### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Machine Design I ME 3650

WMU

GPA 3.52

### View Full Document

## 50

## 0

## Popular in Course

## Popular in Mechanical Engineering

This 64 page Class Notes was uploaded by Mortimer Rutherford on Wednesday September 30, 2015. The Class Notes belongs to ME 3650 at Western Michigan University taught by Judah Ari-Gur in Fall. Since its upload, it has received 50 views. For similar materials see /class/216917/me-3650-western-michigan-university in Mechanical Engineering at Western Michigan University.

## Reviews for Machine Design I

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/30/15

SOLUTION 101 Known A doublethreaded Acme screw of known major diameter is used in a jack having a plain thrust collar of known mean diameter Coef cients of running friction are estimated as 010 for the collar and 013 for the screw Find a Determine the pitch lead thread depth mean pitch diameter and helix angle of the screw b Estimate the starting torque for raising and for lowering a 10000 lb load c If the screw is lifting a 10000 lb load determine the ef ciency of the jack Schematic and Given Data Doublethreaded Acme mad screw 10000 lb 1 1 in l l L l dc 20 in f fC 010 5 fc 2 5 I dm I dc Assumptions 1 The starting friction is about 13 higher than running friction 2 The screw is not exposed to vibration Analysis 1 From Table 103 there are 5 threads per inch p 15 02 in I Because of the doublethreaded screw L 2p 04 in I From Fig 104a Threaded depth 05p 010 in I dm d 05p 090 in I From Eq 101 1 L 1 04 o k tan ndm tan 090 805 I 2 For starting increase the given coef cients of friction by 13 fc 0133f 0173 From Eq 106 an tan1 tan 0L cos x tan1 tan 145O cos 8050 1436quot 101 6 7 8 From Eq 104 szcdc T def1rdm Lcos an 2 ndmcos ocn f L 10000090 l 0173nlt090gt04cos 1436 l 7 lno90cos 1436 017304J 10000013320 2 14775 1330 28075 lbin to raise the load I From Eq 105 de 2 f Ttdm Lcos 1 chdc 2 T ndmcos 0cn f L 10000090 01731t09004cos 1436 2 M090cos 1436 017304J 10000013320 2 16289 1330 149289 lbin to lower the load I From Eq 104 with fC 010f 013 10000090 013m090 04cos 14363 1000001020 2 Tl090oos 1436 01304 j 2 12645 1000 22645 lbin From Eq 104 the friction free torque for raising the load is 10000090 04cos 1436 2 7t090cos 1436 J Ef ciency 636622645 28 I Work input to the screw during one revolution 21tT 21t22645 142288 lbin Work output during one revolution WL 10000202 40001boin Ef ciency Work outWork in 4000142288 28 T T 6366 lbin Comments 1 2 For a double threaded screw the work output during one revolution is WL where L 2p If a small thrust bearing were used so that the collar friction could be neglected the ef ciency would increase to 6366 12645 50 10 2 SOLUTION 102 Known A squarethreaded single thread power screw is used to raise a known load The screw has a mean diameter of 1 in and four threads per inch The collar mean diameter is 175 in The coef cient of friction is estimated as 01 for both the thread and the collar Find a Determine the major diameter of the screw b Estimate the screw torque required to raise the load c If collar friction is eliminated determine the minimum value of thread coef cient of friction needed to prevent the screw from overhauling Schematic and Given Data Load 13750 lb m 1 in l 1 1 1 4 threadsin f f 01 0 fC 9quot 5 dm dC Assumption The screw is not exposed to vibration Analysis a From Fig 104c ddmg1Ol125in l b From Eq 104a T de fndmL chdC 2 mm fL 2 137501 011I1025 I 1375001175 2 11 01025J 2 1245 1b in 1203 lb in 2448 lb in I c From Eq 107a the screw is selflocking if L 025 008 f ndm 10 103 f 2 008 I Therefore the minimum value of thread coef cient of friction needed to prevent the screw from overhauling is 008 SOLUTION 103 Known A doublethreaded Acme screw of known major diameter is used in a jack having a plain thrust collar of known mean diameter Coef cients of running friction are estimated as 009 for the collar and 012 for the screw Find a Determine the pitch lead thread depth mean pitch diameter and helix angle of the screw b Estimate the starting torque for raising and for lowering a 4000 N load c If the screw is lifting a 4000 N load determine the ef ciency of the jack Schematic and Given Data Double threaded Acme Load screw 4000 N d 1 in l I l dc 50 mm f g fc 5 f 012 f 5 l dm I dc gt Assumptions 1 The starting friction is about 13 higher than running friction 2 The screw is not exposed to vibration Analysis 1 From Table 103 there are 5 threads per inch p 15 02 in 00051 m I Because of the doublethreaded screw L 2p 04 in 00102 In I From Fig 104a Threaded depth 05p 010 in 000254 m I dm d 05p 090 in 002286 m I From Eq 101 1 L 1 04 o k tan ndm tan 0901805 I 1 04 2 OOOLII For starting increase the coef cient of friction by 13 fc 012f 016 From Eq 106 an tan1 tan a cos k tan1 tan 145o cos 8050 14360 From Eq 104 T de fndm Lcos and ch dC 2 ndmcos an f L 2 016111002286 00102cos 1436 4000002286 111002286cos 1436quot O160 0102 j 2 4000012005 7 7175 12 19175 Nm to raise the load I From Eq 105 f ndm Lcos an dmcos an f L mm 2 chdc T 2 4000002286 016Tc002286 00102cos 1436quot l 2 m0 02286cos 1436 0 160 0102 4000012005 2 5819 12 618 Nm to lower the load I From Eq 104 with fC 009f 012 T 4000002286 0121c002286 00102Xcos 1436 4000009005 m2 113002286cos 1436 01200102 J 2 700 9 1600 Nm From Eq 104 the friction free torque for raising the load is 4000002286 00102me 1436 2 113002286cos 1436a 649 Nm T Ef ciency 6491600 406 I Work input to the screw during one revolution 21tT 210600 1005 Nm Work output during one revolution 2 WL 4000200051 408 N m Ef ciency Work outWork in 4081005 406 105 Comments 1 For a double threaded screw the work output during one revolution is WL where L 2p 2 If a small thrust bearing were used so that the collar friction could be neglected the ef ciency would increase to 649700 927 SOLUTION 104 Known A square threaded single thread power screw is used to raise a known load The screw has a mean diameter of 1 in and four threads per inch The collar mean diameter is 15 in The coef cient of friction is estimated as 01 for both the thread and the collar Find a Determine the major diameter of the screw b Estimate the screw torque required to raise the load c If collar friction is eliminated determine the minimum value of thread coef cient of friction needed to prevent the screw from overhauling Schematic and Given Data Load 250001b dm 1 in L r J 4 threadsin f g dc 15 In x r rc 01 r 39 f C M 9quot dIn I dC Assumption The screw is not exposed to vibration Analysis a From Fig 104c ddmg1O 22 5 1125in I b From Eq 104a T def1cdmL chdc 2 ndm fL 2 10 6 250001 01mm 025 1 250000115 2 lnlt1 01025J 2 2263 1b in 1875 lb in 4138 1b in I C From Eq 107a the screw is selflocking if gt L 025 O 08 f ndm Tt1 39 f 2 008 I Therefore the minimum value of thread coef cient of friction needed to prevent the screw from overhauling is 008 SOLUTION 105 Known A doublethreaded Acme stub screw of known major diameter is used in a jack having a plain thrust collar of known mean diameter Coef cients of running friction are estimated as 010 for the collar and 011 for the screw Find a Determine the pitch lead thread depth mean pitch diameter and helix angle of the screw b Estimate the starting torque for raising and for lowering a 5000 lb load 0 If the screw is lifting a 5000 lb load at the rate of 4 ftmin determine the screw rpm Also determine the ef ciency of the jack under this steadystate condition d Determine if the screw will overhaul if a ball thrust bearing of negligible friction were used in place of the plain thrust collar Schematic and Given Data Doublethreaded Acme Load stub screw 3500 lb d 2 in LI r l dc 275 in f fc 010 r o 11 Assumptions 1 The starting friction is about 13 higher than running friction 2 The screw is not exposed to vibration 107 Analysis 1 From Table 103 there are 4 threads per inch p 14 025 in Because of the doublethreaded screw L 2p 050 in From Fig 104 b Threaded depth 2 03p 0075 in dm d 03p 1925 in From Eq 101 7 tan 1 L tan1 05 473 dm 19251 2 For starting increase the coef cients of friction by 13 fc 0133f 0147 From Eq 106 an tan 1 tan 01 cos k tan391 tan 145 cos 473 1445quot From Eq 104 Tde Ndm Lcos an 2 1rdmcos 01nfL ch 1C 2 35000 925 01471t192505cos 1445quot l 35000133275 2 111 925cos 1445quot 01470 5 2 7999 6401 1440 lb in to raise the load I From Eq 105 2 1tdmcos 01nfL ch d 2 C 35001925 01471t192505008 1445quot 1 35000133275 2 11925cos 1445quot 014705J 2 2300 6401 8701 lb in to lower the load I 4121nm1n 96 rpm I 05 inrev From Eq 