MECHANICAL PROPERTIES OF MATERIALS

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Ἀθήνη Αγγελίδης
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1 MECHANICAL PROPERTIES OF MATERIALS! Simple Tension Test! The Stress-Strain Diagram! Stress-Strain Behavior of Ductile and Brittle Materials! Hooke s Law! Strain Energy! Poisson s Ratio! The Shear Stress-Strain Diagram! Failure of Materials Due to Creep and Fatigue 1

2 Stress Strain Relationship d 0 P P L 0 σ (MPa) (mm/mm) Upper scale Lower scale 2

3 The Stress-Strain Diagram σ true fracture stress σ f σ u σ f σ Y pl proportional limit elastic limit yield stress ultimate stress fracture stress elastic region yielding strain hardening necking elastic behavior plastic behavior 3

4 Offset Yield Stress σ (MPa) σ Y (mm/mm) (0.2% offset) Offset yield strength for material with no yield points 4

5 σ P A 0 δ L 0 P σ (MPa) 400 σ u 390 MPa (σ y ) u 230 MPa σ fail 295 MPa d 0 L 0 (σ y ) l 220 MPa σ pl 200 MPa E 200x10 3 MPa 200 GPa (mm/mm) Upper scale P Lower scale 5

6 σ (MPa) 400 σ u 360 MPa σ fail 310 MPa σ y 250 MPa 150 σ pl 180 MPa (mm/mm) Upper scale Lower scale E x10 3 MPa 200 GPa 6

7 Stress-Strain Behavior of Ductile and Brittle Materials σ (MPa) Brittle material 400 Ductile material (mm/mm) σ diagrams for a methacrylate plastic 7

8 Elongation Percent Percent reduction elongation of area L A f 0 L 0 A 0 L A 0 f (100%) (100%) 8

9 Temperature Effects: σ (MPa) 10 o C o C Brittle to Ductile 70 o C Ductile to Brittle (mm/mm) σ diagrams for a methacrylate plastic 9

10 Hooke s Law σ σ pl σ E pl pl pl Constant 10

11 Elastic and Plastic Behavior of Materials - Apply load to failure σ EL 1 Failure PL 11

12 - Apply and release load σ EL 1 PL (a) Load is less than proportional limit σ 2 1 EL PL 2 (b) Load is more than proportional limit, but less than elastic limit 12

13 σ PL EL (c) Load is more than elastic limit, and reaply σ PL EL Mechanical hysteresis (d) Repeated load is more than elastic limit loading 13

14 - Comparison σ 1 O σ Apply load once PL EL 3 1 mechanical hysteresis 2 4 Repeated loading n times 14

15 elastic region σ plastic region σ elastic region plastic region A B A O load E A O E unload O mechanical hysteresis permanent set elastic recovery 15

16 Strain Energy Modulus of Resilience Modulus of Toughness σ σ σ pl u t u r pl Modulus of resilience u r Modulus of toughness u t u r 2 1 pl 1 σ pl σ pl 2 2 E 16

17 Modulus of Resilience σ (MPa) σ pl 180 MPa 100 Modulus of resiliency 50 0 (mm/mm) Upper scale Lower scale Modulus of resiliency (u r ) Area under proportional limit u r (1/2)(0.001)(180 MPa) 90 kpa 90 kn/m 2 90 kn m/m 3 90 kj/m 3 Energy per unit volume (90 kn/m 2 )(1 m 3 ) 90 kn m 90 kj 17

18 σ (MPa) EL Modulus of hyper-resiliency (mm/mm) Upper scale Lower scale 18

19 Modulus of Toughness σ (MPa) Failure Modulus of toughness (mm/mm) Upper scale Lower scale 19

20 Example 1 A tension test for a steel alloy results in the stress-strain diagram shown. Calculate the modulus of elasticity and the yield strength based on a 0.2% offset. Identify on the graph the proportional limit, elastic limit, ultimate stress and the fracture stress. σ (MPa) (mm/mm) Upper scale Lower scale 20

