2. Reference solutions#
2.1. Calculation method used for reference solutions#
2.1.1. Linear workpieces#
Isotropic work hardening
For uniaxial traction, the plasticity criterion is written as:
\(∣{\sigma }_{L}∣-R(p)\le 0\)
where \(p\) is the cumulative plastic deformation
\(R(p)=R\text{'}p+{\sigma }^{y}\) and \(R\text{'}=\frac{E{E}_{1}}{E-{E}_{1}}\)
The criterion is then written:
\(∣{\sigma }_{L}∣-R\text{'}p-{\sigma }^{y}\le 0\)
The stress tensor is obtained by:
\(\sigma =A\mathrm{.}(\varepsilon (u)-{\varepsilon }^{p})-\mathrm{3K}\alpha (T-{T}^{\mathrm{ref}})\mathrm{Id}\)
So we deduce the expression for \({\sigma }_{L}\)
\({\sigma }_{L}=E(\varepsilon -\alpha T)-E{\varepsilon }^{p}\) \(({T}^{\mathrm{ref}}=0)\)
In our case, \(\varepsilon =0\) so:
\({\sigma }_{L}=E{\varepsilon }_{L}-E{\varepsilon }^{p}\) with \({\varepsilon }_{L}=-\alpha T\)
So:
If \(∣{\sigma }_{L}∣-R(p)<0\):
p=0 and \({\sigma }_{L}=E{\varepsilon }_{L}\)
If \(∣{\sigma }_{L}∣-R(p)=0\):
\(p=(\frac{∣\sigma ∣-{\sigma }^{y}}{R\text{'}})\)
\({\sigma }_{L}=E{\varepsilon }_{L}-E{\varepsilon }^{p}\)
Application to the loading path
Instant 1:
\(\sigma =E\varepsilon =\mathrm{200MPa}\) and \(R(p)=R\text{'}p+{\sigma }^{y}=\mathrm{100MPa}\) because*p=0*.
We have \({\sigma }_{L}-R(p)\le 0\).
The criterion has not been met, the evolution is elastic: \({\sigma }_{L}=\mathrm{100MPa}\) and \(N=\mathrm{100kN}\)
Instant 2:
The criterion is met:
\(\begin{array}{}{\sigma }_{L}=\frac{E}{E+R\text{'}}(R\text{'}{\varepsilon }_{L}+{\sigma }^{y})=\frac{2\cdot {10}^{11}}{2\cdot {10}^{11}+2.02\cdot \mathrm{10⁹}}(2.02\cdot \mathrm{10⁹}\times 3.5\cdot {10}^{-3}+2\cdot \mathrm{10⁸})=\mathrm{205MPa}\\ N=\mathrm{102.5kN}\end{array}\)
and \(p=2.475\cdot {10}^{-3}\)
Instant 3:
We discharge elastically:
\(\begin{array}{}{\sigma }_{L}=E{\varepsilon }_{L}-E{\varepsilon }^{p}=2\cdot {10}^{11}(1.5\cdot {10}^{-3}-2.475\cdot {10}^{-3})=-195\mathrm{MPa}\\ N=-97.5\mathrm{kN}\end{array}\)
Instant 4:
We laminate again:
Criteria is written as: \(\mid \sigma \mid -R\text{'}p-{\sigma }^{y}=0\) with \(p={p}_{1}+{p}_{2}\) where \({p}_{1}=2.475\cdot {10}^{-3}\)
So we get:
\({p}_{2}=\frac{\mid \sigma \mid -{\sigma }^{y}}{R\text{'}}-{p}_{1}\)
\(\begin{array}{}\sigma =-E{\varepsilon }^{p}=-E({p}_{1}-{p}_{2})\\ \sigma =(\frac{R\text{'}}{R\text{'}+E})(-2E{p}_{1}-\frac{E{\sigma }^{y}}{R\text{'}})=-\mathrm{207.9MPa}\end{array}\)
And so \(N=-\mathrm{103.95kN}\)
Instant 5:
We discharge elastically:
\(\begin{array}{}{\sigma }_{L}=E{\varepsilon }_{L}-E{\varepsilon }^{p}=2\cdot {10}^{11}(2\cdot {10}^{-3}-1.0395\cdot {10}^{-3})=192.1\mathrm{MPa}\\ N=96.05\mathrm{kN}\end{array}\)
Moments 6 and 7:
The reasoning is the same.
