to restrict To properly design a structural it isnecessary stress that the applied load to oneelement thatinisthe less than the load the the stress in the material to a level that wi accidental loadings can occurrestricts thatmember may notorbemechanical accounted for Todoing ensurethis. thisFor safety, it is therefore necessary to choose an necessary to restrict the stress indecay, thesupport. material to a level that will be safe. member canTo fully There are many reasons for design. Atmospheric corrosion, or weathering tend to cause properly design a structural member or mechanical element it is stress that restricts the applied load to one that is less than the To ensure this safety, it is the therefore necessary tosome choose an allowable example, load for which the member is designed may be different materials to deteriorate during service. And lastly, materials, such necessary to restrict the stress in the material to a level that will be safe. member can of fully support. There are many reasons for doing thatconcrete, restricts the applied load to one thaton is lessThe thanintended the load the from loadings measurements asstress wood, or actual fiber-reinforced composites, show high To ensure placed this safety,it. it can is therefore necessary to choose aan allowable example, theorload for which the member is designed may be member in can fullystructure support. are may manynot reasons for due doing this. For orThere machine be exact, to errors in fabrication inthe variability mechanical properties. stress that restricts the applied load to one that is less than load the from actual loadings placed on it. The intended measurem example, the of load which the is designed may be different theforassembly of member its component parts. Unknown vibrations, impact, or One method specifying themember allowable load for a member is to use a can fully support. There are many reasons for doing this. For orinmachine may not be exact, due to errors in fabrica from actual loadings placed onThe it. The intended measurements of astructure accidental loadings can occur that may not be accounted for the number called Emniyet the factor ofGerilmesi safety. factor of safety (F.S.) is a ratio of example, the load for which the member is designed may be different assembly of its component parts. Unknown vibrations, i structure or machine mayallowable not be exact, due to. Here errors or inthetend Atmospheric corrosion, decay, tomeasurements cause the failure load Ffaildesign. to the Fallow Fin isor found failfabrication from load actual loadings placed onweathering it.from The intended of a accidental loadings can occur that may not be accounted f the assembly of its component parts. Unknown vibrations, impact, or materials to deteriorate during service. And lastly, some materials, such experimental testing of the material, and the factor of safety is selected Bir elemanın üzerine gerçek sırasında alınan yüklerden farklı structure gelen or machine mayyükler not betasarım exact, due to errors dikkate in fabrication or in Atmospheric corrosion, decay, or weathering tend accidental loadings can occur that may not be accounted for inare thedesign. as wood, concrete, or fiber-reinforced composites, can show high based on experience so that the above mentioned uncertainties the assembly of its component parts. Unknown vibrations, impact, or materials to deteriorate during service. And lastly, some mate imalatı sırasında ölçüsel gibi elemanın yerine montajı design. Atmospheric corrosion, decay, or weathering tendhatalar to cause variability inaccidental mechanical properties. accounted for olabilir. when theElemanın member is usedloadings under similar conditions of yapılabildiği can occur that may not be accounted for in the as wood, concrete, or fiber-reinforced composites, can sh materials deteriorate during service. And lastly, some materials, such One method of specifying the allowable load for a member is to use a loading and to geometry. Stated mathematically, design. Atmospheric corrosion, decay, Elemanlar or weathering tend to cause umulmayan sırasında hatalar söz ofcomposites, konus tasarım sırasında in mechanical properties. as wood, concrete, or da fiber-reinforced can show highvariability number called the factor safety. olabilmektedir. The factor of safety (F.S.) is asome ratio of materials to deteriorate during service. And lastly, materials, such One method of specifying the allowable load for a member variability in mechanical properties. the failure load F to the allowable load F . Here F is found from fail allow fail yüklere maruzaskalabilirler. Elemanların kullanım sırasında paslanma, wood, concrete, or fiber-reinforced composites, can showçürüme high vb. 1.7etkilere DESIGNofOFsafety SIMPLE(F.S.) CONNEC called the factor of safety. The factor is Ffail One method ofexperimental specifying the allowable load for a member isfactor to useofanumber testing of the material, and the safety is selected variability in mechanical properties. F.S. = (1–8) the failure load F to the allowable load F . Here F is fo number called the factor of safety. The factor of safety (F.S.) is a ratio of based on experience so that the above mentioned uncertainties are fail allow fail F maruz kalabilirler. Ayrıcaof elemanların üretiminde kullanılan allow One method specifying the allowable load for a member beton is to use ve a ahşap gibi testing ofbethe material, and the factor of safety i the failure load Faccounted . Here Fused is found from for whenload theFmember under conditions ofmust In ofisthese equations, theexperimental factor of safety greater fail to the allowable allow number called theany factor offailsafety. The similar factor of safety (F.S.) is a ratio of than 1 in based on experience so gösterebilir. that the