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

[george88b]

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...

 

[Mapes]

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.

 

[george88b]

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

 

[stewartcs]

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

 

[Mapes]

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 (\sqrt{[(\sigma_1-\sigma_2)^2+(\sigma_1-\sigma_3)^2+(\sigma_2-\sigma_3)^2]/2}) is smaller than the Tresca stress value (|\sigma_1-\sigma_3|). 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.

 

[mikex24]

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.

 

[george88b]

the same with me..;p

 

[nvn]

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?

 

[george88b]

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..

 

[Mapes]

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?