# Homework Help: Shear stress question

1. May 12, 2012

### Solidsam

1. The problem statement, all variables and given/known data

2. Relevant 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 12, 2012

### PhanthomJay

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

3. May 13, 2012

### Solidsam

Is Q not =y bar prime * A prime = 4r/3pi * (pi*r^2)/2 ?

4. May 14, 2012

### PhanthomJay

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