on above mentioned uncerta mekanikand özellikleri /oftaşıma büyük değişimler Tüm experimentalmalzemelerin testing of the the factor safety iskapasiteleri selected loading andmaterial, geometry. order to avoid the potential for failure. Specific values depend the the failure Stated load Fmathematically, Ffail is found from fail to the allowable load F allow . Here for 1.7 DESIGN OF SIMPLE CONNECTIONS 47 accounted when the of member is used under similar con based on experience that the above mentioned uncertainties are If the load applied tosothe member is linearly related to the stress types of materials to be used and the intended purpose the structure experimental testing of theyükü material, and the factorondan of safety is selected bunlar alınarak yapıya yüklemek yerine daha düşük izin verilebilir loading geometry. Stated mathematically, accountedwithin for when the member used under similar conditions developed the dikkate member, as inison the case ofgöçme using s example, = P>A and or machine. theof F.S. used and in the design of aircraft based experience soFor that the above mentioned uncertainties are or spaceF loading andthen geometry. mathematically, tavg canStated also express factor ofolacaktır. safety asfail a Genellikle ratio of the vehicle components may close to 1 similar insırasında orderconditions to reduce the weight ofizin P theuygun than 1=inV>A, birweyükleme yapmak tasarım göçmeof yükünün accounted for when is be used under F.S. the = member 1.7 power D(1–8) ESIGN OF SIMPLE CONNECTIONS 47 failure stress s (or t ) to the allowable stress s (or t );∗ that is, the vehicle. Or, in the case of a nuclear plant, the factor safety 1 F nd on the fail OF SIMPLE fail CONNECTIONS allow Stated allow allow 1.7 DESIGN OF SIMPLEof CONNECTIONS 47 loading and geometry. mathematically, 1.7 DESIGN 47 F 1.7 D ESIGN OF SIMPLE CONNECTIONS 47 fail for bilinen some ofbir itsgüvenlik components may /bekatsayısı as high belirlemek as 3 due touygun uncertainties verilebilir yüke oranı olarak faktörü structure F.S. =olacaktır. Ffailthe Fallow in loading orsafety material behavior. In than many1 cases, the factor of safety for aP or spaceIn any of theseF.S. equations, factor of must be greater in = equations, (1–8) (!b)allow P In any of these the factor of safety must be greater than 1 in P s eater than 1 in If the load applied to the member is linearly related to the stress F specific case can be found in design codes and engineering handbooks. fail weight 1 allowfor failure. Specific values order to avoid F.S. the Ppotential depend on the han 1 inof F fail = the potential for 1failure.F.S. order to avoid Specific values depend on the =(1–9) (1–8) public and Assumed uniform 1 within member, asintended in are the intended case of using sstructure =aP>A and These values to the form balance of ensuring sthe of of materials to be used and 1the purpose of depend onsafety theon the typesdeveloped allow F normal stress to allow types of materials bealso used and the intended purpose the structure Ifof load applied to the solution member to is linearly related the structure or machine. tavg = V>A, then weto can express the factorand of as athe ratio of the safety providing aspacereasonable economic ertainties For example, theenvironmental F.S. used in the design of safety aircraft or tructure # in distribution B the or machine. For example, F.S. used the design of aircraft or spacedeveloped within applied to the smember isdesign. linearly totothe orIfspacefailure stress (or tbe )close to the sstress tweight that is, the member, as in the case of using s = fety for athe load failmay fail allow (or allow);∗ of vehicle components toallowable 1 related in orderstress reduce the rraft spaceP ordeveloped (!b)vehicle allow components may be close to 1 in order to reduce the weight of t = V>A, then we can also express the factor of within the member, as in the case of using s = P>A and the weight of avg A "safety as a ra ndbooks. the vehicle. Or,deplasman in the caseload of a applied nuclearoran power theve factor of safety Eğer yük ve arasındaki # plant, # linearly eşitliklerinde olduğu gibi eight of If the to the member is related to the stress ! )allowt ( the vehicle. Or, in the case of a nuclear power plant, the factor of safety b failure stress sfail (or tfail) to the allowable stress sallow (or tof = V>A, thensome we can express the factor of safety a ratio of to theuncertainties actor safety Assumed uniform allow avg ublic and for of also its components may be asmember, highasas 3asdue of safety developed within the in the case of using s = P>A and B normal stress for some of its components may be as high as 3 due to uncertainties s t failure stress s (or t ) to the allowable stress s (or t );∗ that is, uncertainties fail B fail fail fail allow allow olution in loading or material InF.S. the (1–10) factor of safety for a(1–9) rtaintiesto Bmany tor V>A,behavior. then we can also express the factor factor of safety as distribution = cases, ==behavior. avgmaterial B (!bthe )allow in loadingF.S. In many cases, of safety fora aratio of Design ofthe Simple Connections of safety The area of the column tstress s (!b)allow allowin s case can be found design codes and engineering handbooks. allow ety for a for a specific ( ! ) failure (or t ) to the allowable stress s (or t );∗ that is, katsayısı b allow doğrusal ise: # ve yazılabilir. Bu denklemlerde FS from the allowable bea fail # in design fail allow allow emniyet Pallow ( ! ) sfail specific case can be found codes and engineering handbooks. b g handbooks. Assumed uniform AThese " values are intended to form a balance of ensuring public and dbooks. F.S. = Assumed uniform These values are intended to formsimplifying a balance assumptions of ensuring public andnormal !b)allow (uniform Assumed uniform stress sproviding regarding thebüyük behavior of sthe ng failBy making allow Assumed environmental safety and a reasonable economic solution to blicpublic and and nin 1’den büyük olması gerektiği aşikardır, aksi taktirde göçme yükünden bir yüke izin normal stress F.S. = (1–9) orenvironmental stress safety and providing a reasonable economic solution todistribution normalnormal stress s = P>A t = V>A material, the equations and can often be used to ic solution to s avg distribution allow design. ution distribution s *Into some cases, such as columns, the applied load is not linearly related to stress and fail distributiondesign. P analyze or See design or mechanical In F.S.a 13. =simple connection verilmiş olunur. FS değerleri tasarım alanına ve kullanılacak malzemeye göre The area thebe column platenin B the is özel determined P therefore only Eq. 1–8ofcan used determine factor of safety. Chapter A " (1–9)element. or P tobase s allow P A "a section, its from the allowable bearing stress for the concrete. )allow (!bat t particular, if a member is subjected to normal force A " fail A" (!b)allow or F.S.area =uygun (1–10) değişecektir. değerine bakılabilir. (!b)allow(!b)allow İlgili yönetmeliklerden required theFS section is determined from r of the tat allow tfail B is determined or e used to Design of Simple Connections The area of the column base plate F.S. = Design of Simple Connections The area of the column base plate B is determined sment. In The areaTheofarea tfail of the column baseBplate B is determined from the allowable bearingtstress allow for the concrete. the column base F.S. plate P from the allowable bearing stress for the concrete. = is determined (1–10) from the allowable bearing stress for the concrete. A = of the (1–11) tassumptions allowable bearing stress for the concrete. ction, its from the By making simplifying regarding thelinearly allow tbehavior Basit Bağlantıların *In some cases, such as Tasarımı columns, the applied load is not stress fail sto By making simplifying assumptions regarding therelated behavior ofand the allow havior F.S. = (1–10) Assumed unif s = P>A t = V>A the equations and can often be used to of theof the material, therefore only Eq. 1–8 can be used to determine safety. See Chapter 13. tallow tallow = ofV>A material, the equations s = P>A avg andthet factor can often be used to en be to used to analyze or design a simple connection or avg mechanical element. In used *In some cases, such as columns, the applied load is not linearly related t analyze or design a On simple connection or mechanical element. In element. In the linearly other hand, iftoforce the section is subjected to can an average force, therefore onlyits Eq. 1–8 be used to shear determine the factor of safety. particular, if a member isload subjected to normal section, ment. *In In some cases, P See Chap such particular, as columns, the is not related stress at anda l " ————— if applied a member subjected toarea normal force at a section, its P is the a (1–11) section, required tallowpd required area attothe sectiontheisthen determined from tion, its its therefore only Eq. 1–8 can be used determine factor of safety. See Chapterat 13.the section is required area the section is determined from *Inat some Assumed uniform shear stresscases, such as columns, the applied load is not linearly related to stress and tallow therefore only Eq. 1–8 can be used to determine the factor of safety. V See Chapter 13. A = (1–12) P P t(1–11) ear force, P A P= allow P P = (1–11) sA l " ————— (1–11) P allow t" allowpd (1–11) Assumed uniform shear stress " sallow Thestress embedded leng Assumed uniform shear tallow Assumed uniform shear stressAs discussed in Sec. 1.6, the allowable stress used in each of these Assumed uniform shear stress tallow can be determined tallow t equations is determined eithershear by applying of safety to the allow hand, if the section stress of th On the other is subjected to an average force, a factor d hand, if the section P—— (1–12) On the other is subjected to anfailure averagestress shearorforce, ——— l "finding e shear P P material’s normal or shear by these stresses the required area at the section is P ar force, force, then pd t ———— allowl " — ——— the required area at the section is P l— "——— then tallowpd l " ——— directly from an appropriate design code. tallowpdtallowpd P The embedded length l of this rod in concrete Three V examples of where the above equations apply are shown in of these d can be determined using the allowable shear V A = (1–12) d P ty to the tA dglue. Fig. 1–25. = (1–12) stress of the bonding allow (1–12) d P tallow P e (1–12) stresses P " " (Burada üniform kayma gerilmesi varsayımı yapılmıştır.) The embedded length l of this rod in concrete As discussedlength in Sec. 1.6,rodthe allowable stress used in each of these The embedded length l of this rod in concrete ofin this in concrete can be determined using the allowable shear PThe embedded As discussed 1.6, the allowable stress used in each of theseAssumed embedded length l of thisl rod inSec. concrete uniform each ofintheseThe equations can be determined using the allowable shear shown is determined either by applying a factor of safety to the of these can be determined using the allowable shear stress of the bonding glue. can be determined using the allowable sheareither by applying a factor of safety to the shear stress stress of the bonding glue. equations is determined of the bonding tallow normal or shearglue. failure ysafety to theto the material’s stress ofstress the bonding glue. V " Pstress or by finding these stresses material’s normal or shear failure stress or by finding theseP stresses these stresses directly from an appropriate design code. stresses P directly from an appropriate design code. P A" The area of the Three examples of where the above equations apply are shown in P tallow P are shown in PFig. 1–25.Three examples of where the above equations apply are shown in is determined f m hown in which is larges Fig. 1–25. 