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
Stress Strain Relationship d 0 P P L 0 σ (MPa) 400 350 300 250 200 150 100 50 0 (mm/mm) 0.00 0.10 0.20 0.30 0.40 Upper scale 0.0000 0.0010 0.0020 0.0030 0.0040 Lower scale 2
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
Offset Yield Stress σ (MPa) 300 250 200 150 σ Y 100 50 0 (mm/mm) 0.005 0.010 0.002 (0.2% offset) Offset yield strength for material with no yield points 4
σ P A 0 δ L 0 P σ (MPa) 400 σ u 390 MPa (σ y ) u 230 MPa 350 300 250 σ fail 295 MPa d 0 L 0 (σ y ) l 220 MPa 200 150 σ pl 200 MPa 100 200 E 200x10 3 MPa 200 GPa 50 0.001 0 (mm/mm) 0.00 0.10 0.20 0.30 0.40 Upper scale P 0.0000 0.0010 0.0020 0.0030 0.0040 Lower scale 5
σ (MPa) 400 σ u 360 MPa 350 300 σ fail 310 MPa 250 200 σ y 250 MPa 150 σ pl 180 MPa 100 50 0 (mm/mm) 0.00 0.10 0.20 0.30 0.40 Upper scale 0.0000 0.0010 0.0020 0.0030 0.0040 Lower scale E 100 0.0005 200x10 3 MPa 200 GPa 6
Stress-Strain Behavior of Ductile and Brittle Materials σ (MPa) Brittle material 400 Ductile material 300 200 100 0.02 0.04 0.06 (mm/mm) σ diagrams for a methacrylate plastic 7
Elongation Percent Percent reduction elongation of area L A f 0 L 0 A 0 L A 0 f (100%) (100%) 8
Temperature Effects: σ (MPa) 10 o C 400 300 200 100 40 o C Brittle to Ductile 70 o C Ductile to Brittle 0.02 0.04 0.06 (mm/mm) σ diagrams for a methacrylate plastic 9
Hooke s Law σ σ pl σ E pl pl pl Constant 10
Elastic and Plastic Behavior of Materials - Apply load to failure σ EL 1 Failure PL 11
- 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
σ PL EL 1 3 2 (c) Load is more than elastic limit, and reaply σ PL EL 3 1 2 4 Mechanical hysteresis (d) Repeated load is more than elastic limit loading 13
- Comparison σ 1 O σ Apply load once PL EL 3 1 mechanical hysteresis 2 4 Repeated loading n times 14
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
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
Modulus of Resilience σ (MPa) 400 350 300 250 200 150 σ pl 180 MPa 100 Modulus of resiliency 50 0 (mm/mm) 0.00 0.10 0.20 0.30 0.40 Upper scale 0.0000 0.0010 0.0020 0.0030 0.0040 Lower scale Modulus of resiliency (u r ) Area under curve @ 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
σ (MPa) 400 350 300 250 200 EL 1 3 150 100 Modulus of hyper-resiliency 50 0 2 (mm/mm) 0.00 0.10 0.20 0.30 0.40 Upper scale 0.0000 0.0010 0.0020 0.0030 0.0040 Lower scale 18
Modulus of Toughness σ (MPa) 400 350 300 250 200 150 Failure Modulus of toughness 100 50 0 (mm/mm) 0.00 0.10 0.20 0.30 0.40 Upper scale 0.0000 0.0010 0.0020 0.0030 0.0040 Lower scale 19
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) 400 350 300 250 200 150 100 50 0 (mm/mm) 0.00 0.10 0.20 0.30 0.40 Upper scale 0.0000 0.0010 0.0020 0.0030 0.0040 Lower scale 20
σ (MPa) Modulus of Elasticity 400 σ u 100 MPa E 200 GPa 350 0.0005 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) 0.00 0.10 0.20 0.30 0.40 Upper scale 0.0000 0.0010 0.0020 0.0030 0.0040 Lower scale Elastic Limit σ pl 220 MPa Ultimate Stress Fracture Stress σ u 365 MPa σ f 310 MPa 21
Example 2 An aluminum specimen shown has a diameter of d 0 25 mm, a gauge length of L 0 250 mm and is subjected to an axial load of 294.5 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) 750 600 450 d 0 294.5 kn 294.5 kn L 0 300 150 0.01 0.02 0.03 0.04 (mm/mm) 22
The Load is Applied d 0 25 mm 294.5 kn 294.5 kn σ (MPa) L 0 250 mm 750 B F 600 A 450 300 σ - Normal Stress P A 0 3 294.5 10 N ( π / 4)(0.025 m) - The Strain 0.023 mm/mm - The Elongation 2 600 MPa δ L (0.023 mm/mm)(250 mm) 5.75 mm 150 O D (mm/mm) (5.75 mm)/2 (5.75 mm)/2 0.01 0.02 0.03 0.04 0.023 294.5 kn 294.5 kn L 0 250 mm 23
The Load is Removed - Normal Stress d 0 25 mm 294.5 kn 294.5 kn σ P A 0 3 294.5 10 ( π / 4)(0.025 N m) 2 600 MPa σ pl 750 600 450 300 150 σ (MPa) O A OC L 0 250 mm B F E - Permanent Strain 450 MPa 75.0 GPa 0.006 mm / mm CD 0.008 mm/mm The permanent strain, OC 0.023 - CD 600 MPa CD OC 0.023-0.008 0.015 mm/mm - The Permanent Elongation E E CD δ L (0.015 mm/mm)(250 mm) C D (mm/mm) 3.75 mm 0.01 0.02 0.03 0.04 (3.75 mm)/2 (3.75 mm)/2 pl 0.006 0.023 294.5 kn 294.5 kn L 0 250 mm 24
