# Stress-Strain Compression of AA6082-T6 Aluminum Alloy at Room Temperature

Authors: Alexandre da Silva Scari; Bruno Cesar Pockszevnicki; Jánes Landre Junior; Pedro Americo Almeida Magalhaes Junior
Journal: Journal of Structures
 (2014)

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## Abstract

Short cylindrical specimens made of AA6082-T6 aluminum alloy were studied experimentally (compression tests), analytically (normalized Cockcroft-Latham criteria—nCL), and numerically (finite element analysis—FEA). The mechanical properties were determined with the stress-strain curves by the Hollomon equation. The elastic modulus obtained experimentally differs from the real value, as expected, and it is also explained. Finite element (FE) analysis was carried out with satisfactory correlation to the experimental results, as it differs about 1,5% from the damage analysis by the nCL concerning the experimental data obtained by compression tests.


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## Body

## 1. Introduction

Mechanical properties of ductile materials are normally determined by tensile test. However, when the plastic deformation is the aim of the study, the compression test is the most suitable as it allows large deformations without the fracture of the specimen. For this test, the cylindrical specimen is the most adopted. However, in this case, the damage must be evaluated. Numerical analysis (i.e., finite element analysis—FEA) may be used to predict the damage for materials with elastic-plastic behavior. But, first, it is needed to perform classical model (analytical) and material (experimental) analysis.Many studies were carried out concerning aluminum under axial compression; among them, Hopperstad et al. [1], Andrews et al. [2], and Gioux et al. [3] studied the compression of aluminum foams. Ferguson et al. [4] proposed an analytical model for predicting the mechanical properties of bimodal nanoaluminum alloys, concerning the grain size. Pled et al. [5] have done a numerical study of the crushing of circular aluminum tubes, with and without aluminum foam fillers. Han and Kim [6] proposed a new criterion for ductile fracture in sheet metal forming process. Luo et al. [7] performed isothermal compression tests of 7A09 aluminum alloy concerning solid cylindrical specimens. Rees [8] studied aluminum alloy sheets by plane strain compression. Wu et al. [9] determined the flow behavior and constitutive equations of 7050 aluminum alloy in isothermal compression. Kyac
ˇkaj et al. [10] studied the workability of AlMgSi aluminum alloy prepared by powder metallurgy, by means of analytical (damage) and numerical methods.Some researches concerning damage of other metals have also been done, and some are listed here: Stefanik et al. [11] have determined the values of the normalized Cockcroft-Latham (nCL) for multi slight rolling based on tensile test of BSt500S steel. Xia et al. [12] studied the effects of temperature and strain rate on the damage value of AZ80 magnesium alloy. Alexandrov and Jeng [13] proposed a general method for the modified Cockcroft-Latham criterion at elevated temperatures. Quan et al. [14] evaluated the ductile fracture criteria for 42CrMo steel by compressions at different temperatures and strain rates. Landre et al. [15] contributed to the discussion of decoupled ductile fracture criteria and their use with finite element analysis, testing samples of AISI 1040 steel.The goal of this paper is to evaluate and predict the damage of short solid cylindrical specimens made of an aluminum alloy, as just a few researches dealt with it. For this, the parameters of the Hollomon equation were obtained together with the normalized Cockcroft-Latham criterion (nCL) by the experimental data obtained by compression tests. FE analysis was performed and its result differs about 1,5% from the damage analysis by the nCL, which shows that, once validated, FEA can be used to predict damage of elastic-plastic materials.


