Is it a problem if Tresca is larger than Von Mises for yield loci graph?

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In summary, the conversation discusses a problem with calculating the von Mises and Tresca stresses for a shaft and determining which theory is more conservative. The participants suggest that the Tresca theory is more conservative, but one must be careful with the signs in the calculations. They also mention that the von Mises yield boundary is larger than the Tresca yield boundary, and it is expected for the Tresca stress to be larger than the von Mises stress. Additionally, they provide some trivia about the history of the theories.
  • #1
george88b
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hello there

i have a shaft..i found the von mises and the tresca..von mises is 360 and tresca is 430. So when i m trying to produce the loci graph the tresca will be bigger than von mises. I search in internet and there is no examples like that. Is there a problem if the tresca is bigger than von mises?

thnx...
 
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  • #2
Hi george88b, welcome to PF!

The von Mises threshold should be larger, so you probably made a calculation error. If you post your solution, you'll likely get helpful comments.
 
  • #3
thnx for reply..i don't think so that i have error with my solutions because i solved them a 1000 times..:) i have a compressive stress 16.55 MPa and shear stress 95.5 MPa. So the principal stresses are.. σ1=87.5 and σ2=-104.1. Right?;p i think may be my wrong is on principal stresses..(signs). Then i have von mises=365.8 and tresca=422
 
  • #4
george88b said:
thnx for reply..i don't think so that i have error with my solutions because i solved them a 1000 times..:) i have a compressive stress 16.55 MPa and shear stress 95.5 MPa. So the principal stresses are.. σ1=87.5 and σ2=-104.1. Right?;p i think may be my wrong is on principal stresses..(signs). Then i have von mises=365.8 and tresca=422

Perhaps you've made the same error 1000 times. If you don't show your work, we can't help you.

CS
 
  • #5
Also, I think there may be some misunderstanding. The von Mises yield boundary is larger than the Tresca yield boundary. But this means that the von Mises stress value ([itex]\sqrt{[(\sigma_1-\sigma_2)^2+(\sigma_1-\sigma_3)^2+(\sigma_2-\sigma_3)^2]/2}[/itex]) is smaller than the Tresca stress value ([itex]|\sigma_1-\sigma_3|[/itex]). A smaller effective stress means that the von Mises calculation is less conservative with respect to avoiding yield, which is equivalent to having a larger yield boundary.
 
  • #6
I am facing the same problem. The question is: 200mm shaft, Torque= 150KNm, Compressive Load= 520 KN.

Direct stress= -520*10^3/(π*(0.1^2))= -16.55 MPa
Shear Stress=((150*10^3)*0.1)/((π*0.2^4)/32)=95.49 MPa

Principal Stresses:
σ1=0.5(-16.55*10^6)+0.5*sqrt((-16.55*10^6)^2 +4(95.49*10^6)^2)=87*10^6
σ2=0.5(-16.55*10^6)-0.5*sqrt((-16.55*10^6)^2 +4(95.49*10^6)^2)=-104*10^6


to find the tresca i use that formula(i have safety factor 2.2):

|σ1-σ3|=σyield/safety factor and my result is: 421*10^6

to find the von mises:

(σ1^2)+(σ2^2)-(σ1*σ2)=(σyield^2)/Safety factor

and my result is 365*10^6.
 
  • #7
the same with me..;p
 
  • #8
I agree with the second post by Mapes. Well said. The Tresca (maximum shear stress) theory is an overly conservative theory. Ductile materials behave closer to von Mises theory. Therefore, you know the Tresca failure envelope is enclosed within the von Mises failure envelope. Consequently, we know the Tresca effective tensile stress will always be greater than or equal to the von Mises stress. george88b and mikex24, you are observing the correct, expected result.

And here is a bit of trivia. Even though the maximum shear stress theory often bears the name Tresca, did you know Charles-Augustin de Coulomb invented the theory 95 years before Tresca published his paper?
 
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  • #9
so at the yield loci graph the von mises will be smaller than the tresca right? i mean that the circle(von mises) will be inside of the hexagon(tresca) if i m right..
 
  • #10
george88b said:
so at the yield loci graph the von mises will be smaller than the tresca right? i mean that the circle(von mises) will be inside of the hexagon(tresca) if i m right..

No. As stated earlier, the von Mises yield boundary, or yield envelope, surrounds the Tresca yield boundary. When determining the yield boundary, you are setting the von Mises or Tresca effective stress equal to the yield stress, and then asking what combination of principal stresses could give this result. When determining whether a material will yield, you are calculating the effective stress from the principal stresses. Does this make sense?
 

1. What is the difference between Von Mises and Tresca yield criteria?

The Von Mises criterion is a measure of stress that considers the combined effect of normal and shear stresses, while the Tresca criterion only takes into account the maximum shear stress. As a result, the Von Mises criterion is more accurate for predicting the onset of yielding in ductile materials.

2. Why is Von Mises smaller than Tresca in some cases?

This can occur when the material is subjected to a combination of normal and shear stresses that do not align with the principal stress directions. In such cases, the Von Mises criterion accounts for the combined effect of the stresses, while the Tresca criterion only considers the maximum shear stress.

3. What does it mean when Von Mises is smaller than Tresca?

If the Von Mises stress is smaller than the Tresca stress, it indicates that the material is not at risk of yielding under the given loading conditions. This means that the material can withstand the applied stresses without experiencing permanent deformation or failure.

4. How do engineers use the Von Mises and Tresca criteria in design?

Engineers use these yield criteria to determine the maximum allowable stresses for a given material. They can then select a material and design a structure that will not exceed these stress limits, ensuring that the structure will not fail under normal operating conditions.

5. Are there any limitations to using the Von Mises and Tresca criteria?

Yes, these yield criteria are only applicable to ductile materials and do not take into account other factors such as temperature, strain rate, and fatigue. Additionally, they are based on idealized assumptions and may not accurately predict the behavior of real-world materials.

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