Solve Shear Stress Qn: Mx,My,Mz, Mc,I,J, VQ,It

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Solidsam
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Homework Statement



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Homework Equations



Mx=200Nm

My=300Nm

Mz=600Nm


Normal stress caused by bending moment at A = Mc/I = (300*0.02)/((pi*0.02^4)/4)= 47.7MPa. This answer is correct.




Shear stress by Torsional Moment=Tc/J

Polar moment of inertia J=(pi/2)*c^4

So I did (600*0.02)/((pi*0.02^4)/2)= 47.7MPa Is this correct?

&

VQ/It=1000*((4*0.02)/3*pi)*((pi*0.02^2))/(1.257*10^-7*0.04)=10.47 MPa Is this correct?

One of the stress calculations is wrong beacuse when added that should equal 48.8 MPa

So what I'm I doing wrong?
 
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Discussion

PhanthomJay said:
it appears you calculated Q incorrectly, ormade a math error, one or the other. The Q of a semicircle about its base is 2r^3/3

Is Q not =y bar prime * A prime = 4r/3pi * (pi*r^2)/2 ?
 
Solidsam said:
Is Q not =y bar prime * A prime = 4r/3pi * (pi*r^2)/2 ?
certainly, which simplifies to 2r^3/3. So you have made a math error...you
forgot to divide by 2 when determing area of semicircle...and some other calculation error..try again and you should get the correct shear stress as 1.06 MPa