Handout_5 - Classical Dynamics and Fluids P 97 E LASTICITY...

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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 l . 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 Hookes 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 l l where E is Youngs modulus. F F F F For the same extension the force area. For the same force the extension l 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 Hookes law: F A = E l l . This says that the stress (force/unit area) is proportional to the strain (fractional distortion). d l sd l Elastic medium stretched by l due to tension in the 1-direction (axes fixed in the medium). Response in other directions (2 , 3) is a compression a by l . 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 . We will see that the shear modulus G = E 2(1 + ) and the bulk modulus B = E 3(1- 2 ) . a This compression is in the definition of Poissons 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 ( 1 , 2 , 3 ) along the x,y,z-axes. These produce distortions l l ( e 1 ,e 2 ,e 3 ) . Example: tension 1 along the x-axis (i.e. 2 = 3 = 0 ) produces strains E ( e 1 ,e 2 ,e 3 ) = 1 (1 ,- ,- ) N.B. ( 1 , 2 , 3 ) and ( e 1 ,e 2 ,e 3 ) are NOT vectors. The medium responds in a similar way if there are stresses 2 and 3 . Since the strains are linear we have Ee 1 = 1- 2- 3 Ee 2 =- 1 + 2- 3 Ee 3 =- 1- 2 + 3 Thats 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 : 1 = 2 = 3 =- P . Applying the relations Ee 1 = 1- 2- 3 =- P (1- 2 ) etc....
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This note was uploaded on 05/09/2009 for the course DAMTP NST 1B Phy taught by Professor Sfgull during the Spring '07 term at Cambridge.

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Handout_5 - Classical Dynamics and Fluids P 97 E LASTICITY...

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