The Neuber Method is useful in calculating the nonlinear stress at a highly localized stress concentration region in a part. The linear FEA solution has the stress concentration factor, Kt, implicit in the linear solution. The Neuber method would calculate a nonlinear effective stress-strain point which allows the analyst to establish the kt. This paper demonstrate an explicit solution based on the establishment of a linearized stress-strain portion of the material. A related and more accurate method is the ESED

1. Estimates of Notch-Stresses and Strains for Local Yielding using Neuber's Rule
by Julio C. Banks, P. E.
Few closed-form solutions exist for determining notch strains during plastic deformation. Numerical
methods (such as finite elements) can be used, but nonlinear elasto-plastic stress-strain relationships
complicates and increases the cost of the solution. Although nonlinear numerical analysis is sometimes
necessary, various approximate methods have been developed for estimating the notch stresses and
strain. Neuber's rule is the most widely used method and it will be illustrated in this article [1].
Example: Aluminum AMS 4117, T6 used in tubing applications (at T= -400 oF)
Elastic Modulus: E  11.2103ksi Ultimate strain: εu  23%
Yield Strength: Sy  48ksi Ultimate Strength: Su  60ksi
Predicted Using Linear Elastic FEA: σFEA  100.6ksi
Solution
σ = E(ε  0.2%) and Ν = Spεp
2
E
Skt
= where Skt  σFEA
therefore εy
Sy
E
  0.2%  0.6286% and Ν
2
E
Skt
  0.9036ksi
Plastic Stress-Strain linearize relationship: Et
Su  Sy
εu ε y
  53.64ksi b  Sy  Etεy  47.66ksi
2
εp  εyεy  0.001  εu Spεp  Etεp  b SpNεp Skt
E ε p

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0
Neuber Plastic State Point
Sp ε p
ksi
SpN ε p
ksi
εp
%
Julio C. Banks, P.E. MathCAD - Neuber A6061.xmcd page 1 of 5

2. Plastic Stress-Strain linearize relationship: Et
Su  Sy
εu ε y
=
Sp  Sy
εp ε y
=
Initial guess for the plastic stress: Sp  1.2Sy εp  1.2εy
Given
Et
Sp  Sy
εp ε y
= (1)
Ν = Spεp (2)
εy  εp  εu
Sp
εp


 FindSpεp
Sp
ksi
εp
%


48.659
1.857



Post-Process
Skt = KtSp therefore Kt
Skt
Sp
  2.067
Summary
Young's Modulus: E  1.120  104ksi
Tangent Modulus: Et  53.64ksi
Plastic Modulus: Ep
Et
1
  53.90ksi
Et
E

Yield Point: εy  0.63% Sy  48.00ksi
Ultimate Point: εu  23% Su  60.00ksi
Julio C. Banks, P.E. MathCAD - Neuber A6061.xmcd page 2 of 5

3. 2.0 Closed-form solution
The appendix provides the derivation and also the detail closed-form solution. For brevity, only the
essential results are provide in this section
Example
Yield strength: Sy  48.00ksi
Yield strain: εy  0.6286%
Tangent Moduls: Et  53.64ksi
Neuber Constant: Ν  0.9036ksi
B1
1
2

Sy
Et


 εy


   0.4443
B0
 1.685 10 2   
Ν
Et
Λ
B0
B12
8.534 10 2   
εp  B11  1  Λ  1.857% QED
REFERENCE
1. Rolfe & Barson, "Fracture & Fatigue Controls in Structures", Prentice Hall.
Pages 598 through 599.
Julio C. Banks, P.E. MathCAD - Neuber A6061.xmcd page 3 of 5

4. Appendix
Obtain a closed-form solution for the Neuber (plastic) point on the linearized
stress-strain material functions given by equations 1 and 2.
Et
Sp  Sy
εp ε y
= (1)
Ν = Spεp (2)
Solve for Sp from Eq, 2
Sp
Ν
= (3)
εp
Substitute Eq. 3 into Eq. 1
Et
Ν
εp


 Sy
= (4)
εp ε y
Now solver for the plastic strain from Eq. 4
Et
Ν  Syεp
εp
2 εy ε  p
=
1
Ν
Et


Sy
Et


ε  p
εp
2 εy ε  p
=
2 Sy
(5) εp
Et


 εy


 εp
Ν
Et
 = 0
1
2
Let B (6) 1

Sy
Et


 εy


 
B0
Ν
Et
  (7)
Julio C. Banks, P.E. MathCAD - Neuber A6061.xmcd page 4 of 5

5. Substitute equations 6 and 7 into Eq. 5
εp
2 2B1 ε  p  B0 = 0 (8)
The solution of the quadratic equation 8 is
εp B1
1  1  Λ
1  1  Λ


=  (9)
Example
Yield strength: Sy  48.00ksi
Yield strain: εy  0.6286%
Tangent Modulus: Et  53.64ksi
Neuber Constant: Ν  0.9036ksi
Numerical experimentation shows the sign of all pertinent parameter to assist the a-priori selection of
the positive-real root. In order for the solution to be real (not imaginary), 1  Λ  0, Also, Λ  0 and
1  Λ  1, therefore, 1  Λ>1. Since B1  0 and the solution sought is εp  0 then the first root, is
the appropriate answer sought,
B1
1
2

Sy
Et


 εy


   0.4443
B0
 1.685 10 2   
Ν
Et
Λ
B0
B12
8.534 10 2   
εp  B11  1  Λ  1.857%
Julio C. Banks, P.E. MathCAD - Neuber A6061.xmcd page 5 of 5