ue to grvit nd wind lod, the post supporting the sign shown is sujeted simultneousl to ompression, ending, nd torsion. In this hpter ou will lern to determine the stresses reted suh omined lodings in strutures nd mhine omponents. 512
8 T E R rinipl Stresses under Given Loding 513
hpter 8 rinipl Stresses under Given Loding *8.1 Introdution *8.2 rinipl Stresses in em *8.3 esign of Trnsmission Shfts *8.4 Stresses under omined Lodings () Fig. 8.1 () Fig. 8.2 () () ' ' ' *8.1 INTROUTION In the first prt of this hpter, ou will ppl to the design of ems nd shfts the knowledge tht ou quired in hp. 7 on the trnsformtion of stresses. In the seond prt of the hpter, ou will lern how to determine the prinipl stresses in struturl memers nd mhine elements under given loding onditions. In hp. 5 ou lerned to lulte the mimum norml stress s m ourring in em under trnsverse loding (Fig. 8.1) nd hek whether this vlue eeeded the llowle stress s ll for the given mteril. If it did, the design of the em ws not eptle. While the dnger for rittle mteril is tull to fil in tension, the dnger for dutile mteril is to fil in sher (Fig. 8.1). The ft tht s m. s ll indites tht M m is too lrge for the ross setion seleted, ut does not provide n informtion on the tul mehnism of filure. Similrl, the ft tht t m. t ll simpl indites tht V m is too lrge for the ross setion seleted. While the dnger for dutile mteril is tull to fil in sher (Fig. 8.2), the dnger for rittle mteril is to fil in tension under the prinipl stresses (Fig. 8.2). The distriution of the prinipl stresses in em will e disussed in Se. 8.2. epending upon the shpe of the ross setion of the em nd the vlue of the sher V in the ritil setion where M 5 M m, it m hppen tht the lrgest vlue of the norml stress will not our t the top or ottom of the setion, ut t some other point within the setion. s ou will see in Se. 8.2, omintion of lrge vlues of s nd t ner the juntion of the we nd the flnges of W-em or n S-em n result in vlue of the prinipl stress s m (Fig. 8.3) tht is lrger thn the vlue of s m on the surfe of the em. Fig. 8.3 rinipl stresses t the juntion of flnge nd we in n I-shped em. 514 Setion 8.3 will e devoted to the design of trnsmission shfts sujeted to trnsverse lods s well s to torques. The effet of oth the norml stresses due to ending nd the shering stresses due to torsion will e tken into ount. In Se. 8.4 ou will lern to determine the stresses t given point of od of ritrr shpe sujeted to omined loding. First, ou will redue the given loding to fores nd ouples in the setion ontining. Net, ou will lulte the norml nd shering stresses t. Finll, using one of the methods for the trnsformtion of stresses tht ou lerned in hp. 7, ou will determine the prinipl plnes, prinipl stresses, nd mimum shering stress t.
*8.2 RINIL STRESSES IN EM onsider prismti em sujeted to some ritrr trnsverse loding (Fig. 8.4). We denote V nd M, respetivel, the sher nd ending moment in setion through given point. We rell from hps. 5 nd 6 tht, within the elsti limit, the stresses eerted on smll element with fes perpendiulr, respetivel, to the nd es redue to the norml stresses s m 5 MI if the element is t the free surfe of the em, nd to the shering stresses t m 5 VQIt if the element is t the neutrl surfe (Fig. 8.5). 8.2 rinipl Stresses in em Fig. 8.4 Trnsversel loded prismti em. w 515 O Fig. 8.5 Stress elements t seleted points of em. t n other point of the ross setion, n element of mteril is sujeted simultneousl to the norml stresses s 52 M I (8.1) where is the distne from the neutrl surfe nd I the entroidl moment of inerti of the setion, nd to the shering stresses t 52 VQ It (8.2) O in in where Q is the first moment out the neutrl is of the portion of the ross-setionl re loted ove the point where the stresses re omputed, nd t the width of the ross setion t tht point. Using either of the methods of nlsis presented in hp. 7, we n otin the prinipl stresses t n point of the ross setion (Fig. 8.6). The following question now rises: n the mimum norml stress s m t some point within the ross setion e lrger thn the vlue of s m 5 MI omputed t the surfe of the em? If it n, then the determintion of the lrgest norml stress in the em will involve gret del more thn the omputtion of M m nd the use of Eq. (8.1). We n otin n nswer to this question investigting the distriution of the prinipl stresses in nrrow Fig. 8.6 rinipl stresses t seleted points of em.
516 rinipl Stresses under Given Loding retngulr ntilever em sujeted to onentrted lod t its free end (Fig. 8.7). We rell from Se. 6.5 tht the norml nd shering stresses t distne from the lod nd distne ove the neutrl surfe re given, respetivel, Eq. (6.13) nd Eq. (6.12). Sine the moment of inerti of the ross setion is Fig. 8.7 Nrrow retngulr ntilever em supporting single onentrted lod. I 5 h3 12 5 1h21222 5 2 12 3 where is the ross-setionl re nd the hlf-depth of the em, we write nd s 5 I 5 1 3 5 3 2 2 (8.3) t 5 3 2 1 2 2 2 (8.4) Using the method of Se. 7.3 or Se. 7.4, the vlue of s m n e determined t n point of the em. Figure 8.8 shows the results of the omputtion of the rtios s m s m nd s min s m in two setions of the em, orresponding respetivel to 5 2 nd 5 8. In 2 / in / / in / 8 / 0 1.0 0.8 0.6 0 0.010 0.040 1.000 0.810 0.640 0 0.001 0.003 1.000 0.801 0.603 2 8 0.4 0.090 0.490 0.007 0.407 0.2 0.160 0.360 0.017 0.217 0 0.250 0.250 0.063 0.063 0.2 0.360 0.160 0.217 0.017 0.4 0.490 0.090 0.407 0.007 0.6 0.640 0.040 0.603 0.003 0.8 0.810 0.010 0.801 0.001 1.0 1.000 0 1.000 0 Fig. 8.8 istriution of prinipl stresses in two trnsverse setions of retngulr ntilever em supporting single on entrted lod.