21 σ (MPa) Modulus of Elasticity 400 σ u 100 MPa E 200 GPa mm / mm σ fail 300 Yield Strength 250 σ 200 σ y 250 MPa σ y EL 150 Proportional Limit σ pl 100 E σ pl 180 MPa 50 0 (mm/mm) Upper scale Lower scale Elastic Limit σ pl 220 MPa Ultimate Stress Fracture Stress σ u 365 MPa σ f 310 MPa 21

22 Example 2 An aluminum specimen shown has a diameter of d 0 25 mm, a gauge length of L mm and is subjected to an axial load of kn. If a portion of the stress-strain diagram for the material is shown, determine the approximate elongation of the rod when the load is applied. If the load is removed, does the rod return to its original length? Also, compute the modulus of resilience both before and after the load application. σ (MPa) d kn kn L (mm/mm) 22

23 The Load is Applied d 0 25 mm kn kn σ (MPa) L mm 750 B F 600 A σ - Normal Stress P A N ( π / 4)(0.025 m) - The Strain mm/mm - The Elongation MPa δ L (0.023 mm/mm)(250 mm) 5.75 mm 150 O D (mm/mm) (5.75 mm)/2 (5.75 mm)/ kn kn L mm 23

24 The Load is Removed - Normal Stress d 0 25 mm kn kn σ P A ( π / 4)(0.025 N m) MPa σ pl σ (MPa) O A OC L mm B F E - Permanent Strain 450 MPa 75.0 GPa mm / mm CD mm/mm The permanent strain, OC CD 600 MPa CD OC mm/mm - The Permanent Elongation E E CD δ L (0.015 mm/mm)(250 mm) C D (mm/mm) 3.75 mm (3.75 mm)/2 (3.75 mm)/2 pl kn kn L mm 24

25 Modulus of Resilience - Normal Stress d 0 25 mm kn kn σ P A ( π / 4)(0.025 N m) MPa σ pl σ (MPa) L mm 750 B F 600 A (u r ) initial (u r ) final 150 CD C D (mm/mm) O pl ( u ( u - Modulus of Resilience r r ) ) initial final 1 σ pl pl 2 1 (450 MPa)(0.006 mm / mm) MJ / m 1 σ pl pl 2 1 (600 MPa)(0.008 mm / mm) MJ / m 25

26 Example 3 An aluminum rod shown has a circular cross section and is subjected to an axial load of 10 kn. If a portion of the stress-strain diagram for the material is shown, determine the approximate elongation of the rod when the load is applied. If the load is removed, does the rod return to its original length? Take E al 70 GPa. 20 mm 15 mm A B C 10 kn 10 kn σ (MPa) 600 mm 400 mm O (mm/mm) 26

27 The Load is Applied 20 mm 15 mm A B C 10 kn 10 kn 600 mm 400 mm σ (MPa) O BC mm/mm σ σ (mm/mm) P kn A 10 π (0.01 m) AB P kn A 10 π ( m) BC MPa MPa 6 AB Pa AB σ mm / mm Pa E al The elongation of the rod is δ Σ L AB L AB + BC L BC ( )(600 mm) + (0.045)(400 mm) 18.3 mm 27

28 The Load is Removed 20 mm 15 mm A B C 10 kn 10 kn 600 mm 400 mm 56.6 σ pl σ (MPa) O parallel G BC mm/mm OG σ σ (mm/mm) P kn A 10 π (0.01 m) AB P kn A 10 π ( m) BC BC Pa rec σ mm / mm Pa E al The permanent strain, OG mm/mm MPa MPa The elongation of the rod is δ Σ L 0 + OG L BC (400 mm) 17.7 mm 28

29 Poisson s Ratio δ/2 P L δ/2 Original Shape r Tension δ/2 L P Original Shape δ δ/2 Final Shape P δ δ ' long and lat L r ν lat long Final Shape r Compression δ P 29