## 2. Literature Review

### 2.1. Compression Test

The compression test consists of a specimen (normally with cylindrical shape) that is compressed between two parallel plates, with uniform velocity (see Figure1). Mechanical properties such as the yield strength (σ
y) and the elastic modulus (E) can be obtained by this test. However, the ratio of initial length to diameter (L
0
/
D) has great influence on the test results. According to ASTM [17], to accurately determine the elastic modulus in the compression test, it is necessary that L
0
/
D
≥
8 (long specimen). Short specimen presents L
0
/
D
≤
2.Figure 1
Cylindrical specimen in compression test between parallel plates. Source: adapted from [16].Ductile materials in compression may present nonuniform transverse deformation, called barreling. It may be seen from Figure2 that the specimen regions in contact with the plates present almost null plastic deformations (see number 1 on Figure 2), and the number 2 is where the plastic deformations occur. Also, the barreling is represented by the number 3 in Figure 2.Figure 2
Illustration of barreling. Source: adapted from [17].


### 2.2. Plastic Deformation

The experimental data obtained by compression test can be described by the Hollomon equation (see (1)), where σ is the true stress, K is the strength coefficient, ε is the true strain, and n is the strain-hardening exponent:
(1)
σ
=
K
·
ε
n
.
The coefficients in (1) can be obtained by power-law approximation of the experimental data obtained by compression test.With the plate displacements and the corresponding forces, the strain can be calculated by three different ways: normal (2), logarithmic (3), and Lagrange equation (4):
(2)
ε
=
L
-
L
0
L
0
,
(3)
ε
=
ln
⁡
L
L
0
,
(4)
ε
=
L
2
-
L
0
2
2
·
L
0
2
.
Assuming that the areas in contact with the plates (see Figure 1) remain constant, the actuating stress can be obtained with the ratio of the force (F) to initial cross-section area (A
0) as follows:
(5)
σ
=
F
A
0
.


### 2.3. Ductile Fracture Criteria

According to Landre et al. [15], the damage criterion that best fits the workability of elastic-plastic materials was proposed by Cockcroft and Latham (CL). It is based on a critical value of tensile strain per unit of volume and does not take into account the hydrostatic stress (σ
m) because the large principal stress (σ
1
max
⁡) and σ
m normally vary together. The normalized CL criterion (nCL) can be written as
(6)
σ
1
max
⁡
K
·
ε
n
-
1
=
C
,
where (i)
σ
1
max
⁡: large principal stress [MPa];
(ii)
K: strength coefficient [MPa];
(iii)
ε: true strain;
(iv)
n: strain-hardening exponent.The values of K and   ε on (6) are the same for (1).


## 2.1. Compression Test

The compression test consists of a specimen (normally with cylindrical shape) that is compressed between two parallel plates, with uniform velocity (see Figure1). Mechanical properties such as the yield strength (σ
y) and the elastic modulus (E) can be obtained by this test. However, the ratio of initial length to diameter (L
0
/
D) has great influence on the test results. According to ASTM [17], to accurately determine the elastic modulus in the compression test, it is necessary that L
0
/
D
≥
8 (long specimen). Short specimen presents L
0
/
D
≤
2.Figure 1
Cylindrical specimen in compression test between parallel plates. Source: adapted from [16].Ductile materials in compression may present nonuniform transverse deformation, called barreling. It may be seen from Figure2 that the specimen regions in contact with the plates present almost null plastic deformations (see number 1 on Figure 2), and the number 2 is where the plastic deformations occur. Also, the barreling is represented by the number 3 in Figure 2.Figure 2
Illustration of barreling. Source: adapted from [17].


## 2.2. Plastic Deformation

The experimental data obtained by compression test can be described by the Hollomon equation (see (1)), where σ is the true stress, K is the strength coefficient, ε is the true strain, and n is the strain-hardening exponent:
(1)
σ
=
K
·
ε
n
.
The coefficients in (1) can be obtained by power-law approximation of the experimental data obtained by compression test.With the plate displacements and the corresponding forces, the strain can be calculated by three different ways: normal (2), logarithmic (3), and Lagrange equation (4):
(2)
ε
=
L
-
L
0
L
0
,
(3)
ε
=
ln
⁡
L
L
0
,
(4)
ε
=
L
2
-
L
0
2
2
·
L
0
2
.
Assuming that the areas in contact with the plates (see Figure 1) remain constant, the actuating stress can be obtained with the ratio of the force (F) to initial cross-section area (A
0) as follows:
(5)
σ
=
F
A
0
.