eh setion, these rtios hve een determined t 11 different points, nd the orienttion of the prinipl es hs een indited t eh point. It is ler tht s m does not eeed s m in either of the two setions onsidered in Fig. 8.8 nd tht, if it does eeed s m elsewhere, it will e in setions lose to the lod, where s m is smll ompred to t m. ut, for setions lose to the lod, Sint-Vennt s priniple does not ppl, Eqs. (8.3) nd (8.4) ese to e vlid, eept in the ver unlikel se of lod distriuted prolill over the end setion (f. Se. 6.5), nd more dvned methods of nlsis tking into ount the effet of stress onentrtions should e used. We thus onlude tht, for ems of retngulr ross setion, nd within the sope of the theor presented in this tet, the mimum norml stress n e otined from Eq. (8.1). In Fig. 8.8 the diretions of the prinipl es were determined t 11 points in eh of the two setions onsidered. If this nlsis were etended to lrger numer of setions nd lrger numer of points in eh setion, it would e possile to drw two orthogonl sstems of urves on the side of the em (Fig. 8.9). One sstem would onsist of urves tngent to the prinipl es orresponding to s m nd the other of urves tngent to the prinipl es orresponding to s min. The urves otined in this mnner re known s the stress trjetories. trjetor of the first group (solid lines) defines t eh of its points the diretion of the lrgest tensile stress, while trjetor of the seond group (dshed lines) defines the diretion of the lrgest ompressive stress. The onlusion we hve rehed for ems of retngulr ross setion, tht the mimum norml stress in the em n e otined from Eq. (8.1), remins vlid for mn ems of nonretngulr ross setion. owever, when the width of the ross setion vries in suh w tht lrge shering stresses t will our t points lose to the surfe of the em, where s is lso lrge, vlue of the prinipl stress s m lrger thn s m m result t suh points. One should e prtiulrl wre of this possiilit when seleting W-ems or S-ems, nd lulte the prinipl stress s m t the juntions nd d of the we with the flnges of the em (Fig. 8.10). This is done determining s nd t t tht point from Eqs. (8.1) nd (8.2), respetivel, nd using either of the methods of nlsis of hp. 7 to otin s m (see Smple ro. 8.1). n lterntive proedure, used in design to selet n eptle setion, onsists of using for t the mimum vlue of the shering stress in the setion, t m 5 V we, given Eq. (6.11) of Se. 6.4. This leds to slightl lrger, nd thus onservtive, vlue of the prinipl stress s m t the juntion of the we with the flnges of the em (see Smple ro. 8.2). Tensile ompressive 8.2 rinipl Stresses in em Fig. 8.9 Stress trjetories. d e Fig. 8.10 e stress nlsis lotions in I-shped ems. 517 See ro. 8.2, whih refers to progrm tht n e written to otin the results shown in Fig. 8.8. s will e verified in ro. 8.2, s m eeeds s m if # 0.544. rittle mteril, suh s onrete, will fil in tension long plnes tht re perpendiulr to the tensile-stress trjetories. Thus, to e effetive, steel reinforing rs should e pled so tht the interset these plnes. On the other hnd, stiffeners tthed to the we of plte girder will e effetive in preventing ukling onl if the interset plnes perpendiulr to the ompressive-stress trjetories.
518 rinipl Stresses under Given Loding *8.3 ESIGN OF TRNSMISSION SFTS When we disussed the design of trnsmission shfts in Se. 3.7, we onsidered onl the stresses due to the torques eerted on the shfts. owever, if the power is trnsferred to nd from the shft mens of gers or sproket wheels (Fig. 8.11), the fores eerted on the ger teeth or sprokets re equivlent to fore-ouple sstems pplied t the enters of the orresponding ross setions (Fig. 8.11). This mens tht the shft is sujeted to trnsverse loding, s well s to torsionl loding. 3 () 1 2 T 1 1 T 2 T 3 3 () 2 Fig. 8.11 Lodings on ger-shft sstems. The shering stresses produed in the shft the trnsverse lods re usull muh smller thn those produed the torques nd will e negleted in this nlsis. The norml stresses due to the trnsverse lods, however, m e quite lrge nd, s ou will see presentl, their ontriution to the mimum shering stress t m should e tken into ount. For n pplition where the shering stresses produed the trnsverse lods must e onsidered, see ros. 8.21 nd 8.22.
onsider the ross setion of the shft t some point. We represent the torque T nd the ending ouples M nd M ting, respetivel, in horiontl nd vertil plne the ouple vetors shown (Fig. 8.12). Sine n dimeter of the setion is prinipl is of inerti for the setion, we n reple M nd M their resultnt M (Fig. 8.12) in order to ompute the norml stresses s eerted on the setion. We thus find tht s is mimum t the end of the dimeter perpendiulr to the vetor representing M (Fig. 8.13). Relling tht the vlues of the norml stresses t tht point re, respetivel, s m 5 MI nd ero, while the shering stress is t m 5 TJ, we plot the orresponding points X nd Y on Mohr-irle digrm (Fig. 8.14) nd determine the vlue of the mimum shering stress: () 8.3 esign of Trnsmission Shfts M T M () Fig. 8.12 Resultnt loding on the ross setion of shft. T 519 M t m 5 R 5 s m 2 2 1 1t m 2 2 5 M 2I 2 1 T J 2 M Relling tht, for irulr or nnulr ross setion, 2I 5 J, we write T Fig. 8.13 Mimum stress element. t m 5 J 2M2 2 1 T (8.5) It follows tht the minimum llowle vlue of the rtio J for the ross setion of the shft is X J 5 2M2 1 T t ll 2 m (8.6) where the numertor in the right-hnd memer of the epression otined represents the mimum vlue of 2M 2 1 T 2 in the shft, nd t ll the llowle shering stress. Epressing the ending moment M in terms of its omponents in the two oordinte plnes, we n lso write O Y Fig. 8.14 Mohr s irle nlsis. J 5 2M 2 1 M 2 1 T 2 m t ll (8.7) Equtions (8.6) nd (8.7) n e used to design oth solid nd hollow irulr shfts nd should e ompred with Eq. (3.22) of Se. 3.7, whih ws otined under the ssumption of torsionl loding onl. The determintion of the mimum vlue of 2M 2 1 M 2 2 1 T will e filitted if the ending-moment digrms orresponding to M nd M re drwn, s well s third digrm representing the vlues of T long the shft (see Smple ro. 8.3).
160 kn L 375 mm ' SMLE ROLEM 8.1 160-kN fore is pplied s shown t the end of W200 3 52 rolled-steel em. Negleting the effet of fillets nd of stress onentrtions, determine whether the norml stresses in the em stisf design speifition tht the e equl to or less thn 150 M t setion -9. 160 kn 0.375 m SOLUTION Sher nd ending Moment. t setion -9, we hve V M M 5 1160 kn210.375 m2 5 60 kn? m V 5 160 kn 12.6 mm 103 mm 206 mm 206 mm 7.87 mm I 52.9 10 6 m 4 S 511 10 6 m 3 12.6 mm 206 mm 90.4 mm Norml Stresses on Trnsverse lne. Referring to the tle of roperties of Rolled-Steel Shpes in ppendi, we otin the dt shown nd then determine the stresses s nd s. t point : t point : s 5 M S 5 s 5 s 60 kn? m 5 117.4 M 511 3 10 26 3 m 5 1117.4 M2 90.4 mm 103 mm 5 103.0 M We note tht ll norml stresses on the trnsverse plne re less thn 150 M. 103 mm 96.7 mm Y Shering Stresses on Trnsverse lne t point : t point : Q 5 0 t 5 0 Q 5 1206 3 12.62196.72 5 251.0 3 10 3 mm 3 5 251.0 3 10 26 m 3 t 5 V Q It 5 1160 kn21251.0 3 1026 m 3 2 5 96.5 M 152.9 3 10 26 m 4 210.00787 m2 in O R rinipl Stress t oint. The stte of stress t point onsists of the norml stress s 5 103.0 M nd the shering stress t 5 96.5 M. We drw Mohr s irle nd find 2 X s m 5 1 2 s 1 R 5 1 2 s 1 1 2 2 s 2 1 t L 881 mm W200 52 5 103.0 1 103.0 2 2 2 1 196.52 2 s m 5 160.9 M The speifition, s m # 150 M, is not stisfied omment. For this em nd loding, the prinipl stress t point is 36% lrger thn the norml stress t point. For L $ 881 mm, the mimum norml stress would our t point. 520