104w391thfc 01f 011 35000925 01115192505cos 1445quot I 35000 12 75 2 1151925cos 1445 01105J 2 6671 4813 11484 1b in From Eq 13 T Win 175hp 35004 wont Therefore efficiency 9i47 024 24 I 4 From Eq 107 the screw is self locking if gt L cos an 05cos 1445quot f ndm n1925 03908 Thus if f 011 the screw is self locking and not overhauling I SOLUTION 106 Known A squarethreaded single thread power screw is used to raise a known load The screw has a mean diameter of 1 in and four threads per inch The collar mean diameter is 15 in The coef cient of friction is estimated as 01 for both the thread and the collar Find a Determine the major diameter of the screw b Estimate the screw torque required to raise the load c If collar friction is eliminated determine the minimum value of thread coef cient of friction needed to prevent the screw from overhauling Schematic and Given Data Load 12500 lb dm 1 in LEC39 4 threadsin f dc 15 in w r r 01 Assumption The screw is not exposed to vibration Analysis a From Fig 104c ddm 10 22 1125m l 10 9 b From Eq 104a T de fTEdmL chdC 2 ndm fL 2 125001 01n1025 I 12500O1175 39 2 mu 01msz 2 1132 lb in 937 lb in 2069 lb in I C From Eq 10721 the screw is selflocking if 1 M f Mm 1 008 f 2 008 I Therefore the minimum value of thread coef cient of friction needed to prevent the screw from overhauling is 008 SOLUTION 107 Known A doublethreaded Acme stub screw of known major diameter is used in a jack having a plain thrust collar of known mean diameter Coef cients of running friction are estimated as 010 for the collar and 011 for the screw Find a Determine the pitch lead thread depth mean pitch diameter and helix angle of the screw b Estimate the starting torque for raising and for lowering a 5000 lb load 0 If the screw is lifting a 5000 lb load at the rate of 4 ftmin determine the screw rpm Also determine the ef ciency of the jack under this steady state condition d Determine if the screw will overhaul if a ball thrust bearing of negligible friction were used in place of the plain thrust collar Schematic and Given Data Double threaded Acme Load stub screw 50001b d 2 in l I I J dc in y w fc g r011 1010 Assumptions 1 The starting friction is about 13 higher than running friction 2 The screw is not exposed to vibration Analysis 1 From Table 103 there are 4 threads per inch p 14 025 in Because of the double threaded screw L 2p 050 in From Fig 104 b Threaded depth 03p 0075 in dm d 03p 1925 in From Eq 101 A tan1L talrl 473 dm 19251 2 For starting increase the coef crents of frictlon by 13 fC 0133f 0147 From Eq 106 an tan1 tan a cos 1 tan1 tan 145 cos 473 1445quot From Eq 104 de 2 f ndm Lcos 01 ndmcos an f L chdc T 2 50001925 01471t1925 05cos 1445quot 5000013325 2 1t1925c0s 1445 014705 2 11427 8313 19741b in to raise the load I From Eq 105 T de fndm Lcos an ch dC 2 ndmcos an f L 2 50001925 0147n1925 05cos 1445quot 5000013325 2 1t1925cos 1445 014705 2 3285 8313 1160 lb in to lower the load I 05 mfrev From Eq 104 withfc 01f 011 50001925 0111t1925 05cos 1445quot 50000125 2 1t1925cos 1445 011O5 2 953 625 15781b in T 10 11 From Eq 13 Win3 5252 240h 50004 0606 Therefore ef ciency O 025 25 From Eq 107 the screw is self locking if L cos an 05cos 1445 O 08 f 2 ndm n1925 Thus if f 011 the screw is selflocking and not overhauling SOLUTION 108 Known A jack uses a single squarethread screw to raise a known load The major diameter and pitch of the screw and the thrust collar mean diameter are known Running friction coef cients are estimated Find a b C d Determine the thread depth and helix angle Estimate the starting torque for raising and lowering the load Estimate the ef ciency of the jack for raising the load Estimate the power required to drive the screw at a constant 1 revolution per second Schematic and Given Data Assumption The starting friction is about 13 higher than running friction Load Single squarethread screw 50 kN d 36 mm ip 3c 68131 m f r 015 N T re 012 fc 3 I dun dC 1012 Analysis a From Fig 104c Thread depth p2 62 3 m I From Eq 101 2 1 L p 2 tan ndm where dm d 2 33 mm 16 o k tan M33 331 I b For starting increase the coef cients of friction by 13 then f 020fc 016 From Eq 104a zm 2 fndm L ndm fL chdc T 2 500000033 I 020n0033 0006 2 0033 0200006 215 320 535 Nom to raise the load I From Eq 105a 500000160080 2 chdc T 2 z def1tdm L 2 Trdm fL O201t0033 0006 1t0033 0200006 500000033 2 116 320 436 Nam to lower the load I c From Eq 104a withf 015fc 012 500000160080 2 T 500000033 015n0033 0006 1 500000120080 2 1r0033 0150006 2 173 240 413 Nom Work input to the screw during one revolution 21tT 21t413 2595 Nom Work output during one revolution W0 p 500000006 300 Nam Check Torque during load raising with f f c 0 500000033 T 2 478 Nom 0 0006 7t0033 0 Check partial Torque during load raising if collar friction is eliminated 173 Mm Ef ciency screw only 28 173 d From Eq 12 nT 60413 SOLUTION 109 Known An ordinary Cclamp uses a 12 in Acme thread and a collar of 58 in mean diameter Find Estimate the force required at the end of a S in handle to develop a 200 lb clamping force Schematic and Given Data 12 in Acme thread dc 58 in Assumptions 1 Coef cients of running friction are estimated as 015 for both the collar and the screw 2 The screw has a single thread 10 14 Analysis 1 From section 1031 and considering that service conditions may be conducive to relatively high friction estimate f f c z 015 for running friction 2 From Table 103 p 01 in and with a single thread L 01 in 3 From Fig 104a dm mg 05 005 045 in or 1450 From Eq 101 1 L 1 01 7 ta rcdm 39 ta 7c045 From Eq 106 an tan391 tan a cos A tan1 tan 145O cos 4050 14470 Note with k z 4 it is obvious that an z or and well within the accuracy of assumed friction coef cients 4 From Eq 104 4050 T de fndm Lcos an ch dC 2 ndmcos an fL 2 200045 0151r045 01cos 1447quot 39 2000150625 2 Jt045cos 1447 01501 2 1027 937 1964 lb in Use T x 20 lb in At the end of a 5 in handle the clamping force required 2 4 lb I SOLUTION 1010 Known Two identical 3 in major diameter screws single threaded with modi ed square threads are used to raise and lower a SOton sluice gate of a dam An estimated friction coef cient is only 01 for the screw Because of gate friction each screw must provide a lifting force of 26 tons Find Determine the power required to drive each screw when the gate is being raised at the rate of 3 ftmin Also calculate the corresponding rotating speed of the screws 10 15 Schematic and Given Data Motor 1 l f quot111 E 3 in major E nil5 2 diameter 315 quot 5 power screw if139 f01 Q n E l Z V3ftmin 5 iiiziaiiieiiiaiiid 5555 t F Erhulka1pza5513913952139539knit3539 5 391 5139 395 Assumption Collar friction can be neglected Analysis 1 From Table 103 p L 1175 0571 in 2 From Fig 104d B 0571 dm d 2 3 2 271m and or 250 3 Since cos or 0999 use Eq 104a E m 2 frtdm L T ch dc 5200027l Ol1t271 0571 2 1r271 010571 11851 1b in 988 lb ft during load raising 4 To raise the gate 36 inrnin with L 0571 in requires 360571 6305 z 63 rpm 5 From Fig 13 6305988 W 5 j 119 hp say 12 hp Therefore 12 hp are required to drive each screw 1016 0 Check Work output per gate 52000 lb 3 ftmin 156000 lb ftmin 473 hp k tan 1 L tan 1 0571 rcdm 1t271 From Fig 108 Ef ciency 2 40 Thus wreqmd 1674 3 1183 hp 3840 SOLUTION 1011 Known The power screws in Problems a 104 b 107 c 108 d 109 and e 1010 are given Find Calculate the nominal values of torsional axial thread bearing and thread shear stresses Schematic and Given Data Assumptions 1 The length of thread engagement is 15 times the outside diameter of the screw 2 For torsional stress assume effective diameter z dr then T id N r 3 For axial stress where A is not given assume 6 L 11rdr 4 For bearing stress assume d1 z dr then G i W 7rd df 1017 5 For thread shear stress assume d1 z dr and de ne j as the fraction of nut thickness in this problem 15d involved in screw thread shear fracture surface B W W A Of Fig 1011 Then I Analysis a Assume starting friction z 1 running friction or f 0133 From Eq 104a fndm L ch dC ndm fL 2 XV n1 T 2 250001 01331c1 025 25000013315 2 1c1 0133025 2 2686 2494 5180 lb in b From Prob 107 T 19731b in dr d 06p 2 06O25 185 in From Fig 104b j 05p 03p tan 1450p 058 c From Prob 108 T 535 Nm drd p36 630mm d Using starting friction off 53 015 020 in place of running friction f 015 in Prob 109 solution 021t04501cos 14470 l 2 ln045oos1447 0201J 2 948 1250 2198 22 lb in From Fig 104a T 0 j 05p 053 tan 145 063 e Using starting friction off 01 0133 T 52000271 01331t271 0571 2 n271 01330571 14223 1b in From Fig 104d drdp3 T1 243 in 10 18 j 051 05p tan 25 p 052 prob