Course of “Building materials” MS in Building and Architectural Engineering Politecnico di Milano
Exercise n°1: Mechanical properties
1) The results of a tensile test of an aluminium alloy are shown in table 1. The initial length L0 was 50 mm and
original cross section A0 20 mm2. The final length (Lf) was 55.1 mm. Convert the load-elongation data in stress
strain curves and estimate: modulus of elasticity E (GPa); yield strength YS (MPa); tensile strength TS (MPa);
elongation to fracture (%).
Table 1
L (mm)
50.00 50.03 50.06 50.09 50.12 50.15 50.19 50.70 52.00 53.00 55.30
F
0.00 0.80 1.60 2.40 3.20 4.00 5.00 5.20 5.60 5.70 5.40
(kN)
2) In Table 2 mechanical properties of some engineering materials are shown. Compare the materials in term of
stiffness, strength, ductility, toughness, strength to weight and stiffness to weight ratio
Table 2
Density
Material E Yield strength Elongation to fracture (%)
(g/cm3)
(GPa) (MPa)
Quenched and tempered steel 7.85 207 600 20
Al-Zn alloy type 7075 2.7 70 503 11
Polypropilene (PP) 0.90 1.1 30 600
3) Impact test results for a microalloyed pipeline steel are reported in Table 3. Plot the data as impact energy versus
temperature and determine the ductile to brittle transition temperature corresponding to:
a) the average of the maximum and minimum impact energies
b) the temperature at which the impact energy is a required minimum value, i.e. 60 J
Table 3
Temperature (°C) -100°C -80°C -70°C -60°C -40°C -20°C 20°C
Impact energy (J) 20 40 100 150 200 200 200
4) Fracture mechanics allow to take into account the presence of cracks. In Linear Elastic Fracture Mechanics
(LEFM) stress intensity factor K is defined, K a (eq. 1)
where = stress, a = crack length, = geometrical factor.
According to LEFM, the crack can propagate and lead the material to fracture when K = Kc (eq.2). Fracture
toughness (Kc) measures the ability of a material to withstand an applied load in presence of cracks and is a
material property. Consider the following equation, derived from the previous (1 and 2):
a K C (eq. 3)
and answer the following questions (consider the materials in the table 4):
a) A wide plate is subjected to a 100 MPa tensile stress. Estimate the critical length of the crack ac (if the crack is
higher than this value is likely to grow at catastrophic rate).
b) If the maximum length of the crack present in the material is 1 mm, calculate the maximum (critical) stress
that material can withstand and compare with YS or TS.
Table 4
Material YS (MPa) TS (MPa) KIC (MPa·m½)
Steel for metallic construction 300 450 150
Al-Cu alloy 345 480 35
5) High strength steel bar is characterised by yield strength 1000 MPa, tensile strength 1200 MPa and fracture
toughness KC 50 MPa·m½.
a) estimate the critical crack length according to eq. 3 when the material is subjected to 70% of the yield
strength
b) if the maximum length of the crack present in the material is 1 mm, calculate the maximum (critical) stress
that material can withstand and compare with YS or TS.
, Results n.11
Aluminium alloy High strength steel
L (mm) F (kN) (MPa) L (mm) F (kN) (MPa)
50.00 0.00 0.00 0 50.00 0.00 0.00 0
50.03 0.80 0.06 40 50.04 2.40 0.08 165
50.06 1.60 0.12 80 50.11 6.60 0.22 455
50.09 2.40 0.18 120 50.26 15.60 0.52 1076
50.12 3.20 0.24 160 50.30 17.90 0.60 1234
50.15 4.00 0.30 200 50.42 19.30 0.84 1331
50.19 5.00 0.38 250 50.52 19.86 1.04 1370
50.70 5.20 1.4 260 50.68 20.20 1.4 1393
52.00 5.60 4 280 51.58 19.70 3.2 1359
53.00 5.70 6 285 51.84 18.90 3.7 1303
55.30 5.40 10.6 270 52.06 17.60 4.1 1214
52.32 15.70 4.6 1083
52.44 14.40 4.9 993
350 350
tensile strength = 285 MPa
300 300
Engineering stress (MPa)
Engineering stress (MPa)
250 250
200 200
1 = 0,18% Aluminium alloy
150 150
100 100
1 = 120 MPa
Aluminium alloy
50 50
0 0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 2 4 6 8 10 12
Engineering strain (%) Engineering strain (%)
(a) (b)
Figure 1 350
Aluminium alloy 300
Engineering stress (MPa)
yield strength 0,2% = 250 MPa
(a) Young’s modulus (E) 66.6 GPa 250
(b) tensile strength (TS) 285 MPa 200
(c) yield strength (YS) 250 MPa 150
100
Aluminium alloy
50
0
