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

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Discussion Overview

The discussion revolves around calculating shear stress and normal stress in a mechanical context, specifically related to bending moments and torsional moments. Participants are examining the correctness of various equations and calculations, including the polar moment of inertia and the Q factor for a semicircular cross-section.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents calculations for normal stress and shear stress, questioning the correctness of their results.
  • Another participant suggests that there may be a calculation error in determining the Q value, indicating that the Q for a semicircle about its base is 2r^3/3.
  • A further reply reiterates the Q calculation, proposing an alternative expression for Q as y bar prime times A prime, which simplifies to 2r^3/3.
  • Another participant points out a potential error in the area calculation for the semicircle, suggesting that the shear stress should be recalculated to yield a different value of 1.06 MPa.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correctness of the initial calculations, with multiple competing views on the proper calculation of Q and the resulting shear stress. The discussion remains unresolved as participants continue to challenge and refine each other's claims.

Contextual Notes

Limitations include potential errors in mathematical calculations and assumptions regarding the geometry of the semicircular cross-section. Specific steps in the calculations remain unresolved.

Solidsam
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Homework Statement



26eyH.png


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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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
 
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
 

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