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Cantilever Structure (Posted on 2010-02-26) Difficulty: 3 of 5

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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re: Am interested in learning Comment 6 of 6 |
(In reply to Am interested in learning by Dej Mar)

Dej Mar,

The typical engineering approach is to not worry about whether any particular member is a "tie" or a "strut" (building on your nomenclature).  Instead, approach the problem by realizing that each and every point on the structure must be in equilibrium (i.e ” forces in the horizontal and vertical = 0), and that the entire structure + weight must also be in equilibrium.  Since the latter is not a point, but exists in two dimensions, equilibrium for the structure means both ”Forces = 0 and ”Moments=0 as well.  A shortcut can be realized if you also remember that one consequence of 'struts" and "ties" is that they are classical "two-force members".   This means that the net force must always be acting along their length.  Thus the "sign" of the force (compression or tension) is worked out automatically by the algebra.

If you roll this all together, you can conclude that the only reaction force at the wall at C is a horizontal force (no vertical component).  The magnitude can be found by summing moments for the entire structure.  It follows that the equal and opposite force to the Weight, W, must be a vertical force up (acting on the structure) at B.  This is allowed only because of the angular (i.e. non-horizontal) member there.  This force W then must represent the vertical component of the force in this member, and since the total force must be along the member, this force can be solved for.  Then, for equilibrium to hold, the horizontal member at B must provide the balance of the horizontal force required to react against the horizontal force at C.  This is the basis for the solution I supplied in my earlier comment.

Bractals has created an interesting and classically formulated problem.  It's beauty lies in that there is really no redundancy (i.e. extra members) that would force one to consider cross section properties of the struts and ties in order to balance the real world deflections along their lengths that occur in the real world - something a lot harder to do without modern computers and code.  In fact, as a practicing engineer, I have often marvelled at the structures one can see in very old truss bridges - many are elegantly simple in their design - in part because they represent solutions that could be calculated by hand (or with slide rules).

Edited on February 28, 2010, 9:25 pm
  Posted by Kenny M on 2010-02-28 21:21:46

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