236 CHAPTER 3 Torsion. Strain Energy in Torsion

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1 36 CHAPER 3 orsion Srain Energy in orsion Problem A solid circular bar of seel (G psi) wih lengh 3 in. and diameer d 1.75 in. is subjeced o pure orsion by orques acing a he ends (see figure). d (a) Calculae he amoun of srain energy U sored in he bar when he maximum shear sress is 5 psi. (b) From he srain energy, calculae he angle of wis (in degrees). Soluion Seel bar d G psi 3 in. d 1.75 in. max 5 psi max 16 d 3 d 3 max 16 I P d 3 (Eq. 1) (a) SRAIN ENERGY U d 3 max 3 GI P 16 G d d max 16G Subsiue numerical values: U 3. in.-lb (b) ANGE OF WIS U f f U Subsiue for and U from Eqs. (1) and (): f max Gd Subsiue numerical values: f.1353rad.775 (Eq. ) (Eq. 3) Problem 3.9- A solid circular bar of copper (G 5 GPa) wih lengh.75 m and diameer d mm is subjeced o pure orsion by orques acing a he ends (see figure). (a) Calculae he amoun of srain energy U sored in he bar when he maximum shear sress is 3 MPa. (b) From he srain energy, calculae he angle of wis (in degrees)

2 SECION 3.9 Srain Energy in orsion 37 Soluion 3.9- Copper bar d (A) SRAIN ENERGY U d3 max 3 GI P 16 G d G 5 GPa.75 m d mm max 3 MPa max 16 d 3 d 3 max 16 (Eq. 1) d max 16G Subsiue numerical values: U 5.36 J (b) ANGE OF WIS U f f U (Eq. ) I P d 3 Subsiue for and U from Eqs. (1) and (): f max Gd Subsiue numerical values: f.6667 rad 1.53 (Eq. 3) Problem A sepped shaf of solid circular cross secions (see figure) has lengh 5 in., diameer d 1. in., and diameer d 1 1. in. he maerial is brass wih G psi. Deermine he srain energy U of he shaf if he angle of wis is 3.. d d1 Soluion Sepped shaf d d1 SRAIN ENERGY U a 16 () 16 () GI P Gd Gd 1 8 G 1 d 1 d 1 (Eq. 1) d 1 1. in. d 1. in. 5 in. G psi (brass) rad Also, U f (Eq. ) Equae U from Eqs. (1) and () and solve for : Gd 1 d f 16(d 1 d ) U f Gf d 3 1 d d f radians 1 d SUBSIUE NUMERICA VAUES: U.6 in.-lb

3 38 CHAPER 3 orsion Problem 3.9- A sepped shaf of solid circular cross secions (see figure) has lengh.8 m, diameer d mm, and diameer d 1 3 mm. he maerial is seel wih G 8 GPa. Deermine he srain energy U of he shaf if he angle of wis is 1.. Soluon 3.9- Sepped shaf d d1 d 1 3 mm d mm.8 m G 8 GPa (seel) rad Equae U from Eqs. (1) and () and solve for : G d 1 d f 16(d 1 d ) U f Gf d 3 1 d d f radians 1 d SUBSIUE NUMERICA VAUES: U 1.8 J SRAIN ENERGY U a 16 () 16 () GI P Gd Gd 1 8 G 1 d 1 d 1 (Eq. 1) Also, U f (Eq. ) Problem A canilever bar of circular cross secion and lengh is fixed a one end and free a he oher (see figure). he bar is loaded by a orque a he free end and by a disribued orque of consan inensiy per uni disance along he lengh of he bar. (a) Wha is he srain energy U 1 of he bar when he load acs alone? (b) Wha is he srain energy U when he load acs alone? (c) Wha is he srain energy U 3 when boh loads ac simulaneously? Soluion Canilever bar wih disribued orque G shear modulus I P polar momen of ineria orque acing a free end orque per uni disance

4 SECION 3.9 Srain Energy in orsion 39 (a) OAD ACS AONE (EQ. 3-51a) U 1 GI P (c) BOH OADS AC SIMUANEOUSY (b) OAD ACS AONE From Eq. (3-56) of Example 3-11: U 3 6GI P A disance x from he free end: (x) x [(x)] U 3 dx 1 GI P GI P ( x) dx dx GI P GI P 3 6GI P NOE: U 3 is no he sum of U 1 and U. x Problem Obain a formula for he srain energy U of he saically indeerminae circular bar shown in he figure. he bar has fixed suppors a ends A and B and is loaded by orques and a poins C and D, respecively. Hin: Use Eqs. 3-6a and b of Example 3-9, Secion 3.8, o obain he reacive orques. A C D B Soluion Saically indeerminae bar A A B B C D REACIVE ORQUES From Eq. (3-6a): ( 3 ) A B 3 A 5 INERNA ORQUES AC 7 CD 7 DB 5 SRAIN ENERGY (from EQ. 3-53) U a n i1 1 B 7 GI P U 19 3GI P i i G i I Pi 1 B GI AC p CD DB R 5 R