We find:
\(\begin{array}{}{N}_{(\mathrm{inst.6})}=\mathrm{105.87kN}\\ {N}_{(\mathrm{inst.7})}=-\mathrm{44.13kN}\end{array}\)
Kinematic work hardening
The calculation method is the same, but in this case, the plasticity criterion is written as:
\(\sigma -{X}_{\mathrm{eq}}-{\sigma }^{y}\le 0\) with \({X}_{\mathrm{eq}}=C{({\varepsilon }^{p})}_{\mathrm{eq}}=\frac{3}{2}C{\varepsilon }^{p}=\frac{E{E}_{t}}{E-{E}_{t}}{\varepsilon }^{p}\)
With the previous notations, the criterion is written:
\(∣{\sigma }_{L}-R\text{'}{\varepsilon }^{p}∣-{\sigma }^{y}\le 0\) from where \({\sigma }_{L}=R\text{'}{\varepsilon }^{p}\pm {\sigma }^{y}\) (depending on the direction of flow).
Application to the loading path
Instant 1:
The criterion has not been met, the evolution is elastic: \({\sigma }_{L}=\mathrm{100MPa}\) and \(N=\mathrm{100kN}\)
Instant 2:
The criterion is met: \(\mid {\sigma }_{L}-R\text{'}{\varepsilon }^{p}\mid -{\sigma }^{y}=0\)
\({\sigma }_{L}=R\text{'}{\varepsilon }^{p}+{\sigma }^{y}=2.02\cdot \mathrm{10⁹}\times 2.475\cdot {10}^{-3}+2\cdot \mathrm{10⁸}=\mathrm{205MPa}\)
Instant 3:
We discharge elastically:
\(\begin{array}{}{\sigma }_{L}=E{\varepsilon }_{L}-E{\varepsilon }^{p}=2\cdot {10}^{11}(1.5\cdot {10}^{-3}-2.475\cdot {10}^{-3})=-\mathrm{195MPa}\\ N=-\mathrm{97.5kN}\end{array}\)
Instant 4:
\(\mid \sigma -R\text{'}{\varepsilon }^{p}\mid -{\sigma }^{y}=0\) with \({p}_{1}=2.475\cdot {10}^{-3}\)
\(\begin{array}{}{\varepsilon }^{p}={p}_{1}-{p}_{2}\\ {p}_{2}={p}_{1}-\frac{\mid \sigma +{\sigma }^{y}\mid }{R\text{'}}\\ \sigma =-E{\varepsilon }^{p}=-E({p}_{1}-{p}_{2})\\ \sigma =-E(\frac{\mid \sigma +{\sigma }^{y}\mid }{R\text{'}})=-\mathrm{198MPa}\\ N=-\mathrm{99kN}\end{array}\)
Instant 5:
We discharge elastically:
\(\begin{array}{}{\sigma }_{L}=E{\varepsilon }_{L}-E{\varepsilon }^{p}=2\cdot {10}^{11}(2\cdot {10}^{-3}-9.9\cdot {10}^{-4})=202\mathrm{MPa}\\ N=\mathrm{101kN}\end{array}\)
Moments 6 and 7:
The reasoning is the same. We find:
\(\begin{array}{}{N}_{(\mathrm{inst.6})}=\mathrm{103kN}\\ {N}_{(\mathrm{inst.7})}=-\mathrm{47kN}\end{array}\)
2.1.2. Pinto-Menegotto model#
This model is described in the Code_Aster Reference Manual [R5.03.09] [bib1]. The constitutive law of steels is composed of two distinct parts: monotonic loading composed of three successive zones (linear elasticity, plastic level and work hardening) and cyclic loading where the path between two points of reversal (half-cycle) is described by an analytical expression curve of the type \(\sigma =f(\varepsilon )\).
