Strongest Possible Beam From a Log
Date: 07/11/2002 at 22:37:41 From: Bill Gemberling Subject: Calculus - Optimization Rectangular Beam The strength of a rectangular beam is directly proportional to the product of its width and the square of the depth of a cross section. Find the dimensions of the strongest beam that can be cut from a cylindrical log of radius 'a'. There is a drawing included with the problem in the book that makes the corners of the rectangle come to the edge of the log. I know I'm missing some formula to figure this out. I understand the product to be: wd^2 = S, where w = width, d = depth, S = strength of a rectangular beam. Thank you for your help. Bill Gemberling firstname.lastname@example.org
Date: 07/13/2002 at 12:58:07 From: Doctor Fwg Subject: Re: Calculus - Optimization Rectangular Beam Dear Bill, Here is one way to treat this problem: If the width and depth of a rectangular beam are W and D, respectively, and the strength of the beam (S), in terms of W and D, is S = kWD^2 where k is a constant of proportionality, find the dimensions of the strongest rectangular beam that can be cut from a uniform cylindrical log having a radius of value A. Start with: 1) S = kWD^2. But if the distance between the center of the cut beam and one corner is A, and the horizontal distance between the center of the cut beam and one side is X, and the vertical distance between the center of the cut beam and its top side is Y, then, let W = 2X and D = 2Y. This changes S to: 2) S = k(2X)(2Y)^2 = 8k(X)(Y)^2. These stipulations allow the following: 3) X^2 + Y^2 = A^2. So: 4) Y^2 = A^2 - X^2 Now using the right hand side (RHS) of Equation 4 in Equation 2 (to eliminate Y^2), one gets: 5) S = 8k(X)(A^2 - X^2) = 8k(A^2X - X^3). Now, if one takes the first derivative of S with respect to (WRT) X and sets that result to zero, one can find the value of X (in terms of A, which is a constant) which causes S to be a maximum (because the second derivative of S WRT X is negative). One should get the result that X = [Sqrt(1/3)]A and Y = [Sqrt(2/3)]A. Remember that W = 2X and D = 2Y. It will also be possible to show that Y/X = D/W = sqrt(2). I hope this helps. With Best Regards, - Doctor Fwg, The Math Forum http://mathforum.org/dr.math/
Date: 07/14/2002 at 15:54:40 From: Bill Gemberling Subject: Thank you (Calculus - Optimization Rectangular Beam) Doctor Fwg, Thank you for your help with the Optimization Rectangular Beam problem. I am taking Calculus for the first time and this has been helpful. Once again thank you for your help. Bill Gemberling
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