MECHANICS OF MATERIALS Poisson s Ratio

Transcription

1 Poisson s Raio For a slender bar subjeced o axial loading: ε x x y 0 The elongaion in he x-direcion i is accompanied by a conracion in he oher direcions. Assuming ha he maerial is isoropic (no direcional dependence), ε y ε 0 Poisson s raio is defined as laeral srain ε y ε ν axial srain ε ε x x 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-1

2 Generalied Hooke s Law For an elemen subjeced o muli-axial loading, he normal srain componens resuling from he sress componens may be deermined from he principle of superposiion. This requires: 1) srain is linearly relaed o sress 2) deformaions are small Wih hese resricions: ν x y ν ε x + ν x y ν ε y + ν ν x y ε The McGraw-Hill Companies, Inc. All righs reserved. 2-2

3 Dilaaion: Bulk Modulus Relaive o he unsressed sae, he change in volume is e 1 ( 1+ ε )( 1+ ε )( 1+ ε ) 1 1+ ε + ε + ε [ ] [ ] x y x y ε x + ε y + ε 1 2ν ( + + ) x y dilaaion (change in volume per uni volume) For elemen subjeced o uniform hydrosaic pressure, ( 1 2ν ) p e p k 1 ( 2ν ) k bulk modulus Subjeced o uniform pressure, dilaaion mus be negaive, herefore 0 <ν < The McGraw-Hill Companies, Inc. All righs reserved.

4 Shearing Srain A cubic elemen subjeced o a shear sress will deform ino a rhomboid. The corresponding shear srain is quanified in erms of he change in angle beween he sides, τ xy ( ) f γ xy A plo of shear sress vs. shear srain is similar il o he previous plos of normal sress vs. normal srain excep ha he srengh values are approximaely half. For small srains, τ Gγ τ Gγ τ Gγ xy xy y y where G is he modulus of rigidiy or shear modulus. x x 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-4

5 xample 2. A recangular block of maerial wih modulus of rigidiy G 90 ksi is bonded o wo rigid horional plaes. The lower plae is fixed, while he upper plae is subjeced o a horional force P. Knowing ha he upper plae moves hrough 0.04 in. under he acion of he force, deermine a) he average shearing srain in he maerial, and b) he force P exered on he plae. SOLUTION: Deermine he average angular deformaion or shearing srain of he block. Apply Hooke s law for shearing sress and srain o find he corresponding shearing sress. Use he definiion of shearing sress o find he force P The McGraw-Hill Companies, Inc. All righs reserved. 2-5

6 Deermine he average angular deformaion or shearing srain of he block. γ xy 0.04in. an γ xy γ xy rad 2in. Apply Hooke s law for shearing sress and srain o find he corresponding shearing sress. τ xy Gγ xy ( 90 psi)( rad) 1800psi Use he definiion of shearing sress o find he force P. P τ xy A ( 1800 psi )( 8in. )( 2.5in. ) 6 lb xy P 6.0kips 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-6

7 Relaion Among, ν, and G An axially loaded slender bar will elongae in he axial direcion and conrac in he ransverse direcions. An iniially cubic elemen oriened as in op figure will deform ino a recangular parallelepiped. The axial load produces a normal srain. If he cubic elemen is oriened as in he boom figure, i will deform ino a rhombus. Axial load also resuls in a shear srain. Componens of normal and shear srain are relaed, 2G ( 1+ν ) 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-7

8 Sample Problem 2.5 A circle of diameer d 9 in. is scribed on an unsressed aluminum plae of hickness /4 in. Forces acing in he plane of he plae laer cause normal sresses x 12 ksi and 20 ksi. For x 6 psi and ν 1/, deermine he change in: a) he lengh of diameer AB, b) he lengh of diameer CD, c) he hickness of he plae, and d) he volume of he plae The McGraw-Hill Companies, Inc. All righs reserved. 2-8

9 SOLUTION: Apply he generalied Hooke s Law o valuae he deformaion componens. find he hree componens of normal δ d ( in./in. )( 9in. ) B A ε srain. x ε ε ε x y + x 1 ν y ν 1 ( 12ksi) 0 ( 20ksi) 6 psi ν x + y ν x in./in. ν in./in. ν y + in./in. δ C ε d D δ ε y δ B A in. ( in./in. )( 9in. ) δ C D ( in./in. )( 0.75in. ) δ Find he change in volume e ε V x + ε ev y + ε V in /in in. in. ( ) in in 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-9

10 Composie Maerials Fiber-reinforced composie maerials are formed from lamina of fibers of graphie, glass, or polymers embedded in a resin marix. Normal sresses and srains are relaed by Hooke s Law bu wih hdirecionally i dependen d moduli of elasiciy, x y x y ε ε ε x Transverse conracions are relaed by direcionally dependen values of Poisson s raio, e.g., ν xy ε y ε y ν x ε x ε x Maerials wih direcionally dependen mechanical properies are anisoropic The McGraw-Hill Companies, Inc. All righs reserved. 2 -