5 CHAPER 3 orsion Problem A saically indeerminae sepped shaf ACB is fixed a ends A and B and loaded by a orque a poin C (see figure). he wo segmens of he bar are made of he same maerial, have lenghs A and B, and have polar momens of ineria I PA and I PB. Deermine he angle of roaion of he cross secion a C by using srain energy. Hin: Use Eq. 3-51b o deermine he srain energy U in erms of he angle. hen equae he srain energy o he work done by he orque. Compare your resul wih Eq. 3-8 of Example 3-9, Secion 3.8. A I PA A C B I PB B Soluion Saically indeerminae bar A I PA SRAIN ENERGY (FROM EQ. 3-51B) U a n i1 GI Pi f i A GI PAf GI PBf i A B Gf I PA A I PB B C B I PB B WORK DONE BY HE ORQUE W f EQUAE U AND W AND SOVE FOR Gf I PA I PB f A B A B f G( B I PA A I PB ) (his resul agrees wih Eq. (3-8) of Example 3-9, Secion 3.8.) Problem Derive a formula for he srain energy U of he canilever bar shown in he figure. he bar has circular cross secions and lengh. I is subjeced o a disribued orque of inensiy per uni disance. he inensiy varies linearly from a he free end o a maximum value a he suppor.

6 SECION 3.9 Srain Energy in orsion 1 Soluion Canilever bar wih disribued orque (x) = x x disance from righ-hand end of he bar dx d x EEMEN d Consider a differenial elemen d a disance from he righ-hand end. d exernal orque acing on his elemen d ()d j dj d d SRAIN ENERGY OF EEMEN dx du [(x)] dx 1 GI P GI P x dx SRAIN ENERGY OF ENIRE BAR U du 8 GI P x dx U 3 GI P 8 GI P x dx 8 GI P 5 5 EEMEN dx A DISANCE x (x) (x) dx (x) inernal orque acing on his elemen (x) oal orque from x o x x x x (x) d j dj x

7 CHAPER 3 orsion Problem A hin-walled hollow ube AB of conical shape has consan hickness and average diameers d A and d B a he ends (see figure). A B (a) Deermine he srain energy U of he ube when i is subjeced o pure orsion by orques. (b) Deermine he angle of wis of he ube. Noe: Use he approximae formula I P d 3 / for a hin circular ring; see Case of Appendix D. d A d B Soluion hin-walled, hollow ube A x (x) hickness d A average diameer a end A d B average diameer a end B d(x) average diameer a disance x from end A d(x) d A d B d A x POAR MOMEN OF INERIA I P d 3 I P (x) [d(x)]3 (a) SRAIN ENERGY (FROM EQ. 3-5) U dx GI P (x) From Appendix C: G dx dx (a bx) 1 3 b(a bx) dx B d A d B d A x R 3 B B d A d B d A 3 x R (Eq. 1) herefore, (d B d A ) (d B d A )(d B ) (d B d A )(d A ) (d A d B ) d A d B Subsiue his expression for he inegral ino he equaion for U (Eq. 1): U G (d A d B ) d A d B G d A d B d A d B (b) ANGE OF WIS Work of he orque : W f W U f (d A d B ) G d A d B Solve for : f (d A d B ) G d A d B 1 dx B d A d 3 B d A x R B d A d B d A x R

8 SECION 3.9 Srain Energy in orsion 3 **Problem A hollow circular ube A fis over he end of a solid circular bar B, as shown in he figure. he far ends of boh bars are fixed. Iniially, a hole hrough bar B makes an angle wih a line hrough wo holes in ube A. hen bar B is wised unil he holes are aligned, and a pin is placed hrough he holes. When bar B is released and he sysem reurns o equilibrium, wha is he oal srain energy U of he wo bars? (e I PA and I PB represen he polar momens of ineria of bars A and B, respecively. he lengh and shear modulus of elasiciy G are he same for boh bars.) I PA ube A I PB Bar B ube A Bar B Soluion Circular ube and bar I PA I PB ube A Bar B ube A Bar B UBE A COMPAIBIIY A B A FORCE-DISPACEMEN REAIONS f A GI PA f B GI PB orque acing on he ube A angle of wis Subsiue ino he equaion of compaibiliy and solve for : I PAI PB bg I PA I PB BAR B B SRAIN ENERGY U a GI P GI PA GI PB G 1 I PA 1 I PB Subsiue for and simplify: U b G I PA I PB I PA I PB orque acing on he bar B angle of wis