1.7 1.7 1.7 " w niform ess w P " t allow P P " The area of the bolt for this lap joint is determined from the shear stress, V " is P largest between the plates. which V"P P P P Fig. 1–25 The area of the bolt for this lap joint The area of the bolt for this lap joint is determined from thestress, shear stress, is determined from the shear which is largest between the plates. Assumed uniform shearAssumed stress uniform tallowshear stress tallow P P A" t P allow A" t allow Fi #17 P P The area of the bolt for this lap joint The area of the bolt for this lap joint is determined from the shear stress, determined from shear stress, which isislargest between thethe plates. which is largest between the plates. 1.7 DESIGN SIMPLEOFCS ONNECTIONS 1.7 OF DESIGN IMPLE CONNECTIONS 49 49 1.7 DESIGN OF SIMPLE DESIGN OF SIMPLE CONNECTIONS 49 1.7 D ESIGN OF SIMPLE CO 1.7 DESIGN OF SIMPLE CONNECTIONS 49 1 EXAMPLE 1.13 1 1.7 DESIGN OF SIMPLE CONNECTIONS 49 1.13 he control arm isEXAMPLE subjected to the loading shown in Fig. in 1–26a. DESIGN OF SIMPLE CO The control is subjected to the loading shown Fig.arm 1–26a. 1 The control is subjected to the loading shown in Fig.1.71–26a. " arm 1.7 XAMPLE 1.13 1.13 EXAMPLE 1 EXAMPLE 1.13 1 etermine to the nearest diameter of the steel pinsteel EXAMPLE 1.13 in. the Determine to the nearest required diameter of the 4 in. the4 required Determine to thepin nearest 14 in. the required diameter of the steel pin The control arm is subjected to the loading shown in Fig. 1–26a. = 8 ksi. C if the allowable shear stress for the steel is t allow is tat = the 8 ksi. at C if the allowable shear stress for the steel = 8Fig. ksi.1–26a. C if allowable shear stress for the steel is tallow 1 Çelik için kayma emniyet gerilmesi # allow ise isCshown mesnedindeki perçinin çapını belirleyiniz. EXAMPLE 1.13 The control arm subjected to pin the loading shown in in. Determine to the nearest the required diameter of The control arm is subjected to the loading insteel Fig. 1–26a. 1 EXAMPLE 1.13 the 4 1 1 in. Determine to the nearest the required diameter of the steel pin at CDetermine if the allowable stress forthe therequired steel is tdiameter 4 the steel pin in. to theshear nearest of allow = 8 ksi. 4 A to the FAB The control arm Bisatsubjected inthe Fig. 1–26a. A loading shown FAB stress forAthe steel is tallow = 8 ksi. at C for ifcontrol the allowable shear F = C ifB the allowable shear stress steel tallow The armis is subjected B 8 ksi.to the loading shown in Fig. 1–26a. AB Determine to the nearest 14 in. the required diameter of the steel pin 1 Determine diameter of the steel pin A = 8 ksi.to the nearest 4 in. the Frequired AB CESIGN if theOFallowable shear stress for theBsteel 1.7 at D SIMPLE CONNECTIONS 49 is tallow at C if the allowable shear stress for A the steel is tallow = 8 ksi. FAB A B 8 in. B 1.7 1–26a. teel pin 8 in. 1 DESIGN OF SIMPLE CONNECTIONS 8 in. C C C 49 8 in. 8 in. C FAB B 8 in. FAB 8 in. 8 in. B A Cx C Cx 8 in. 1 FAB A 8 in. 8 in. 8 in. 8 in. C C Cx C 5 5 8 in. 3 3 8 in. 5 4 3 3 in. 2 in. 4 3 C3 in. Cx 2 in. C 3 in. 3Cin. 2 3in.in. 5 kip 2 4in. 5 kip C 4 3 in. 2 in. 5 kip4 5 Cx C C hown in Fig. 1–26a. 5 kip 3 C C C C 3 kip 5 kip 5 x y Cy 3 kip F Cy 3 kip 3k 4 3 kip " 2 in. 3 4 5 5 meter ofABthe steel " pin 3 kip3 in. 3 52 in. C 3 in. 35 kip (b) Fig. 1–26 (b) 3 in. 2(b C (a) Fig. 1–26 Cx35 kip C C (a) 4 33 in. 2 in. 4Fig. 1–26 2Cin. allow = 8 ksi. (a) 3 in. kip C 4 C y 5 in. 2 in. x 5C kip 5 kip Çözüm: Serbest cisim diyagramı: 5 33 kip 5 kip 3 Cy 3 kip 3 in. 2 in. 3 kip 4 Cy 3(b)5 3 kip (a) 3 4 3 kip Fig. 1–26 OLUTION 3 in. 3 in. 2 in. 5 kip SOLUTION F (b) 2 Fig.4 1–26 SOLUTION (a) 3 in. 2 in. (b) Fig. 1–26 AB (a) 5 kip 8 in. C 3 kip 5 kip y nternalInternal Shear Force. A free-body diagram of the arm is shown 3diagram kip Cy Shear Force. A free-body ofInternal the arm Shear is in shown in 3 kip Force. A free-body 3diagram of the arm is shown in kip SOLUTION (b) g. 1–26b. For equilibrium we have Fig. 1–26 (a) Fig. 1–26b. For equilibrium we have (b) Fig. 1–26b. For equilibrium we have Fig. 1–26 SOLUTION (a) in SOLUTION Internal Shear Force. A free-body diagram of the arm is shown 8 in. 3 5C 5 1 C + ©MCd +=©M 0; C = 0; FAB18 in.2 3in.2 kip -133in.2 kip in.2 0 Force. A 35 d+ FAB-18For kip -135we in.2 5B 15 kip A 35Shear B=15 Internal diagram armA 3is shown in ©M = in.2 0; =of0the FABA 18free-body in.2 - 3 kip 5 kip Fig. Internal 1–26b. equilibrium have Shear Force. A free-body arm is shown in 13 in.2of-the Cdiagram 5 B 15 in.2 = 0 SOLUTION 8 in. Cx C Fig. 1–26b. For equilibrium we have SOLUTION = F 3For kipequilibrium Fig.F1–26b. we have 3 AB 3 kip AB = d+ ©M =3 0; FAB 18 in.2 kiparm 13 in.2 - 5 kip = 0= 3 kip 5 C A Internal Shear Force. free-body diagram of 3the is shown in A 5 B 15 in.2FAB 4 + Internal Shear Force. A the arm shown 3 free-body 4 d+ ©M = 0; F 18 in.2 3 kip diagram 13 in.2 - of5 kip = 0 in A 35 B 15isin.2 + 4 : ©F = 0; -3 kip C + 5 kip = 0 C = 1 kip C AB 3 in. 2 in. A B d+-3 ©M 0; in.2 kip - 5kip in.2 +3 3kip x: xC x0 A 5 B 15 ©Fx For = 0;equilibrium kip -5have C + 5F kip == Cxx 13 == 1in.2 Fig. 1–26b. we= ABA18 : ©F 0;kip 3 kip - have Cx += 50kip A 45 B = 0 Cx = 1 kip 5 BAB F kipx 5 Fig. 1–26b. For equilibrium we FAB = 3 kip c ©F =c ©F 0; y=Cy=0;0; C3+ 3C kip - 35kip = in.2 0 A 3 B -=C =c=6A 3C kip A 35 5B 13 4=in.2 3 15 kip ykip y0 -0; kip --3 kip 60;kip d +y+©M F 18y=in.2 3 kip 5+ kip 6.082 kip AB 3 C ABx +5©M ©F =0;= 3 kip kip kipin.2 C = 6 kip ©F kip -5 CxF kip 0= 0 FCxC18 =y 6.082 1 kip A3B = 0 A B 5 By y= 5 d+ in.2 3 -kip5 13 C: 4 5 - 5 kip A 5 By15 in.2 = 0 x(b) C C AB + ©F = 0; 4 he pinThe at Cpin resists the resultant force at C, which is : 3 kip C + 5 kip = 0 Cx = 1 kip + A B x x at C resists the resultant force at C, which is 5 5 : ©F = 0; -3 kip C + 5 kip = 0 C = 1 kip 3 A B The pin at C5 resistsCthe=resultant force at C, which is x x6 kip kip 5 + c ©Fy = x0;FAB = 3 C 6.082 kip y3 - 3 kip - 5 kip A 5 B = 0 y FAB = 3 kip3.041 3 3 kip 23 in. 4 2 2kip2 2in. 2 = 6.082 3 c FC = 211 kip2 + 16 kip + ©F = 0; C 3 kip 5 kip = 0 Cy kip = 6 kip 3.041 kip 4 A B 4 + 2 2 y y FC -3 = + the 16 kip2 kip c211 54 = 6.082 + ©F = C -== 306.082 kip -C kip1which = is 0 = 211 Cy kip2 = 6 kip A5B F 6.082 : ©Fx = 0; The kip C0; kip kip + 16 kip2 kip pin at C resists resultant force at C, A 5y5B kip y-kip2 x + 5 x5 = + ©F C 5 kip : = 0; 3 kip C + 5 kip = 0 C = 1 kip A B 3.041 kip x x x C 3 kip 5 3.041 kip y nce the pin the is subjected to double shear, ashear, shear 3.041 Theof pin at2 at C resists the force at C, which is Since pin is subjected to Cdouble aforce force ofkip 3.041 kipisresultant 3 resultant The pin at resists the force C, which 2 shear Since the pin is subjected to double a shear force of 3.041 3.041kipkip Pin atkip C shear, c + ©F = 0; C 3 kip 5 kip = 0 C = 6 kip F = 211 kip2 + 16 kip2 = 6.082 kip A B 3 6.082 Pin at C hown in y y y C 5 (b) between cts over its over cross-sectional area between the armthe andarm each supporting cand Fig. 1–26 +2acts ©F 0; Cy -kip2 3 kip -between 5kip2 kip 2A 5the = 0 and Ceach kip 2 B acts its" cross-sectional area each supporting y = its y = 6supporting over cross-sectional area arm 2F = 211 + 16 = 6.082 kip " (c) 3.041 kip 3.041 kip (c) FC =at 211 kip2 is +shear, 16 kip2 =C 6.082 The C resists the1–26c. resultant force C,towhich Since the pin is subjected double a shear forcekip of 3.041 kip af for thepin pin, Fig. 1–26c. leaf foratthe pin, Fig. The C resists the resultant force at C,awhich is leafpin foratpin the pin, Fig. 1–26c. Pin at Since the is subjected to double shear, shear force ofC3.041 kip 3.041 kip actsSince over its cross-sectional area the arm each supporting the tobetween double shear, a and shear force of 3.041 kip 2 equired Area. Area. We 3.041 kip .2 = 0Required FChave = We 211have kip2pin + is16subjected kip22 = 6.082 kip (c) 2 2 Pin at C Required Area. We have acts over its cross-sectional area between the arm and each supporting F = 211 kip2 + 16 kip2 = 6.082 kip for the pin, 1–26c. " leafVacts C over its Fig. cross-sectional area between the arm and each supporting3.041 kip 3.041 kip 3.041 kip a shear V double (c) 2leafforce pinAisin subjected of pin, 3.041 kip1–26c. V forthe Fig. f theSince arm isthe shown 3.041 kip shear force 2the =Required =forto =shear, 0.3802 in leaf the = pin,2We Fig. 1–26c. Since pin is subjected to double shear, Area. have A = = 0.3802 in Pin at Ca A = kesilmeye = = 0.3802 in2 of 3.041 kip tallow tallow 2 tesirlidir. 8area kip>in acts over its cross-sectional between the armRequired and each supporting C mesnedindeki perçin çift Yani iki noktadan çalışılmaktadır. 2 8 kip>in t Area. We haveallow 8 kip>in acts over cross-sectional area between (c)the arm and each supporting kip Area. WeVhave 3.041 kip its 2 leaf for the pin, Fig. Required 1–26c. 2 = d 2 A = 2 leaf for2the = pin, 0.3802 in1–26c. 2 kip d 3.041 Fig. 2 V 5 kip A 35 B 15 in.2 = 0 2 3.041 kip tallow Vin 8 kip>in pa b pa = 0.3802 in b =A 0.3802 kip Required Area. A = in2 pa =d = 0.3802 6.082We kip have = 0.3802 in2 in = = = 0.3802 2 2 tallow 28bkip>in 2 2 Area. We have 2Required tallow 8 kip>in 2 d 3.041 kip Vd = 0.696 d =in. 0.696 in. 2 d = kip 0.696 in. pa b =in20.3802 in V d 23.041 A = = = 0.3802 2 2 d =b = 0.3802 = 0.3802 in2 Use C a xpin diameter of 3.0418of = having 1 kip 2 A = pa allow kip>in2 kip Use a pin ahaving a tdiameter pad =b 0.696 = 0.3802 in Use ain. pin having a diameter tallow 2 8ofkip>in2 in 3 2 30.750 in. d =3.041 Ans. = 2kip d= Cy = 6 kip dkip in.a diameter = 0.7502in. 4 in. 041 kip in. in. 6.082 d d=d342 =in.0.696 Ans. = 0.750 4 = d = 0.696 in. Ans. Use a pin having of pa b 0.3802 in Pin at C " " " pa of b = 0.3802 in2 2 Use a pin having a diameter hporting is 3 2 Use a pin having a diameter of (c) Ans. d = 0.696din.= 4 in. = 0.750 in. d == 0.696 d = 34 in. Ans. 0.750in. in. 3.041dkip 82 kipUse a pin having a diameter of = 34 in. = 0.750 in. Ans. Use a pin having a diameter of 3.041 kip ar force of 3.041 kip d = 34 in.Pin=at0.750 Ans. in. C d = 34 in. = 0.750 in. Ans. and each supporting (c) 02 in2 Ans. Ans. #18 6.08 6.082 k 6.082 k disk as shown in Fig. 1–27a. If the rod passes through a 40-mm-diameter hole, determine the minimum required diameter of the rod and The suspender rod is supported at its end by a fixed-connected circular The suspender rod is supported at its end thickness by a fixed-connected circular to support the 20-kN load. the minimum of the disk needed 50 C H Adisk P T E as R 1shown S T R EinS SFig. 1–27a. If the rod passes through a 40-mm-diameter TER 1 STRESS disk as shown in Fig. 1–27a. If the rod passes through a 40-mm-diameter 50 CHAPTER 1 STRESS allowable normal for the STRESS hole, determine the minimum The required diameter ofstress the rod androd is sallow = 60 MPa, and the the minimum required diameter of disk the rod and= 35 MPa. allowable shear stress for the is tallow 50 C H A P Tthe E R 1 hole, S T R Edetermine SS minimum thickness of the disk needed to support the 20-kN load. the minimum thickness of the disk needed to support the 20-kN load. EXAMPLE The 1.14 allowable normal stress for the rod is sallow = 60 MPa, and the 1.14 1EXAMPLE 1.14 = 60 MPa,t and the The allowable normal stress for the rod is sallow 40 mm 40 mm 1 allowable shear stress for the disk is tallow = 35 MPa. = 35 MPa. allowable shear stress for the