## 2.3. Ductile Fracture Criteria

According to Landre et al. [15], the damage criterion that best fits the workability of elastic-plastic materials was proposed by Cockcroft and Latham (CL). It is based on a critical value of tensile strain per unit of volume and does not take into account the hydrostatic stress (σ
m) because the large principal stress (σ
1
max
⁡) and σ
m normally vary together. The normalized CL criterion (nCL) can be written as
(6)
σ
1
max
⁡
K
·
ε
n
-
1
=
C
,
where (i)
σ
1
max
⁡: large principal stress [MPa];
(ii)
K: strength coefficient [MPa];
(iii)
ε: true strain;
(iv)
n: strain-hardening exponent.The values of K and   ε on (6) are the same for (1).


## 3. Materials and Methods

### 3.1. Experimental Tests

The cylindrical specimens were made of AA6082-T6 aluminum alloy. Its chemical composition was determined by laboratory (chemical analysis) and is presented on Table1, which also presents the reference values of MATWEB [19]. According to MATWEB [19], this alloy presents yield stress σ
y = 250 MPa, ultimate stress σ
u = 290 MPa, elastic modulus E = 70 GPa, and hardness Brinell HB = 90.Table 1
Chemical composition (%) of the aluminum alloy studied and the MATWEB reference.
Element
Si
Mg
Mn
Fe
Cu
Zn
Ti
Cr
Pb
Sn
Al
Specimen
1.12
0.24
0.46
0.21
0.017
0.002
0.023
0.005
0.001
0.000
97.89
MATWEB
0.70–1.3
0.60–1.2
0.40–1.0
≤0.50
≤0.10
≤0.20
≤0.10
≤0.25
≤0.05
≤0.05
95.2–93.3It may be seen from Table1 that the magnesium concentration (0.24%) of the specimens is a little lower than the minimum indicated by MATWEB (0.60%–1.20%). Magnesium is added to an aluminum alloy in order to improve its strength.For the compression tests, the plate displacement velocity was constant and equal to 2 mm/min. Ten specimens were tested, all of them with diameterD = 15.78mm and average initial length L
0 = 28.7 mm. Thus, L
0
/
D
=
1.82, which represents short specimens. The displacements adopted for each specimen are presented in Table 2.Table 2
Displacements applied on the specimens.
Displacement [mm]
Specimens
0.50
1a
1b
0.75
2a
2b
1.50
3a
3b
2.50
4a
4b
4.00
5a
5bAfter the tests, with the flange displacements and the respective forces applied for each instant of time, some information may be calculated (i.e., strains and tensions) and the yield stress can be determined. Also, the stress-strain curves for each specimen were obtained and, with the power-law approximation, the coefficients of the Hollomon equation (see (1)) were determined along with the critical values of the nCL.Figure3 presents one of the specimens before the compression test.Figure 3
Cylindrical specimen before compression test.


### 3.2. Finite Element Analysis (FEA)

The finite element (FE) model consists of hexahedral elements with 1.0 mm length. The software adopted was Hypermesh (preprocessor), Abaqus (solver), and Hyperview (postprocessor). Assuming that the lower and upper faces present only axial displacement in the compression test, in the FE model, the lower face was restricted (all degrees of freedom (DOFs)) and a plate with contact elements was used to apply the compression force in the upper face. Its value is 75526 N according to specimen4b (see Table 3). The elastic modulus considered was the real one (70 GPa) Figure 4 presents the FE model analyzed.Table 3
Experimental results summary.
Specimen
Δ
L
[mm]
F
[N]
Stress
[MPa]
ε
normal
ε
Lagrange
ε
logarithmic
E
[MPa]
σ
y
[MPa]
1a
0.078
44525
227.7
0.01750
0.01735
0.01766
13008.2
227.7
1b
0.027
49714
254.2
0.01464
0.01453
0.01475
17368.0
252.2
2a
0.187
56542
289.1
0.02380
0.02352
0.02409
13323.5
245.5
2b
0.198
57570
294.4
0.02341
0.02313
0.02369
17006.2
228.7
3a
0.866
64384
329.2
0.04957
0.04834
0.05084
15697.4
217.4
3b
0.921
70400
360.0
0.05116
0.04985
0.05252
18178.4
250.4
4a
1.753
76888
393.1
0.08245
0.07905
0.08605
16920.9
253.0
4b
1.612
75526
386.2
0.07799
0.07495
0.08120
14040.0
261.4
5a
3.317
77916
398.4
0.13820
0.12865
0.14873
17049.7
239.7
5b
3.012
84035
429.7
0.12996
0.12152
0.13922
15654.3
249.2Figure 4
FE model.