0 0.2 0.4 0.6 0.8 1
Engineering strain (%)
(c)
1
In the correction of the exercise also another material (high strength steel) is reported for comparison purposes
Course of “Building materials” (F. Bolzoni) Exercise n.1 pag. 2
, 1600 1600
tensile strength = 1393 MPa
1400 High strength
High strenght steelsteel 1400
Emgineering stress (MPa)
Engineering stress (MPa)
1200 1200
1000 1000
800 800
600 2 = 0,22% 600
High High
strength steel
strenght steel
400 400
= 455 MPa
200 200
0 0
0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 5
Engineering strain (%) Engineering strain (%)
(a) (b)
Figure 2 1600
High strength steel 1400
yield strength 0,2% = 1320 MPa
Emgineering stress (MPa)
a) Young’s modulus (E) 207 GPa 1200
1000
b) tensile strength (TS) 1393 MPa
800
c) yield strength (YS) 1320 MPa
600
400
High strenght steel
200
High strength steel
0
0 0.2 0.4 0.6 0.8 1
Engineering strain (%)
(c)
Results n.2
Density
Material E Yield Tensile Elongation to YS/ E/
(g/cm3)
(GPa) strength strength fracture (%)
(MPa) (MPa)
Quenched and tempered steel 7.85 207 600 700 20 44.2 26.4
Al-Zn alloy type 7075 2.7 70 503 572 11 186.3 25.9
Polypropilene (PP) 0.90 1.1 30 35 600 34.4 1.2
(*) for YS/and E/unit of measurement is not indicated
Stiffness (Young’s modulus E) Quenched and tempered steel > Al-Zn 7075 alloy > Polypropilene
Strength (yield) Quenched and tempered steel > Al-Zn alloy 7075 > Polypropilene
Ductility (elongation to fracture) Polypropilene > Quenched and tempered steel > Al-Zn alloy 7075
Toughness Is a measure of absorbed energy and correspond to the area under the curve. With the
available data only a rough estimation can be made, according to the values of YS, TS and E%.
Polypropilene (E% = 600, YS 30 MPa, TS 35 MPa) ≥ Quenched and tempered steel (E% = 20, YS 600
MPa, TS 700 MPa) > Aluminium alloy 7075 (E% = 11, YS 503 MPa, TS 572 MPa)
Strength to weight ratio (YS/) Al-Zn alloy 7075 > quenched and tempered steel > polypropilene
Stiffness to weight ratio (E/Quenched and tempered steel > Al-Zn alloy 7075 > polypropilene
Course of “Building materials” (F. Bolzoni) Exercise n.1 pag. 3
, Results n.3
In the figure below the curve energy vs temperature is shown. The ductile-brittle transition temperature,
defined in the two proposed ways, is about -70°C or -75°C. This evaluation is conventional, because the
ductile brittle transition really occur in a temperature range. The second criterium (minimum impact
energy) is easier to verify in acceptance tests.
250
200
Energy (J)
150
100
50
0
‐120 ‐100 ‐80 ‐60 ‐40 ‐20 0 20 40
Temperature (°C)
Figure 3
Results n.4 (results are also reported for a brittle material only for comparison purposes)
Question a. Critical value of the crack length ac can be evaluated by rearranging equation 3 of the text.
Steel, very tough material (KC 150 MPa·m1/2) shows a very high critical crack length (716 mm) when the
applied stress is 100 MPa; lower value (39 mm) is found for Al-Cu alloy. Critical crack length is very low
(0.29 mm) for a brittle material like alumina.
2
1K
ac IC
Question b. Critical stress (max) calculated by means of equation 3 for crack length 1 mm can be
compared with YS (or TS). For steel and Al-Zn alloy the critical stress calculated with fracture mechanics
is higher than yield and tensile strength, while for brittle material (alumina), here reported only for
comparison purposes, the critical stress is much lower than tensile strength.
Table 5
=100 MPa amax= 1 mm
½
Material YS (MPa) TS (MPa) KIC (MPa•m ) acritical (mm) max (MPa)
Mild steel 300 450 150 716 2676
Al-Cu alloy 345 480 35 39 624
Al2O3 (alumina) 210 3 0.29 54
Results n.5
Using the same formulas of the previous exercise, the results are presented in the Table 6. In presence of
only 1 mm crack, the material fail with applied stress 892 MPa, lower than the yield (89% of YS). On the
other hand, if a design stress 70% of the yield strength is considered, the critical crack length is only 1.6
mm long. These results, even if calculated by a simplified approach, highlight the possible critical
conditions arising when a material with low fracture toughness is used.