41 kips 9 ft 239.4 kip ft 20 kips 20 ft 3.2 kips/ft 59 kips 5 ft V 9 ft 11 ft 41 kips 5 ft 16 kips ( 239.4) 12.2 kips 7.8 kips ( 279.4) (40) 43 kips M d 21 in. 10.5 in. 9.88 in. 20 kips t f 0.615 in. X 1.45 ksi 3.2 kips/ft t w 0.400 in. W21 62 40 kip ft S 127 in 3 we t w d 8.40 in 2 1.45 ksi 21.3 ksi 21.3 ksi 21.4 ksi 22.6 ksi 21.3 ksi O Y SMLE ROLEM 8.2 The overhnging em supports uniforml distriuted lod of 3.2 kips/ ft nd onentrted lod of 20 kips t. nowing tht for the grde of steel to e used s ll 5 24 ksi nd t ll 5 14.5 ksi, selet the wide-flnge shpe tht should e used. SOLUTION Retions t nd. We drw the free-od digrm of the em. From the equilirium equtions SM 5 0 nd SM 5 0 we find the vlues of R nd R shown in the digrm. Sher nd ending-moment igrms. Using the methods of Ses. 5.2 nd 5.3, we drw the digrms nd oserve tht ƒ M ƒ m 5 239.4 kip? ft 5 2873 kip? in. ƒ V ƒ m 5 43 kips Setion Modulus. For M m 5 2873 kip? in. nd s ll 5 24 ksi, the minimum eptle setion modulus of the rolled-steel shpe is 2873 kip? in. S min 5 ƒ M ƒ m 5 5 119.7 in 3 s ll 24 ksi Seletion of Wide-Flnge Shpe. From the tle of roperties of Rolled-Steel Shpes in ppendi, we ompile list of the lightest shpes of given depth tht hve setion modulus lrger thn S min. We now selet the lightest shpe ville, nmel W21 3 62 Shering Stress. Sine we re designing the em, we will onservtivel ssume tht the mimum sher is uniforml distriuted over the we re of W21 3 62. We write t m 5 V m 43 kips 5 5 5.12 ksi, 14.5 ksi (O) 2 we 8.40 in rinipl Stress t oint. We hek tht the mimum prinipl stress t point in the ritil setion where M is mimum does not eeed s ll 5 24 ksi. We write s 5 M m S Shpe S (in 3 ) W24 3 68 154 W21 3 62 127 W18 3 76 146 W16 3 77 134 W14 3 82 123 W12 3 96 131 2873 kip? in. 5 5 22.6 ksi 127 in 3 s 5 s 9.88 in. 5 122.6 ksi2 5 21.3 ksi 10.50 in. onservtivel, t 5 V 12.2 kips 5 5 1.45 ksi 2 we 8.40 in We drw Mohr s irle nd find 2 s m 5 1 21.3 ksi 21.3 ksi 2 s 1 R 5 1 1 11.45 ksi2 2 2 2 s m 5 21.4 ksi # 24 ksi (O) 521
200 G 200 r 60 r 80 imensions in mm 200 200 r 0.060 m r 0.080 m E r E 160 M F 6.63 kn F 2.49 kn r E 0.160 m E T 199 N m T 398 N m F E 3.73 kn E F E 3.73 kn F 2.49 kn TE 597 N m F 6.63 kn F E 3.73 kn 0.932 kn 0.6 m 2.80 kn 0.2 m M 373 N m 186 N m 560 N m E E SMLE ROLEM 8.3 The solid shft rottes t 480 rpm nd trnsmits 30 kw from the motor M to mhine tools onneted to gers G nd ; 20 kw is tken off t ger G nd 10 kw t ger. nowing tht t ll 5 50 M, determine the smllest permissile dimeter for shft. SOLUTION Torques Eerted on Gers. Oserving tht f 5 480 rpm 5 8, we determine the torque eerted on ger E: T E 5 2pf 5 30 kw 2p18 2 5 597 N? m The orresponding tngentil fore ting on the ger is F E 5 T E 5 597 N? m 5 3.73 kn r E 0.16 m similr nlsis of gers nd ields 20 kw T 5 2p18 2 5 398 N? m F 5 6.63 kn 10 kw T 5 2p18 2 5 199 N? m F 5 2.49 kn We now reple the fores on the gers equivlent fore-ouple sstems. ending-moment nd Torque igrms T 398 N m 6.22 kn 0.2 m 2.90 kn T 199 N m 0.4 m E F 2.49 kn T E 597 N m F 6.63 kn 398 N m 597 N m T E M 580 N m 1244 N m 1160 N m E M T M 2 2 ritil Trnsverse Setion. omputing 2M 1 M 1 T 2 t ll potentill ritil setions, we find tht its mimum vlue ours just to the right of : 2 2 2 2M 1 M 1 T m 5 2111602 2 1 13732 2 1 15972 2 5 1357 N? m imeter of Shft. For t ll 5 50 M, Eq. (7.32) ields J 5 2M 2 2 2 1 M 1 T m t ll 5 1357 N? m 50 M 5 27.14 3 10 26 m 3 For solid irulr shft of rdius, we hve J 5 p 2 3 5 27.14 3 10 26 5 0.02585 m 5 25.85 mm imeter 5 2 5 51.7 mm 522
ROLEMS 8.1 W10 3 39 rolled-steel em supports lod s shown. nowing tht 5 45 kips, 5 10 in., nd s ll 5 18 ksi, determine () the mimum vlue of the norml stress s m in the em, () the mimum vlue of the prinipl stress s m t the juntion of the flnge nd we, () whether the speified shpe is eptle s fr s these two stresses re onerned. 10 ft 8.2 Solve ro. 8.1, ssuming tht 5 22.5 kips nd 5 20 in. Fig. 8.1 8.3 n overhnging W920 3 449 rolled-steel em supports lod s shown. nowing tht 5 700 kn, 5 2.5 m, nd s ll 5 100 M, determine () the mimum vlue of the norml stress s m in the em, () the mimum vlue of the prinipl stress s m t the juntion of the flnge nd we, () whether the speified shpe is eptle s fr s these two stresses re onerned. 8.4 Solve ro. 8.3, ssuming tht 5 850 kn nd 5 2.0 m. 8.5 nd 8.6 () nowing tht s ll 5 24 ksi nd t ll 5 14.5 ksi, selet the most eonomil wide-flnge shpe tht should e used to support the loding shown. () etermine the vlues to e epeted for s m, t m, nd the prinipl stress s m t the juntion of flnge nd the we of the seleted em. Fig. 8.3 2 kips/ft 12.5 kips 15 kips 10 kips 9 ft 3 ft 3 ft 6 ft 6 ft 12 ft Fig. 8.5 Fig. 8.6 8.7 nd 8.8 () nowing tht s ll 5 160 M nd t ll 5 100 M, selet the most eonomil metri wide-flnge shpe tht should e used to support the loding shown. () etermine the vlues to e epeted for s m, t m, nd the prinipl stress s m t the juntion of flnge nd the we of the seleted em. 275 kn 1.5 m 3.6 m 1.5 m 275 kn 40 kn 2.2 kn/m 4.5 m 2.7 m Fig. 8.7 Fig. 8.8 523
524 rinipl Stresses under Given Loding 8.9 through 8.14 Eh of the following prolems refers to rolled-steel shpe seleted in prolem of hp. 5 to support given loding t miniml ost while stisfing the requirement s m # s ll. For the seleted design, determine () the tul vlue of s m in the em, () the mimum vlue of the prinipl stress s m t the juntion of flnge nd the we. 8.9 Loding of ro. 5.73 nd seleted W530 3 66 shpe. 8.10 Loding of ro. 5.74 nd seleted W530 3 92 shpe. 8.11 Loding of ro. 5.77 nd seleted S15 3 42.9 shpe. 8.12 Loding of ro. 5.78 nd seleted S12 3 31.8 shpe. 8.13 Loding of ro. 5.75 nd seleted S460 3 81.4 shpe. 8.14 Loding of ro. 5.76 nd seleted S510 3 98.2 shpe. 8.15 The vertil fore 1 nd the horiontl fore 2 re pplied s shown to disks welded to the solid shft. nowing tht the dimeter of the shft is 1.75 in. nd tht t ll 5 8 ksi, determine the lrgest permissile mgnitude of the fore 2. 6 in. 2 8 in. 1 3 in. 10 in. 10 in. Fig. 8.15 8.16 The two 500-l fores re vertil nd the fore is prllel to the is. nowing tht t ll 5 8 ksi, determine the smllest permissile dimeter of the solid shft E. 60 mm 90 mm Q 100 mm 7 in. 500 l 7 in. 4 in. 6 in. 7 in. 7 in. 4 in. E 4 kn 80 mm 140 mm 500 l Fig. 8.16 8.17 For the ger-nd-shft sstem nd loding of ro. 8.16, determine the smllest permissile dimeter of shft E, knowing tht the shft is hollow nd hs n inner dimeter tht is 2 3 the outer dimeter. Fig. 8.18 8.18 The 4-kN fore is prllel to the is, nd the fore Q is prllel to the is. The shft is hollow. nowing tht the inner dimeter is hlf the outer dimeter nd tht t ll 5 60 M, determine the smllest permissile outer dimeter of the shft.