## 3.1. Experimental Tests

The cylindrical specimens were made of AA6082-T6 aluminum alloy. Its chemical composition was determined by laboratory (chemical analysis) and is presented on Table1, which also presents the reference values of MATWEB [19]. According to MATWEB [19], this alloy presents yield stress σ
y = 250 MPa, ultimate stress σ
u = 290 MPa, elastic modulus E = 70 GPa, and hardness Brinell HB = 90.Table 1
Chemical composition (%) of the aluminum alloy studied and the MATWEB reference.
Element
Si
Mg
Mn
Fe
Cu
Zn
Ti
Cr
Pb
Sn
Al
Specimen
1.12
0.24
0.46
0.21
0.017
0.002
0.023
0.005
0.001
0.000
97.89
MATWEB
0.70–1.3
0.60–1.2
0.40–1.0
≤0.50
≤0.10
≤0.20
≤0.10
≤0.25
≤0.05
≤0.05
95.2–93.3It may be seen from Table1 that the magnesium concentration (0.24%) of the specimens is a little lower than the minimum indicated by MATWEB (0.60%–1.20%). Magnesium is added to an aluminum alloy in order to improve its strength.For the compression tests, the plate displacement velocity was constant and equal to 2 mm/min. Ten specimens were tested, all of them with diameterD = 15.78mm and average initial length L
0 = 28.7 mm. Thus, L
0
/
D
=
1.82, which represents short specimens. The displacements adopted for each specimen are presented in Table 2.Table 2
Displacements applied on the specimens.
Displacement [mm]
Specimens
0.50
1a
1b
0.75
2a
2b
1.50
3a
3b
2.50
4a
4b
4.00
5a
5bAfter the tests, with the flange displacements and the respective forces applied for each instant of time, some information may be calculated (i.e., strains and tensions) and the yield stress can be determined. Also, the stress-strain curves for each specimen were obtained and, with the power-law approximation, the coefficients of the Hollomon equation (see (1)) were determined along with the critical values of the nCL.Figure3 presents one of the specimens before the compression test.Figure 3
Cylindrical specimen before compression test.


## 3.2. Finite Element Analysis (FEA)

The finite element (FE) model consists of hexahedral elements with 1.0 mm length. The software adopted was Hypermesh (preprocessor), Abaqus (solver), and Hyperview (postprocessor). Assuming that the lower and upper faces present only axial displacement in the compression test, in the FE model, the lower face was restricted (all degrees of freedom (DOFs)) and a plate with contact elements was used to apply the compression force in the upper face. Its value is 75526 N according to specimen4b (see Table 3). The elastic modulus considered was the real one (70 GPa) Figure 4 presents the FE model analyzed.Table 3
Experimental results summary.
Specimen
Δ
L
[mm]
F
[N]
Stress
[MPa]
ε
normal
ε
Lagrange
ε
logarithmic
E
[MPa]
σ
y
[MPa]
1a
0.078
44525
227.7
0.01750
0.01735
0.01766
13008.2
227.7
1b
0.027
49714
254.2
0.01464
0.01453
0.01475
17368.0
252.2
2a
0.187
56542
289.1
0.02380
0.02352
0.02409
13323.5
245.5
2b
0.198
57570
294.4
0.02341
0.02313
0.02369
17006.2
228.7
3a
0.866
64384
329.2
0.04957
0.04834
0.05084
15697.4
217.4
3b
0.921
70400
360.0
0.05116
0.04985
0.05252
18178.4
250.4
4a
1.753
76888
393.1
0.08245
0.07905
0.08605
16920.9
253.0
4b
1.612
75526
386.2
0.07799
0.07495
0.08120
14040.0
261.4
5a
3.317
77916
398.4
0.13820
0.12865
0.14873
17049.7
239.7
5b
3.012
84035
429.7
0.12996
0.12152
0.13922
15654.3
249.2Figure 4
FE model.