9 CHAPER 3 orsion **Problem A heavy flywheel roaing a n revoluions per minue is rigidly aached o he end of a shaf of diameer d (see figure). If he bearing a A suddenly freezes, wha will be he maximum angle of wis of he shaf? Wha is he corresponding maximum shear sress in he shaf? (e lengh of he shaf, G shear modulus of elasiciy, and I m mass momen of ineria of he flywheel abou he axis of he shaf. Also, disregard fricion in he bearings a B and C and disregard he mass of he shaf.) Hin: Equae he kineic energy of he roaing flywheel o he srain energy of he shaf. A d B C n (rpm) Soluion Roaing flywheel Shaf Flywheel d diameer of shaf U Gd f 6 UNIS: G (force)/(lengh) d diameer n rpm KINEIC ENERGY OF FYWHEE K.E. 1 I mv v n 6 n rpm K.E. 1 I m n 6 UNIS: n I m 18 I m (force)(lengh)(second) radians per second K.E. (lengh)(force) SRAIN ENERGY OF SHAF (FROM EQ. 3-51b) U GI Pf I P 3 d I P (lengh) radians lengh U (lengh)(force) EQUAE KINEIC ENERGY AND SRAIN ENERGY K.E. U n I m 18 Gd f 6 Solve for : f n I m 15d B G MAXIMUM SHEAR SRESS (d) f I P GI P Eliminae : Gdf max Gd n I m 15d B G max n GI m 15d B

10 SECION 3.1 hin-walled ubes 5 hin-walled ubes Problem A hollow circular ube having an inside diameer of 1. in. and a wall hickness of 1. in. (see figure) is subjeced o a orque 1 k-in. Deermine he maximum shear sress in he ube using (a) he approximae heory of hin-walled ubes, and (b) he exac orsion heory. Does he approximae heory give conservaive or nonconservaive resuls? 1. in. 1. in. Soluion Hollow circular ube APPROXIMAE HEORY (EQ. 3-63) 1. in. 1. in. 1 k-in. 1. in. r radius o median line r 5.5 in. d ouside diameer 1. in. d 1 inside diameer 1. in. 1 r 1 k-in. 631 psi (5.5 in.) (1. in.) approx 631 psi EXAC HEORY (EQ. 3-11) (d ) I P 16(1 k-in.)(1. in.) [(1. in.) (1. in.) ] 6831 psi exac 683 psi d 3 d d 1 Because he approximae heory gives sresses ha are oo low, i is nonconservaive. herefore, he approximae heory should only be used for very hin ubes. Problem 3.1- A solid circular bar having diameer d is o be replaced by a recangular ube having cross-secional dimensions d d o he median line of he cross secion (see figure). Deermine he required hickness min of he ube so ha he maximum shear sress in he ube will no exceed he maximum shear sress in he solid bar. d d d Soluion 3.1- Bar and ube SOID BAR A m (d)(d) d (Eq. 3-6) d max 16 d 3 (Eq. 3-1) max A m d (Eq. 3-61) EQUAE HE MAXIMUM SHEAR SRESSES AND SOVE FOR RECANGUAR UBE d 16 d 3 d min d 6 d If min, he shear sress in he ube is less han he shear sress in he bar.