8.19 Negleting the effet of fillets nd of stress onentrtions, determine the smllest permissile dimeters of the solid rods nd. Use t ll 5 60 M. rolems 525 200 mm 500 N 180 mm 160 mm 1250 N Fig. 8.19 nd 8.20 8.20 nowing tht rods nd re of dimeter 24 mm nd 36 mm, respetivel, determine the mimum shering stress in eh rod. Neglet the effet of fillets nd of stress onentrtions. 90 M 8.21 It ws stted in Se. 8.3 tht the shering stresses produed in shft the trnsverse lods re usull muh smller thn those produed the torques. In the preeding prolems their effet ws ignored nd it ws ssumed tht the mimum shering stress in given setion ourred t point (Fig. 8.21) nd ws equl to the epression otined in Eq. (8.5), nmel, O () T t 5 J 2M2 2 1 T V M Show tht the mimum shering stress t point (Fig. 8.21), where the effet of the sher V is gretest, n e epressed s t 5 J 1M os 22 1 2 3 V 1 T 2 90 O T where is the ngle etween the vetors V nd M. It is ler tht the effet of the sher V nnot e ignored when t $ t. (int: Onl the omponent of M long V ontriutes to the shering stress t.) Fig. 8.21 () 8.22 ssuming tht the mgnitudes of the fores pplied to disks nd of ro. 8.15 re, respetivel, 1 5 1080 l nd 2 5 810 l, nd using the epressions given in ro. 8.21, determine the vlues of t nd t in setion () just to the left of, () just to the left of. M 8 in. 4 in. 3.5 in. E 8.23 The solid shfts nd EF nd the gers shown re used to trnsmit 20 hp from the motor M to mhine tool onneted to shft EF. nowing tht the motor rottes t 240 rpm nd tht t ll 5 7.5 ksi, determine the smllest permissile dimeter of () shft, () shft EF. 6 in. F 8.24 Solve ro. 8.23, ssuming tht the motor rottes t 360 rpm. Fig. 8.23
526 rinipl Stresses under Given Loding 8.25 The solid shft rottes t 360 rpm nd trnsmits 20 kw from the motor M to mhine tools onneted to gers E nd F. nowing tht t ll 5 45 M nd ssuming tht 10 kw is tken off t eh ger, determine the smllest permissile dimeter of shft. 0.2 m M 0.2 m 0.2 m F E 120 mm 120 mm Fig. 8.25 8.26 Solve ro. 8.25, ssuming tht the entire 20 kw is tken off t ger E. 8.27 The solid shft nd the gers shown re used to trnsmit 10 kw from the motor M to mhine tool onneted to ger. nowing tht the motor rottes t 240 rpm nd tht t ll 5 60 M, determine the smllest permissile dimeter of shft. 100 mm M 90 mm E Fig. 8.27 8.28 ssuming tht shft of ro. 8.27 is hollow nd hs n outer dimeter of 50 mm, determine the lrgest permissile inner dimeter of the shft.
8.29 The solid shft E rottes t 600 rpm nd trnsmits 60 hp from the motor M to mhine tools onneted to gers G nd. nowing tht t ll 5 8 ksi nd tht 40 hp is tken off t ger G nd 20 hp is tken off t ger, determine the smllest permissile dimeter of shft E. 8.4 Stresses under omined Lodings 527 M 4 in. 6 in. F 8 in. 6 in. 3 in. G 4 in. E 4 in. Fig. 8.29 8.30 Solve ro. 8.29, ssuming tht 30 hp is tken off t ger G nd 30 hp is tken off t ger. *8.4 STRESSES UNER OMINE LOINGS In hps. 1 nd 2 ou lerned to determine the stresses used entri il lod. In hp. 3, ou nled the distriution of stresses in lindril memer sujeted to twisting ouple. In hp. 4, ou determined the stresses used ending ouples nd, in hps. 5 nd 6, the stresses produed trnsverse lods. s ou will see presentl, ou n omine the knowledge ou hve quired to determine the stresses in slender struturl memers or mhine omponents under firl generl loding onditions. onsider, for emple, the ent memer E of irulr ross setion tht is sujeted to severl fores (Fig. 8.15). In order to determine the stresses produed t points or the given lods, we first pss setion through these points nd determine the fore-ouple sstem t the entroid of the setion tht is required to mintin the equilirium of portion. This sstem represents the internl fores in the setion nd, in generl, onsists of three F 1 F 2 F 3 F 4 F 5 Fig. 8.15 Memer E sujeted to severl fores. E F 6 The fore-ouple sstem t n lso e defined s equivlent to the fores ting on the portion of the memer loted to the right of the setion (see Emple 8.01).
528 rinipl Stresses under Given Loding M F 1 V F 3 V M T F 2 M () M V () V Fig. 8.17 Internl fores seprted into () those using norml stresses () those using shering stresses. () Fig. 8.18 Norml stresses nd shering stresses. () Fig. 8.19 omined stresses. p p Fig. 8.20 rinipl stresses nd orienttion of prinipl plnes. T Fig. 8.16 etermintion of internl fores t the setion for stress nlsis. fore omponents nd three ouple vetors tht will e ssumed direted s shown (Fig. 8.16). The fore is entri il fore tht produes norml stresses in the setion. The ouple vetors M nd M use the memer to end nd lso produe norml stresses in the setion. The hve therefore een grouped with the fore in prt of Fig. 8.17 nd the sums s of the norml stresses the produe t points nd hve een shown in prt of Fig. 8.18. These stresses n e determined s shown in Se. 4.14. On the other hnd, the twisting ouple T nd the shering fores V nd V produe shering stresses in the setion. The sums t nd t of the omponents of the shering stresses the produe t points nd hve een shown in prt of Fig. 8.18 nd n e determined s indited in Ses. 3.4 nd 6.3. The norml nd shering stresses shown in prts nd of Fig. 8.18 n now e omined nd displed t points nd on the surfe of the memer (Fig. 8.19). The prinipl stresses nd the orienttion of the prinipl plnes t points nd n e determined from the vlues of s, t, nd t t eh of these points one of the methods presented in hp. 7 (Fig. 8.20). The vlues of the mimum shering stress t eh of these points nd the orresponding plnes n e found in similr w. The results otined in this setion re vlid onl to the etent tht the onditions of ppliilit of the superposition priniple (Se. 2.12) nd of Sint-Vennt s priniple (Se. 2.17) re met. This mens tht the stresses involved must not eeed the proportionl limit of the mteril, tht the deformtions due to one of the lodings must not ffet the determintion of the stresses due to the others, nd tht the setion used in our nlsis must not e too lose to the points of pplition of the given fores. It is ler from the first of these requirements tht the method presented here nnot e pplied to plsti deformtions. Note tht our present knowledge llows ou to determine the effet of the twisting ouple T onl in the ses of irulr shfts, of memers with retngulr ross setion (Se. 3.12), or of thin-wlled hollow memers (Se. 3.13).
