graywolfscvm
Mechanical
- Jul 19, 2006
- 2
I'm writing a set of calculations that will show the allowable stress for a cylindrical tank with a cone bottom head. Usually we would use ASME code to determine this but the half apex angle of the cone head is greater than 30 degrees which means, according to ASME code, we have to present an analysis proving the discontinuity stress are within limits. I'm using the example from "Roark's Formulas for Stress & Strain", sixth edition, article 12.4, (page 579 - 582). Ok, here is my problem and I apologize in advance if this question is elementary.
I understand most of the example and I see where each equation is comming from. It's just the last part of the example that I can't seem to understand.
On page 581, the example equation shows the summation of the radial deflections for the end of the cone and equating it to the sum of the radial deflection for the cylinder.
0.00944+12.59E-6Qo+4.677E-6Mo+=1.00796-20.16E-6Vo+6.689E-6Mo
Then the next equation is the same thing but for the meridian rotations.
0.000674+4.677E-6Qo+3.395E-6Mo=0+6.698E-6Vo-4.452E-6Mo
Finally it shows the equation of the radial forces.
Qo+5091cos45=Vo
My problem is I can not seem to determine how the forces were solved, (Qo=-2110, Vo=1490, Mo=2443). I know it's a simple solution but I think I've looked at this too long to see the answer right in front of my face.
If anyone has seen this or has this book and can give me a hint that would be great.
Thank you for taking the time to read my question.
-Steve CVM
I understand most of the example and I see where each equation is comming from. It's just the last part of the example that I can't seem to understand.
On page 581, the example equation shows the summation of the radial deflections for the end of the cone and equating it to the sum of the radial deflection for the cylinder.
0.00944+12.59E-6Qo+4.677E-6Mo+=1.00796-20.16E-6Vo+6.689E-6Mo
Then the next equation is the same thing but for the meridian rotations.
0.000674+4.677E-6Qo+3.395E-6Mo=0+6.698E-6Vo-4.452E-6Mo
Finally it shows the equation of the radial forces.
Qo+5091cos45=Vo
My problem is I can not seem to determine how the forces were solved, (Qo=-2110, Vo=1490, Mo=2443). I know it's a simple solution but I think I've looked at this too long to see the answer right in front of my face.
If anyone has seen this or has this book and can give me a hint that would be great.
Thank you for taking the time to read my question.
-Steve CVM