stiffness of a beam question

Hi, I'm having a hard time with this question.

the stiffness of a beam is proportional to it's width times the cube of it's depth. What are the dimensions of the stiffest beam you an produce from a beam 12" in diameter.

So I draw a rectangle inside a circle, and the diagonal inside the rectangle is 12. The width (w) times the depth cubed equals the stiffness.

S=wd^3

d^2+w^2=144 w=sqr(144-d^2)

S=sqr(144-d^2)(d^3)

S'=sqr(144-d^2)(3d^2)+d^3(1/2(144-d^2)^-1/2)

boils down to -d^2(6d^2+d-864)=0, to find critical points.

d=0, or 6d^2+d-864=0

d=(sqr(20737)-1)/2

w= sqr(144-(sqr(20737)-1729)/12)

did I do this right?

The teacher, tried to get us started on this problem, but following through from where he started, I run into a problem which I don't understand.

The teacher, started by trying to find ds/dw.

Here is where he left us off, S=W(144-w^2)^(3/2)

I found it's derivative, 3w^2sqr(144-w^2)+(144-w^2)^3/2

when I solve for 0, I get w=-12, or 12.

Makes no sense because if the width is 12, the depth is zero. Where did I go wrong.

Re: stiffness of a beam question

nobody can help?

I feel like I'm going crazy or something. I cannot find a problem, I have calculated the dw/ds like 5 times and set it to zero, and I get, 12, -12. The correct answer should be 6.

Can anyone find the derivative of s=w(144-w^2)^1/2? Set it equal to zero, what do you get?

I am wondering if my calculator is just unable to find the 6?

Re: stiffness of a beam question

your first derivative for S'(d) is incorrect, it should be:

$\displaystyle (3d^2)\sqrt{144-d^2} + \frac{1}{2}\frac{d^3}{\sqrt{144-d^2}}(-2d)$

$\displaystyle =(3d^2)\sqrt{144-d^2} - \frac{d^4}{\sqrt{144-d^2}}$

setting this equal to 0, and multiplying by the square root in the denominator (since d = 12 makes no sense) you get:

$\displaystyle (-d^2)(3d^2 + d^2 - 432) = 0$ which simplifies to:

$\displaystyle d^2 - 108 = 0$ , giving $\displaystyle d = 6\sqrt{3},\ w = 6$.

your second derivation is also incorrect, you should have:

$\displaystyle \frac{dS}{dw} = (-3w^2)\sqrt{144-w^2} + (144-w^2)^{\frac{3}{2}}$ ,

setting that equal to 0, and taking out the factor under the root (as before), we get:

$\displaystyle 144 - 4w^2 = 0$ which leads directly to $\displaystyle w = 6,\ d = 6\sqrt{3}$, as it should.

(and i can add from personal experience, that a 6x10 beam is a perfectly reasonable size)

Re: stiffness of a beam question