MHB What Is the Optimal Geometry for a Concrete Column Supporting 1250 Tonnes?

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The discussion focuses on determining the optimal geometry for a concrete column designed to support a 1250-tonne load while considering the maximum allowable stress of 15 MPa. Participants explore the relationship between the column's height, cross-sectional area, and material volume to minimize costs. One participant suggests treating the cross-section as variable along the height of the column rather than constant, which could lead to a more efficient design. The stress at any height can be expressed as a function of the cross-sectional area, prompting further analysis on how to achieve minimal material use while maintaining structural integrity. The conversation emphasizes the importance of innovative thinking in engineering design, particularly in the absence of standard textbooks.
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When building a tall support, often the self weight of the support must be considered. For an optimal support, the volume of material, and hence the cost, will be a minimum. If the maximum allowable stress in concrete is 15 MPa, determine the optimal geometry of a column 100 metres tall made of concrete to support a mass of 1250 tonnes at its top. Hint: think of the shape of the CN tower
There's no textbook for this course, so we're expected to use common sense to answer these questions...
$$\sigma = \frac{F}{A}=\frac{\pi D^2/4\cdot H\cdot W+1250\cdot 10^3 \cdot 9.81}{\pi D^2/4} $$
W=25 Kn/m, and putting all the information in, I get diameter is 1.1m. Is that right, and I was told that the radius is different at the top, not sure what to do.
 
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Rido12 said:
When building a tall support, often the self weight of the support must be considered. For an optimal support, the volume of material, and hence the cost, will be a minimum. If the maximum allowable stress in concrete is 15 MPa, determine the optimal geometry of a column 100 metres tall made of concrete to support a mass of 1250 tonnes at its top. Hint: think of the shape of the CN tower
There's no textbook for this course, so we're expected to use common sense to answer these questions...
$$\sigma = \frac{F}{A}=\frac{\pi D^2/4\cdot H\cdot W+1250\cdot 10^3 \cdot 9.81}{\pi D^2/4} $$
W=25 Kn/m, and putting all the information in, I get diameter is 1.1m. Is that right, and I was told that the radius is different at the top, not sure what to do.

Hey Rido! (Smile)

It appears you have already made the assumption that the support is a massive cylinder.
Suppose we make no assumptions about the shape yet, and just assume a cross section of $A$.
And suppose we do not make the assumption that the cross section is constant, but instead is a function of height. That is $A=A(y)$.

What formula would yet get for the stress at height $y$? (Wondering)

And since we want to use a minimal amount of material, suppose this stress is equal to the maximum stress at each $y$.
What can you deduce from that?
 
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