The truss shown below is constructed entirely of steel members (E = 200 GPa) all
ID: 1820008 • Letter: T
Question
The truss shown below is constructed entirely of steel members (E = 200 GPa) all of which have cross-sectional areas of 3250 mm2. Estimate how far the point "R" moves horizontally in response to the indicated loads.Explanation / Answer
By taking moments about one of the supports we see that the reaction forces at A and E are 50 N upward. Figured thusly: about A -100(6)+Fe(12) = 0 => Fe= 600/12 = 50 N summing all vertical forces Fe+Fa -100 =0 =>100-50 = 50 N =Fa Lets work from the A side If we imagine we break sides AG and AB and ignore the remainder of the truss to the right of our break, we can only maintain static equilibrium if we apply forces on our stubs equal to the original internal stress on the link. Let's assume that AG is compressive(C) and AB is tensile (T). We then sum forces in the horizontal direction so that Fab-Fag(3/5)=0 (3/5 being the cosine of the angle the force Fag works at). When we sum forces in the vertical direction we see that the reaction force Fa must equal the vertical portion the the compressive force in Fag. Fab has no vertical component. Fa-Fag(sin(angle))=0 50=Fag(4/5) => Fag =50(5/4) Fag = 62.5N plugging back in to our first equation and solving for Fab Fab=62.5(3/5) Fab = 37.5 N as our signs are positive, we guessed correctly on Fab being (T) and Fag being (C) Now let's analyze about point B breaking our links and replacing the rest of the bridge with the resultant force vectors Summing vertical forces we see that the only vertical component we have is Fbg so we set this to 0 Assuming the links have no weight, Fbg =0 Summing horizontal we see that Fbc-Fab=0 so Fbc=Fab Fbc=37.5 N (T)
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