11 6 CHAPER 3 orsion Problem A hin-walled aluminum ube of recangular cross secion (see figure) has a cenerline dimensions b 6. in. and h. in. he wall hickness is consan and equal o.5 in. (a) Deermine he shear sress in he ube due o a orque 15 k-in. (b) Deermine he angle of wis (in degrees) if he lengh of he ube is 5 in. and he shear modulus G is. 1 6 psi. b h Soluion hin-walled ube Eq. (3-6): A m bh. in. h Eq. (3-71)wih 1 :J b h b h J 8.8 in. b b 6. in. h. in..5 in. 15 k-in. 5 in. G. 1 6 psi (a) SHEAR SRESS (EQ. 3-61) 15 psi A m (b) ANGE OF WIS (EQ. 3-7) f.651 rad GJ.373 Problem 3.1- A hin-walled seel ube of recangular cross secion (see figure) has cenerline dimensions b 15 mm and h 1 mm. he wall hickness is consan and equal o 6. mm. (a) Deermine he shear sress in he ube due o a orque 165 N m. (b) Deermine he angle of wis (in degrees) if he lengh of he ube is 1. m and he shear modulus G is 75 GPa. Soluion 3.1- hin-walled ube b 15 mm h 1 mm h 6. mm 165 N m 1. m b G 75 GPa Eq. (3-6): A m bh.15 m Eq. (3-71)wih 1 :J b h b h J m (a) SHEAR SRESS (EQ. 3-61) 9.17 MPa A m (b) ANGE OF WIS (EQ. 3-7) f. rad GJ.1

12 SECION 3.1 hin-walled ubes 7 Problem A hin-walled circular ube and a solid circular bar of he same maerial (see figure) are subjeced o orsion. he ube and bar have he same cross-secional area and he same lengh. Wha is he raio of he srain energy U 1 in he ube o he srain energy U in he solid bar if he maximum shear sresses are he same in boh cases? (For he ube, use he approximae heory for hin-walled bars.) ube (1) Bar () Soluion hin-walled ube (1) A m r J r 3 A r r U 1 GJ (r max ) G(r 3 ) r max G Bu r A U 1 A max G max A m r r max SOID BAR () U (r 3 max ) GI P 8G r Bu r A RAIO r U 1 U A r max r I P U A max G I P r r r 3 max 3 r max G Problem Calculae he shear sress and he angle of wis (in degrees) for a seel ube (G 76 GPa) having he cross secion shown in he figure. he ube has lengh 1.5 m and is subjeced o a orque 1 kn m. r = 5 mm = 8 mm r = 5 mm b = 1 mm Soluion Seel ube = 8 mm r = 5 mm r = 5 mm G 76 GPa 1.5 m 1 kn. m b = 1 mm A m r br (5 mm) (1 mm)(5 mm) 17,85 mm m br (1 mm)(5 mm) 51. mm J A m (8 mm)(17,85 mm ) m 51. mm mm SHEAR SRESS 1 kn. m A m (8 mm)(17,85 mm ) ANGE OF WIS 35. MPa f GJ (1 kn. m) (1.5 m) (76 GPa)( mm ).995 rad.57

13 8 CHAPER 3 orsion Problem A hin-walled seel ube having an ellipical cross secion wih consan hickness (see figure) is subjeced o a orque 18 k-in. Deermine he shear sress and he rae of wis (in degrees per inch) if G psi,. in., a 3 in., and b in. (Noe: See Appendix D, Case 16, for he properies of an ellipse.) b a Soluion Ellipical ube b a 18 k-in. G psi consan. in a 3. in. b. in. FROM APPENDIX D, CASE 16: A m ab (3. in.)(. in.) in. m p[1.5(a b) ab] [1.5(5. in.) 6. in. ] in. J A m (. in.)(18.85 in. ) m in in. SHEAR SRESS 18 k in. A m (. in.)(18.85 in. ) 39 psi ANGE OF WIS PER UNI ENGH (RAE OF WIS) u f GJ 18 k-in. (1 1 6 psi)(17.9 in. ) u radin..8in. Problem A orque is applied o a hin-walled ube having a cross secion in he shape of a regular hexagon wih consan wall hickness and side lengh b (see figure). Obain formulas for he shear sress and he rae of wis. b

14 SECION 3.1 hin-walled ubes 9 Soluion Regular hexagon b engh of side hickness m 6b b SHEAR SRESS 3 A m 9b ANGE OF WIS PER UNI ENGH (RAE OF WIS) J A m m d S A m 9b3 m u GJ G(9b 3 ) 9Gb 3 (radians per uni lengh) FROM APPENDIX D, CASE 5: 6 n 6 A m nb co b 6b co 3 33b Problem Compare he angle of wis 1 for a hin-walled circular ube (see figure) calculaed from he approximae heory for hin-walled bars wih he angle of wis calculaed from he exac heory of orsion for circular bars. (a) Express he raio 1 / in erms of he nondimensional raio r/. (b) Calculae he raio of angles of wis for 5, 1, and. Wha conclusion abou he accuracy of he approximae heory do you draw from hese resuls? r C Soluion APPROXIMAE HEORY f 1 GJ EXAC HEORY f GI P hin-walled ube r C J r 3 f 1 Gr 3 From Eq. (3-17): I p r (r ) f GI P Gr(r ) RAIO f 1 f r r 1 r e b r f 1 f 1 1 b 1 / As he ube becomes hinner and becomes larger, he raio 1 / approaches uniy. hus, he hinner he ube, he more accurae he approximae heory becomes.

