Structural Analysis/beams/deflection

Question

Question :

The cantilever beam AB of length L carries a uniformly distributed W load , which includes the weight of the beam.

(Note that the left end end A is free , while right end has a fixed support)

(i) Derive the equation of the elastic curve.

(ii) Compute the maximum displacement of the beam.

My work :

First I considered a section of beam of length x from left end A and derived an equation for bending moment(M) which gave,

M + (Wx/L)(x/2) = 0

M = -(W*(x^2))/2L) which is equal to EIy' = -(W*(x^2))/2L)

==> EIy' = -(W*(x^3))/6L) + C1

==> EIy = -(W*(x^4))/24L) + C1L + C2 , where C1, C2 are arbitrary constants and y' & y' denote second & first derivatives of y, respectively.

EI is constant.

I found C1 = (w*(L^2))/6 and C2 = -(w*(L^3))/8

==> EI*y' = -(W*(x^3))/6L) + (w*(L^2))/6

==> EI*y = -(W*(x^4))/24L) + (w*(L^3))/6 + -(w*(L^3))/8

For the 2nd part,

I tried calculating max. displacement manually by equating y' = 0 and substituting the x value I got, to y.

y' = -(W*(x^3))/6L) + (w*(L^2))/6 = 0

==> -(W*(x^3))/6L) + (w*(L^3))/6L = 0

==> (w*(L^3))/6L = (W*(x^3))/6L)

==> (w*(L^3)) = (W*(x^3))

==> L^3 = x^3 ==> x^3 - L^3 = 0

==> (x-L)(x^2 + Lx + L^2) = 0

==> x = L or x^2 + Lx + L^2 = 0

But at x=L, y= 0

Hence, x^2 + Lx + L^2 = 0

==> x =[ -L + /- sqrt(L^2 - 4L^2)]/2

==> x = [ -L + /- sqrt(-3L^2)]/2 ] , which are not real values

Following a different process, then I differentiated y w.r.t.x and equated it to 0

EI*y = -(W*(x^4))/24L) + (w*(L^3))/6 + -(w*(L^3))/8

y' = w/24EI(4L^2 - ((3x^3)/L) )

and equating y' = 0 to find x such that max.y can be found,

w/24EI(4L^2 - ((3x^3)/L) ) = 0

4L^2 - ((3x^3)/L) ) = 0

3x^3 = 4L^3 x^3 = 4/3(L^3) ,

which also doesn't seem to give the expected answer of x = 0

Could you help find my mistake?

Thanks!

Answer 1

Answer 2

Hey Ashley. The good news is your methodology is ok, but there are several mathematical errors. Be careful to distinguish between x and L...X is the variable you are differentiating/integrating with respect to, while L is a constant.

When you integrate EIy' you get

EIy = -(W*(X^4))/24L) + C1X (not L) + C2

This explains why the second method did not yield the same result as the first method.

There is actually nothing wrong with the first method you used. You just have to recall one thing. From calculus, setting the derivative of a function to zero gives you the local maximima/minima. You still have to check the boundaries of your interval to find the global maximum. The equation you calculated for y is valid for 0<x<L. So in order to find the global max you have to check x = 0 and x = L against whatever values you get when you plug in your critical points. In this particular case your critical point is at x=L, so that point is already taken care of. When you plug in x=L you get y = 0, which is correct, since your beam is fixed at x = L (this also happens to be your minimum deflection). When you plug in x = 0 you get y = -(w*(L^3))/8. This is your max deflection.

The Bottom line is that in order to find the max or minimum of a function you need to test both the critical points and the boundaries of your interval. For some beam deflection problems you can skip checking the boundaries if your beam is supported at the ends (since you already know wherever you have a support the deflection is 0). However if you have a free end, such as one side of a cantilevered beam, you have to remember to check that boundaries! Hope this helps. Best of luck!