Plasticity – Yield Criteria

It’s relatively easy to measure when a material plastically yields in uniaxial tension or compression, but in most in-service applications there is usually a more complicated multi-axial stress state loading a structure.

Uniaxial Stress Multiaxial Stress

While this question is simple to pose the answer to it can be quite complicated. What we would like is giventhe yield strength of a material in a uniaxial stress state is there some “criteria” for evaluating yieldingunder a multiaxial stress state?

Plasticity – Yield Criteria

Maximum Shear Stress Theory: The theory assumes that yielding will occur in a multiaxial stress state when the maximum shear stress in the multiaxial stress state equals the shear stress in simple tension at yield.

If yielding in simple tension occurs at a normal stress of !", the shear stress is 1/2!"

σ 3 −σ 1( ) = ±σ o

σ 3 −σ 2( ) = ±σ o

σ 2 −σ 1( ) = ±σ o

±1/ 2 σ 3 −σ 1( ) = 1/ 2σ o

−σ o

−σ o

σ o

σ o

σ 2 =σ 1 +σ o

σ 2 =σ 1 −σ o

σ 1

σ 2

Plasticity – Yield Criteria

Distortional Energy Theory: The theory assumes that yielding will occur in a multiaxial stress state when the distortional energy in the multiaxial stress state equals the distortional energy in simple tension at yield.

Distortional Energy:

In terms of principal stresses

In terms of a general stress state

In simple tension σ 1 =σ o; σ 2 =σ 3 = 0

I2d = 1/ 6 σ o2 +σ o

2⎡⎣ ⎤⎦ = 1/ 3σ o2

1/ 3σ o2 = 1/ 6 σ 1 −σ 2( )2 + σ 2 −σ 3( )2 + σ 3 −σ 1( )2⎡

⎣⎢⎤⎦⎥

σ o =12

σ 1 −σ 2( )2 + σ 2 −σ 3( )2 + σ 3 −σ 1( )2⎡⎣⎢

⎤⎦⎥1/2

σ 12 +σ 2

2 −σ 1σ 2 =σ o2

σ 1

σ 2

I2d = 1/ 6 σ 1 −σ 2( )2 + σ 2 −σ 3( )2 + σ 3 −σ 1( )2⎡⎣⎢

⎤⎦⎥

I2d = 1/ 6 σ 11 −σ 22( )2 + σ 22 −σ 33( )2 + σ 33 −σ 11( )2 + 6 σ 122 +σ 23

2 +σ 312( )⎡

⎣⎢⎤⎦⎥

Plasticity – Yield Criteria

Importantly, in order to use the maximum shear stress criteria you need to have the principle stresses. This is notthe case for the distortional energy criteria.

Example: Calculate when yielding will occur under a state of pure shear:

σ 12σ 12

σ 12

σ 12 σ 1

σ 1σ 2 = −σ 1

σ 2 = −σ 1 Max. Shear Stress:

σ 3 −σ 1( ) = ±σ o

σ o =12

σ 1 −σ 2( )2 + σ 2 −σ 3( )2 + σ 3 −σ 1( )2⎡⎣⎢

⎤⎦⎥1/2

Distortional energy:

σ 1 = ±σ o

3

σ 1 = ±σ o

2

s3 (= s1) is the largest principal stresss1 (= s2) is the smallest principal stress

13σ 02 =σ 12

2

σ 12 =σ o

3

Examples of Yield criteria problems

A solid is under a state of stress such that !" = $!% = %!$. It starts to plastically yield when !% = 140 MPa.1. What is the yield stress of this solid in uniaxial tension? 2. If the stress state is changed to '( = − '* +,- '. = 0,at what value of '*will yielding occur?

1. !" = $!% = %!$: !% = 140 Mpa!" = 420, !$ = 210 MPa

Using Tresca:

σ 3 −σ 1( ) = ±σ o

σ 3 −σ 2( ) = ±σ o

σ 2 −σ 1( ) = ±σ o

420 − 140 = '3 = 280 56+

Using von Mises:

σ o2 = 12

σ 1 −σ 2( )2 + σ 2 −σ 3( )2 + σ 3 −σ 1( )2⎡⎣⎢

⎤⎦⎥

σ o2 = 12280( )2 + −70( )2 + −210( )2⎡

⎣⎢⎤⎦⎥= 63700

σ o = 252.4MPa

2. If the stress state is changed to !" = − !% &'( !) = 0,at what value of !%will yielding occur?

Using Tresca:

420 − 140 = !. = 280 01&

σ 3 −σ 1( ) = ±σ o

σ 3 −σ 2( ) = ±σ o

σ 2 −σ 1( ) = ±σ o

± σ 3 −σ 1( ) = ±σ o

!" = − !%±σ 3 = ±140MPa

Using von Mises:

σ o2 = 12

σ 1 −σ 2( )2 + σ 2 −σ 3( )2 + σ 3 −σ 1( )2⎡⎣⎢

⎤⎦⎥

σ o2 = 12

−σ 3( )2 + σ 3( )2 + 2σ 3( )2⎡⎣⎢

⎤⎦⎥= 6σ 3

2

σ o = 3σ 3

σ 3 =σ o

3= 252.4

3= 145.7 MPa

Examples of Yield criteria problems

Examples of Yield criteria problems

A solid has a yield strength in simple tension of 23 MPa. Determine if yielding will occur under the multiaxialstress state using both the Tresca and von Mises criteria

σ ij =0 0 00 10 100 10 −5

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

MPa

Find the principal stresses as 15, 0 and -10 MPa.Using Tresca:

± σ 3 −σ 1( ) = ±σ o

15− −10( ) = 25 > 23 MPa: Yielding will occur

Using von Mises:

σ o2 = 12

σ 1 −σ 2( )2 + σ 2 −σ 3( )2 + σ 3 −σ 1( )2⎡⎣⎢

⎤⎦⎥

1215( )2 + 10( )2 + −25( )2⎡

⎣⎢⎤⎦⎥1/2

= 21.8

21.8 < 23 MPa: Yielding will NOT occur