## 4. Results and Discussion

### 4.1. Experimental Tests

Table3 presents the experimental results summary for each specimen subjected to compression test and also the calculation results (strain, stress, and elastic modulus). It can be seen from Table 3 that ε
normal(2) present intermediate values among ε
logarithmic(3) and ε
Lagrange(4). So, ε
normal was used to calculate the elastic modulus.The stress-strain curves obtained experimentally are presented in Figure5.Figure 5
Stress-strain curves from compression tests.According to Table3 and Figure 5,(i)
specimen 1a did not enter the plasticity region;
(ii)
the average elastic modulus is 16 GPa, lower than the real value (70 GPa). This lower value was expected and is due to theL
0
/
D
=
1.82 ratio, according to [17];
(iii)
the yield stress varied from 217.4 MPa to 261.4 MPa, consistent with the real value of 250 MPa; also, these values agree with the results obtained by Hopperstad et al. [1];
(iv)
strain calculated with the Lagrange equation (see (4)) leads to the lowest results.With the stress-strain curves presented in Figure 5, power-law approximations were done in order to determine the coefficients of the Hollomon equation (1). The average values are presented in Table 4. Yet, the power-law approximation results are of high quality, as R
2
=
0.999 for each regression made.Table 4
Average coefficients of the Hollomon equation obtained experimentally.
Strain
Strength coefficient,K [MPa]
Strain-hardening exponent,n
ε
normal
587.5
0.203
ε
logarithmic
593.1
0.208
ε
Lagrange
585.5
0.205According to Table4,(i)
the average value of the strength coefficient (K) is 588.7 MPa;
(ii)
the average value of the strain-hardening coefficient (n) is 0.205.Comparing the values presented in Table4 with the AA2024-T4 properties presented by Dowling [18] and Han and Kim [6] (see Table 5), the strain-hardening coefficient (n) is almost the same (2.5% different). Also, the strength coefficient (K) increases with the strength of the aluminum alloy (or hardness).Table 5
Comparison among aluminum alloys properties.
Aluminum alloy
K
[MPa]
n
Ultimate stress,σ
u [MPa]
AA6082 T6
588.7
0.205
290
AA2024 T4a
806
0.200
476
AA6111b
504
0.270
272
*Source: aDowling [18]; bHan and Kim [6].Yet, according to Hosford [20], the strength coefficient (K) lies between 400 MPa and 550 MPa. Also, common values of the strain-hardening coefficient (n) remain between 0.20 and 0.30. Comparing with Table 5, the values of K are lower with respect to AA6082-T6 and AA2124-T4 alloys but the values of n are in accordance.Rees [8] obtained E
= 75 GPa for AC 120 and HE30TF aluminum alloys with the compression of sheets instead of cylindrical specimens. Although this difference is in the elastic modulus, the strain-hardening coefficient (n) for the AC 120 aluminum alloy obtained by the same author was 0.218, very similar to the value obtained in this work (0.205, see Table 5).