Load Torque d dr Torsion Axial Bearing Shear part w T 1 stress stresso stresso stressr a 25000 5180 1125quot 0875quot 05 394 ksi 416 ksi 636 ksi 108 ksi b 5000 1993 2quot 185quot 058 16 ksi 19 ksi 110 ksi 049 ksi c 50 kN 535 Nm36 mm 30 mm 05 101 MPa 707 MP 1607 MPa 196 MPa 1 200 22 05quot 040quot 063 18 ksi 16 ksi 28 ksi 034 ksi e 5200014223 3quot 243quot 052 50 ksi 112 ksi 214 ksi 29 ksi Comment In the above table the units of load W are lb and the units of torque T are lbin unless speci ed otherwise SOLUTION 1012D Known Several different types of fasteners are used that prevent or resist unauthorized removal Find Examine and sketch several fasteners used in vending machines computers and other items that prevent or resist unauthorized removal Analysis 1 The article Twenty Tamper proof Fasteners by Federico Strasser appearing in Product Engineering Design Manual edited by Douglas C Greenwood Mc Graw Hill 1959 p 7273 illustrates twenty tamper proof fasteners The abstract of the article states Ways to prevent or indicate unauthorized removal of fasteners in vending machines instruments radios TV sets and other units Included are positively retained fasteners to prevent loss where retrieval would be dif cult The author provides the following descriptions 1a b 2a Wax or other suitable material lls recess above screw Wax ush with plate hides screw position if surface is painted Cupped sleeve riveted in screw hole provides cavity for wax when plate is too thin for recessing Pin prevents rotation of square cup which would allow screw to be removed without disturbing wax Lead seal crimped over twisted ends of wire passing through screw allows only limited slackening of nut Two or more screws strung through heads with wire are protected against unauthorized removal by only one seal Code or other signet can be embossed on seals during crimping Sheetmetal disk pressed into groove can only be removed with dif culty and discourages tampering 1019 4a Spannerhead screws are available in all standard heads and sizes from US manufacturers Special driver is required for each screw size except l4in diameter and above Left handed screw thread is sometimes suf cient to prevent unauthorized loosening Special head lets screw be driven but not unscrewed Tapped cover and casing allows screw with reduced shank diameter to be completely unscrewed from casing yet retained positively in cover For thin sheetmetal covers split ring on reduced shank is preferable Snap ring in groove or transverse pin are effective on unreduced shank Simple and cheap method is ber washer pushed over thread Openended slot in sliding cover allows screw end to be staked or burred so screw cannot be removed once assembled Nut is retained on screw by staking or similar method but if removal of nut is occasionally necessary coaxial bindinghead screw can be used Where screw end must be ush with nut pin through nut tangential to undercut screw limits nut movement Rotatable nut or screw should have suf cient lateral freedom to accommodate slight differences in location when two or more screws are used SOLUTION 1013D Known There are various types of commercially available locknuts Find Give examples of locknuts for 1 pins keys tabs safety wire 2 deformed threads 3 secondary spring elements 4 frictional interference and 5 free spinning until seated Ana 1 lysis The article The Fundamentals of Selecting Locknuts by CC Feroni in Product Design Engineering Manual edited by Douglas C Greenwood McGrawHill 1959 p 47 49 lists and illustrates the basic classes of commercial locknuts The author provides the following discussion for each basic class CLASS I Pins Keys Tabs Safety Wire Assembly is more costly than for other types and there is the possibility of forgetting the locking member at assembly Also the very vibration which the locknuts are intended to withstand may cause fatigue failure of the pin tab or wire On the other hand these types of nuts are most satisfactory where some relative motion exists between the respective part of the joint CLASS 11 Deformed Threads Widely used for moderate service conditions because of low original cost Several factors limit their performance 1 Locking friction is a result of high pressure in a few localized places 2 Performance is usually erratic 3 Cyclic loading and vibration can materially wear the interference points Often merely loading the nut causes local yielding and a loss of locking torque Also deformed threads do not lend themselves to frequent reuse 1020 CLASS III Secondary Spring Elements These often give an attractive balance between cost and performance Initial cost is usually low and reliability is adequate for many applications Limitations 1 Spring member may fail from vibrations 2 Locking effectiveness is reduced or even lost if bolt stretches or mating surfaces wear 3 Most nuts of this type tend to score the surfaces on which they bear CLASS IV Frictional Interference Highest performance but more expensive than any of the other classes Locking action comes from plastic deformation of elements of the nut itself These elements are either non metallic inserts or slotted collars With either spring back tendency which produces action is uniform and does not depend solely upon bolt tension Also these types need not be seated to lock Reusability is high Operating temperature is limited CLASS V Free Spinning These include types that are free to spin until seated Advantages Easy and inexpensive installation easily removed and replaced Disadvantages Any loosening of joint or loss of bolt tension converts them into free running plain nuts Also since the clamping force is in addition to bolt stress and is usually con ned to the lower threads where load is already high combined stress in these threads lowers the fatigue and impact resistance SOLUTION 1014D Known There are various types of commercially available locknuts Find Developa list of ten factors that should be considered in selecting the class of locknut that should be used Analysis The article The Fundamentals of Selecting Locknuts by CC Feroni in Product Design Engineering Manual edited by Douglas C Greenwood McGrawHill 1959 p 47 49 lists and illustrates the basic classes of commercial locknuts The de nitions for the basic classes of commercial locknuts are given in Solution10l3D l The author presents the following discussion for some of the factors that determine what class of locknut to use 2 Can the risk be run that occasionally something may accidentally be forgotten in assembly Class I nuts depend on a secondary element such as a lock washer or a cotter pin Not only might this be overlooked in assembly but lock wires or cotter pins can break under severe vibration Without the secondary element these nuts are as free to spin as plain nuts There seems to be little justi cation for the popularity that still exists for this class Probably it is partially inertia in resisting a change from practice and partially a reluctance to trust what seems like less positive methods Actually many types often perform better Can the nut be seated ght against the work Class V types lock only when they are tightly seated against one of the surfaces being joined If any relaxation occurs they are free to turn off Such relaxation can occur due to creep of the bolt wear or corrosion of mating surfaces 1021 3 Is locking pressure spread evenly or concentrated on a few threads If vibration is severe the friction which prevents a nut from loosening should be spread over as large an area as possible This requires expensive precise manufacture Some nuts lock with interference between a few threads or load some threads more than others 4 Are all nuts of a given type equally reliable Some classes of nuts are not uniform in their locking ability The amount of distortion of shape which causes locking may vary from nut to nut Also these types cannot adjust to the normal variations of bolt diameter within normal tolerance limits Thus on bolts with diameters near the low limit some nuts may not lock at all 5 Will the joint be exposed to high temperatures Some types of Class IV locknuts have plastic or other nonmetallic inserts to obtain locking action and are not recommended for use above 250 F 6 Will the nut be frequently removed and reused Often bolted joints must be broken periodically for inspection repairs access or maintenance The types of nuts which jam a few threads together for locking can damage bolt threads to the extent that the joint cannot be remade unless bolts are replaced 7 Is