The stiffness S of a rectangular beam is proportional to its width (w) times the cube of its depth/height (h). find the dimensions (i.e.: w and h) of stiffest beam that can be cut from a log which has a circular cross-section with diameter of 12-in.
You would use the relation of the circular cross-section of (w/2)²+(h/2)²=(12/2)² to relate h and w.
After multiplying by 4 on both sides and rearranging, we have w=√(12²-h²).
Now the stiffness has been defined as
S=k(w)(h³) where k is a constant.
If you substitute w from above then S is now a function of h only.
Can you take it from here?
h^3=s/kw
h^3=s/k(12-h)
sorry, I'm still confused how to get the function of h. please help me to complete the assignment. thank you
You need to use the relation
w=√12²-h²) to eliminate w from the function S.
S(h)=k w h³
=k √12²-h²) h³
To find the maximum/minimum stiffness, you will proceed normally to equate
S'(h) = 0
to solve for h.
yes i understand now, thanks for your help
You're welcome!
To find the dimensions of the stiffest beam that can be cut from a log with a circular cross-section, you need to maximize the stiffness function. The stiffness (S) is proportional to the width (w) times the cube of the depth/height (h).
Given that the log has a circular cross-section with a diameter of 12 inches, we can relate the width (w) and depth/height (h) as follows:
w = 12 - 2h
Now, we can express the stiffness (S) in terms of h:
S = w * h^3
= (12 - 2h) * h^3
= 12h^3 - 2h^4
To find the dimensions (w and h) of the stiffest beam, we need to find the maximum value of S. We can do this by finding the critical points of the function, where its derivative equals zero.
Let's differentiate the stiffness function with respect to h:
dS/dh = 36h^2 - 8h^3
Now, set the derivative equal to zero and solve for h:
36h^2 - 8h^3 = 0
Factor out h^2:
h^2(36 - 8h) = 0
From this equation, we have two possibilities:
h^2 = 0 -> h = 0 (discard this solution as it does not make sense in our context)
36 - 8h = 0 -> h = 4.5
So, we have found that h = 4.5 is a critical point.
To determine if this critical point is a maximum or minimum, we can examine the second derivative of the function. If the second derivative is positive, it is a minimum, and if it is negative, it is a maximum.
Let's find the second derivative of the stiffness function:
d^2S/dh^2 = 72h - 24h^2
Evaluate the second derivative at h = 4.5:
d^2S/dh^2 = 72(4.5) - 24(4.5)^2
= 72 * 4.5 - 24 * 20.25
= 324 - 486
= -162
Since the second derivative is negative, it indicates that h = 4.5 is the maximum point.
Now, substitute this value of h back into the expression for w:
w = 12 - 2h
= 12 - 2(4.5)
= 12 - 9
= 3
Therefore, the dimensions of the stiffest beam that can be cut from the log are w = 3 inches and h = 4.5 inches.