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

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SUMMARY

The discussion focuses on calculating shear and normal stresses in a beam subjected to bending and torsional moments. The bending moments are given as Mx = 200 Nm, My = 300 Nm, and Mz = 600 Nm. The correct normal stress at point A is calculated as 47.7 MPa using the formula Mc/I. The shear stress calculation using the torsional moment resulted in an error due to incorrect computation of Q, which should be 2r^3/3 for a semicircle. The final corrected shear stress is determined to be 1.06 MPa.

PREREQUISITES
  • Understanding of shear and normal stress calculations
  • Familiarity with polar moment of inertia (J) and area moment of inertia (I)
  • Knowledge of bending and torsional moments in mechanics
  • Ability to apply the Q calculation for semicircular cross-sections
NEXT STEPS
  • Study the derivation and application of the Q calculation for different shapes
  • Learn about the significance of the polar moment of inertia in torsional stress analysis
  • Explore advanced topics in beam theory, including shear and moment diagrams
  • Investigate the effects of varying cross-sectional shapes on stress distribution
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Mechanical engineers, civil engineers, and students studying structural mechanics who need to understand stress analysis in beams under various loading conditions.

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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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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