speed of assembly important It obviously takes more assembly time for nuts which require extra motions to lock such as insertion of cotter pins or lock wires With large volume production this is an important factor 8 Is ease of assembly important When a locknut is required for a relatively inaccessible location the free spinning type Class V may be preferred 9 Will the nut damage39the bolt or the work surfaces When a joint design is critical and maximum strength is required stress raisers which jammedon nuts can cause should be noted Is there relative motion between parts bolted together If this is so a castellated nut with pin or lock wire may be best since repeated rotary motion might loosen other types 10 SOLUTION 1015D Known The web site httpwwwnuttycom lists different kinds of a nuts b bolts and c washers Find Review the web site and list the different kinds of a nuts b bolts and c washers Comment on how to evaluate the products promoted on a web site Analysis 3 b nuts hex nut nylon insert lock nut grade 5 hex nut jam nylon insert lock nut grade 8 hex nut grade 8 torque lock nut jam nut grade 8 ange torque lock nut heavy hex nut 2way lock nut machine screw nut left hand nut acorn cap nut tee nut acme nut rod coupling nut slotted nut reducer rod coupling nut castle nut k lock keps nut 2H heavy hex nut ange lock nut wing nut square nut wood insert nut cage nut I bolts grade 2 hex head cap screw carriage bolt grade 5 hex head cap screw lag bolt grade 8 hex head cap screw tap full thread bolt grade 8 ange frame bolt grade 5 shaker screen bolt square head bolt grade 8 shaker screen bolt grade 5 1022 c carriage bolt grade 8 plow bolt hanger bolt step bolt elevator bolt A325 structural bolt L anchor bolt A490 structural bolt J bolt I washers USS at washer grade 8 USS at washer SAE at washer grade 8 SAE at washer shim washer fender washer split lock washer grade 8 split lock washer nishing cup washer galvanizedrubber bonding dock washer square bevel washer internal tooth lock washer external tooth lock washer high collar split lock washer I SOLUTION 1016D Known The web site httpwwwboltsciencecom provides information related to bolt technology Find Review the web site information related to bolted joint technology and answer the following questions a Is vibration the most frequent cause of boltnut loosening If not what is the most frequent cause of loosening b What are three common causes of relative motion in threads c Can conventional spring lock washers be used to prevent self loosening when bolts without lock washers would loosen because of relative motion d What is prevailing torque e What are direct tension indicators Analysis The web site provides the following answers a b c d 6 It is widely believed that vibration causes bolt loosening By far the most frequent cause of loosening is side sliding of the nut or bolt head relative to the joint resulting in relative motion occurring in the threads If this does not occur then the bolts will not loosen even if the joint is subjected to severe vibration I There are three common causes of the relative motion occurring in the threads 1 Bending of parts which results in forces being induced at the friction surface If slip occurs the head and threads will slip which can lead to loosening 2 Differential thermal effects caused as a result of either differences in temperature or differences in clamped materials 3 Applied forces on the joint can lead to shifting of the joint surfaces leading to bolt loosening I Conventional spring lock washers are no longer speci ed because it has been shown that they actually aid self loosening rather than prevent it I The prevailing torque is the torque required to run a nut down a thread on certain types of nuts designed to resist vibration loosening The resistance can be provided by a plastic insert or a non circular head Threads coated with an adhesive also exhibit a prevailing torque I Direct Tension Indicators DTI s is a term sometimes used to describe load indicating washers Projections on the face of the washer usually on the face abutting the bolt head or nut that deform under loading as the bolt is tensioned An indication of the tension in the bolt can be made by measuring the gap 1023 between the washer face and the nut or bolt head The smaller the gap the greater the tension in thebolt Load indicating washers are commonly used in civil rather than mechanical engineering applications I Comment The task of developing a procedure for evaluating the integrity of information provided by a web site is left as an exercise for a bushy tailed student We suggest starting by de ning the word quotintegrityquot As with most all problem solving we rst need to understand andor de ne the problem SOLUTION 1017 Known The bolt shown is made from cold drawn steel The load uctuates continuously between 0 and 8000 lb Find a The minimum required value of initial load to prevent loss of compression of the plates b The minimum force in the plates for the uctuating load when the preload is 8500 lb Schematic and Given Data F 1A1 I Fe 0 to 800011 kc 6kb v Fe D Assumption The bolt nut and plate materials do not yield Analysis 1 Compression of the plates is lost when Fe 0 when maximum load is applied From Eq 1013 EEhle 6kb Q O kb 6kb 8000 7 8000 6857 lb I 2 Minimum force in plates occurs when uctuating load is maximum From Eq 1013 EEim e 1024 8500 6kb 8000 8500 g 8000 1643 lb I kb 6kb SOLUTION 1018 Known The bolt shown is made from cold drawn steel The load uctuates continuously between 0 and 8000 lb Find a The minimum required value of initial load to prevent loss of compression of the plates b The minimum force in the plates for the uctuating load when the preload is 8500 lb Schematic and Given Data Fe Fe0to 8000 lb kc4kb V Fe Assumption The bolt nut and plate materials do not yield Analysis 1 Compression of the plates is lost when Fc 0 when maximum load is applied From Eq 1013 L F Fckbkc Fe i 8000 5 8000 6400 lb I kb 4kb 5 Minimum force in plates occurs when uctuating load is maximum From Eq 1013 0 kc F F kt 8500 41 8000 8500 6400 2100 lb I kb 4kb 1025 SOLUTION 1019 Known In a given assembly two parts are clamped together by a bolt The ratio of the clamped member stiffness and the bolt stiffness is given the initial bolt tension and the range of the uctuating external load are also given Find Draw a graph plotting force vs time showing three or four external load uctuations and corresponding curves showing the uctuations in total bolt load and total joint clamping force Schematic and Given Data F l T I Fe0to 6000 lb kc6kb F1 11001b l C m ILJIIII 11 Assumption The bolt size and material are such that the bolt load remains within the elastic range Analysis 1 The total bolt load when an external load is applied is from Eq 1013 Fb Fi kb JFC 1100 1 6000 kb k 1 6 1957 lb Fc Fi C 1100 Q 6000 kc k 7 4043 lb since 4043 lt 0 FC 0 lb and Pb 6000 lb When Fc 0 separation takes place 1100 6 Fe Fc 0 and thus Fe g 1100 1283 lb 3 With no external load Fb PC Fi 1026 Fb 6000 l 7 I E V 1100 393 0 52 Fe 3 C SOLUTION 1020 Known In a given assembly two parts are clamped together by a bolt The ratio of the clamped member stiffness and the bolt stiffness is given the initial bolt39tension and the range of the uctuating external load are also given Find Draw a graph plotting force vs time showing three or four external load uctuations and corresponding curves showing the uctuations in total bolt load and total joint clamping force Schematic and Given Data Fe0to 6000 lb 5 kc 31 3 J Fi 1100 lb Assumption The bolt size and material are such that the bolt load remains within the elastic range Analysis 1 Using Eq 1013 for Fe 6000 lb 1027 kb 1 2600 lb kc 3 PC F1 Fe 1100 6000 kc kb 4 3400 lb Since 3400 lt 0 Fc 0 and Pb 6000 lb 2 When Fc 0 separation takes place 1100 Fe FC 0 and thus Fe 3141100 1467 lb 3 ForFe01bFbFcFi Total force 1b SOLUTION 1021 Known The cylinder head of a pistontype air compressor is held in place by ten bolts Total joint stiffness is four times total bolt stiffness Each bolt is tightened to an initial tension of 5000 N The total external force acting to separate the joint uctuates between 0 and 20000 N Find Draw a graph plotting force vs time showing three or four external load uctuations and correSponding curves showing the uctuations in total bolt load and total joint clamping force 1028 Schematic and Given Data Fc Ot0 20000N Assumption The bolt size and material are such that the bolt load remains within the elastic range Analysis 1 The total bolt