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Unformatted text preview: Classical Dynamics and Fluids P 97 E LASTICITY — F UNDAMENTALS d l F F l • Revision : An elastic wire of length l , cross-section A is subject to a stretching force F and extends by δl . d l F elastic region fracture • In the elastic region the wire returns to its original length when the force is removed. • Stretching beyond this point causes plastic deformation and fracture. • In the linear elastic region the material obeys Hooke’s Law. (1678 a ) ‘Ut tensio, sic vis’ , which means ‘as the extension, so the force’, i.e. the extension is proportional to the force . • Definition : for small elastic displacements F = EA δl l where E is Young’s modulus. F F F F • For the same extension the force ∝ area. • For the same force the extension δl ∝ l . a Hooke actually published it in 1676 as ‘ceiiinosssttuv’ Classical Dynamics and Fluids P 98 Y OUNG ’ S M ODULUS AND P OISSON ’ S R ATIO • Hooke’s law: F A = E δl l . • This says that the stress (force/unit area) is proportional to the strain (fractional distortion). d l sd l • Elastic medium stretched by δl due to tension in the 1-direction (axes fixed in the medium). • Response in other directions (2 , 3) is a compression a by σδl . • σ is the Poisson ratio . • For an isotropic solid (i.e. one with no preferred directions) the elastic properties are completely determined by the Young modulus E and the Poisson ratio σ . • We will see that the shear modulus G = E 2(1 + σ ) and the bulk modulus B = E 3(1- 2 σ ) . a This “compression” is in the definition of Poisson’s ratio — some materials actually expand (i.e. their Poisson ratio is negative.) Classical Dynamics and Fluids P 99 S TRESS AND S TRAIN • For an linear elastic medium STRESS ∝ STRAIN (force/unit area trying (fractional distortions to distort the medium) of the medium) t t t t 1 2 t 3 2 1 • For a unit cube of material apply tensions ( τ 1 ,τ 2 ,τ 3 ) along the x,y,z-axes. These produce distortions δl l ≡ ( e 1 ,e 2 ,e 3 ) . • Example: tension τ 1 along the x-axis (i.e. τ 2 = τ 3 = 0 ) produces strains E ( e 1 ,e 2 ,e 3 ) = τ 1 (1 ,- σ,- σ ) • N.B. ( τ 1 ,τ 2 ,τ 3 ) and ( e 1 ,e 2 ,e 3 ) are NOT vectors. • The medium responds in a similar way if there are stresses τ 2 and τ 3 . Since the strains are linear we have Ee 1 = τ 1- στ 2- στ 3 Ee 2 =- στ 1 + τ 2- στ 3 Ee 3 =- στ 1- στ 2 + τ 3 • That’s all there is to elasticity in isotropic media. . . Classical Dynamics and Fluids P 100 B ULK M ODULUS P P + • Consider a medium under isotropic pressure P : τ 1 = τ 2 = τ 3 =- P . • Applying the relations Ee 1 = τ 1- στ 2- στ 3 =- P (1- 2 σ ) etc....
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- Spring '07
- Force, classical dynamics, Classical Dynamics and Fluids