Two fores 1 nd 2, of mgnitude 1 5 15 kn nd 2 5 18 kn, re pplied s shown to the end of r, whih is welded to lindril memer of rdius 5 20 mm (Fig. 8.21). nowing tht the distne from to the is of memer is 5 50 mm nd ssuming tht ll stresses remin elow the proportionl limit of the mteril, determine () the norml nd shering stresses t point of the trnsverse setion of memer loted t distne 5 60 mm from end, () the prinipl es nd prinipl stresses t, () the mimum shering stress t. Internl Fores in Given Setion. We first reple the fores 1 nd 2 n equivlent sstem of fores nd ouples pplied t the enter of the setion ontining point (Fig. 8.22). This sstem, whih represents the internl fores in the setion, onsists of the following fores nd ouples: 1. entri il fore F equl to the fore 1, of mgnitude F 5 1 5 15 kn 2. shering fore V equl to the fore 2, of mgnitude V 5 2 5 18 kn 3. twisting ouple T of torque T equl to the moment of 2 out the is of memer : T 5 2 5 118 kn2150 mm2 5 900 N? m 4. ending ouple M, of moment M equl to the moment of 1 out vertil is through : M 5 1 5 115 kn2150 mm2 5 750 N? m 5. ending ouple M, of moment M equl to the moment of 2 out trnsverse, horiontl is through : M 5 2 5 118 kn2160 mm2 5 1080 N? m The results otined re shown in Fig. 8.23.. Norml nd Shering Stresses t oint. Eh of the fores nd ouples shown in Fig. 8.23 n produe norml or shering stress t point. Our purpose is to ompute seprtel eh of these stresses, nd then to dd the norml stresses nd dd the shering stresses. ut we must first determine the geometri properties of the setion. Geometri roperties of the Setion We hve 5 p 2 5 p10.020 m2 2 5 1.257 3 10 23 m 2 I 5 I 5 1 4p 4 5 1 4p10.020 m2 4 5 125.7 3 10 29 m 4 J 5 1 2p 4 5 1 2p10.020 m2 4 5 251.3 3 10 29 m 4 We lso determine the first moment Q nd the width t of the re of the ross setion loted ove the is. Relling tht 5 43p for semiirle of rdius, we hve nd Q 5 5 1 2 p2 4 3p 5 2 3 3 5 2 10.020 m23 3 5 5.33 3 10 26 m 3 t 5 2 5 210.020 m2 5 0.040 m Norml Stresses. We oserve tht norml stresses re produed t the entri fore F nd the ending ouple M, ut tht the ouple M EXMLE 8.01 Fig. 8.21 M Fig. 8.22 60 mm M V T 900 N m M Fig. 8.23 T F V 18 kn 50 mm 2 18 kn M 750 N m 1 15 kn 4 3 F 15 kn 529
530 52.5 M Fig. 8.24 107.4 M (M) 107.4 53.7 53.7 O Y Fig. 8.25 E F 2 s 2 p 128.8 M in 21.4 M Fig. 8.26 Fig. 8.27 53.7 M 75.1 M X 18 kn 52.5 p 22.2 s 22.8 15 kn does not produe n stress t, sine is loted on the neutrl is orresponding to tht ouple. etermining eh sign from Fig. 8.23, we write s 52 F 1 M 1750 N? m210.020 m2 5211.9 M 1 I 125.7 3 10 29 m 4 5211.9 M 1 119.3 M s 51107.4 M Shering Stresses. These onsist of the shering stress (t ) V due to the vertil sher V nd of the shering stress (t ) twist used the torque T. Relling the vlues otined for Q, t, I, nd J, we write 1t 2 V 51 VQ I t 51118 3 103 N21 5.33 3 10 26 m 3 2 1125.7 3 10 29 m 4 210.040 m2 5119.1 M 1t 2 twist 52 T 1900 N? m2 10.020 m2 52 5271.6 M J 251.3 3 10 29 m 4 dding these two epressions, we otin t t point. t 5 1t 2 V 1 1t 2 twist 5119.1 M 2 71.6 M t 5252.5 M (M) In Fig. 8.24, the norml stress s nd the shering stresses nd t hve een shown ting on squre element loted t on the surfe of the lindril memer. Note tht shering stresses ting on the longitudinl sides of the element hve een inluded.. rinipl lnes nd rinipl Stresses t oint. We n use either of the two methods of hp. 7 to determine the prinipl plnes nd prinipl stresses t. Seleting Mohr s irle, we plot point X of oordintes s 5 1107.4 M nd 2t 5 152.5 M nd point Y of oordintes s 5 0 nd 1t 5 252.5 M nd drw the irle of dimeter XY (Fig. 8.25). Oserving tht O 5 5 1 2 1107.42 5 53.7 M X 5 52.5 M we determine the orienttion of the prinipl plnes: tn 2u p 5 X 5 52.5 53.7 5 0.97765 2u p 5 44.4 i 15 kn u p 5 22.2 i 18 kn We now determine the rdius of the irle, R 5 2153.72 2 1 152.52 2 5 75.1 M nd the prinipl stresses, 18 kn 15 kn s m 5 O 1 R 5 53.7 1 75.1 5 128.8 M s min 5 O 2 R 5 53.7 2 75.1 5221.4 M The results otined re shown in Fig. 8.26.. Mimum Shering Stress t oint. This stress orresponds to points E nd F in Fig. 8.25. We hve t m 5 E 5 R 5 75.1 M Oserving tht 2u s 5 908 2 2u p 5 908 2 44.48 5 45.68, we onlude tht the plnes of mimum shering stress form n ngle u p 5 22.8 l with the horiontl. The orresponding element is shown in Fig. 8.27. Note tht the norml stresses ting on this element re represented O in Fig. 8.25 nd re thus equl to 153.7 M.
