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Problems and Solutions: Problem 2.1 Find the reactions acting at A and B in terms of P and C the dimensions shown (xp/L). Isolate member BC and draw a free body diagram that will enable you to determine the forces acting on this member. Find those forces, again in terms of P and the dimensions are shown. Find the force in the horizontal member of the structure.
Problem 2.2 P 2L L 60o x y x p A B For the truss shown below, i) Isolate the full truss structure and replace the applied loads with an equivalent load (no moment) acting at some distance, b, from the left end. What is b? ii) Determine the reactions at f and l. iii) Find the force in member ch with but a single additional free body diagram
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Problem 2.3 Find the force in the member CD of the structure F shown in terms of P. All members, save CF are of equal length. In this, use method of joints starting from node B. Note: You must determine the reaction at B to get started. G (Prof. 3.11 text)
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Problem 2.4 The rigid, weight-less, beam carries a load P at its right end and is supported at the left end by two (frictionless pins). What can you say about the reactions acting at A and B? E.g., “they are equivalent to...”
Problem 2.1 First, we need to find the reaction forces at point A and B, we get that
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Now consider the horizontal member, the only possible way to have equilibrium condition is that the forces action at point D and E are both horizontal forces FD FE (i.e., no vertical forces acting on the points). You can simply take a moment at either point D or E and you will find D E that if there is any vertical forces at either D or E, the member DE cannot be in the equilibrium condition (summation of moment is not equal to zero).
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Figure 1.1 shows you the member FBD of the member BC. From the FBD, we get that
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Problem 2.2 From figure 2.1 (a) and (b), we can find the equivalent force by
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Figure 2.1 (a), (b), and (c) for top, middle, and bottom, respectively To find the forces in the member ch with a single FBD, we use method of section and cut the truss as shown in the figure 2.2. We get that
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The negative sign for Fch indicates that the direction that I assumed in the figure above is not cor2W rect. (If the angle θ is 60 degree, the Fch is equal to F = –---------- )And the correct direction is opposite of ch 33 that in the figure 2.2. It is important to note the correct Fch in this case is compression.
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Problem 2.3 First, we need to find the reaction at points A and B. Assuming each member length is L (except the member CF which is longer than other members), we get that
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However, since AC is a two force member, Ray must be 0 to maintain the equilibrium condition for 5P that member.. Therefore, we get that Rby = P and .
Now consider cutting the joint B, we get that
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Problem 2.4 The reactions acting at point A and B are equivalent to PL/ D h. They are equal but opposite in the direction. Also, the total vertical reaction at the point A and B are equal to P (acting upward). S
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