### 4.2. FEA Results

The FEA results for the specimen4b are presented in Figure 6 and in Table 6. The minimum principal stress obtained is −458.9 MPa. Also, the plastic equivalent strain is 0.09822. Comparing these results with those presented in Table 3, it may be seen that(i)
the minimum principal stress obtained (−458.9 MPa) is 6.8% greater than the specimen5b result (−429.5 MPa);
(ii)
the plastic strain obtained (0.09822) is 14.1% greater than the logarithmic strain presented by the specimen4a   (0.08605);
(iii)
Figure6 is in accordance with Figure 2.Table 6
FEA results.
Minimum principal stress [MPa]
Plastic equivalent strain
−458.9
0.09822FEA results: (a) minimum principal stress; (b) plastic equivalent strain-specimen4b.
(a)
(b)


### 4.3. Damage Results

Table7 presents the results concerning the nCL criterion for each specimen subjected to compression test. The nCL curves obtained experimentally are presented in Figure 7.Table 7
nCL criterion results summary.
Specimen
nCL [—]
1a
0.015512
1b
0.015027
2a
0.025153
2b
0.025272
3a
0.051318
3b
0.057543
4a
0.091840
4b
0.086313
5a
0.140321
5b
0.144128Figure 7
nCL curves from compression tests.According to the above results,(i)
as the compression tests were stopped before fracture, the curves on Figure7 did not present asymptotic behavior;
(ii)
specimen5b presented the highest value of nCL (0.144128, see Table 7) as it was subjected to the highest compression force (84035N, see Table 3);
(iii)
the orange value of plastic equivalent strain (PEEQ) on Figure 6(b) varies from 0.07706 to 0.08764. Comparing this maximum (0.08764) with the maximum nCL obtained for specimen4b (0.086313, see Table 7), the numerical value is 1.54% greater than the experimental result. As FE formulation has some known intrinsic errors, the difference of 1.54% between the mentioned results shows a satisfactory correlation between experimental and numerical analysis.Figure8 illustrates the orange region of Figure 6(b).Figure 8
Orange regions highlighted from Figure6(b).


## 4.1. Experimental Tests

Table3 presents the experimental results summary for each specimen subjected to compression test and also the calculation results (strain, stress, and elastic modulus). It can be seen from Table 3 that ε
normal(2) present intermediate values among ε
logarithmic(3) and ε
Lagrange(4). So, ε
normal was used to calculate the elastic modulus.The stress-strain curves obtained experimentally are presented in Figure5.Figure 5
Stress-strain curves from compression tests.According to Table3 and Figure 5,(i)
specimen 1a did not enter the plasticity region;
(ii)
the average elastic modulus is 16 GPa, lower than the real value (70 GPa). This lower value was expected and is due to theL
0
/
D
=
1.82 ratio, according to [17];
(iii)
the yield stress varied from 217.4 MPa to 261.4 MPa, consistent with the real value of 250 MPa; also, these values agree with the results obtained by Hopperstad et al. [1];
(iv)
strain calculated with the Lagrange equation (see (4)) leads to the lowest results.With the stress-strain curves presented in Figure 5, power-law approximations were done in order to determine the coefficients of the Hollomon equation (1). The average values are presented in Table 4. Yet, the power-law approximation results are of high quality, as R
2
=
0.999 for each regression made.Table 4
Average coefficients of the Hollomon equation obtained experimentally.
Strain
Strength coefficient,K [MPa]
Strain-hardening exponent,n
ε
normal
587.5
0.203
ε
logarithmic
593.1
0.208
ε
Lagrange
585.5
0.205According to Table4,(i)
the average value of the strength coefficient (K) is 588.7 MPa;
(ii)
the average value of the strain-hardening coefficient (n) is 0.205.Comparing the values presented in Table4 with the AA2024-T4 properties presented by Dowling [18] and Han and Kim [6] (see Table 5), the strain-hardening coefficient (n) is almost the same (2.5% different). Also, the strength coefficient (K) increases with the strength of the aluminum alloy (or hardness).Table 5
Comparison among aluminum alloys properties.
Aluminum alloy
K
[MPa]
n
Ultimate stress,σ
u [MPa]
AA6082 T6
588.7
0.205
290
AA2024 T4a
806
0.200
476
AA6111b
504
0.270
272
*Source: aDowling [18]; bHan and Kim [6].Yet, according to Hosford [20], the strength coefficient (K) lies between 400 MPa and 550 MPa. Also, common values of the strain-hardening coefficient (n) remain between 0.20 and 0.30. Comparing with Table 5, the values of K are lower with respect to AA6082-T6 and AA2124-T4 alloys but the values of n are in accordance.Rees [8] obtained E
= 75 GPa for AC 120 and HE30TF aluminum alloys with the compression of sheets instead of cylindrical specimens. Although this difference is in the elastic modulus, the strain-hardening coefficient (n) for the AC 120 aluminum alloy obtained by the same author was 0.218, very similar to the value obtained in this work (0.205, see Table 5).


