Deflection of Cantilever Beam Arrangement

[phiska]

If two cantilever beams are fixed at Left Hand Side (LHS) and joined to each other 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)
Where P= load in prop
EI=constant

 

[Speed]

Can we see the full question?

 

[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?

 

[phiska]

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

 

[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

For your case:

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

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

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

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

 

[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 as well as getting the correct answer.

 

[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).

 

[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.

 

[phiska]

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