Asked by Creating a formula from 2 variables
Hi, I'm having trouble with this problem...
"If the strength of a rectangular beam of wood varies as its breadth and the square of its depth, find the dimensions of the strongest beam that can be cut out of a round log, diameter d. "
I can't come up with an equation with only 1 variable.
Help me, Please!!!
Many thanks, Stu.
"If the strength of a rectangular beam of wood varies as its breadth and the square of its depth, find the dimensions of the strongest beam that can be cut out of a round log, diameter d. "
I can't come up with an equation with only 1 variable.
Help me, Please!!!
Many thanks, Stu.
Answers
Answered by
levy
Im not really sure but i think its
360(d)(120)
360(d)(120)
Answered by
Creating a formula from 2 variables
Hi Levy, any chance you could explain how you came up with that solution?
I believe I should find values for Breath and Depth with relation to diameter, but I am stuck for a way to do it.
Many thanks.
I believe I should find values for Breath and Depth with relation to diameter, but I am stuck for a way to do it.
Many thanks.
Answered by
Damon
rectangle must fit in circle of diameter d
B =chord1 of circle = d sin (theta1/2)
H = chord 2 of circle = d sin( [pi-theta]/2)
because the angles subtended by the inscribed breadth and depth add up to a straight line. pi is straight line angle in radians.
S = k B H^2
S = k d sin (theta/2)*d^2 sin^2 ([pi-theta]/2)
let A = theta/2 to make it all easier
S = k d^3 sin A sin^2 (pi/2-A)
but sin (pi/20-A) = cos(A)
so
S = k d^3 sin(A) cos^2(A)
take derivative and set to 0
0 = dS/dA = k d^3[ -2 sin^2(A)cos(A)+ cos^3(A)]
2 sin^2 (A) = cos^2(A)
but cos^2(A) = 1 - sin^2(A)
2 sin^2 A = 1 - sin^2(A)
3 sin^2 (A) = 1
sin(A) = sqrt(3)/3 = about .577
so A is about 35.2 degrees
so
B = d sin (A) = .577 d
H = d sin(90-A) =d sin 54.8deg = .817 d
B =chord1 of circle = d sin (theta1/2)
H = chord 2 of circle = d sin( [pi-theta]/2)
because the angles subtended by the inscribed breadth and depth add up to a straight line. pi is straight line angle in radians.
S = k B H^2
S = k d sin (theta/2)*d^2 sin^2 ([pi-theta]/2)
let A = theta/2 to make it all easier
S = k d^3 sin A sin^2 (pi/2-A)
but sin (pi/20-A) = cos(A)
so
S = k d^3 sin(A) cos^2(A)
take derivative and set to 0
0 = dS/dA = k d^3[ -2 sin^2(A)cos(A)+ cos^3(A)]
2 sin^2 (A) = cos^2(A)
but cos^2(A) = 1 - sin^2(A)
2 sin^2 A = 1 - sin^2(A)
3 sin^2 (A) = 1
sin(A) = sqrt(3)/3 = about .577
so A is about 35.2 degrees
so
B = d sin (A) = .577 d
H = d sin(90-A) =d sin 54.8deg = .817 d
Answered by
Creating a formula from 2 variables
Damon, you're a God-send!
If you're ever in North Wales I'll buy you a pint!
1x10^3 Thanks.
Stu.
If you're ever in North Wales I'll buy you a pint!
1x10^3 Thanks.
Stu.
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