Course of “Building materials” (F. Bolzoni) Exercise n.1 pag. 4
Exercise n°1: Mechanical properties
1) The results of a tensile test of an aluminium alloy are shown in table 1. The initial length L0 was 50 mm and
original cross section A0 20 mm2. The final length (Lf) was 55.1 mm. Convert the load-elongation data in stress
strain curves and estimate: modulus of elasticity E (GPa); yield strength YS (MPa); tensile strength TS (MPa);
elongation to fracture (%).
Table 1
L (mm)
50.00 50.03 50.06 50.09 50.12 50.15 50.19 50.70 52.00 53.00 55.30
F
0.00 0.80 1.60 2.40 3.20 4.00 5.00 5.20 5.60 5.70 5.40
(kN)
2) In Table 2 mechanical properties of some engineering materials are shown. Compare the materials in term of
stiffness, strength, ductility, toughness, strength to weight and stiffness to weight ratio
Table 2
Density
Material E Yield strength Elongation to fracture (%)
(g/cm3)
(GPa) (MPa)
Quenched and tempered steel 7.85 207 600 20
Al-Zn alloy type 7075 2.7 70 503 11
Polypropilene (PP) 0.90 1.1 30 600
3) Impact test results for a microalloyed pipeline steel are reported in Table 3. Plot the data as impact energy versus
temperature and determine the ductile to brittle transition temperature corresponding to:
a) the average of the maximum and minimum impact energies
b) the temperature at which the impact energy is a required minimum value, i.e. 60 J
Table 3
Temperature (°C) -100°C -80°C -70°C -60°C -40°C -20°C 20°C
Impact energy (J) 20 40 100 150 200 200 200
4) Fracture mechanics allow to take into account the presence of cracks. In Linear Elastic Fracture Mechanics
(LEFM) stress intensity factor K is defined, K a (eq. 1)
where = stress, a = crack length, = geometrical factor.
According to LEFM, the crack can propagate and lead the material to fracture when K = Kc (eq.2). Fracture
toughness (Kc) measures the ability of a material to withstand an applied load in presence of cracks and is a
material property. Consider the following equation, derived from the previous (1 and 2):
a K C (eq. 3)
and answer the following questions (consider the materials in the table 4):
a) A wide plate is subjected to a 100 MPa tensile stress. Estimate the critical length of the crack ac (if the crack is
higher than this value is likely to grow at catastrophic rate).
b) If the maximum length of the crack present in the material is 1 mm, calculate the maximum (critical) stress
that material can withstand and compare with YS or TS.
Table 4
Material YS (MPa) TS (MPa) KIC (MPa·m½)
Steel for metallic construction 300 450 150
Al-Cu alloy 345 480 35
5) High strength steel bar is characterised by yield strength 1000 MPa, tensile strength 1200 MPa and fracture
toughness KC 50 MPa·m½.
a) estimate the critical crack length according to eq. 3 when the material is subjected to 70% of the yield
strength
b) if the maximum length of the crack present in the material is 1 mm, calculate the maximum (critical) stress
that material can withstand and compare with YS or TS.
, Results n.11
Aluminium alloy High strength steel
L (mm) F (kN) (MPa) L (mm) F (kN) (MPa)
50.00 0.00 0.00 0 50.00 0.00 0.00 0
50.03 0.80 0.06 40 50.04 2.40 0.08 165
50.06 1.60 0.12 80 50.11 6.60 0.22 455
50.09 2.40 0.18 120 50.26 15.60 0.52 1076
50.12 3.20 0.24 160 50.30 17.90 0.60 1234
50.15 4.00 0.30 200 50.42 19.30 0.84 1331
50.19 5.00 0.38 250 50.52 19.86 1.04 1370
50.70 5.20 1.4 260 50.68 20.20 1.4 1393
52.00 5.60 4 280 51.58 19.70 3.2 1359
53.00 5.70 6 285 51.84 18.90 3.7 1303
55.30 5.40 10.6 270 52.06 17.60 4.1 1214
52.32 15.70 4.6 1083
52.44 14.40 4.9 993
350 350
tensile strength = 285 MPa
300 300
Engineering stress (MPa)
Engineering stress (MPa)
250 250
200 200
1 = 0,18% Aluminium alloy
150 150
100 100
1 = 120 MPa
Aluminium alloy
50 50
0 0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 2 4 6 8 10 12
Engineering strain (%) Engineering strain (%)
(a) (b)
Figure 1 350
Aluminium alloy 300
Engineering stress (MPa)
yield strength 0,2% = 250 MPa
(a) Young’s modulus (E) 66.6 GPa 250
(b) tensile strength (TS) 285 MPa 200
(c) yield strength (YS) 250 MPa 150
100
Aluminium alloy
50
0
0 0.2 0.4 0.6 0.8 1
Engineering strain (%)
(c)
1