## 4.2. FEA Results

The FEA results for the specimen4b are presented in Figure 6 and in Table 6. The minimum principal stress obtained is −458.9 MPa. Also, the plastic equivalent strain is 0.09822. Comparing these results with those presented in Table 3, it may be seen that(i)
the minimum principal stress obtained (−458.9 MPa) is 6.8% greater than the specimen5b result (−429.5 MPa);
(ii)
the plastic strain obtained (0.09822) is 14.1% greater than the logarithmic strain presented by the specimen4a   (0.08605);
(iii)
Figure6 is in accordance with Figure 2.Table 6
FEA results.
Minimum principal stress [MPa]
Plastic equivalent strain
−458.9
0.09822FEA results: (a) minimum principal stress; (b) plastic equivalent strain-specimen4b.
(a)
(b)


## 4.3. Damage Results

Table7 presents the results concerning the nCL criterion for each specimen subjected to compression test. The nCL curves obtained experimentally are presented in Figure 7.Table 7
nCL criterion results summary.
Specimen
nCL [—]
1a
0.015512
1b
0.015027
2a
0.025153
2b
0.025272
3a
0.051318
3b
0.057543
4a
0.091840
4b
0.086313
5a
0.140321
5b
0.144128Figure 7
nCL curves from compression tests.According to the above results,(i)
as the compression tests were stopped before fracture, the curves on Figure7 did not present asymptotic behavior;
(ii)
specimen5b presented the highest value of nCL (0.144128, see Table 7) as it was subjected to the highest compression force (84035N, see Table 3);
(iii)
the orange value of plastic equivalent strain (PEEQ) on Figure 6(b) varies from 0.07706 to 0.08764. Comparing this maximum (0.08764) with the maximum nCL obtained for specimen4b (0.086313, see Table 7), the numerical value is 1.54% greater than the experimental result. As FE formulation has some known intrinsic errors, the difference of 1.54% between the mentioned results shows a satisfactory correlation between experimental and numerical analysis.Figure8 illustrates the orange region of Figure 6(b).Figure 8
Orange regions highlighted from Figure6(b).


## 5. Conclusions

The ratioL
0
/
D = 1,82 is not suitable for determining the elastic modulus (E), as the average value of E obtained was 16 GPa and right value is 70 GPa. According to ASTM [17], the ideal ratio for that purpose is L
0
/
D
=
8. Yet, different shapes in compression (instead of cylindrical specimens) may lead to more accurate values of the elastic modulus.The average yield stress obtained (242.5 MPa) is 3.0% lower than the value informed by MATWEB [19] (250 MPa). This difference may be due to the magnesium concentration on this aluminum alloy (0.24%) which is lower than the minimum specified for this alloy (0.60%).The coefficients of the Hollomon equation obtained by power-law approximation are in accordance with the literature for the AA6082-T6 aluminum alloy. Also, the values ofK and n were used to calculate nCL for each specimen.FEA results presented satisfactory correlation to the experimental tests as the difference, concerning specimen4b, is about 1,5% from the damage analysis by the nCL criterion, which shows that, once validated, FEA can be used to predict damage of elastic-plastic materials.



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*Source: 387680-2014-11-13.xml*