As before, the deformations imposed are thermal deformations: \(\varepsilon =-\alpha T\)
2.1.2.1. Case without buckling#
First load
Linear elasticity: \(\sigma =E\varepsilon\)
Instant 1:
\(N=E\varepsilon S=2\cdot {10}^{11}\times 1\cdot {10}^{-3}\times 5\cdot {10}^{-4}=\mathrm{100kN}\)
Plastic bearing: \(\sigma ={\sigma }^{y}\)
Degree 4 polynomial: \(\sigma ={\sigma }_{u}-({\sigma }_{\mathrm{su}}-{\sigma }_{y}^{0}){(\frac{{\varepsilon }_{u}-\varepsilon }{{\varepsilon }_{u}-{\varepsilon }_{h}})}^{4}\)
Instant 2:
\(\varepsilon =3.5\cdot {10}^{-3}>{\varepsilon }_{h}=2.3\cdot {10}^{-3}\), we use the degree 4 polynomial:
\(\sigma =\mathrm{209.416MPa}\) and \(N=\mathrm{104.708kN}\)
Cycles
Half cycle 1:
We determine \({\zeta }_{p}^{0}\):
\({\zeta }_{p}^{0}={\varepsilon }_{r}^{0}-{\varepsilon }_{y}^{0}=3.5\cdot {10}^{-3}-{1.10}^{-3}=2.5\cdot {10}^{-3}\) because \({\varepsilon }_{r}^{0}={\varepsilon }_{(\mathrm{inst.2})}\)
Then \(\Delta {\sigma }^{0}\):
\(\Delta {\sigma }^{0}={E}_{h}{\zeta }_{p}^{0}=2\cdot {10}^{9}\times 2.5\cdot {10}^{-3}=\mathrm{5MPa}\)
Where \({\sigma }_{y}^{1}={\sigma }_{y}^{0}\cdot \mathrm{sign}(-{\zeta }_{p}^{0})+\Delta {\sigma }^{0}=-200+5=-\mathrm{195MPa}\)
We then calculate \({\varepsilon }_{y}^{1}\):
\({\varepsilon }_{y}^{1}={\varepsilon }_{r}^{0}+\frac{{\sigma }_{y}^{1}-{\sigma }_{r}^{0}}{E}=3.5\cdot {10}^{-3}+\frac{(-195-209.416)\mathrm{10⁶}}{2.0\cdot {10}^{11}}=1.477\cdot {10}^{-3}\)
We thus determine \(\sigma *=f(\varepsilon*)\), defined by:
\({\sigma }^{}=b{\varepsilon }^{}+(\frac{1-b}{{(1+{({\varepsilon }^{})}^{R})}^{(1/R)}}){\varepsilon }^{}\), with \(b=\frac{{E}_{h}}{E}\)
\(\begin{array}{}{\varepsilon }^{}=\frac{\varepsilon -{\varepsilon }_{r}^{0}}{{\varepsilon }_{y}^{1}-{\varepsilon }_{r}^{0}}\\ \\ {\sigma }^{}=\frac{\sigma -{\sigma }_{r}^{0}}{{\sigma }_{y}^{1}-{\sigma }_{r}^{0}}\end{array}\)
\({\xi }_{p}^{0}=\frac{{\zeta }_{p}^{0}}{\mid {\varepsilon }_{y}^{1}-{\varepsilon }_{r}^{0}\mid }\) and \(\mathrm{R¹}={R}_{0}-\frac{{A}_{1}\cdot {\xi }_{p}^{0}}{{A}_{2}+{\xi }_{p}^{0}}\)
We get \({\xi }_{p}^{0}=-1.23\) and \(\mathrm{R¹}=3.51\)
We can then calculate the value of \(\sigma\) at times 3 and 4:
Instant 3:
\({\varepsilon }^{}=\frac{{\varepsilon }_{(\mathrm{inst.3})}-{\varepsilon }_{r}^{0}}{{\varepsilon }_{y}^{1}-{\varepsilon }_{r}^{0}}=\frac{1.5\cdot {10}^{-3}-3.5\cdot {10}^{-3}}{1.477\cdot {10}^{-3}-3.5\cdot {10}^{-3}}=0.988\)
\({\sigma }^{}=b{\varepsilon }^{}+(\frac{1-b}{{(1+{({\varepsilon }^{})}^{R})}^{(1/R)}}){\varepsilon }^{}=0.01\times 0.988+(\frac{1-0.01}{{(1+{(0.988)}^{3.51})}^{(1/3.51)}})=0.82\)
and \(\sigma ={\sigma }^{}({\sigma }_{y}^{1}-{\sigma }_{r}^{0})+{\sigma }_{r}^{0}=0.82\times (-195-209.416)+209.416=-122\mathrm{MPa}\)
Hence \(N=-\mathrm{61kN}\)
Instant 4:
We use the same method, with \(\varepsilon =0\).
