By Louis L. Bucciarelli
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Extra info for Engineering Mechanics for Structures
38 Chapter 2 Furthermore, we require the three forces to sum to zero to satisfy force equilibrium, so: F + W + N = 0. This is shown graphically in the figure at the far right above. From this we obtain what we were asked to show. (A bit of analytical geometry leads to both relationships). Now from this we can state another rule, namely • For an isolated system subject to but three forces, a three force system, the three forces must be concurrent. That is their lines of action must all run together and intersect at a common point.
This time there are no special tricks, no special effects hidden in subsystems, that would enable us to go further. That’s it. We can not solve the problem. Rather, we have solved the problem in that we have shown that the equations of equilibrium are insufficient to the task. Observe • That the forces in the members might depend upon how well a machinist has fabricated the additional member CD. Say he or she made it too short. Then, in order to assemble the structure, you are going to have to pull the node D down toward point C in order to fasten the new member to the others at D and to the ground at C.
Construct an expression for the force component which compresses the block at the bottom in terms of the weight of the block W and the dimensions shown. Express you result in terms of nondimensional factors. F=W a θ A θ B ha C hb D W E 2b If θ = 30 o a = b = h a = h b, What must the coefficient of friction be to ensure the block does not slide out of the grip? 1 Internal Forces in Members of a Truss Structure We are ready to start talking business, to buy a loaf of bread. Up until now we have focused on the rudimentary basics of the language; the vocabulary of force, moment, couple and the syntax of static equilibrium of an isolated particle or extended body.