In the correction of the exercise also another material (high strength steel) is reported for comparison purposes
Course of “Building materials” (F. Bolzoni) Exercise n.1 pag. 2
, 1600 1600
tensile strength = 1393 MPa
1400 High strength
High strenght steelsteel 1400
Emgineering stress (MPa)
Engineering stress (MPa)
1200 1200
1000 1000
800 800
600 2 = 0,22% 600
High High
strength steel
strenght steel
400 400
= 455 MPa
200 200
0 0
0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 5
Engineering strain (%) Engineering strain (%)
(a) (b)
Figure 2 1600
High strength steel 1400
yield strength 0,2% = 1320 MPa
Emgineering stress (MPa)
a) Young’s modulus (E) 207 GPa 1200
1000
b) tensile strength (TS) 1393 MPa
800
c) yield strength (YS) 1320 MPa
600
400
High strenght steel
200
High strength steel
0
0 0.2 0.4 0.6 0.8 1
Engineering strain (%)
(c)
Results n.2
Density
Material E Yield Tensile Elongation to YS/ E/
(g/cm3)
(GPa) strength strength fracture (%)
(MPa) (MPa)
Quenched and tempered steel 7.85 207 600 700 20 44.2 26.4
Al-Zn alloy type 7075 2.7 70 503 572 11 186.3 25.9
Polypropilene (PP) 0.90 1.1 30 35 600 34.4 1.2
(*) for YS/and E/unit of measurement is not indicated
Stiffness (Young’s modulus E) Quenched and tempered steel > Al-Zn 7075 alloy > Polypropilene
Strength (yield) Quenched and tempered steel > Al-Zn alloy 7075 > Polypropilene
Ductility (elongation to fracture) Polypropilene > Quenched and tempered steel > Al-Zn alloy 7075
Toughness Is a measure of absorbed energy and correspond to the area under the curve. With the
available data only a rough estimation can be made, according to the values of YS, TS and E%.
Polypropilene (E% = 600, YS 30 MPa, TS 35 MPa) ≥ Quenched and tempered steel (E% = 20, YS 600
MPa, TS 700 MPa) > Aluminium alloy 7075 (E% = 11, YS 503 MPa, TS 572 MPa)
Strength to weight ratio (YS/) Al-Zn alloy 7075 > quenched and tempered steel > polypropilene
Stiffness to weight ratio (E/Quenched and tempered steel > Al-Zn alloy 7075 > polypropilene
Course of “Building materials” (F. Bolzoni) Exercise n.1 pag. 3
, Results n.3
In the figure below the curve energy vs temperature is shown. The ductile-brittle transition temperature,
defined in the two proposed ways, is about -70°C or -75°C. This evaluation is conventional, because the
ductile brittle transition really occur in a temperature range. The second criterium (minimum impact
energy) is easier to verify in acceptance tests.
250
200
Energy (J)
150
100
50
0
‐120 ‐100 ‐80 ‐60 ‐40 ‐20 0 20 40
Temperature (°C)
Figure 3
Results n.4 (results are also reported for a brittle material only for comparison purposes)
Question a. Critical value of the crack length ac can be evaluated by rearranging equation 3 of the text.
Steel, very tough material (KC 150 MPa·m1/2) shows a very high critical crack length (716 mm) when the
applied stress is 100 MPa; lower value (39 mm) is found for Al-Cu alloy. Critical crack length is very low
(0.29 mm) for a brittle material like alumina.
2
1K
ac IC
Question b. Critical stress (max) calculated by means of equation 3 for crack length 1 mm can be
compared with YS (or TS). For steel and Al-Zn alloy the critical stress calculated with fracture mechanics
is higher than yield and tensile strength, while for brittle material (alumina), here reported only for
comparison purposes, the critical stress is much lower than tensile strength.
Table 5
=100 MPa amax= 1 mm
½
Material YS (MPa) TS (MPa) KIC (MPa•m ) acritical (mm) max (MPa)
Mild steel 300 450 150 716 2676
Al-Cu alloy 345 480 35 39 624
Al2O3 (alumina) 210 3 0.29 54
Results n.5
Using the same formulas of the previous exercise, the results are presented in the Table 6. In presence of
only 1 mm crack, the material fail with applied stress 892 MPa, lower than the yield (89% of YS). On the
other hand, if a design stress 70% of the yield strength is considered, the critical crack length is only 1.6
mm long. These results, even if calculated by a simplified approach, highlight the possible critical
conditions arising when a material with low fracture toughness is used.
Course of “Building materials” (F. Bolzoni) Exercise n.1 pag. 4