\(\begin{array}{}{\varepsilon }^{}=1.73\\ {\sigma }^{}=0.56\\ \sigma =-20\mathrm{MPa}\\ N=-\mathrm{10kN}\end{array}\)
Half cycle 2: Instant 5 and 6:
The calculation method is identical, we determine:
\({\zeta }_{p}^{1}\), \({\sigma }_{y}^{2}\), \({\varepsilon }_{y}^{2}\), \({\xi }_{p}^{1}\), \(\mathrm{R²}\), then \({\sigma }_{(\mathrm{inst.5})}^{}=f({\varepsilon }_{(\mathrm{inst.5})}^{})\) and \({\sigma }_{(\mathrm{inst.6})}^{}=f({\varepsilon }_{(\mathrm{inst.6})}^{})\)
and finally \({\sigma }_{(\mathrm{inst.5})}\) and \({\sigma }_{(\mathrm{inst.6})}\).
Half-cycle 3: Instant 7: Same
2.1.2.2. Case with buckling#
First load
Same as the previous case.
Cycles
Half cycle 1 (compression) :
The calculation method is the same, but the value of the asymptote slope is changed:
A new coefficient \({b}_{c}\) is calculated:
\({b}_{c}=a(5.0-L/D)e(b\xi \text{'}\frac{E}{{\sigma }^{y}-{\sigma }^{\infty }})=0.006\times (5.0-5.9){e}^{(0.01\times 1.477\cdot {10}^{-3}\frac{2\cdot {10}^{11}}{2\cdot {10}^{8}-1.36\cdot {10}^{8}})}=-0.0057\)
Next, as in the model without buckling, we must determine \({\sigma }_{y}^{n}\). The reasoning is the same, but an additional constraint \({\sigma }_{s}^{}\) is added in order to correctly position the curve in relation to the asymptote.
\({\sigma }_{s}^{}={\gamma }_{s}bE\frac{b-{b}_{c}}{1-{b}_{c}}=0.028\times 0.01\times 2\cdot {10}^{11}\times \frac{0.01+0.0057}{1+0.0057}=0.87\mathrm{MPa}\)
where \({\gamma }_{s}\) is given by: \({\gamma }_{s}=\frac{11.0-L/D}{10({e}^{c(L/D)}-1.0)}=0.028\)
Half cycle 2 (traction) :
In traction, a reduced Young’s modulus is adopted:
\({E}_{r}=E({a}_{5}+(1.0-{a}_{5}){e}^{-{a}_{6}\xi {"}^{2}})=2\cdot {10}^{11}\times (\mathrm{0,88}+(1-0.88){e}^{(-620\times 1.473\cdot {10}^{-6})})=1.99\cdot {10}^{11}\mathrm{MPa}\)
with \({a}_{5}=1.0+(5.0-L/D)/7.5=0.88\)
The rest of the method is the same.
2.2. Benchmark results#
Constant \(N\) normal effort on the bar
2.3. Uncertainty about the solution#
None, the solution is analytical
2.4. Bibliographical references#
[1] Code_Aster reference manual [R5.03.09].
[2] S. ANDRIEUX: TD 1 Three perfect Von Mises thermoelastoplastic bars. In « Introduction to thermoplasticity in Code_Aster », HI-74/96/013 November 1996 (course reference manual).