disk is t 50 C H A P T E R C H A P T E R 1 S T R E S S 1 S EXAMPLE RESS allow 1.14 The rod is supported at its end by a fixed-connected circular 1 " The suspender rod is supported at its end bysuspender a fixed-connected circular The suspender supported atthe itsrod endpasses by a fixed-connected circular t rodinis 40 mm The suspender rod is supported at its end by a fixed-connected circular disk as shown Fig. 1–27a. If through a 40-mm-diameter tallow 40 mm disk as shown in Fig. 1–27a. If the roddisk passes through at 40-mm-diameter 40 mm A 40amm as shown in Fig. 1–27a. If the rod passes through a 40-mm-diameter The suspender rod is supported at its end by fixed-connected circular Şekildeki askı çubuğu ucundaki dairesel disk vasıtasıyla mesnetlidir. Eğer çubuk 40 mm disk as shown in Fig. 1–27a. If the rod passes through a 40-mm-diameter hole, determine the minimum required diameter of the rod and hole, determine the minimum required diameter the of the rod and EXAMPLE 1.14 PLE 1.14 hole, hole, determine minimum required the rod and disk as shown inen Fig. 1–27a. If rod passes through a of 40-mm-diameter determine the thickness minimum diameter of the rod and çapı the minimum thickness ofthe theve disk needed to support theiçin 20-kN load. çapında bir delikten geçiyorsa için gerekli küçük 20 kNdiameter yükü taşımak the minimum ofrequired the çubuk disk the needed to support the 20-kN load. tallow minimum thickness of the disk needed to support the 20-kN load. hole, determine the minimum required diameter of the rod andthe A t thediskin minimum thickness of the disk needed to support the 20-kN load. s = 60 MPa, The allowable normal stress for the rod is and allow allow A send = its MPa, The normal stress the and the enallowable küçük kalınlığını belirleyiniz. Çubuk için normal emniyet gerilmesi # = 60the allow The suspender rod isrod supported at end by athe fixed-connected The suspender rod is for supported atisits by a60and fixed-connected circular scircular MPa, allowable normal stress for the rod is to andload. the the minimum thickness offor disk needed support 20-kN allow s = 35 60 MPa. MPa, The allowable normal stress for the rod isisThe the allowable shear stress the disk allow dis tallow = 35 MPa. = allowable shear stress for the disk t 20 kN allow disk as shown in Fig. 1–27a. If the rod passes through a 40-mm-diameter ve disk için kayma emniyet gerilmesi # olarak verilmektedir. disk as shown in Fig. 1–27a. If the rod passes through a 40-mm-diameter = 35 MPa. allowable shear stress for the disk is t s = 60 MPa, The allowable normal stress for the rod is and the allow allow allowable shear stress for the disk is tallow = 35 MPa. hole, determine the minimum required diameter of the rod and hole, determine the minimum required diameter of the and= 35 MPa. the disk rod is ttallow 40 mm 40 mm t allowable shear stress for 40 mm (b) 40 to mmsupport t 40 mm thickness the disktoneeded the 20-kN d ofneeded the40minimum thickness of the disk support the 20-kN load. 40 mm 20 kN load. t 20 kN mm the minimum d 40 mm 20 kN sallow =t and 60 MPa, The allowable normal foristhe rod 40 mm sallow =is 60 MPa, The allowable normal stress forstress the rod the and the 40 mm tallow (b) (a) = 35 MPa. allowable shear stress for the disk is t t MPa. allowable shear stress for the disk is tallow = 35allow A allow (b) A t 20 kN 20 kN A t t 40 mm (a)40 mm (a) end by a fixed-connected circular passes through a 40-mm-diameter quired diameter of the rod and needed to support the 20-kN load. " MPa, and the e rod is sallow = 60 tallow = 35 MPa.Çözüm: d tallow 40 mm Fig. 1–27 40 mm allow A tallow A SOLUTION Fig. 1–27 20 kN Fig. 1–27 d 20 dkN tallow tallow A 20 the kN rod is 20 kN. Diameter 20 of kN Rod.A By inspection, the axial force in d (b)kN Thus the required (b) cross-sectional area of the rod is 20 d SOLUTION SOLUTION 20 kN (b) 20 kN (b) kN in the rod is 20 kN. Diameter of Rod. By inspection, the axial20force 20 kN 3 (b) Diameter of Rod. By inspection, the axial force in the rod is 20 kN. 20110 2 N P the 20 p 2 (a) kNis Thus the cross-sectional rod (a)required d Aarea ; = of = d d cross-sectional 20 kN Thus the required area of the rod is sallow(a) 420 kN 6011062 N>m2 (a) Fig. 1–27 3 (a) Fig. 1–27 20110 2 N 3 P Fig. 1–27 p 2 Fig. (b) 1–27 (b) 20110 2 N P p ; A = d = so 2 that 2 ; 20 kN 4 20 kNd60110 A = = 62 N>m Fig. 1–27 sallow 40 mm s 4 SOLUTION 6011062 N>m2 allow SOLUTION SOLUTION (a) (a) SOLUTION ofinRod. By inspection, axial in mm the rod is 20 kN. Ans. = the 0.0206 m force = 20.6 so that By inspection,Diameter Diameter of Rod. theDiameter axial force theBy rodinspection, is 20 kN. dthe SOLUTION ofrequired Rod. axial force in is the rod is 20 kN. so that t allow By inspection, the axial force Diameter of the Rod. in the rod is 20 kN. Thus the cross-sectional area of the rod Fig. 1–27 A Thus Fig. 1–27 required cross-sectional area ofthe therequired rod is cross-sectional area of the rod is Thus Diameter of Rod. Byofinspection, the axial force the free-body rod is 20 kN. Thus the required cross-sectional area of the rod Thickness on inthe diagram in d = is0.0206 m = 20.6 mmDisk. As shown 3 Ans. Ans. d = 0.0206 m = 20.6 mm Thus required cross-sectional area of the rod is 3 the 20110 2 N P p Fig. 1–27b, the material at the sectioned area of the disk must resist 20110 2 N 3 PSOLUTION p 2 3 SOLUTION ; d2 =20110 26N pprevent 20110 ; A = d = 2 NA 6=A =P s2 ;shear 2the disk through the hole. If this 24 P p 2 stress to movement of 60110 2 N>m =rodin dthe allow on the 3rod Thickness Disk. shown theaxial free-body diagram in kN. 4 of of 60110 2As N>m Diameter Rod. By inspection, force ; sallow A = dThickness 6 kN. 20110 2 diagram Nis2 20 Diameter of=Rod. inspection, the axial force isthe 20 PAs sallow 4passumed Disk. shown onisin