15 5 CHAPER 3 orsion *Problem A hin-walled recangular ube has uniform hickness and dimensions a b o he median line of he cross secion (see figure). How does he shear sress in he ube vary wih he raio a/b if he oal lengh m of he median line of he cross secion and he orque remain consan? From your resuls, show ha he shear sress is smalles when he ube is square ( 1). a b Soluion Recangular ube,, and m are consans. b e k (1 b) consan k m b k min hickness (consan) a, b Dimensions of he ube b a b m (a b) consan consan SHEAR SRESS A A m ab bb m m b(1 b) consan m b m (1 b) A m a m b (1 b) A m b B (1 b) R ()(1 b) (1 b) b m m b min 8 m b a b From he graph, we see ha is minimum when 1 and he ube is square. AERNAE SOUION m (1 b) B R b d db B b()(1 b) (1 b) (1) R m b or (1)(1) 1 hus, he ube is square and is eiher a minimum or a maximum. From he graph, we see ha is a minimum.

16 SECION 3.1 hin-walled ubes 51 *Problem A ubular aluminum bar (G 1 6 psi) of square cross secion (see figure) wih ouer dimensions in. in. mus resis a orque 3 lb-in. Calculae he minimum required wall hickness min if he allowable shear sress is 5 psi and he allowable rae of wis is.1 rad/f. in. in. Soluion Square aluminum ube Ouer dimensions:. in.. in. G 1 6 psi 3 lb-in. allow 5 psi in. u allow.1 radf.1 1 radin. e b ouer dimension. in. Cenerline dimension b A m (b ) m (b ) J A m m (b ) (b )3 (b ) in. HICKNESS BASED UPON SHEAR SRESS A m UNIS: in. b in. lb-in. psi (. in. ) 3 lb-in. (5 psi) 1 3 in.3 3( ) 1 Solve for :.915 in. HICKNESS BASED UPON RAE OF WIS u GJ G(b ) (b 3 )3 Gu UNIS: in. G psi rad/in. (. in. ) 3 1( ) 3 9 Solve for :.1 in. A m 9 1 ANGE OF WIS GOVERNS min.1 in. (b ) 3 lb-in ( 1 6 psi)(.11 radin.)

17 5 CHAPER 3 orsion *Problem A hin ubular shaf of circular cross secion (see figure) wih inside diameer 1 mm is subjeced o a orque of 5 N m. If he allowable shear sress is MPa, deermine he required wall hickness by using (a) he approximae heory for a hin-walled ube, and (b) he exac orsion heory for a circular bar. 1 mm Soluion hin ube (b) EXAC HEORY 1 mm 5, N m d 1 inner diameer 1 mm allow MPa is in millimeers. r Average radius 5 mm r 1 Inner radius 5 mm r Ouer radiusa m r 5 mm r I p I p (r r 1 ) [(5 ) (5) ] MPa (5, N. m)(5 ) [(5 ) (5) ] (5 ) (5) 5 Solve for : 7. mm (5 N. m)() ()( MPa) mm3 he approximae resul is 5% less han he exac resul. hus, he approximae heory is nonconservaive and should only be used for hin-walled ubes. (a) APPROXIMAE HEORY A m (r ) r MPa 5, N. m or 5 5 5, N. m ( MPa) Solve for : 6.66 mm mm3

18 SECION 3.1 hin-walled ubes 53 Problem A long, hin-walled apered ube AB of circular cross secion (see figure) is subjeced o a orque. he ube has lengh and consan wall hickness. he diameer o he median lines of he cross secions a he ends A and B are d A and d B, respecively. Derive he following formula for he angle of wis of he ube: G d A d d B A d B A B Hin: If he angle of aper is small, we may obain approximae resuls by applying he formulas for a hin-walled prismaic ube o a differenial elemen of he apered ube and hen inegraing along he axis of he ube. d A d B Soluion hin-walled apered ube d A A x d(z) hickness d A average diameer a end A d B average diameer a end B orque d(x) average diameer a disance x from end A. d(x) d A d B d A x J r 3 d 3 J(x) [d(x)]3 For elemen of lengh dx: df dx GJ(x) dx dx B db B d A d B d A G B d A d 3 B d A x R 3 x R For enire ube: f G From able of inegrals (see Appendix C): dx B d A d 3 B d A x R dx (a bx) 1 3 b(a bx) f G C 1 d B d A d A d B d A G B (d B d A )d B (d B d A )d R A f G d A d B d A d B S x

