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
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.
To find the shape of the stiffest beam that can be cut from a cylindrical log, we need to maximize the stiffness of the beam.
Given that the stiffness of a beam is proportional to the product of the width and the cube of the depth, let's denote the width of the beam as w and the depth as d. The stiffness, S, can be expressed as:
S = k * w * d^3,
where k is the constant of proportionality.
The cross-sectional area of a beam can be calculated using the formula for the area of a circle:
A = π * (D/2)^2 = (π/4) * D^2,
where D is the diameter of the log.
Since the beam is cut from the log, the width and the depth of the beam cannot exceed the diameter of the log, i.e., w ≤ D and d ≤ D.
To maximize the stiffness, we need to find the values of width and depth that yield the largest value for S, while still satisfying the constraints w ≤ D and d ≤ D.
To find the maximum value, we can take the partial derivative of S with respect to both w and d, and set them equal to zero:
∂S/∂w = k * d^3 = 0,
∂S/∂d = 3 * k * w * d^2 = 0.
From the first equation, we have d = 0. From the second equation, we have w = 0 or d = 0.
However, d = 0 is not a meaningful solution since depth cannot be zero. Therefore, we are left with w = 0.
Next, we need to consider the constraints w ≤ D and d ≤ D. Since w = 0, we have d ≤ D.
Therefore, the stiffest beam that can be cut from the cylindrical log has a width of zero and a depth equal to the diameter of the log, i.e., w = 0 and d = D.
In terms of shape, this means that the stiffest beam is a rectangular cross-section with negligible width (essentially a line) and a depth equal to the diameter of the log.