Finding the shear centre of a non-homogeneous cross section cross

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The discussion focuses on finding the shear center of a non-homogeneous cross section with a given ratio of material properties. The user calculates the equivalent cross section and identifies the shear center's location based on symmetry. They derive the shear stress and moment equations but arrive at a different answer than expected, prompting them to question their calculations and the textbook's accuracy. Adjustments to the cross section's dimensions are considered, yet the user still struggles to reconcile their results with the textbook answer. The conversation highlights the complexities of calculating shear centers in non-homogeneous materials and the potential for discrepancies in provided solutions.
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for the following cross section
Capture.JPG

the ratio Eb/Ep=30

find the shear centre of the cross section
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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)*\intdt\intQds

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??
 
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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"?
 
o yes, that's correct, meant to be moment, but it still won't 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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