^2x +C_2$$
Using the B.C.'s @ pt. B again, we have Y=0 and x=L we get:

$$C_2 = -\frac{1}{3}PL^3$$ and that now leads to:

$$EI Y =-\frac{1}{6}Px^3 +\frac{1}{2}PL^2x -\frac{1}{3}PL^3$$

We use that equation for the situation of the bottom beam, x=L/2:

$$EI Y =-\frac{1}{6}P(\frac{L}{2})^3 +\frac{1}{2}PL^2(\frac{L}{2}) -\frac{1}{3}PL^3$$

After a little algebra...
$$EI Y = (-\frac{1}{48}+\frac{1}{4}-\frac{1}{3})PL^3$$

$$EI Y = (-\frac{1}{48}+\frac{12}{48}-\frac{16}{48})PL^3$$

$$Y = -\frac{5PL^3}{48EI}$$

You can now go back and substitute x=0 for the top beam to get the second part of the deflection for the top beam.

Sorry I took so long.

Last edited: Mar 16, 2005
9. Mar 16, 2005

### phiska

Thank you so much.... you've been really helpful!