load when an external load is applied is from Eq 1013 Fb F1 kb Fe 500010 4 20000 kb k l 4 54000 N kc 4 PC 2 F1 Fe 50000 20000 kc kb 5 34000 N 2 60000 54000 50000 40000 34 000 30000 Total force N 20000 10000 Time D 1 029 SOLUTION 1022 Known The cylinder head of a pistontype air compressor is held in place by ten bolts Total joint stiffness is four times total bolt stiffness Each bolt is tightened to an initial tension of 5000 N The total external force acting to separate the joint uctuates between 10000 and 20000 N Find Draw a graph plotting force vs time showing three or four external load uctuations and draw corresponding curves showing the uctuations in total bolt load and total joint clamping force Schematic and Given Data Fc 10000 to 20000 N Assumption The bolt size and material are such that the bolt load remains within the elastic range Analysis 1 Using Eq 1013 for Fe 20000 N kb 1 Fb F F 5000 10 20000 x kbk e 1 14 54000 N F F k F 50000amp20 000 C quot 1 kckb e quot 3 5 9 34000 N 2 For Fe 10000 N Fb 50000 1 00000 52000 N 1 4 Fc 50000 3 10000 42000 N 1030 Total force kN SOLUTION 1023 Known Two parts of a machine are held together by bolts that are initially tightened to provide a total initial clamping force of 10000 N The elasticities are such that kc Zkb Find a Determine the external separating force that would cause the clamping force to be reduced to 1000 N b If this separating force is repeatedly applied and removed determine values of mean and alternating force acting on the bolts Schematic and Given Data Initial Fe 210000 N 3 Assumption The stress on the bolt is within the elastic limit 10 31 Analysis a From Eq 1013 kc kc kb FcFi Fe 1000 10000 2 2 Fe 9000 3 Fe 1 3 Hence Fe 13500 N I b Load off Fb Pi 10000 N Load on Fb 10000 13500 14500 N 1 4 Fm w 12250 N 14500 1 000 a 2 9 2250 N I SOLUTION 1024 Known Two parts of a machine are held together by bolts that are initially tightened to provide a total initial clamping force of 2000 lb The elasticities are such that kc Skb Find a Determine the external separating force that would cause the clamping force to be reduced to 500 lb b If this separating force is repeatedly applied and removed determine values of mean and alternating force acting on the bolts Schematic and Given Data I Initiach20001b E kc5kb 10 32 Assumption The stress on the bolt is within the elastic limit Analysis a From Eq 1013 FcFi kc Fe kckb 500 2000 Fe 1500 24e Hence Fe 1800 lb I b Load off Fb F1 2000 lb Load on Fb 2000 1800 2300 lb Fm 2000 3 2300 2150 lb I Fa 23OO 2000 1501b I SOLUTION 1025 Known Two parts of a machine are held together by bolts that are initially tightened to provide a total initial clamping force of 2000 lb The elasticities are such that kc 6kb Find a Determine the external separating force that would cause the clamping force to be reduced to 500 1b b If this separating force is repeatedly applied and removed determine values of mean and alternating force acting on the bolts Schematic and Given Data Fe 1A1 Initiach2000N h5kb 1033 Assumption The bolt stress is less than the elastic limit of the bolt material Analysis 21 From Eq 1013 Fc Fi 39 L Fe I kc kb 5002000 621ng15007613e Hence Fe 1750 lb I b Load off Fb Pi 2000 lb Load on Fb 2000 1750 22501b Fm 29amp2 22 5921251b I F 2250 2000 a 2 1251b I SOLUTION 1026 Known Drawing 1 and 2 are identical except for placement of the spring washer The bolt and the clamped members are quotin nitelyquot rigid in comparison with the spring washer In each case the bolt is initially tightened to a force of 10000 N before two known external loads are applied Find a For both arrangements draw block A as a freebody in equilibrium b For both arrangements draw a bolt force vs time plot for the case involving repeated application and removal of the external loads Schematic and Given Data rm I s i i i A l 1 1 L1 J i 1000N 1000N 1000N 1000N spring washer 1 2 Assumption The bolt stress is within the elastic limit of the bolt material 1034 Analysis a 1 10000 N 2 8000 N I M39K I A q Will lMl 1000N12000N 1000N 1000N10000N IOOON Time gt Time gt 1 2 Comment Note that in neither case does the 10 kN force of the exible spring washer change SOLUTION 1027 Known Drawing 1 and 2 are identical except for placement of the spring washer The bolt and the clamped members are quotin nitelyquot rigid in comparison with the spring washer In each case the bolt is initially tightened to a force of 10000 N before two known external loads are applied Find Plot Fb and Fe versus Fe for drawings 1 and 2 Schematic and Given Data 1000 N 1000 N 51mg 1 2 1035 Assumption The bolt stress is within the elastic limit of the bolt material Analysis at Drawing 1 Drawing 2 10000 N 8000 N F A 1000N12000N 1000N 1000N10000N 1000N Cb Drawing 1 A Fb 12 kN K Fe O 2kN Fe Drawing 2 ii Fb Fi 10 kN 8 kN Fe O 7 2 kN Fe Comment Note that in neither case does the 10 kN force exerted by the exible spring washer change with change in the external load Fe 1036 SOLUTION 1028D Known An idea is given for reducing the uctuating load in a connecting rod bolt Find Explain why this idea will or will not work Schematic and Given Data Flexible Sheeve Assumption The bolt and sleeve remain within their elastic range Analysis This idea will work But other considerations may render it impractical Any increase in force applied to the bolt head requires shortening of the quot exiblequot added sleeve N o appreciable sleeve shortening is possible without separation at the rod cap interface SOLUTION 1029D Known A known static tensile load is applied to two metric ISO screws Find Select appropriate screws and specify a tightening torque Schematic and Given Data 33kN Rotating shaft Metric ISO screw 1037 AssumptionsDecisions 1 2 3 4 The load of 33 kN is shared equally by each screw No bending of the machine screws bolts takes place ie the bolt is in axial tension Use a SF of 4 A relatively inexpensive class 58 steel is chosen for the screw material Analysis 1 From Table 105 choose class 58 steel with proof strength of 380 MPa Nominal load on each screw 3 35qu 165 kN Using a safety factor of 4 design overload for each bolt 165 kN X 4 66 kN 66900 N gt At 1737 mmz t From Table 102 use screw size M18 X 25 coarse threads with At 192 mm2 From Eq 1011a F1 09 At Sp 09192380 657 kN From Eq 1012 T 02 F1 d 0265718 2365 Nm ogt380MPa Comments In the analysis step 1 the solution for area A is independent of the stiffness ratio kab and also independent of the initial tension F3 Regardless of these quantities static failure of the bolts will occur only when the overload is suf cient to yield the entire bolt cross section with the pillow block pulled away from the mating xed surface ie FC 0 The optimal initial tension would be the highest value that does not yield the bolts enough to damage them after taking them out and reinstalling them many times The procedure to obtain At 1737 mm2 was based on slight yielding of the entire bolt cross section when the design overload is reached With the M18 X 25 screws At 192 m2 a design overload of 73 kN would cause slight yielding A small additional overload would distort the bolts so that they would not be reusable However a substantially higher overload would be needed to bring the material to its ultimate strength and fracture the bolts ratio SSp 520 380 137 In some situations the design overload might be based on using the ultimate strength of the bolt material rather than its proof or yield strength 1038 SOLUTION 1030 Known A known steel bolt clamps three steel plated and is loaded in double shear Find Determine the force capacity of the joint Schematic and Given Data Normal Load carried by Overload causing Friction forces shear failure 1 l f l l E E 39 2 2 F I 39gt F E E 2 2 a b 1 in12 UNF grade 5 steel bol Assumptions 1 The bolt is tightened to its full proof load that is F1 SpAt 2 The bolt fails in double shear 3 The bolt and plates have adequate strength to prevent other failure modes 4 The wrenchtorque variation is roughly 30 percent 5 There is a 10 percent initial loss in tension during the rst few weeks of service Analysis 1 For 1 in12 UNF grade 5 steel bolt Table 101 gives At 0663 in2 and Table 104 shows that Sp 85 ksi Therefore the initial tension is Fi SpA 850000663 56355 lb With i30 torquewrench variation and 10 initial tension loss during the rst few weeks of service a conservative assumption of working value of Fi is about 35500 lb 2 The coef cient for semipolished steel is approximately 03 and for sand or grit blasted steel approximately 05 For this case a friction of 