Homework 6 Solutions 1. The components of the uniaxial tension test system are shown in Figure 1. The load (force) applied to the specimen was measured by a loadcell. A digital extensometer clamped onto the specimen gauge section measured the strain. The loadcell and the extensometer were connected to a computer to record their values every 0.5 sec. (4 POINTS)
Loading direction
Loadcell
x2
0
Specimen Extensometer x1 Clamps
Fixed End Figure 1. Setup used to conduct uniaxial tension measurements of stress and strain.
2. From the experimental data recorded in the lab, the stress in the specimens was computed by using the relation (6 POINTS)
Stress
Force Area
; EngineeringStress
P
2 initial 4 d
Example engineering stress vs. strain curves for the cold rolled steel (1045 CR) and the aluminum alloy (6061-T6) specimens are shown in Figures 2(a-d).
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1200
1200
1000
1000
) a800 P M ( s 600 s e r t S 400
) a800 P M ( 600 s s e r t S400
200
200
0
0 0
3
6
9
12
15
Strain (%)
0
0.4 0.6 Strain (%)
(a)
(b)
350
350
300
300
) 250 a P M200 ( s s 150 e r t S 100
) 250 a P 200 M ( s s150 e r t S 100
50
50
0
0.2
0.8
1
0 0
5
10
15
20
0
0.5
Strain (%)
Strain (%)
(c)
(d)
1
Figure 2. Engineering stress vs. strain for: (a-b) Cold rolled Steel (1045-CR), (c-d) Al 6061-T6.
3. The stress and strain computed using the experimental measurements correspond to
and
σ11
,
ε11
respectively, since the load and strain measured by the loadcell and the extensometer were along the x1 direction. (2 POINTS) 4. The elastic moduli of the steel and Al alloy were calculated from the slope of the initially straight segment of the stress vs. strain curves. Note that in Figure 3, the strain was plotted as percent strain. Hence, the elastic modulus was obtained by multiplying slope of this curve with 100. You can apply linear regression directly in MS Excel or Matlab to find the slope of the curves in Figures 3(a,b).
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The elastic modulus of cold rolled steel is E steel = 2077.1×100 MPa = 208 GPa. Similarly, the modulus of aluminum 6061-T6 is found to be E Al = 65.49 GPa. The expected values for the elastic moduli for cold rolled steel and aluminum alloy are 201 GPa, and 73.1 GPa, respectively. The difference between the experimentally measured and expected values could be due to different material composition or errors in strain measurement since the extensometer could slide during specimen loading. (6 POINTS)
350
1200
y = 2077.1x + 7.9437 1000 ) a 800 P M ( 600 s s e r t S 400
) 250 a P M200 ( s s 150 e r t S 100
200
50
0
0 0
0.2
0.4
0.6
0.8
y = 654.92x + 4.356
300
1
0
0.2
Strain (%)
0.4
0.6
0.8
1
Strain (%)
(a)
(b)
Figure 3. Section of the stress vs. strain (below 1% strain) curves for (a) cold rolled steel, (b) Al 6061-T6
to calculate the Young’s moduli.
5. Proportional limit or elastic limit is the highest stress at which the stress in the specimen is still linearly proportional to the strain. The proportional limits of cold rolled steel and aluminum alloy were 420 MPa and 220 MPa, respectively. (see Figure 4) (2 POINTS) Yield stress is the stress at which the material has fully begun to deforming irreversibly (plastically).
We usually report the 0.2% yield stress, which is the stress at which a line drawn passing through the 0.2% strain point and parallel to the linear part of the stress vs. strain response intersects the stress vs. strain curve (Figure 4). Hence, the yield stresses of cold rolled steel and the Al alloy are 800 MPa and 245 MPa, respectively. The expected values of the yield stress for cold rolled steel and Al 6061-T6 are 530 MPa and 324 MPa, respectively. The experimentally measured values could differ because of material processing and composition. Therefore, we should never use literature quoted values when
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we use materials for performance critical or safety sensitive applications, instead, we should take our own precise measurements with the materials at hand. (2 POINTS)
Ultimate strength is the maximum stress that a material can carry before failure. For the particular
materials, changes in the cross-section are disregarded when calculating ultimate strength. Using Figure 2 we find that the ultimate strength cold rolled steel and Al 6061-T6 are 998 MPa and 291 MPa respectively. (2 POINTS)
Failure stress is the stress in the material at the time of failure. Changes in the cross section are
disregarded in calculating failure strength. From Figure 2, we see that the failure stress for cold rolled steel and Al 6061-T6 are 771 MPa and 245 MPa, respectively. (2 POINTS)
1200
350 y = 2077.1x + 7.9437
1000
300
) 800 a P M ( 600 s s e r t S 400
) 250 a P M200 ( s s 150 e r t S 100
Proportional limit Yield Point
200
y = 654.92x + 4.356
50
0
0 0
0.2
0.4
0.6
0.8
1
0
Strain (%)
0.2
0.4
0.6
0.8
1
Strain (%)
(a)
(b)
Figure 4. Section of the stress vs. strain curves (below 1% strain) showing the elastic limit and yield
stress for (a) cold rolled steel, and (b) Al 6061-T6.