4.5 in. 0.90 in. 1.8 in. E G 4.5 in. J 2.5 in. T SMLE ROLEM 8.4 horiontl 500-l fore ts t point of rnkshft whih is held in stti equilirium twisting ouple T nd retions t nd. nowing tht the erings re self-ligning nd eert no ouples on the shft, determine the norml nd shering stresses t points, J,, nd L loted t the ends of the vertil nd horiontl dimeters of trnsverse setion loted 2.5 in. to the left of ering. 500 l 4.5 in. 4.5 in. 2.5 in. T 250 l 500 l 1.8 in. 250 l M 625 l in. E V 250 l L J T 900 l in. 0.9-in. dimeter G 6290 psi 6290 psi L () J 6290 psi 6290 psi 524 psi J L () 524 psi 0 0 8730 psi () J L 8730 psi 0 5770 psi 6290 psi J L 8730 psi 6810 psi 6290 psi 8730 psi SOLUTION Free od. Entire rnkshft. 5 5 250 l 1l M 5 0: 21500 l211.8 in.2 1 T 5 0 T 5 900 l? in. Internl Fores in Trnsverse Setion. We reple the retion nd the twisting ouple T n equivlent fore-ouple sstem t the enter of the trnsverse setion ontining, J,, nd L. V 5 5 250 l T 5 900 l? in. M 5 1250 l212.5 in.2 5 625 l? in. The geometri properties of the 0.9-in.-dimeter setion re 5 p10.45 in.2 2 5 0.636 in 2 I 5 1 4p10.45 in.2 4 5 32.2 3 10 23 in 4 J 5 1 2p10.45 in.2 4 5 64.4 3 10 23 in 4 Stresses rodued Twisting ouple T. Using Eq. (3.8), we determine the shering stresses t points, J,, nd L nd show them in Fig. (). t 5 T J 5 1900 l? in.210.45 in.2 64.4 3 10 23 in 4 5 6290 psi Stresses rodued Shering Fore V. The shering fore V produes no shering stresses t points J nd L. t points nd we first ompute Q for semiirle out vertil dimeter nd then determine the shering stress produed the sher fore V 5 250 l. These stresses re shown in Fig. (). Q 5 1 2 p2 4 3p 5 2 3 3 5 2 3 10.45 in.23 5 60.7 3 10 23 in 3 t 5 VQ It 5 1250 l2160.7 3 1023 in 3 2 5 524 psi 132.2 3 10 23 in 4 210.9 in.2 Stresses rodued the ending ouple M. Sine the ending ouple M ts in horiontl plne, it produes no stresses t nd. Using Eq. (4.15), we determine the norml stresses t points J nd L nd show them in Fig. (). s 5 0M 0 I 5 1625 l? in.210.45 in.2 32.2 3 10 23 in 4 5 8730 psi Summr. We dd the stresses shown nd otin the totl norml nd shering stresses t points, J,, nd L. 531
130 mm 50 kn 75 kn 100 mm 30 kn 25 mm 200 mm E G F 40 mm 20 mm 70 mm 140 mm V 30 kn 50 kn V 75 kn M 8.5 kn m G E F M 3 kn m 0.020 m M 8.5 kn m G 0.025 m 0.140 m M 3 kn m E F 0.045 m 0.025 m 0.040 m t 0.040 m 1 V (M) 66.0 M in O Z 33.0 33.0 R 2 p Y 1 0.0475 m 17.52 M (M) 13.98 in SMLE ROLEM 8.5 Three fores re pplied s shown t points,, nd of short steel post. nowing tht the horiontl ross setion of the post is 40 3 140-mm retngle, determine the prinipl stresses, prinipl plnes nd mimum shering stress t point. SOLUTION Internl Fores in Setion EFG. We reple the three pplied fores n equivlent fore-ouple sstem t the enter of the retngulr setion EFG. We hve V 5230 kn 5 50 kn V 5275 kn M 5 150 kn210.130 m2 2 175 kn210.200 m2 528.5 kn? m M 5 0 M 5 130 kn210.100 m2 5 3 kn? m We note tht there is no twisting ouple out the is. The geometri properties of the retngulr setion re 5 10.040 m210.140 m2 5 5.6 3 10 23 m 2 I 5 1 12 10.040 m210.140 m2 3 5 9.15 3 10 26 m 4 I 5 1 12 10.140 m210.040 m2 3 5 0.747 3 10 26 m 4 Norml Stress t. We note tht norml stresses s re produed the entri fore nd the ending ouples M nd M. We determine the sign of eh stress refull emining the sketh of the foreouple sstem t. s 51 1 0M 0 2 0 M 0 I I 50 kn 13 kn? m210.020 m2 18.5 kn? m210.025 m2 5 1 2 5.6 3 10 23 2 m 0.747 3 10 26 m 4 9.15 3 10 26 m 4 s 5 8.93 M 1 80.3 M 2 23.2 M s 5 66.0 M Shering Stress t. onsidering first the shering fore V, we note tht Q 5 0 with respet to the is, sine is on the edge of the ross setion. Thus V produes no shering stress t. The shering fore V does produe shering stress t nd we write Q 5 1 1 5 310.040 m210.045 m2410.0475 m2 5 85.5 3 10 26 m 3 t 5 V Q 5 175 kn2185.5 3 1026 m 3 2 t I t 19.15 3 10 26 m 4 5 17.52 M 210.040 m2 rinipl Stresses, rinipl lnes, nd Mimum Shering Stress t. We drw Mohr s irle for the stresses t point tn 2u p 5 17.52 33.0 2u p 5 27.96 u p 5 13.98 R 5 2133.02 2 1 117.522 2 5 37.4 M s m 5 O 5 O 1 R 5 33.0 1 37.4 s min 5 O 5 O 2 R 5 33.0 2 37.4 t m 5 37.4 M s m 5 70.4 M s min 527.4 M 532
ROLEMS 8.31 6-kip fore is pplied to the mhine element s shown. nowing tht the uniform thikness of the element is 0.8 in., determine the norml nd shering stresses t () point, () point, () point. 8 in. 8 in. 6 kips 35 8 in. 1.5 in. 1.5 in. d e f Fig. 8.31 nd 8.32 18 mm 8.32 6-kip fore is pplied to the mhine element s shown. nowing tht the uniform thikness of the element is 0.8 in., determine the norml nd shering stresses t () point d, () point e, () point f. 8.33 For the rket nd loding shown, determine the norml nd shering stresses t () point, () point. 8.34 through 8.36 Memer hs uniform retngulr ross setion of 10 3 24 mm. For the loding shown, determine the norml nd shering stresses t () point, () point. 20 mm 60 100 mm Fig. 8.33 4 kn 60 mm 9 kn 60 mm 9 kn 60 mm 9 kn G 30 60 mm 12 mm 30 G 60 mm 12 mm 30 G 60 mm 12 mm 12 mm 40 mm 12 mm 40 mm 12 mm 40 mm Fig. 8.34 Fig. 8.35 Fig. 8.36 533
534 rinipl Stresses under Given Loding 8.37 Severl fores re pplied to the pipe sseml shown. nowing tht the pipe hs inner nd outer dimeters equl to 1.61 nd 1.90 in., respetivel, determine the norml nd shering stresses t () point, () point. 200 l 150 l t 8 mm 50 mm 20 mm 4 in. 4 in. 150 l 10 in. 225 mm Fig. 8.37 6 in. 50 l Fig. 8.38 60 E 8.38 The steel pile hs 100-mm outer dimeter nd n 8-mm wll thikness. nowing tht the tension in the le is 40 kn, determine the norml nd shering stresses t point. 8.39 The illord shown weighs 8000 l nd is supported struturl tue tht hs 15-in. outer dimeter nd 0.5-in. wll thikness. t time when the resultnt of the wind pressure is 3 kips, loted t the enter of the illord, determine the norml nd shering stresses t point. 6 ft 3 ft 9 ft 8 kips l 3 kips 2 ft 3 ft 3 ft 8 ft Fig. 8.39 Fig. 8.40 F 8.40 thin strp is wrpped round solid rod of rdius 5 20 mm s shown. nowing tht l 5 100 mm nd F 5 5 kn, determine the norml nd shering stresses t () point, () point.
8.41 vertil fore of mgnitude 60 l is pplied to the rnk t point. nowing tht the shft E hs dimeter of 0.75 in., determine the prinipl stresses nd the mimum shering stress t point loted t the top of the shft, 2 in. to the right of support. 1 in. rolems 535 8.42 13-kN fore is pplied s shown to the 60-mm-dimeter st-iron post. t point, determine () the prinipl stresses nd prinipl plnes, () the mimum shering stress. E 2 in. 60 8 in. 5 in. Fig. 8.41 13 kn 300 mm 150 mm 100 mm E 125 mm 1.4 kn m 10 kn Fig. 8.42 240 mm 8.43 10-kN fore nd 1.4-kN? m ouple re pplied t the top of the 65-mm dimeter rss post shown. etermine the prinipl stresses nd mimum shering stress t () point, () point. Fig. 8.43 8.44 Fores re pplied t points nd of the solid st-iron rket shown. nowing tht the rket hs dimeter of 0.8 in., determine the prinipl stresses nd the mimum shering stress t () point, () point. 1 in. 50 kips 0.9 in. 2 kips 2.4 in. 0.9 in. 2 in. 2500 l 2.5 in. 3.5 in. h 10.5 in. 1.2 in. 1.2 in. 6 kips 600 l Fig. 8.44 8.45 Three fores re pplied to the r shown. etermine the norml nd shering stresses t () point, () point, () point. 8.46 Solve ro. 8.45, ssuming tht h 5 12 in. Fig. 8.45 4.8 in. 1.8 in.