the in 6Byof 2 60110 2 N>m 2 free-body shear stress to be uniformly distributed over the sectioned sallow 20 kN 4Thus 60110 2 N>m ; A = = d Fig. 1–27b, the material at the sectioned area of the disk must resist the required cross-sectional area of the rod is Thus the required cross-sectional area of the rod is 6 2 Fig. 1–27b, the material at the sectioned area of the disk must resist s 4 V = 60110 # 2 N>m we have 20 kN, allowarea, then, since so that shear stress to prevent movement of the disk through the hole. If this so that so that movement shear stress to prevent of the 3 disk through the hole. If this so that (b) 3 shear stressPis assumed to be20110 uniformly distributed over the sectioned 20110 2distributed N 2 N p P shear passumed 2 stress is to be uniformly overmthe 2 that d = so ; V =dm Ans. d = 0.0206 = sectioned 20.6 mm ; = dsince A = area,Athen, ==kN, 1.7 D ESIGN OF SIMPLE CONNECTIONS we have 20 2011032 N 51 6 2 Ans. = 0.0206 20.6 mm 6 we 2V 2 N>m sthen, 4 60110 sallow 4 area, since have V = 20 kN, Ans. d = 0.0206 m = 20.6 mm allow 60110 2 N>m A = ; 2p10.02 m21t2 = Ans. d = 0.0206 m = 20.6 mm Çubuk çapı hesabı: # # tallow # d = 0.0206 m = 20.6 mm 3511062 N>m2 Ans. in Thickness of Disk. As shown on the free-body diagram 3 Thickness of Disk. As shownThickness on the free-body in on the free-body 1–27 so that 20110diagram 2As N shown so that 3 V of Disk. diagram 20110 2 N Thickness of Disk. As onsectioned the free-body diagram in1.7atresist Fig.m21t2 1–27b, the disk material the sectioned area of the disk must in resist EXAMPLE = shown ;atVthe 2p10.02 =the Fig.1.15 1–27b, theAmaterial area of D ESIGN OF SIMPLE CONNECTIONS 1 in51 6 must 2shown A = ; 2p10.02 m21t2 = Fig. 1–27b, the material at the sectioned area of through the4.55 diskmm must resist t -3disk Thickness of Disk. As on the free-body diagram allow 35110 2 N>m 6 2 4.55110 Fig. 1–27b, the material at the sectioned area of the disk must resist shear stress to prevent movement of the the hole. If this Ans. t = 2 m = t allow # d of stresshesabı: to prevent movement the stress disk the hole. IfN>m 35110 2mm Diskshear kalınlığı # # this Ans. d = 0.0206 m = 20.6 Ans. =shear 0.0206 mthrough =the20.6 mm to prevent movement of the disk through hole. If this Fig. 1–27b, material attothe sectioned area of the the disk must resist to prevent movement of the disk through the hole. If this shear stress is assumed be uniformly distributed over the sectioned The shear shaft stress shown in Fig. 1–28a is supported by the collar at C, which is shear stress is assumed to be uniformly distributed over thetosectioned shear stress isthe assumed bekN, uniformly distributed overthe thehole. sectioned shear stress prevent the disk through If this the axial force in theto rod kN. shear stress is20 assumed toVbe uniformly distributed over sectioned area, then, since weofhave V 20 -3to attached theisshaft and located on the right side of the bearing B. =movement area, then, since we have = 20 kN, EXAMPLE 1.15 t = 4.55110 Ans. inover the sectioned 2 m =at=free-body 4.55 mm -3 Thickness of Disk. As shown on20 the free-body diagram Thickness of Disk. As shown on the diagram in area, then, since we have V kN, 1 shear stress is assumed to be uniformly distributed # t = 4.55110 Ans. 2 m = 4.55 mm ea of the rod is area, then, have V value = 20 kN, Determine thesince largest of Pwe for the axial forces at E and F so Fig. 1–27b, the at material at the sectioned area of must the disk must resist Fig. 1–27b, the material the sectioned area of the disk resist area, then, since we have V = 20 kN, 3 that the bearing stress oninthe collar does not exceed allowable 3an The shaft shown Fig. 1–28a is to supported byofthe C,disk which is 20110 2N Şekildeki şaft B mesnedinin sağ tarafında bulunan noktasında şafta bitişik yaka ile3 mesnetlidir. E ve VC shear stress prevent movement ofat the through the hole. If this shear stress to prevent movement disk through the hole. If this 20110 2the Ncollar 11032 N stress of V75 3 A stress 20110 2 N 1s 2 = MPa and the average normal in the shaft = ; 2p10.02 m21t2 = V b allow attached toshear the shaft and located on the right side of the bearing at B. 20110 2 N A = ; 2p10.02 m21t2 = 6 2 V F noktalarında etki eden taşınabilecek en büyük P kuvvetini hesaplayınız. Yakadaki emniyet gerilmesi shear stress is assumed to be uniformly distributed over the sectioned t stress is assumed to be uniformly over the sectioned 6 allow 2 35110 2 N>m A 35110 =55 MPa. ; distributed 2p10.02 m21t2 = tallow allowable stress of =1s = exceed ; an 2p10.02 m21t2 6 32 N 2 062 N>m2 doesAnot 20110 t2 allow6 = tallow V2 N>m the largest value P the axial forces at E and F so 2kN, 35110 2 N>m area, then, have VA =we 75 tDetermine MPa emniyet gerilmesi 55 olarak area, then, since V of =since 20for kN, allow ve şafttaki 35110 2MPa N>m =20have ;weverilmektedir. 2p10.02 m21t2 = that the bearing stress on the collar doestnot allow exceed an allowable 3511062 N>m2 -3 -3 t = 4.55110 Ans. 2 m = 4.55 mm stress of 1sAb2allow = 75tMPa and normal stress in the 3shaft = -3 4.55110 Ans. 2average m =204.55 60 mmtheB -3 3 t 20110 mm mm = 4.55110 Ans. 2 m = 4.55 mm 2 N 20110 2 N V V P P t = 4.55110 Ans. 2 m = 4.55 mm 2allow = 55 MPa. does not exceed stress of 1s t80 A; = ; 2p10.02 -3 3P mm m21t2 = t = 4.55110 A = an allowable 2P Ans. 2 62 N>m2 2 m = 4.55 mm Ans. m = 20.6 mm tallow2p10.02 m21t2 = 35110622PN>m t 35110 allow F E C # A(a) on the free-body diagram in Axialmust 2resist P oned area of the disk Force the disk through the hole. If thisF mly distributed over the sectioned 3P e Axial 2P 20110 2 N 3511062 N>m2 3P 2P B 20 mm m = = 80 4.55 mm mm =-3 4.55110-32 t 2m t = 4.55110 E (b) 4.55 2 Pmm C Ans. Ans.P 3P (b) (a) Force 3 P 60 mm Position (c) Fig. 1–28 Position SOLUTION (c) Ans.we will