11 Sain-Venan s Principle Loads ransmied hrough rigid plaes resul in uniform disribuion of sress and srain. Concenraed loads resul in large sresses in he viciniy of he load applicaion poin. Sress and srain disribuions become uniform a a relaively shor disance from he load applicaion poins. Sain-Venan s Principle: Sress disribuion may be assumed independen of he mode of load applicaion excep in he immediae viciniy of load applicaion poins The McGraw-Hill Companies, Inc. All righs reserved. 2-11

12 Sress Concenraion: Hole Disconinuiies of cross secion may resul in high localied or concenraed sresses. K max ave 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-12

13 Sress Concenraion: Fille 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-1

14 xample 2.12 SOLUTION: Deermine he larges axial load P ha can be safely suppored by a fla seel bar consising i of wo porions, boh mm hick, and respecively 40 and 60 mm wide, conneced by filles of radius r 8 mm. Assume an allowable normal sress of 165 MPa. Deermine he geomeric raios and find he sress concenraion facor from Fig. 2.64b. Find he allowable average normal sress using he maerial allowable normal sress and he sress concenraion facor. Apply he definiion of normal sress o find he allowable load The McGraw-Hill Companies, Inc. All righs reserved. 2-14

15 Deermine he geomeric raios and find he sress concenraion i facor from Fig. 2.64b. D 60mm 1.50 d 40mm K 1.82 r d 8mm 40mm 0.20 Find he allowable average normal sress using he maerial allowable normal sress and he sress concenraion facor. K 165 MPa 1.82 max ave 90.7 MPa Apply he definiion of normal sress o find he allowable load. P A ( 40mm)( mm)( 90.7 MPa) ave The McGraw-Hill Companies, Inc. All righs reserved N P 6.kN

16 lasoplasic Maerials Previous analyses based on assumpion of linear sress-srain relaionship, i.e., sresses below he yield sress Assumpion is good for brile maerial which h rupure wihou yielding If he yield sress of ducile maerials is exceeded, hen plasic deformaions occur Analysis of plasic deformaions is simplified by assuming an idealied elasoplasic maerial Deformaions of an elasoplasic maerial are divided ino elasic and plasic ranges Permanen deformaions resul from loading beyond he yield sress 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-16

17 Plasic Deformaions A P A max ave K P A K lasic deformaion while maximum sress is less han yield sress Maximum sress is equal o he yield sress a he maximum elasic loading A loadings above he maximum elasic load, a region of plasic df deformaions develop near he hole hl As he loading increases, he plasic P U A region expands unil he secion is a K P a uniform sress equal o he yield sress 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-17

18 Residual Sresses When a single srucural elemen is loaded uniformly beyond is yield sress and hen unloaded, i is permanenly deformed bu all sresses disappear. This is no he general resul. Residual sresses will remain in a srucure afer loading and unloading if - only par of he srucure undergoes plasic deformaion - differen pars of he srucure undergo differen plasic deformaions Residual sresses also resul from he uneven heaing or cooling of srucures or srucural elemens 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-18

19 xample 2.14, 2.15, 2.16 A cylindrical rod is placed inside a ube of he same lengh. The ends of he rod and ube are aached o a rigid suppor on one side and a rigid plae on he oher. The load on he rod-ube assembly is increased from ero o 5.7 kips and decreased back o ero. a) draw a load-deflecion diagram for he rod-ube assembly b) deermine he maximum elongaion c) deermine he permanen se d) calculae he residual sresses in he rod and ube. A r r 0.075in. 0 6 ( ) 6ksi r 2 psi A 0.0in ( ) 45ksi 2 psi 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-19

20 xample 2.14, 2.15, 2.16 a) Draw a load-deflecion diagram for he rod-ube assembly 2 ( P ) ( ) A ( 6ksi)( 0.075in ) r r r 2.7 kips ( ) r 6 psi ( δ ) ( ε ) L L ( 0in. ) r r 6 - in. r 0 2 ( P ) ( ) A ( 45ksi)( 0.0in ) ( δ ) ( ε ) 90 L L - 6 psi 4.5kips ( ) 45 psi ( 0in. ) in psi P P r + P δ δ r δ 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-20

21 xample 2.14, 2.15, 2.16 b,c) deermine he maximum elongaion and permanen se A a load of P 5.7 kips, he rod has reached he plasic range while he ube is sill in he elasic range Pr ( Pr ) 2.7 kips P P P ( ) P A ε δ ε L r.0 kips 0ksi 2 0.1in kips.0 kips 0 L 15 6 psi 0in. psi δ max δ 60 in. The rod-ube assembly unloads along a line parallel o 0 r 4.5 kips m 125 kips - 6 in. P δ m in. slope 5.7 kips kips in. max in. δ p δ + δ max ( ) in. δ p 14.4 in The McGraw-Hill Companies, Inc. All righs reserved. 2-21

22 xample 2.14, 2.15, 2.16 Calculae he residual sresses in he rod and ube. Calculae he reverse sresses in he rod and ube caused by unloading and add hem o he maximum sresses. δ ε L 45.6 in in. 0in. in. 6 ( 1.52 )( 0 psi) 45.6ksi 6 ( 1.52 )( 15 psi) 22.8ksi ε r r ε residual, ( ) residual, r r + r. + ksi 9 6ksi ( ) ksi 7.2ksi 2006 The McGraw-Hill Companies, Inc. All righs reserved. 2-22