The stiffness of a rectangular beam varies as its breadth...  TOPIC_SOLVED

Limits, differentiation, related rates, integration, trig integrals, etc.

The stiffness of a rectangular beam varies as its breadth...

Postby Luke53 on Sun Feb 09, 2014 10:04 am

Hi, can someone help me getting started with this problem?

The stiffness of a rectangular beam varies as its breadth and as the cube of its hight. Find the dimensions of the stiffest beam which can be cut from a circular log 12 inches in diameter.

Thanks.
Luke.
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Re: The stiffness of a rectangular beam varies as its breadt  TOPIC_SOLVED

Postby buddy on Mon Feb 10, 2014 11:08 am

Luke53 wrote:Hi, can someone help me getting started with this problem?

The stiffness of a rectangular beam varies as its breadth and as the cube of its hight. Find the dimensions of the stiffest beam which can be cut from a circular log 12 inches in diameter.

to get started try here.

For this one, I think you have to use the pythagorean thm to relate the diameter to the breadth (b) & height (h). The diameter is d = sqrt{b^2 + h^2} so d^2 = b^2 + h^2 = 12^2 = 144. Solve this for b to get b = sqrt{144 - h^2} If stiffness is S then S = bh^3 = sqrt{144 - h^2}*h^3. Do the derivative to find the max/min.
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Re: The stiffness of a rectangular beam varies as its breadt

Postby Luke53 on Tue Feb 11, 2014 8:20 am

Thank you for putting me on the right track.
Solving the derivative of: (144-h^2)^0.5 * h^3 = 0, this gives: 3 h^2(144-h^2)^0.5 - h^4(144-h^2)^-0.5 = 0 (by the poduct rule).
Solving this eqn: h = -6*sqrt(3) and h = 6*sqrt(3), (only the positive value is valid).
So the height should be 10.4 inches and the width 6 inches for max. stiffness of this beam.
S = bh^3 gives a maximum value for these results, so I assume this must be the right answer.
Thanks again.
Luke.
Luke53
 
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am


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