04 is assumed Therefore the force required to slip each of two interface 35500 X 04 14200 lb Hence the value of F required to overcome friction is in the range of 28000 lb 3 For the two shear planes involved the larger value of force that can be transmitted through the bolt itself is F 2 SysA where A the area of the bolt at the shear plane 1t124 0785 in2 and Sys 058 Sy 05892 53 ksi Thus for yielding of the two shear planes 1039 F 2530000785 83210 lb I 4 The estimated 83210 lb would increase the shear stress to the shear yield strength over the entire cross section of the shear planes A further increase in load would cause total shear yield failure The estimated load for complete failure is F 2ASus where from Eq 1016 Sus z 74 ksi I Therefore F 2078574 X 103 116180 lb I SOLUTION 1031 Known Three SAE class 98 steel bolts with a speci ed safety factor are used to attach a bracket of known geometry that supports a known vertical load Find Determine an appropriate bolt size Schematic and Given Data See Figure 1032 of the textbook The safety factor SF 10 Assumptions 1 The clamped members are rigid and do not de ect with load 2 The load tends to rotate the bracket about point A 3 The shear loads are carried by friction Design Analysis With the assumption of rigid clamped members and shear loads carried by friction the eccentricity of the applied load has no effect on bolt loading With the bracket tending to rotate about about point A the strain and hence the load imposed upon the two bolts D is four times that imposed upon bolt E Let FD and FE denote the tensile loads carried by bolts D and E Summation of moments about point A for the design overload of 24 kN10 240 kN gives 500240 IOOFE 4OOFD 4OOFD 25FD 400FD 400FD or FD 14545 kN 2 Class 98 steel has a proof strength of 650 MPa Hence the required tensile stress area 1s A 1935 53 224 mm2 Reference to Table 102 indicates the required thread size to be M20 X 25 Comments 1 Because of appearance and to provide additional safety a larger bolt size might be selected 2 As in Sample Problem 102 the bolt size required is independent of kb kc and F1 except for the fact that F1 must be large enough to justify the assumption that shear forces are transmitted by friction With an assumed coef cient of friction of 04 and an initial tension after considering tightening variations and initial relaxation of at least 0558pAl compare the available shear friction force using 16mm bolts with the applied shear overload 1040 Available friction force 3 bOItSOSSSpAf 3055650 MPa245 mm204 105105 N which represents a margin of safety with respect to the 24kN applied overload plus the rotational tendency caused by the overload eccentricity The second effect is dealt with in Sample Problem 105 SOLUTION 1032 Known Three SAE class 98 steel bolts having a speci ed safety factor are used to attach a bracket of known geometry that supports a known vertical load Find Select an appropriate bolt size Schematic and Given Data See Sample Problem 104 and Fig 1032 of the textbook The safety factor SF 10 Assumptions The shear forces caused by the eccentric vertical load are carried completely by the bolts The vertical shear load is distributed equally among the three bolts The tangential shear force carried by each bolt is proportional to its distance from the center of gravity of the group of bolts Design Analysis 1 Neglecting friction has no effect on bolt stresses in the threaded region where attention was focused in Sample Problem 104 and in Problem 1031 For this problem attention is shifted to the bolt shear plane at the interface between bracket and xed plate This plane experiences the tensile force of 14545 kN calculated in Problem 1031 in addition to the shear force calculated in the following step 2 The applied eccentric shear force of 24 kN10 240 kN tends to displace the bracket downward and also rotate it clockwise about the center of gravity of the bolt group cross section For three bolts of equal size the center of gravity corresponds to the centroid of the triangular pattern as shown in the following gure This gure shows the original applied load dotted vector replaced by an equal load applied at the centroid solid vector plus a torque that is equal to the product of the force and the distance it was moved As assumed each bolt carries onethird of the vertical shear load plus a tangential force with respect to rotation about the center of gravity that is proportional to its distance from the center of gravity Calculations on the gure show this tangential force to be 62 kN for each of the top bolts The vector sum of the two shear forces is obviously greatest for the upper right bolt Routine calculation shows V 136 kN 10 41 39 240lcN applied overload 439 V l 240 kN 150 mm CG of bolt group 36 kNm cross section 200 I FM 1 8O 80 chg 0 240 kN 150 mm F180 mm F180 mm 139200 mm 39F 62kN The critical upperright bolt is thus subjected to a tensile stress 039 145450A and a shear stress I 136000A Substitution in the distortion energy equation gives an equivalent tensile stress of c 102 312 391454502 31360002 Equating this to the proof stress gives 276846 A Therefore A 426 mm2 Finally A m 113 or dq ampE Q233mm Thus a shank diameter of 24 mm is required 276846 A sp 650 MPa Comment In comparing this solution with that of Problem 1031 note that for this particular case shear plus tension in the bolt shear plane proved to be more critical than tension alone in the threads 1 042 SOLUTION 1033 Known Two parts of a machine each carrying a static load are held together by bolts The factor of safety and the ratios of yield strength and proof strength of the nuts to the yield strength and proof strength of the bolts are known Find a Determine the size of class 58 coarsethread metric bolts required b Determine the least number of threads that must be engaged for the thread shear strength to be equal to the bolt tensile strength Schematic and Given Data TilT 31 kN Assumptions 1 For steel Sys z 058Sy 2 The loads are equally distributed among the threads 3 Synut 07Sybolt Analysis 21 From Table 105 Sp z 380 MPa ForceSF 31004 N Sp 380 MPa From Table 102 select M8 X 125 with AI 366 mm2 I b Bolt tensile strength z A sy 366 mm2 o Sy Nut shear strength 1td075tSys 1t8 mm075t05807Sy where Sy pertains to the bolt material Equating the strengths gives t 478 mm For pitch 125 mm this corresponds to 383 z 39 threads I At 326 mm2 10 43 SOLUTION 1034 Known The bolts that attach a bracket to an industrial machine must each carry a static tensile load of 4 kN The safety factor is 5 The nuts are made of a steel with 23 the yield strength and proof strength of the bolt steel Find a Determine the size of class 58 coarsethread metric bolts required b Determine the least number of threads that must be engaged for the thread shear strength to be equal to the bolt tensile strength Schematic and Given Data 1 T HI 4kN Assumptions 1 For steel Sys z 0588y 2 The loads are equally distributed among the threads 3 Synut 23 Syboit Analysis a From Table 105 Sp 380 MPa ForceSF 40005 N 2 At Sp 380 MPa 526 m From Table 102 select M10 X 15 I b Bolt tensile strength z At 39101t 58 mm2 symt Nut shear strength z 1td075tSys 100 mm075t05823Sybou where Sys pertains to the nut material and Sybolt to the bolt material Equating the strengths gives t 637 mm For pitch 150 mm this corresponds to 425 z 43 thread I 10 44 SOLUTION 1035 Known A UNF bolt made from SAE grade 5 steel carries a static tensile load of 3000 lb The bolt is used with a nut made of steel corresponding to SAE grade 2 speci cations The safety factor is 4 based on the proof strength Find a Determine the size of the UNF bolt b Determine the least number of threads that must be engaged for the thread shear strength to be equal to the bolt tensile strength Schematic and Given Data 1 T UNF bolt SAE grade 5 SF 4 Nut SAE grade 2 3000 lb Assumptions 1 For steel Sys 0588 2 The loads are equally distributed among the threads Analysis a For the 3000 lb load it is obvious that drequired lt 1 in Hence from Table 104 select Sp 85 ksi 30004 At required 85 0 06 014 1n2 From Table 101 select 12 in20 thread with A 01599 in2 I b Tensile strength z At Sy 01599 in292000 psi 14710 lb Nut shear strength 2 1td075tSys 1t05075t05857000 Equating the strengths gives t 038 in For pitch 120 in this corresponds to 755 say 76 threads I SOLUTION 1036 Known A UNF bolt made from SAE grade 5 steel carries a static tensile load of 2001b The bolt is used with a nut made of steel corresponding to SAE grade 1 speci cations The safety