# finding the shear centre of a non-homogeneous cross section cross

by Dell
Tags: centre, cross, nonhomogeneous, shear
 P: 590 for the following cross section the ratio Eb/Ep=30 find the shear centre of the cross section ------------------------------------------- first of all i need to find the equivalent cross section, which will be the same except that the horizontal parts will be of length n*L=30*1=30cm since this cross section has an axis of symmetry, i know that the shear centre passes through that axis, now all i need to find is the distance "eo" i know that i can move the force to the shear centre and the cross section must feel the same moment, i calculated the moment about a point that passes through the Vertical portion so that only the sums of the horizontal shearing stresses have an effect on the moment, i know that Q1y is the 1st area moment of each of the horizantal portions, Q1y=(0.15*s)*1.075 =0.16125*s I=10.6cm^4 shearing stress=$$\int$$$$\int$$(V*Q/I/b)da*2.15 where 2.15 is the distance between the horizontal forces V*e=$$\int$$$$\int$$(V*Q/I/b)da*2.15 e=2.15/(I*0.15)*$$\int$$dt$$\int$$Qds e=[2.15/(I*0.15)]*0.15*$$\int$$(0.16125*s)ds -->from 0 to 30 [2.15/(10.6*0.15)]*0.15*0.16125*302/2=14.7cm the correct answer is somehow meant to be 0.408cm, can anyone see where i have gone wrong??
 P: 667 your line: "shearing stress=LaTeX Code: \\int LaTeX Code: \\int (V*Q/I/b)da*2.15 where 2.15 is the distance between the horizontal forces" comment: I'm not sure what all those integral signs are doing. I thought shearing stress was just V*Q/I/b. (That is, the result of integration). OK, so you multiply by area to get force, and then by 2.15 to obtain moment. So should your "shearing stress" read as "moment"?
 P: 590 o yes, thats correct, meant to be moment, but it still wont change the answer, im beginning to think that the way i expanded the cross section was incorrect, i just made the 2 horizontals 30 times longer ( to the right) maybe i was meant to expand them to both sides, 14.5 to each side, doing this would still not give me a correct answer, but much closer, by doing this i get 0.49cm is it possible that the answer i was given in the textbook (0.408) is incorrect, could someone please check this for me

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