Modulus of Resilience - Normal Stress d 0 25 mm 294.5 kn 294.5 kn σ P A 0 3 294.5 10 ( π / 4)(0.025 N m) 2 600 MPa σ pl σ (MPa) L 0 250 mm 750 B F 600 A 450 300 (u r ) initial (u r ) final 150 CD 0.008 C D (mm/mm) O 0.01 0.02 0.03 0.04 pl 0.006 0.023 ( u ( u - Modulus of Resilience r r ) ) initial final 1 σ pl pl 2 1 (450 MPa)(0.006 mm / mm) 2 3 1.35 MJ / m 1 σ pl pl 2 1 (600 MPa)(0.008 mm / mm) 2 3 2.40 MJ / m 25
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 60 50 40 30 20 10 O 0.02 0.04 0.06 0.08 0.10 0.12 (mm/mm) 26
The Load is Applied 20 mm 15 mm A B C 10 kn 10 kn 600 mm 400 mm 56.6 31.83 σ (MPa) 60 50 40 30 20 10 O BC 0.045 mm/mm 0.02 0.04 0.06 0.08 0.10 0.12 σ σ (mm/mm) P kn A 10 π (0.01 m) AB 31. 83 2 P kn A 10 π (0.0075 m) BC 56. 6 2 MPa MPa 6 AB 31.83 10 Pa AB σ 0.0004547 mm / mm 9 70 10 Pa E al The elongation of the rod is δ Σ L AB L AB + BC L BC (0.0004547)(600 mm) + (0.045)(400 mm) 18.3 mm 27
The Load is Removed 20 mm 15 mm A B C 10 kn 10 kn 600 mm 400 mm 56.6 σ pl 31.83 σ (MPa) 60 50 40 30 20 10 O parallel G BC 0.045 mm/mm 0.02 0.04 0.06 0.08 0.10 0.12 OG σ σ (mm/mm) P kn A 10 π (0.01 m) AB 31. 83 2 P kn A 10 π (0.0075 m) BC 56. 6 2 6 BC 56.6 10 Pa rec σ 0.000808 mm / mm 9 70 10 Pa E al The permanent strain, OG 0.0450-0.000808 0.0442 mm/mm MPa MPa The elongation of the rod is δ Σ L 0 + OG L BC 0.0442(400 mm) 17.7 mm 28
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
P y t x L b P z x ν z y z Assumption: Homogeneous Isotropic Elastic 30
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 0.32. P 80 kn y 50 mm x P 80 kn z 1.5 m 100 mm 31
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) 16 200(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 )] -25.6 µ δ x - x L x -[25.6(10-6 )(0.1 m) -2.56 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) -1.28 µm 32
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
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 400 300 200 100 100 mm d γ 50 mm V O 0.008 0.54 0.73 γ (rad) 34
(a) The shear modulus, proportional limit, and the ultimate shear stress. τ (MPa) 400 300 τ u 370 τ pl 270 200 100 G O γ pl 0.008 γ (rad) 0.73 - Proportional limit ; τ pl 270 MPa - Shear Modulus ; 270 MPa G 33. 75 GPa 0.008 rad - Ultimate shear stress; τ u 370 MPa 35
(b) The maximum distance d and the magnitude of V if the material behaves elastically τ (MPa) 400 300 200 τ avg 180 MPa 100 Oγ 180 0.00533 75 mm 100 mm 1.35 MN d γ γ 180 0.00533 50 mm τ avg γ (rad) 0.73 - Shear stress τ τ V avg A 1.35MN τ avg (0.1 m 0.075 m) - Shear strain γ - The distance d 180 MPa τ avg γ G 180MPa γ (33.75GPa) 0.005333 tan( 0.00533 rad) 0.00533 rad d 0.267 mm d 50 mm 36
Example 6 An aluminum specimen shown has a diameter of d 0 25 mm and a gauge length of L 0 250 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 0 165 kn 37
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 0 ----------(1) d 0 25 mm d f L 0 250 mm - Stress and Strain Relation ship P 165 kn σ A ( π / 4)(0.025 m) 2 336.1 MPa Since σ < σ Y 440 MPa, the material behaves elastically. The modulus of elasticity is E al σ long 165 kn 9 336.1 10 70.0 10 Pa long 0.0048 long 6 Pa 38
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 0 250 mm ν 0.346 - Poisson s Ratio ν lat long 165 kn 0.346 lat 0.00480 mm / mm From lat -0.00166 mm/mm δ' (0.00166)(25 mm) 0.0415 mm eq.( 1) : d f d0 + latd0 25 mm + ( 0.00166)(25 mm) 24. 96 mm 39
Failure of Materials due to Stress Relaxation, Creep, and Fatigue Stress Relaxation σ (MPa) 200 150 100 50 200 400 600 800 1000 σ τ diagram for stainless steel at 1200 o F and creep strain at 1% t (hrs.) 40
Creep 10-3 (µ) 2.0 1.5 1.0 0.5 200 400 600 800 1000 σ τ diagram, typical aluminum t (hrs.) 41
Fatigue Fatigue limit (endurance limit) σ (MPa) 400 (σ el ) st 210 (σ fs ) al 130 300 200 100 (σ el ), Endurance limit 0.1 (σ fs ), Fatigue strength @ 500(10 6 ) cycles 1 10 100 1000 Structural steel aluminum S-N diagram for steel and aluminum alloys (N axis has a logarithmic scale) N (10 6 ) 42