536 rinipl Stresses under Given Loding 8.47 Three fores re pplied to the r shown. etermine the norml nd shering stresses t () point, () point, () point. 60 mm 24 mm 15 mm 180 mm 32 mm 750 N 40 mm 30 mm 16 mm 50 mm 50 mm 375 mm 120 kn 75 mm 75 mm 50 kn 30 Fig. 8.47 8.48 Solve ro. 8.47, ssuming tht the 750-N fore is direted vertill upwrd. 8.49 For the post nd loding shown, determine the prinipl stresses, prinipl plnes, nd mimum shering stress t point. 500 N 10 kn 8.50 For the post nd loding shown, determine the prinipl stresses, prinipl plnes, nd mimum shering stress t point. Fig. 8.49 nd 8.50 8.51 Two fores re pplied to the smll post s shown. nowing tht the vertil portion of the post hs ross setion of 1.5 3 2.4 in., determine the prinipl stresses, prinipl plnes, nd mimum shering stress t point. 6000 l 500 l 4 in. 1.5 in. 2.4 in. 6 in. 1 in. 3.25 in. 1.75 in. Fig. 8.51
8.52 Solve ro. 8.51, ssuming tht the mgnitude of the 6000-l fore is redued to 1500 l. 8.53 Three steel pltes, eh 13 mm thik, re welded together to form ntilever em. For the loding shown, determine the norml nd shering stresses t points nd. rolems 537 d e 400 mm 60 mm 30 mm 60 mm 75 mm 9 kn 150 mm t 13 mm 13 kn Fig. 8.53 nd 8.54 8.54 Three steel pltes, eh 13 mm thik, re welded together to form ntilever em. For the loding shown, determine the norml nd shering stresses t points d nd e. 8.55 Two fores re pplied to W8 3 28 rolled-steel em s shown. etermine the prinipl stresses nd mimum shering stress t point. 90 kips W8 28 4 in. 24 in. 20 kips Fig. 8.55 nd 8.56 8.56 Two fores re pplied to W8 3 28 rolled-steel em s shown. etermine the prinipl stresses nd mimum shering stress t point.
538 rinipl Stresses under Given Loding 8.57 Two fores 1 nd 2 re pplied s shown in diretions perpendiulr to the longitudinl is of W310 3 60 em. nowing tht 1 5 25 kn nd 2 5 24 kn, determine the prinipl stresses nd the mimum shering stress t point. 75 mm 1 0.6 m 2 1.2 m W310 60 Fig. 8.57 nd 8.58 8.58 Two fores 1 nd 2 re pplied s shown in diretions perpendiulr to the longitudinl is of W310 3 60 em. nowing tht 1 5 25 kn nd 2 5 24 kn, determine the prinipl stresses nd the mimum shering stress t point. 8.59 vertil fore is pplied t the enter of the free end of ntilever em. () If the em is instlled with the we vertil ( 5 0) nd with its longitudinl is horiontl, determine the mgnitude of the fore for whih the norml stress t point is 1120 M. () Solve prt, ssuming tht the em is instlled with 5 38. l 1.25 m W250 44.8 h Fig. 8.59 l Fig. 8.60 8.60 fore is pplied to ntilever em mens of le tthed to olt loted t the enter of the free end of the em. nowing tht ts in diretion perpendiulr to the longitudinl is of the em, determine () the norml stress t point in terms of,, h, l, nd, () the vlues of for whih the norml stress t is ero.
*8.61 5-kN fore is pplied to wire tht is wrpped round r s shown. nowing tht the ross setion of the r is squre of side d 5 40 mm, determine the prinipl stresses nd the mimum shering stress t point. rolems 539 d 2 d Fig. 8.61 *8.62 nowing tht the struturl tue shown hs uniform wll thikness of 0.3 in., determine the prinipl stresses, prinipl plnes, nd mimum shering stress t () point, () point. 3 in. *8.63 The struturl tue shown hs uniform wll thikness of 0.3 in. nowing tht the 15-kip lod is pplied 0.15 in. ove the se of the tue, determine the shering stress t () point, () point. 2 in. 10 in. 4 in. 6 in. 0.15 in. 3 in. 9 kips Fig. 8.62 1.5 in. 2 in. 4 in. 15 kips 10 in. Fig. 8.63 *8.64 For the tue nd loding of ro. 8.63, determine the prinipl stresses nd the mimum shering stress t point.
REVIEW N SUMMRY This hpter ws devoted to the determintion of the prinipl stresses in ems, trnsmission shfts, nd odies of ritrr shpe sujeted to omined lodings. We first relled in Se. 8.2 the two fundmentl reltions derived in hps. 5 nd 6 for the norml stress s nd the shering stress t t n given point of ross setion of prismti em, s 52 M I t 52 VQ It (8.1, 8.2) where V 5 sher in the setion M 5 ending moment in the setion 5 distne of the point from the neutrl surfe I 5 entroidl moment of inerti of the ross setion Q 5 first moment out the neutrl is of the portion of the ross setion loted ove the given point t 5 width of the ross setion t the given point rinipl plnes nd prinipl stresses in em O in Fig. 8.28 in Using one of the methods presented in hp. 7 for the trnsformtion of stresses, we were le to otin the prinipl plnes nd prinipl stresses t the given point (Fig. 8.28). We investigted the distriution of the prinipl stresses in nrrow retngulr ntilever em sujeted to onentrted lod t its free end nd found tht in n given trnsverse setion eept lose to the point of pplition of the lod the mimum prinipl stress s m did not eeed the mimum norml stress s m ourring t the surfe of the em. While this onlusion remins vlid for mn ems of nonretngulr ross setion, it m not hold for W-ems or S-ems, where s m t the juntions nd d of the we with the flnges of the em (Fig. 8.29) m eeed the vlue of s m ourring t points nd e. Therefore, the design of rolled-steel em should inlude the omputtion of the mimum prinipl stress t these points. (See Smple ros. 8.1 nd 8.2.) d e 540 Fig. 8.29
In Se. 8.3, we onsidered the design of trnsmission shfts sujeted to trnsverse lods s well s to torques. Tking into ount the effet of oth the norml stresses due to the ending moment M nd the shering stresses due to the torque T in n given trnsverse setion of lindril shft (either solid or hollow), we found tht the minimum llowle vlue of the rtio J for the ross setion ws J 5 2M 2 1 T 2 m t ll (8.6) In preeding hpters, ou lerned to determine the stresses in prismti memers used il lodings (hps. 1 nd 2), torsion (hp. 3), ending (hp. 4), nd trnsverse lodings (hps. 5 nd 6). In the seond prt of this hpter (Se. 8.4), we omined this knowledge to determine stresses under more generl loding onditions. Review nd Summr esign of trnsmission shfts under trnsverse lods Stresses under generl loding onditions 541 M F 1 F 5 E F 6 F 1 F 3 V V M T F 2 F 3 F 4 F 2 Fig. 8.30 Fig. 8.31 For instne, to determine the stresses t point or of the ent memer shown in Fig. 8.30, we pssed setion through these points nd repled the pplied lods n equivlent fore-ouple sstem t the entroid of the setion (Fig. 8.31). The norml nd shering stresses produed t or eh of the fores nd ouples pplied t were determined nd then omined to otin the resulting norml stress s nd the resulting shering stresses t nd t t or. Finll, the prinipl stresses, the orienttion of the prinipl plnes, nd the mimum shering stress t point or were determined one of the methods presented in hp. 7 from the vlues otined for s, t, nd t.