factor is 5 based on the proof strength Find a Determine the size of the UNF bolt b Determine the least number of threads that must be engaged for the thread shear strength to be equal to the bolt tensile strength 10 45 Schematic and Given Data T T UNF bolt l l SAE grade 5 J Nut SAE grade 1 200 lb Assumptions 1 For steel Sys z 0583y 2 The loads are equally distributed among the threads Analysis a For the 200 lb load it is obvious that drequifed lt 1 in Hence from Table 104 select Sp z 85 ksi 2005 2 m 1n From Table 101 select 0164 in 36 thread with At 001474 in2 b Tensile strength z A Sy 001474 in292000 psi 1356 lb Nut shear strength 2 1td075tSys 1t0164075t05836000 where Sy is obtained from Table 104 Equating the strengths gives t 0168 in For pitch 136 in this corresponds to 605 say 61 threads At required 1 046 SOLUTION 1037 Known A gear reducer of known weight is lifted using a steel eyebolt of SAE grade 5 The housing into which the bolt is threaded has only half the yield strength of the bolt steel Find a Select a suitable bolt size for a safety factor of 10 b Determine the minimum number of threads that should be engaged Schematic and Given Data Tensile failure l Stripping along surface quotAquot of Fig 1011 Assumption The loads are equally distributed among the threads Analysis a From Table 104 Sp 85 ksi WSF 20001b 10 2 02 5 A sp 85000 psi 3 m From Table 101 sizes 35 in18 UNF or in 10 UNC would be appropriate Arbitrarily choosing in10 UNC a SF of 10 14 is provided b For balanced tensile and quotstrippingquot strength Eq 1 in Sec 1045 gives t 047d 047075 in 0353 in But if threaded member strength is only half that of the bolt use t 0706 in With threads per in 10 Number of threads 070610 706 threads engaged I 10 47 SOLUTION 1038D Known An industrial motor weighing 22 kN is to be provided with a class 88 steel eye bolt for use when it is lifted The housing into which the bolt is threaded has only half the yield strength of the bolt steel Find a Select a suitable bolt size Explain brie y your choice of safety factor b Determine the minimum number of threads that should be engaged Schematic and Given Data Tensile failure Stripping along surface quotAquot of Fig 1011 Assumption The loads are equally distributed among the threads Analysis a Select safety factor This case corresponds to 3 in Section 612 except that an impact factor is involved 8 in Section 612 Impact upon a chain hoist becoming taut can exceed the factor of 2 associated with a suddenly applied load In a normal situation one might select say SF 25 multiplied by an impact factor of 4 to get an overall SF of 10 But this is a case where consequences of failure could be catastrophic If the motor is lifted over someone39s head and the cost of an oversize bolt is minimal Let us arbitrarily select SF z 20 From Table 105 Sp 600 MPa WSF 22000 N 20 sp quot 600 MPa From Table 102 sizes M33X2 or M36X4 would be appropriate A 733 mm2 Arbitrarily choosing M36X4 a SF of 20 22 is provided I b For balanced tensile and quotstrippingquot strength Eq d in Sec 1045 gives t 047d 04736 mm 1692 mm 1048 But if threaded member strength is only half that of the bolt use t 3384 mm with p 4 m Number of threads 33844 846 threads engaged Specifying an integral number say 9 threads engaged minimum I SOLUTION 1039 Known The internal pressure of a pressure vessel with a gasketed end plate is suf ciently uniform that the bolt loading can be considered static A gasket clamping pressure of at least 13 MPa is needed Find a For 12 16 and 20m bolts with coarse threads and made of SAE class 88 or 98 steel determine the number of bolts needed b If the ratio of bolt circle circumference to number of bolts should not exceed 10 nor be less than 5 state which of the bolt sizes considered gives a satisfactory bolt spacing Schematic and Given Data Assumption When calculating gasket area the bolt hole area is negligible Analysis E 4 Diquot pressure a Clamping force required 192802 1402 13 600358 N 4 Clamping force At Sp 600 kN Number of Bolt dla bolt 90 roof b 102 P 1quot a Tab 10 5 10a d At 093 forcebolt bolts required 12 mm 343 m2 650 MPa 493 kN 1217 13 16 157 650 918 654 7 20 245 600 1323 454 5 b For 12 mm bolts spacing 2301t13 5558 mm 463d 1049 For 16 mm bolts spacing 230n7 1032 mm 645d For 20 mm bolts spacing 230n5 1445 mm 723d The 16 and 20 mm bolts satisfy the given guidelines The 12 mm bolts are a little too close together SOLUTION 1040 Known The internal pressure of a pressure vessel with a gasketed end plate is suf ciently uniform that the bolt loading can be considered static A gasket clamping pressure of at least 10 MPa is needed Find a For 12 16 and 20 m bolts with coarse threads and made of SAE class 88 or 98 steel determine the number of bolts needed b If the ratio of bolt diameter should not exceed 10 nor be less than 5 state which of the bolt sizes considered gives a satisfactory bolt spacing Schematic and Given Data Assumption When calculating gasket area the bolt hole area is negligible Analysis a Clamping force required D3 D3 pressure 1t 2 2 Z 280 140 10 461814 N z 460 kN Clam in force At sp P g 460 kN Number of Bolt dia bolt 90 proof Tab 102 Tab 1051 m d At 09 SP forcebolt bolts required 12 mm 843 m2 650 MPa 493 kN 933 10 16 157 650 918 501 6 20 245 600 1323 347 4 10 50 b For 12 mm bolts spacing 230n10 7226 mm 602d For 16 mm bolts spacing 230116 1204 mm 753d For 20 mm bolts spacing 2301t4 1806 mm 903d The 12 16 and 20 mm bolts all satisfy the given guidelines I SOLUTION 1041 Known A bolt made of steel is initially tightened to its full yield strength of 80 kN Elasticities are such that kc 3k A plot of load vs time is given Find a Complete the curves for bolt force and clamping force b Make a simple drawing showing the shape of the residual stress curve in a 39 threaded section of the bolt Schematic and Given Data l Sy80kN Assumption 1 Use an idealized stressstrain curve for the bolt 2 The stress concentration factor at the notch is K 3 Analysis a With ii L the elastic load division is 14 to the bolt and 34 to the kb kc 1 3 clamped members But for the initial load application kb 0 1051 Bolt yielding Time gt b With the assumption that K 3 we have 2sy sy 0 sy 28y 38y 4 I L l 1 Answer SOhd me E 139 Elastic stress for tightening load A Actual stress when tightened K 3 notch SOLUTION 1042 Known A grade 5 l in diameter UNF bolt has rolled threads and is used to clamp two rigid members together such that kc 4kb There is a force tending to separate the members that uctuates rapidly between 0 and 20000 lb There is a possibility of slight bolt bending Two values of initial tension are to be investigated 1 the normal value that a mechanic might tend to apply Eq e in Section 107 and 2 the theoretical limiting value of AtSy Find a Make a force vs time plot for each of the two values of initial tension b Estimate the safety factor corresponding to each of the values of initial tension where failure is considered to be either eventual fatigue fracture or joint opening clamping force dropping to zero 1052 Schematic and Given Data 1 Grade 5 lin UNF bolt rolled threads Mi kc4kb Fc 0 to 20000 lb Assumption The material has an idealized stressstrain curve with the change from elastic to plastic occurring at the yield strength Analysis a Force vs time plot for two values of initial tension 1 From Table 101 At 0663 in2 From Table 104 Sy 92 ksi From Table 106 Kf 3 Eq e tightening Fi 16000d 16000 lb quotFull tighteningquot F1 AtSy 066392000 61000 lb kb 15 kbkC 45 of load to clamped member Elastic response kb 0 kbkc all of load to clamped member When bolt is plastic 1053 b 5 10 Estimate the safety factor for joint opening For Fe 2 20000 lb Fc 0 when F1 16000 lb Thus for F1 16000 lb SF 1 and for F1 2 61000 lb SF 6100016000 38 In estimating the safety factor for fatigue Sn 3 Sn CLCGCs Sn 05Su Fig 85 Su 120 ksi Table 104 CL 1 Table 81 CG O 8 some bendingTable 81 Cs Table 106 Sn O51201081 48 ksi For F1 16000 lb 16000 6 0663 and an overload to failure proceeds as shown on the 6m Ga plot to Ca 18 ksi For Pi 61000 lb the thread root stresses proceed from Gm Sy 03 0 to the same limiting value of 63 18 ksi 3 724 ksi at thread root 40 6 Limiting co life ksi 30 fatigue point 20 10 l l l I l l 1 J J I 60 724 80 92 100 6m kSi 120 Since Fa 2000 lb for the bolt for either value of Fi 2000 0663 hence SF 189 2 For F1 16000 lb SF 1 limited by separation For Pi 610001b SF 2 limited by fatigue Ga 3 9 ksi 10 54 SOLUTION 1043 Known A grade 7 34 in UNF bolt with rolled threads is used in a joint such that the clamped