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 Cantilever Structure (Posted on 2010-02-26)

The cantilever structure shown in the figure consists of 4n-1 struts of the same length plus one that is half that length. Each strut can handle a maximum tension force T before it will snap and a maximum compression force C before it will buckle. The structure is connected to a wall at points B and C. A weight W is attached at point A. The weight W is increased until two struts fail - one from tension and the other from compression.

What is the value of the ratio C/T if n = 25?

Consider the struts as weightless.

 See The Solution Submitted by Bractals Rating: 3.0000 (1 votes)

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 Solution | Comment 5 of 6 |
I've just spotted that the solution has been posted - never mind..

Let T be the tension in AD and S be the compression in the horizontal member with one end at A.
Since the joint at A is in equilibrium, resolving horizontal components of forces on that joint gives:                S = Pcos(60) = P/2                                    (1)
At every other joint (not on the wall), only the slanting members have vertical force components, so for equilibrium of all joints these components must be equal and opposite. Since they’re all at the same angle, it follows that all the slanting members have equal forces (P) alternating as tensions and compressions from the left-most member.
Where each pair of slanting members meet, together they therefore exert a horizontal force of Pcos(60) + Pcos(60) = P on that joint, and this must be balanced by the difference between the forces exerted by the horizontal members on either side of the joint. So, starting from the result in (1), the lower horizontal members have compressions of P/2, 3P/2, 5P/2,..........,(2n - 1)P/2, while the upper horizontal members have tensions P, 2P, 3P,.............,nP.

It follows that the members with the greatest compression and tension are the horizontal ones jointed to the wall, and if these are both critical, then the ratio is given by:
C/T  = [(2n - 1)P/2]/[nP] = (2n - 1)/(2n)
When n = 25, this will be 49/50

 Posted by Harry on 2010-02-28 00:01:00

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