REVIEW ROLEMS 8.65 () nowing tht s ll 5 24 ksi nd t ll 5 14.5 ksi, selet the most eonomil wide-flnge shpe tht should e used to support the loding shown. () etermine the vlues to e epeted for s m, t m, nd the prinipl stress s m t the juntion of flnge nd the we of the seleted em. 1.5 kips/ft 12 ft 6 ft Fig. 8.65 8.66 etermine the smllest llowle dimeter of the solid shft, knowing tht t ll 5 60 M nd tht the rdius of disk is r 5 80 mm. r 150 mm 8.67 Using the nottion of Se. 8.3 nd negleting the effet of shering stresses used trnsverse lods, show tht the mimum norml stress in irulr shft n e epressed s follows: Fig. 8.66 150 mm T 600 N m s m 5 J 31M2 1 M 2 2 1 2 1 1M 2 1 M 2 1 T 2 2 1 2 4 m 8.68 The solid shft rottes t 450 rpm nd trnsmits 20 kw from the motor M to mhine tools onneted to gers F nd G. nowing tht t ll 5 55 M nd ssuming tht 8 kw is tken off t ger F nd 12 kw is tken off t ger G, determine the smllest permissile dimeter of shft. 150 mm F 225 mm 225 mm 60 mm M 150 mm 100 mm 60 mm E G 542 Fig. 8.68
8.69 Two 1.2-kip fores re pplied to n L-shped mhine element s shown. etermine the norml nd shering stresses t () point, () point, () point. Review rolems 543 1.8 in. 0.5 in. 1.2 kips 12 in. 1.2 kips 6 in. 1.0 in. 45 mm 45 mm 1500 N 1200 N Fig. 8.69 1.0 in. 75 mm 8.70 Two fores re pplied to the pipe s shown. nowing tht the pipe hs inner nd outer dimeters equl to 35 nd 42 mm, respetivel, determine the norml nd shering stresses t () point, () point. 20 mm 8.71 lose-oiled spring is mde of irulr wire of rdius r tht is formed into heli of rdius R. etermine the mimum shering stress produed the two equl nd opposite fores nd 9. (int: First determine the sher V nd the torque T in trnsverse ross setion.) Fig. 8.70 R R 8.72 Three fores re pplied to 4-in.-dimeter plte tht is tthed to the solid 1.8-in. dimeter shft. t point, determine () the prinipl stresses nd prinipl plnes, () the mimum shering stress. 6 kips 2 in. 2 in. 6 kips r V T 2.5 kips ' Fig. 8.71 8 in. Fig. 8.72
544 rinipl Stresses under Given Loding 8.73 nowing tht the rket hs uniform thikness of 5 8 in., determine () the prinipl plnes nd prinipl stresses t point, () the 3 kips mimum shering stress t point. 30 5 in. 2 in. 2.5 in. 8.74 Three fores re pplied to the mhine omponent s shown. nowing tht the ross setion ontining point is 20 3 40-mm retngle, determine the prinipl stresses nd the mimum shering stress t point. Fig. 8.73 50 mm 20 mm 150 mm 40 mm 0.5 kn 3 kn 2.5 kn 160 mm Fig. 8.74 8.75 nowing tht the struturl tue shown hs uniform wll thikness of 0.25 in., determine the norml nd shering stresses t the three points indited. 3 in. 6 in. 600 l 600 l 1500 l 5 in. 1500 l 2.75 in. 0.25 in. 3 in. 20 in. 300 mm 40 mm 60 mm 600 N Fig. 8.76 Fig. 8.75 8.76 The ntilever em will e instlled so tht the 60-mm side forms n ngle etween 0 nd 908 with the vertil. nowing tht the 600-N vertil fore is pplied t the enter of the free end of the em, determine the norml stress t point when () 5 0, () 5 908. () lso, determine the vlue of for whih the norml stress t point is mimum nd the orresponding vlue of tht stress.
OMUTER ROLEMS The following prolems re designed to e solved with omputer. 8.1 Let us ssume tht the sher V nd the ending moment M hve een determined in given setion of rolled-steel em. Write omputer progrm to lulte in tht setion, from the dt ville in ppendi, () the mimum norml stress s m, () the prinipl stress s m t the juntion of flnge nd the we. Use this progrm to solve prts nd of the following prolems: (1) ro. 8.1 (Use V 5 45 kips nd M 5 450 kip? in.) (2) ro. 8.2 (Use V 5 22.5 kips nd M 5 450 kip? in.) (3) ro. 8.3 (Use V 5 700 kn nd M 5 1750 kn? m) (4) ro. 8.4 (Use V 5 850 kn nd M 5 1700 kn? m) 8.2 ntilever em with retngulr ross setion of width nd depth 2 supports single onentrted lod t its end. Write omputer progrm to lulte, for n vlues of nd, () the rtios s m s m nd s min s m, where s m nd s min re the prinipl stresses t point (, ) nd s m the mimum norml stress in the sme trnsverse setion, () the ngle u p tht the prinipl plnes t form with trnsverse nd horiontl plne through. Use this progrm to hek the vlues shown in Fig. 8.8 nd to verif tht s m eeeds s m if # 0.544, s indited in the seond footnote on pge 517. in Fig. 8.2 p 8.3 isks 1, 2,..., n re tthed s shown in Fig. 8.3 to the solid shft of length L, uniform dimeter d, nd llowle shering stress t ll. Fores 1, 2,..., n of known mgnitude (eept for one of them) re pplied to the disks, either t the top or ottom of its vertil dimeter, or t the left or right end of its horiontl dimeter. enoting r i the rdius of disk i nd i its distne from the support t, write omputer progrm to lulte () the mgnitude of the unknown fore i, () the smllest permissile vlue of the dimeter d of shft. Use this progrm to solve ro. 8.18. i 1 L r i n 1 2 2 i i n Fig. 8.3 545
546 rinipl Stresses under Given Loding 8.4 The solid shft of length L, uniform dimeter d, nd llowle shering stress t ll rottes t given speed epressed in rpm (Fig. 8.4). Gers G 1, G 2,..., G n re tthed to the shft nd eh of these gers meshes with nother ger (not shown), either t the top or ottom of its vertil dimeter, or t the left or right end of its horiontl dimeter. One of these gers is onneted to motor nd the rest of them to vrious mhine tools. enoting r i the rdius of disk G i, i its distne from the support t, nd i the power trnsmitted to tht ger (1 sign) or tken off tht ger (2 sign), write omputer progrm to lulte the smllest permissile vlue of the dimeter d of shft. Use this progrm to solve ros. 8.27 nd 8.68. i L r i G 1 G 2 G i G n Fig. 8.4 8.5 Write omputer progrm tht n e used to lulte the norml nd shering stresses t points with given oordintes nd loted on the surfe of mhine prt hving retngulr ross setion. The internl fores re known to e equivlent to the fore-ouple sstem shown. Write the progrm so tht the lods nd dimensions n e epressed in either SI or U.S. ustomr units. Use this progrm to solve () ro. 8.45, () ro. 8.47. M V h V M Fig. 8.5