19 5 CHAPER 3 orsion Sress Concenraions in orsion he problems for Secion 3.11 are o be solved by considering he sress-concenraion facors. Problem A sepped shaf consising of solid circular segmens having diameers. in. and D. in. (see figure) is subjeced o orques. he radius of he fille is R.1 in. If he allowable shear sress a he sress concenraion is 6 psi, wha is he maximum permissible orque max? D R Soluion Sepped shaf in orsion D R USE FIG. 3-8 FOR HE SRESS-CONCENRAION FACOR R.1 in.. in..5 D. in. 1.. in.. in. D. in. R.1 in. allow 6 psi K 1.5 max K mom K 16 max 3 D 1 max 3 max 16K (. in.)3 (6 psi) 16(1.5) max 6 lb-in. 6 lb-in. Problem A sepped shaf wih diameers mm and D 6 mm is loaded by orques 11 N m (see figure). If he allowable shear sress a he sress concenraion is 1 MPa, wha is he smalles radius R min ha may be used for he fille? Soluion Sepped shaf in orsion mm D 6 mm 11 N m allow 1 MPa D R USE FIG. 3-8 FOR HE SRESS-CONCENRAION FACOR max K nom K 16 3 K 3 max 16 D 6 mm 1.5 mm ( mm)3 (1 MPa) 16(11 N # m) D 1.37 From Fig. (3-8) wih 1.5 and K 1.37, R we ge.1 R min.1( mm). mm

20 SECION 3.11 Sess Concenraions in orsion 55 Problem A full quarer-circular fille is used a he shoulder of a sepped shaf having diameer D 1. in. (see figure). A orque 5 lb-in. acs on he shaf. Deermine he shear sress max a he sress concenraion for values as follows:.7,.8, and.9 in. Plo a graph showing max versus. Soluion Sepped shaf in orsion D R (in.) D / R (in.) R/ K max (psi) D 1. in. 5 lb-in..7,.8, and.9 in. Full quarer-circular fille (D R) R D.5 in. 1, max (psi) 5 USE FIG. 3-8 FOR HE SRESS-CONCENRAION FACOR max K nom K (5 lb-in.) K 3 56 K (in.) Noe ha max ges smaller as ges larger, even hough K is increasing.

21 56 CHAPER 3 orsion Problem he sepped shaf shown in he figure is required o ransmi 6 kw of power a rpm. he shaf has a full quarer-circular fille, and he smaller diameer 1 mm. If he allowable shear sress a he sress concenraion is 1 MPa, a wha diameer D will his sress be reached? Is his diameer an upper or a lower limi on he value of D? Soluion Sepped shaf in orsion D R P 6 kw 1 mm n rpm allow 1 MPa Full quarer-circular fille POWER P n (Eq. 3- of secion 3.7) 6 P was n rpm Newon meers 6P n 6(6 13 W) 1,3 N ( rpm) # m USE FIG. 3-8 FOR HE SRESS-CONCENRAION FACOR max K nom K 16 3 K max( 3 ) 16 Use he dashed line for a full quarer-circular fille. R.75R.75D.75 (1 mm) mm D R 1 mm (7.5 mm) 115 mm D 115 mm his value of D is a lower limi (If D is less han 115 mm, R/ is smaller, K is larger, and max is larger, which means ha he allowable sress is exceeded.) (1 MPa)()(1 mm)3 16(1,3 N # 1.37 m)

22 SECION 3.11 Sess Concenraions in orsion 57 Problem A sepped shaf (see figure) has diameer D 1.5 in. and a full quarer-circular fille. he allowable shear sress is 15, psi and he load 8 lb-in. Wha is he smalles permissible diameer? Soluion Sepped shaf in orsion D R D 1.5 in. allow 15, psi 8 lb-in. Full quarer-circular fille D R R D.75 in. USE FIG. 3-8 FOR HE SRESS-CONCENRAION FACOR Use rial-and-error. Selec rial values of (in.) R (in.) R/ K max (psi) , , , max (psi) max K nom K 16 3 K 3,5 16(8 lb-in.) B R K 3 16, 15, 1, allow =1.31in. 13, From he graph, minimum 1.31 in.