Derivation of Deflection from Euler-Bernoulli Beam Equation

[bill nye scienceguy!]

Homework Statement

I want to derive a formula for deflection w(x) from the Euler-Bernoulli beam equation. It's essentially only four integrations but I'm not sure about my boundary conditions, particularly wrt shear. The beam is a cantilever with a point load at the unsupported end.

And apologies in advance for the clumsy latex...

Homework Equations

P = load

w = deflection = 0 when x =0

$$\frac{dw}{dx}$$= slope = 0 when x =0

EI$$\frac{d^{2}w}{dx^{2}}$$ = bending moment = 0 when x = L

-EI$$\frac{d^{3}w}{dx^{}3}$$ = shear force = 0 when?

I guess my question is: what boundary condition do I need to get rid of the C$$_{1}$$ after the first integration and I suppose if this is the right way to go about this at all!

The Attempt at a Solution

Here's what I've done so far:

EI$$\frac{d^{4}w}{dx^{4}}$$=P

EIEI$$\frac{d^{3}w}{dx^{3}}$$=Px + C$$_{1}$$

I've left C$$_{1}$$ here and carried it through since I don't have a clue about the shear BC.

EI$$\frac{d^{2}w}{dx^{2}}$$=P$$\frac{x^{2}}{2}$$ + C$$_{1}$$x +C$$_{2}$$

EI$$\frac{d^{2}w}{dx^{2}}$$ = bending moment = 0 when x = L, so

C$$_{2}$$=-$$\frac{PL^{2}}{2}$$-C$$_{1}$$L

EI$$\frac{d^{2}w}{dx^{2}}$$=P$$\frac{x^{2}}{2}$$ + C$$_{1}$$x - $$\frac{PL^{2}}{2}$$-C$$_{1}$$L

EI$$\frac{dw}{dx}$$=$$\frac{Px^{3}}{6}$$+C$$_{1}$$$$\frac{x^{2}}{2}$$-[$$\frac{PL^{2}}{2}$$-C$$_{1}$$L]x + C$$_{3}$$

$$\frac{dw}{dx}$$ = 0 when x = 0 so C$$_{3}$$=0

and finally

EIw=$$\frac{Px^{4}}{24}$$+$$\frac{C_{1}x^{3}}{6}$$-[$$\frac{PL^{2}}{2}$$-C$$_{1}$$L]$$\frac{x^{2}}{2}$$ + C$$_{4}$$

w=0 when x=0 so C$$_{4}$$=0

so:

EIw=$$\frac{Px^{4}}{24}$$+$$\frac{C_{1}x^{3}}{6}$$-[$$\frac{PL^{2}}{2}$$-C$$_{1}$$L]$$\frac{x^{2}}{2}$$

 

[Mapes]

Why did you set $EI(d^4 w/dx^4)$ equal to P? The units don't match up, for one. Isn't P a force? (Know what I'm getting at? )

 

[pongo38]

Why has your first eqn got EI in it? What is your definition of shear force? That should tell you the constant you are looking for. It doesn't have to be zero.

 

[bill nye scienceguy!]

I don't have a good text on beam theory to hand so this is all coming from a mixture of wikipedia and efunda. So from the statement on the wiki page (http://en.wikipedia.org/wiki/Euler–Bernoulli_beam_equation) that:

$$\frac{d^{2}}{dx}$$(EI$$\frac{d^{2}u}{dx^{2}}$$)=P

[where I've called deflection u and load P]

I was trying to get to the statement of deflection as a function of length, distance from supported end, load and constant EI from efunda (http://www.efunda.com/formulae/soli...e=cantilever_endload&search_string=cantilever)

w(x)=-$$\frac{Px^{2}}{6EI}$$$$\left(3L-x\right)$$

I've looked at the units in the first statement and I'm not sure what the function

$$\frac{d^{4}u}{dx^{4}}$$

actually is? Apart from being the fourth derivative of displacement, but what is it physically?

 

[Mapes]

So from the statement on the wiki page (http://en.wikipedia.org/wiki/Euler–Bernoulli_beam_equation) that:

$$\frac{d^{2}}{dx}$$(EI$$\frac{d^{2}u}{dx^{2}}$$)=P

[where I've called deflection u and load P]

Where are you seeing that equation on that page?

I see

$$\frac{d^{2}}{dx^2}\left(EI\frac{d^{2}u}{dx^{2}}\right)=w(x)$$

for a distributed load and

$$-\frac{d}{dx}\left(EI\frac{d^{2}u}{dx^{2}}\right)=F\,(\mathrm{or~}P)$$

for a point load.

 

[bill nye scienceguy!]

Where are you seeing that equation on that page?

I see

$$\frac{d^{2}}{dx^2}\left(EI\frac{d^{2}u}{dx^{2}}\right)=w(x)$$

for a distributed load and

$$-\frac{d}{dx}\left(EI\frac{d^{2}u}{dx^{2}}\right)=F\,(\mathrm{or~}P)$$

for a point load.

I totally missed that, that makes things a lot easier. So now:

1st integration

-EI($$\frac{d^{2}u}{dx^{2}}$$)=Px +C1

$$\frac{d^{2}u}{dx^{2}}$$=0 when x=L so C1=-PL

2nd integration

-EI$$\frac{du}{dx}$$=$$\frac{Px^{2}}{2}$$-PLx+C2

$$\frac{du}{dx}$$=0 when x=0 so C2=0

3rd integration

-EIu=$$\frac{Px^{3}}{6}$$-$$\frac{PLx^{2}}{2}$$+C3

u=0 when x=0 so C3=0

which leaves me with:

-EIu=$$\frac{Px^{3}}{6}$$-$$\frac{PLx^{2}}{2}$$

u=$$\frac{-Px^{2}}{2}$$(L-$$\frac{x}{3}$$)

multiplying through by 3 gives me

3u=$$\frac{-3Px^{2}}{2}$$(3L-x)

and then

u=$$\frac{-Px^{2}}{2}$$(3L-x)

which still isn't quite right. Can you spot where I've gone wrong in the maths?

 

[pongo38]

For your first integration, how do you get Px for the moment at x? Is x measured from the support or from the point load?

 

[Mapes]

You multiplied the right side by 3 twice.

 

[bill nye scienceguy!]

For your first integration, how do you get Px for the moment at x? Is x measured from the support or from the point load?

I'm measuring x from the support and Mapes, of course you're right. Thanks very much for your help!