STANDARD SAMPLE. Reduced section " Diameter. Diameter. 2" Gauge length. Radius

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Transcription:

MATERIAL PROPERTIES

TENSILE MEASUREMENT F l l 0 A 0 F

STANDARD SAMPLE Reduced section 2 " 1 4 0.505" Diameter 3 4 " Diameter 2" Gauge length 3 8 " Radius

TYPICAL APPARATUS Load cell Extensometer Specimen Moving crosshead

ENGINEERING STRESS F F A 0 l l 0 A 0 F

COMPRESSIVE LOAD F A 0 l 0 l F A 0 F ( b) T

SHEAR AND TORSIONAL A 0 F T F θ T F (c)

ENGINEERING STRESS F A 0

ENGINEERING STRAIN l i l 0 l 0 l l 0

HOOKES LAW Unload Stress Slope = modulus of elasticity Load 0 0 Strain

2 = Tangent modulus (at 2 ) Stress 1 = Secant modulus (between origin and 1 ) Strain

STRESS Force/Area

STRAIN Elongation

HOOKE S LAW E

Young s Modulus of Copper 110 x 10 3 MPa EXAMPLE A piece of copper originally 305 mm long is pulled in tension with a stress of 276 MPa. If the deformation is entirely elastic, what will be the resultant elongation?

EXAMPLE 2 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa and an original diameter of 3.8 mm. will experience only elastic deformation when a tensile load of 2000 N is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.42 mm 250 mm

WHAT ARE WE ACTUALLY DOING HERE + Attractive force F A Attraction Unload Force F Repulsion 0 r 0 Repulsive force F R Net force F N Interatomic separation r Stress Slope = modulus of elasticity + (a) Repulsive energy E R Load Net energy E N Interatomic separation r 0 E 0 0 Strain Attractive energy E A We can see that the gradient is similar for a small region above and below the zero force (b) position (equilibrium). Potential energy E Attraction Repulsion 0

0.04 0.03 0.02 0.01 0-0.01 Force (N), Potential (J) -0.02-0.03-0.04-0.05-0.06 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 Distance (arb)

ON AN ATOMIC LEVEL

Strongly bonded df dr r 0 Force F 0 Separation r Weakly bonded

U 0 r r 0 U 0 r r 0 U 0 ~ 20 kj/mol U 0 ~ 400 kj/mol 12 10 Slope = E 1.5 0.4 8 6 4 2 1.0 0.5 0.3 0.2 0.1 Slope = E 40 20 s 109 N m 2 0 2 4 6 8 r r 0 0 0.5 1.0 s (10 6 psi) s 109 N m 2 0 0.1 0.2 0.3 r r 0 0 20 40 (10 3 psi) 10 12 Figure 5.3 1.5 0.4 0 1 2 3 0 1 2 3 4 (b) (a) Schematic diagrams showing potential energy (top) andappliedstress(bottom)

Temperature ( F) 400 0 400 800 1200 1600 70 Modulus of elasticity (GPa) 400 300 200 100 Steel Aluminum Tungsten 60 50 40 30 20 10 Modulus of elasticity (10 6 psi) 0 0 200 0 200 400 600 800 Temperature ( C)

Poisson s Ratio for Various Materials Modulus of Elasticity Shear Modulus Poisson s Material GPa 10 6 psi GPa 10 6 psi Ratio Metal Alloys Tungsten 407 59 160 23.2 0.28 Steel 207 30 83 12.0 0.30 Nickel 207 30 76 11.0 0.31 Titanium 107 15.5 45 6.5 0.34 Copper 110 16 46 6.7 0.34 Brass 97 14 37 5.4 0.34 Aluminum 69 10 25 3.6 0.33 Magnesium 45 6.5 17 2.5 0.35 Ceramic Materials

POISSONS RATIO ν = d/d 0 L/L 0 σ t L = L L 0 d d L 0 L s t

MATERIAL PROPERTIES Poisson s Ratio for Various Materials Modulus of Elasticity Shear Modulus Poisson s Material GPa 10 6 psi GPa 10 6 psi Ratio Metal Alloys Tungsten 407 59 160 23.2 0.28 Steel 207 30 83 12.0 0.30 Nickel 207 30 76 11.0 0.31 Titanium 107 15.5 45 6.5 0.34 Copper 110 16 46 6.7 0.34 Brass 97 14 37 5.4 0.34 Aluminum 69 10 25 3.6 0.33 Magnesium 45 6.5 17 2.5 0.35 Ceramic Materials

NEGATIVE POISSON RATI0

NEGATIVE POISSON RATI0

Elastic Plastic y Stress P 0.002 Strain

Upper yield point Stress y Lower yield point Strain