member stiffness is only half the bolt stiffness The bolt initial tension corresponds to Eq e in Section 107 During operation there is an external separating force that uctuates between 0 and P The bending of the bolts is negligible Find a Estimate the maximum value of P that would not cause eventual bolt failure b Estimate the maximum value of P that would not cause joint separation Schematic and Given Data Grade 7 34 in UNF bolt rolled threads 2kc kb Assumption The material has an idealized stress strain curve with the change from elastic to plastic occurring at the yield strength Analysis at 1 From Eq e Pi 16000d 16000 120001b 2 Using Eq 1013 Maximum Forcebolt F1 kb J P kb k 12000 2 P 12000 23 P 1b 1 2 Minimum Forcebolt Fj 12000 lb Thus Fm 12000 13 P 1b Fa 13 P lb 3 From Table 101 At 0373 in2 From Table 106 Kf 3 From Table 104 Su 133 ksi Sy 115 ksi 2 1P E 439 a A K 3 0373 3 0373 5 Sn Sn CLCGCs Sn CL 1 Table 81 CG 09 Table 81 10 55 CS 1 Table 106 3n 051331091 5985 ksi 6 For F1 12000 lb 12000 oi 0373 3 965 k31 7 5985 60 4O 6 Eventual ksi fatigue failure 20293lik i OIZIOJZOlt iOI O IOOIIZIO om ksi sy115 Su133 P 8 Ga 0373 15000 Thus P 5595 z 5600 lb for eventual fatigue failure I b 9 From Eq 1013 FC 12000 L P 1 2 At separation PC 0 0 12000 P Thus P 36000 lb for separation I Comment The value of P for separation would be somewhat less because of yielding in the thread root region which reduces the value of kb SOLUTION 1044 Known The cap of an automotive connecting rod is secured by two class 109 M8X125 bolts with rolled threads The grip and unthreaded length can both be taken as 16 mm The connecting rod cap the clamped member has an effective cross section area of 250 mm2 per bolt The initial tension and the maximum external load are known 1056 Find a Estimate the bolt tightening torque required b Determine the maximum total load per bolt during operation c Construct free body diagrams when the maximum load of 18 kN is pushing downward on the center of the cap d Determine the safety factor for fatigue Schematic and Given Data Class 109 M8 x 125 bolts Rolled threads F1 kN Max Fe 18 kN Assumption The material follows an idealized stress straincurve based on Sy Analysis a From Eq 1012 T 02Fi d T 02220000008 352 Nm b Ab 82 5027 mm2 Note use full 8 mm diameter to nd kb Since Eb E and Lb LC then k39s are directly proportional to A39s kb Ab 5027 0167 kb kC Ab Ac 5027 250 From Eq 1013 Fb max 22 kN 01679 kN 235 kN Thus 1057 d 1 d2 d3 d4 145 235 J J 235 145 145 18 235 235 From Table 105 Su 1040 MPa Sy 940 MPa From Table 106 Kf 3 Sn Sn CLCGCs Eq 81 n 05Su Fig 85 CL 1 Table 81 Co 07 Table 81 Cs 1 Table 106 Sn 0510401071 364 MPa nn5 2 2275 kN Fm 235 22 2 52 Kr 3 186 GPa At 3679 obviously relieved by yielding F a 075 ya AJKf 36795 0061 GPa 61 MPa Fa 075 kN Cm 1058 d5 61 MPa yielding l 040 0 200 400 600 800 1000 cm MPa 940 d6 The drawing shows that Ga 60 MPa is as high as Ga can be without eventual fatigue failure Hence SF 2 10 I SOLUTION 1045 Known Two grade 88 bolts with M20 X 25 rolled threads are used to attach a pillow block The bolts are initially tightened in accordance with Eq 1011a Joint stiffness is estimated to be three times bolt stiffness The external load tending to separate the pillow block from its support varies rapidly between 0 and P Find a Estimate the maximum value of P that would not cause eventual fatigue failure of the bolts b Show on a mean stress altemating stress diagram points representing thread root stresses 1 just after initial tightening 2 during operation with the load uctuating between 0 and P2 and 3 with the machine shut down after operating with the 0 to P2 load Schematic and Given Data P Grade 88 bolts M20 X 25 Rolled threads min I i I l l 1 n 10 59 Assumptions 1 2 3 Bolt bending is negligible The material behaves as predicted by an idealized stress strain curve based on Sy The safety factor is 2 Analysis 1 From Eq 1011a F1 09AtSp where A 245 mm2 Table 102 Sp 600 MPa Table 105 F1 09245600 1323 kN Since Kf 30 Table 106 fl 132300 a At Kf 245 3 1620 MPa Therefore the bolt will yield Alternating bolt force Alternating bolt stress Fa P AKf 16245 3 Sn Sn39CLCGCs EQ 81 Sn Su 830 MPa Table 105 CL 1 Table 81 CG 09 Table 81 C5 1 Table 106 n 058301091 3735 MPa 63 3735 400 l 0 200 400 600 800 cm MPa 660 From the drawing Ga 140 MPa P 140 m 3 Thus P 183 kN The points 1 2 and 3 are shown in the above gure 1 0 60 SOLUTION 1046 Known Two aluminum plates are held together by a grade 7 12 in UNF bolt The effective area of the aluminum plates in compression is estimated to be 12 times the crosssectional area of the steel bolt The bolt is initially tightened to 90 of its proof strength Gust loads varying from zero to P tend to pull the plates apart The safety factor is 13 Find a Determine the maximum value of P that will not cause eventual bolt fatigue failure b Determine the clamping force that will remain when this value of P acts Schematic and Given Data Grade 7 12 in UNF bolt Assumptions 1 Bolt bending is negligible 2 Threads are rolled Analysis 1 From Eq 1011a F1 09AISp At 01599 in Table 101 Sp 105 ksi Table 104 F1 0901599105000 15100 lb With this initial tightening load the threads roots are yielded 2 SnSn CLCGCs Eq 81 n Su 133 ksi Table 104 CL 1 Table 81 CG 09 Table 81 C5 1 Table 106 Sn 051331091 60 ksi 1061 Eventual fatigue failure is limited by this point With rolled threads as assumed from Table 106 Kf 30 Fa F3 53 t Kr 15000 01599 3 Fa 8001b e ee 3112 14 Ec Ac 14 Hence kb kbkC 14 1 Thus Fa 0 IP 800 0 1P P 800013 6150 lb If the bolt did not yield at all when initially tightened the clamping force remaining would be F1 08P 15100 4900 10200 lb If the bolt was fully yielded when initially tightened the clamping force remaining would be F1 P 15100 6150 89501b The answer is between these values 8950 lb lt FC lt 10200 lb 10 62 SOLUTION 1047 Known Solutions to problems a 1042 b 1043 c 1044 d 1045 and e 1046 are given as the information in Section 1012 and Table 107 Find Comment on the probable accuracy of the fatigue results If previous designs had been made based on these earlier results state whether or not it is important to specify that the bolt threads be rolled after heat treatment Analysis Base Thread root Nominal Table 107 alt strength problem S 0 a 5 a roll before roll after a 1033 120 ksi 18 ksi 6 ksi 10 ksi 21 ksi b 1034 133 ksi 15 ksi 5 ksi 10 ksi 21 ksi C 1035 1040 MPa 61 MPa 203 MPa 69 MPa 145 MPa d 1036 830 MPa 140 MPa 467 MPa 69 MP3 145 MPa 6 1037 133 ksi 15 ksi 5 ksi 10 ksi 21 ksi quotdividing outquot Kf 3 in each case Comment In each case the earlier results appear satisfactory with threads rolled before as well as after heattreatment SOLUTION 1048 Known A critical application requires the smallest possible bolt for resisting a dynamic separating force varying from 0 to 100 kN It is estimated that by using an extra high strength bolt steel with Sp 1200 MPa and using special equipment to control initial tightening to the full AtSp a stiffness ratio of kdkb 6 can be realized Any of the bolt threads and nishes listed in Table 107 may be selected The safety factor is 13 Find a With respect to eventual fatigue failure determine the smallest size metric bolt that can be used b State the thread and nish selected c With this bolt tightened as speci ed determine the clamping force that will remain at least initially when the lOOkN load is applied Schematic and Given Data F 1A1 F0t0100kN Sp1200MPa 6 1Q SF 13 1063 Assumption The reduction in clamping force approaches the full applied load of 100 kN Analysis 1 Alternating component of separating force 50 kN Since Fa felt by bolt 579 714 kN 2 From Table 107 nd Sa 179 MPa with SF 13 The nominal value of 63 can be 17313 138 MPa 5anom K 9 138 MPa 3 t Hence At 517 mm2 3 Tentatively choose M10X15 thread with A1 580 mmz But this is unsatisfactory because of separation when the bolt is tightened to full proof load kb becomes small We have conservatively assumed that the reduction in clamping force approaches the full applied load of 100 kN This exceeds the initial clamping force of Atsp 580 mm21200 MPa 69600 N To provide SF 13 against separation and assuming kb 2 0 the required F1 is equal to 13100 kN 130000 N spat 1200 MPa At or At 108 mm2 Use M14X2 thread This gives Fi 1200 MPa115 m2 138000 N I Even with kb 0 this is a minimum clamping force of 138 kN 100 kN 38 kN chm 1064

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I made $350 in just two days after posting my first study guide."

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.