30 P y t x L b P z x ν z y z Assumption: Homogeneous Isotropic Elastic 30

31 Example 4 A bar made of A-36 steel has the dimensions shown. If an axial force of P 80 kn is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically. Take E 200 GPa and ν st P 80 kn y 50 mm x P 80 kn z 1.5 m 100 mm 31

32 The change in the bar s length (δ) P 80 kn y 50 mm x P 80 kn z 1.5 m 100 mm - z direction P 80 kn σ z A (0.1 m)(0.05 m) (10 9 E st σ z z 6 16(10 ) Pa ) Pa 6 80(10 ) mm / mm 80µ z z MPa - x and y direction Poisson' s Ratio : ν st lat long x z y x y - ν st z -0.32[(80(10-6 )] µ δ x - x L x -[25.6(10-6 )(0.1 m) µm z δ z z L z [80(10-6 )(1.5 m)] 120 µm δ y - y L y -[25.6(10-6 )(0.05 m) µm 32

33 The Shear Stress-Strain Diagram y y τ xy γ xy /2 x γ xy /2 π/2 - γ xy x τ τ u τ f τ Gγ τ pl G γ pl γ u γ f γ G E 2(1 + ν ) 33

34 Example 5 A specimen of titanium alloy is tested in torsion and the shear stress-strain diagram is shown. (a) Determine the shear modulus G, the proportional limit, and the ultimate shear stress. (b ) Determine the distance d that the top of a block of this material, shown, could be displaced horizontally by a shear force V of 135 MN. τ (MPa) 75 mm mm d γ 50 mm V O γ (rad) 34

35 (a) The shear modulus, proportional limit, and the ultimate shear stress. τ (MPa) τ u 370 τ pl G O γ pl γ (rad) Proportional limit ; τ pl 270 MPa - Shear Modulus ; 270 MPa G GPa rad - Ultimate shear stress; τ u 370 MPa 35

36 (b) The maximum distance d and the magnitude of V if the material behaves elastically τ (MPa) τ avg 180 MPa 100 Oγ mm 100 mm 1.35 MN d γ γ mm τ avg γ (rad) Shear stress τ τ V avg A 1.35MN τ avg (0.1 m m) - Shear strain γ - The distance d 180 MPa τ avg γ G 180MPa γ (33.75GPa) tan( rad) rad d mm d 50 mm 36

37 Example 6 An aluminum specimen shown has a diameter of d 0 25 mm and a gauge length of L mm. If a force of 165 kn applies to the specimen shown, determine the diameter of the specimen. Take E 70 GPa,G al 26 GPa and σ Y 440 MPa. 165 kn d 0 L kn 37

38 The diameter of the specimen (d f ) if a force 165 kn applies 165 kn d f d 0 + δ d 0 + e lat d (1) d 0 25 mm d f L mm - Stress and Strain Relation ship P 165 kn σ A ( π / 4)(0.025 m) MPa Since σ < σ Y 440 MPa, the material behaves elastically. The modulus of elasticity is E al σ long 165 kn Pa long long 6 Pa 38

39 165 kn - G and E Relationship G E 2(1 + ν ) d f 26 GPa 70 GPa 2(1 +ν ) d 0 25 mm L mm ν Poisson s Ratio ν lat long 165 kn lat mm / mm From lat mm/mm δ' ( )(25 mm) mm eq.( 1) : d f d0 + latd0 25 mm + ( )(25 mm) mm 39

40 Failure of Materials due to Stress Relaxation, Creep, and Fatigue Stress Relaxation σ (MPa) σ τ diagram for stainless steel at 1200 o F and creep strain at 1% t (hrs.) 40

41 Creep 10-3 (µ) σ τ diagram, typical aluminum t (hrs.) 41

42 Fatigue Fatigue limit (endurance limit) σ (MPa) 400 (σ el ) st 210 (σ fs ) al (σ el ), Endurance limit 0.1 (σ fs ), Fatigue 500(10 6 ) cycles Structural steel aluminum S-N diagram for steel and aluminum alloys (N axis has a logarithmic scale) N (10 6 ) 42

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