Asked by manjeet
The stiffness of a given length of beam is proportional to the product of the width and the cube of the depth. Find the shape of the stiffest beam which can be cut from a cylindrical log (of the given length) with cross-sectional diameter of D
Answers
Answered by
bobpursley
stiffeness=k*w*d^3
Log: diameter D. so in a circle, find the relationship with depth.
well, if you sketch a circle, and draw in a d depth and W width, you see they are 90 deg with each other, and if you connect the other ends, you have a hypoenuse that runs thru the center (you need to think that out item 1). Then, the length of the hypotensue is D.
D^2=d^2+W^2 or
W= sqrt(D^2-d^2).
stiffness= d^3*sqrt(D^2-d^2)
ds/ddepth= 3d^2*sqrt( ) + d^3 (-2d/sqrt( ) ) =0 for max
or
sqrt ( )=2d^2/sqrt( ) or
D^2-d^2=2d^2
3d^2=D^2
or d=D/3 and then w= sqrt(D^2-d^2)=sqrt(D^2-D^2/9)...
check my math. I am in a hurry.
Log: diameter D. so in a circle, find the relationship with depth.
well, if you sketch a circle, and draw in a d depth and W width, you see they are 90 deg with each other, and if you connect the other ends, you have a hypoenuse that runs thru the center (you need to think that out item 1). Then, the length of the hypotensue is D.
D^2=d^2+W^2 or
W= sqrt(D^2-d^2).
stiffness= d^3*sqrt(D^2-d^2)
ds/ddepth= 3d^2*sqrt( ) + d^3 (-2d/sqrt( ) ) =0 for max
or
sqrt ( )=2d^2/sqrt( ) or
D^2-d^2=2d^2
3d^2=D^2
or d=D/3 and then w= sqrt(D^2-d^2)=sqrt(D^2-D^2/9)...
check my math. I am in a hurry.
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