6. The maximum shear stress in the specimen at yield is calculated using the Mohr’s circle shown in Figure 5. For cold rolled steel, direction, σ22 and σ12 are 0. T =
σ11
at yield point was 800 MPa. Since load was applied in the x 1
=800 MPa. (4 POINTS)
σy
From the Mohr’s circle, we see that the maximum shear stress is T/2 = 400 MPa for cold rolled steel. Similarly, for Al 6061-T6, the maximum shear stress was 123 MPa.
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Figure 5. Calculation of maximum shear stress in a uniaxially loaded specimen using Mohr’s
circle.
7. The true normal stress is the maximum normal stress that a material experiences just before failure. The true cross section is considered (just before failure) to calculate the failure stress. For cold rolled steel the load and the diameter of the neck at failure were 30.01 kN and 5.93 mm. The true failure stress was 30.01×10 3/(π×5.932/4) MPa = 1,087 MPa. Similarly, the true failure stress of Al 6061-T6 was 371 MPa. (2 POINTS) The true shear stress is half of the true normal stress as we have seen using Mohr’s circle in Figure
5. So, the true shear stress at failure for cold rolled steel and Al 6061-T6 was 544 MPa and 186 MPa, respectively. (2 POINTS)
8. The fracture cross sections of the Al alloy exhibited minor cup and cone shapes with the material undergoing limited necking. This tells us that the Al alloy was not very ductile. On the contrary, the fracture cross section of the steel specimen was that of a cup and cone indicating decent ductility for the steel specimen. The cup and cone fracture surfaces indicate that the material clearly failed due to shear stresses. (2 POINTS)
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Additional Notes: Sometimes
it is not possible to obtain good cup and cone fracture surface due to
the presence of defects or micro cracks in the specimen can cause premature failure, not good initial specimen alignment in the test setup, and loading rate; when specimens are loaded fast they fail as brittle materials and may not have cup and cone fracture surfaces.
9. Specific elastic modulus is the ratio between elastic modulus of a material and its density. The specific yield stress and specific maximum stress are defined similarly. The densities of Al and steel are 2.7 g/cm3 and 7.8 g/cm 3, respectively. The specific elastic modulus, specific yield and maximum stresses are tabulated in the Table 1. (2 POINTS)
Table 1. Specific mechanical properties of steel and aluminum alloys Material
Cold rolled steel (1045-CR) Aluminum 6061-T6
Specific elastic
Specific yield stress
2 -2
Specific maximum
2 -2
2 -2
modulus (m s )
(m s )
stress (m s )
26.6×106
0.10×106
0.13×106
24.3×10 6
0.09×106
0.11×106
Steel has marginally higher specific properties but Al is much lighter.
10. The material with higher specific yield stress is chosen as an appropriate for aircraft components to design for reliable flight. From Table 1, we see that the specific moduli for Al and steel are comparable. Considering the density of both materials, it would be much more appropriate to use Al to design for energy efficiency. Hence, the Al alloy is recommended for airplane components. We do not consider the specific strength as much because we do not want to design close to it. Instead, we design based on the value of the yield strength at which the material is still “one piece”. To be safe we also apply safety margins that reduce even further the stress envelop we can use to design and we design for a maximum stress that is much less than the yield stress. (2 POINTS) Materials with high specific elastic modulus (stiffness) are preferred in design aircraft because they provide better flight stability. This ensures that aircraft components such as the wings are stiff and maintain critical design parameters. From Table 1, we see that the specific elastic moduli of the Al and the steel alloys are comparable. It would be way more appropriate to use the Al alloy to design for energy efficiency since Al is much lighter. Hence, Al 6061-T6 is the choice between the two materials for airplane structures. (2 POINTS)
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Finally, criteria (a) and (b) are the most important for reliable flight.
11. The mechanical properties of Al 6061-T6, cold rolled steel ar e summarized below: (2 POINTS) 12. Property
Cold Rolled Steel
Aluminum Alloy
(1045-CR)
(6061-T6)
Elastic modulus (GPa)
208
65.5
Expected elastic modulus (GPa)
201
73
Proportional limit (MPa)
420
220
Yield stress (MPa)
800
245
Shear stress at yield (MPa)
400
123
Expected yield stress (MPa)
530
324
Ultimate stress (MPa)
998
291
Failure stress (MPa)
771
245
True normal stress (MPa)
1087
371
True shear stress (MPa)
544
186
Specific elastic modulus (m 2s-2)
26.6 × 106
24.3 × 106
Specific yield stress (m 2s-2)
0.10 × 106
0.09 × 106
Specific maximum stress (m 2s-2)
0.13 × 106
0.11 × 106
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