determine P for each possible failure m = 4.55 mm To solve the problem Fig. 1–28 condition. Then we will choose the smallest value. Why? Normal Stress. Using the method of sections, the axial load within SOLUTION region FETo of solve the shaft 2P, whereas the determine largest axialPforce, 3P, occurs the isproblem we will for each possible failure within region EC, Fig. 1–28b. Thechoose variation the internal loading is condition. Then we will the of smallest value. Why? clearly shown on the normal-force diagram, Fig. 1–28c. Since the cross- #19 80 mm 2P P B 160 20mm mm B 80 mm 20 mm 2P 1 2 P E B 60 mm C B P 60 mm A P F 20 mm P A 20 mm (b) 3P 80 mm E P P 2 P the collar at C, which is 2 P 80 mm C P 2 P 2 P B 60 mm 3P 80 mm byisthe A shown2 P 2 Pcollar at 80 C, mmwhich is The shaft in FFig. 1–28a is supported by the collar at 2P 2 P20 mm The shaft in Fig. 1–28a is (a)supported E Fshown C C, which E side of the bearing at PB. P C FP FE and 3P(b)at B. 80 mm CtheEright Axial (a)the Cbearing 2of P the on (b) 2P attached to the shaft and located on the right side bearing at B. attached to the shaft located side of 3P al forces atFE and F so (b) (b) Axial E Determine Cthe (a) Determine the largest value ofForce P the for(a) axial forces E and F so forces at E and F so largest value of at P for the axial ot exceed an allowable (a) Force (b) (a) Axial Axial (b) that theshaft bearing stressthat on Force collar does noton exceed an allowable the bearing stress the collar does not exceed an allowable Axial Axial Force normal stress in the (a) 3P Force Force 1s 2 = 75 MPa stress of and the average normal stress in the shaft 1s 2 = 75 MPa stress of and the average normal stress in the shaft Axial b allow Çözüm: 2P b allow w = 55 MPa. 3P Force does not exceed an allowable = 55 MPa. stress of 1s 2 does not exceed ant2allowable stress of 1s allow 2P t allow = 55 MPa. Position 3P 3P 2P 3P Serbest cisim diyagramı: Position 2P 3P (c) 2P B 2P 20 Position (c) Fig. 1–28 3P mm Position B 60 mmA P 60 mm B Position A 20 mm 2P Position 20 mm 3P (c) 80 mm (c) P 2P P Fig. 1–28 P P (c) 3P Position 80 mm 80 mm (c) 2P # 2P 2P C Fig. 1–282 P Fig. 1–28 SOLUTION E F (b) (c)F EFig. 1–28 C C Fig. 1–28 SOLUTION (b) To solve the problem we will determine P for each possible failure (b) Normal kuvvet diyagramı: Fig. 1–28 SOLUTION SOLUTION To solve the problem we will determine P for each possible failure (a) (a) condition. Then we will choose the smallest value. Why? SOLUTION SOLUTION 1.7 P DESIGN OF SIMPLE CONNECTIONS 51 Why? ToAxial solve the To problem weAxial will determine forThen each possible failure condition. we the smallest value. solve the problem we will determine Pwill forchoose each possible failure To solve the problem we will determine P for each possible failure Normal Stress. Using the method of sections, the axial load within OLUTION To solve the problem we will determine P for each possible failure Force Forcewethe condition. Then we will Then choose smallest Why? value. Why? condition. will choosevalue. the smallest 3P 2P EXAMPLE 1.15 EXAMPLE 1.1560 mm A F A P 3P P (b) 3P 3P Normal Stress. Using theaxial method of3P, sections, Then we will choose the smallest value. region ofeach theThen shaft is 2P, whereas largest force, occursthe axial load within o solve the problemcondition. we will determine PFEfor possible failure condition. we will chooseWhy? the smallest value. Why? ure Normal Stress. Using the method of sections, the axial load within region FE of the shaft is 2P, whereas the largest axial force, 3P, occurs Normal Stress. Using the method of sections, the axial load within within region EC, Fig. of 1–28b. The variation ofsections, the internal loading is ndition. Then we will chooseStress. the smallest value. LE 1.15 Normal Using theWhy? method sections, the axialof load within Normal Stress. Using the method the axial load within 3P 3P 1 internal loading is region FE of the shaft is 2P, whereas the largest axial force, 3P, occurs within region EC, Fig. 1–28b. The variation of the region FE of the shaft is 2P, whereas the largest axial force, 3P, occurs 2P clearly shown onof the normal-force diagram, Fig. 1–28c. Sinceforce, the cross2P region FE of the shaft is 2P, whereas the largest axial force, 3P, occurs n ormal Stress. Using the method of sections, the axial load within region FE the shaft is 2P, whereas the largest axial 3P, occurs hin within regionwithin EC, Fig. 1–28b. variation of the internal is diagram, clearly shown on thePosition normal-force Fig. 1–28c. Since the crossregion EC,The 1–28b. variation of loading the internal loading is to Position sectional area ofFig. the entire shaft region subjected ft shown in the Fig.shaft 1–28a supported by collar at C, which isThe within region EC, Fig. 1–28b. The variation of is theconstant, internal loading is is internal # clearly gion FE of isis2P, whereas thethe largest axial force, 3P, occurs within region EC, Fig. 1–28b. The variation ofEC the loading is is subjected to urs shown onthe the normal-force diagram, Fig. 1–28c. Since the crosssectional area of the entire shaft is constant, region EC clearly shown on the normal-force diagram, Fig. 1–28c. Since the cross(c) maximum average normal stress.Applying Eq.Fig. 1–11, we have (c) Fig. dthin to the shaft and Fig. located on the right side of the bearing at B. clearly shown on the normal-force diagram, 1–28c. Since the crossregion EC, 1–28b. The variation of the internal loading is Fig. 1–28 clearly shown on the normal-force diagram, 1–28c. Since the crossg is sectional area of the gerilmeler entire shaft is dikkate constant, region EC istaşınabilecek subjected tois 1–28 theFmaximum average normal stress.Applying we have sectional area ofatthe shaft is1–28 constant, region EC subjected toEq. 1–11, Fig. P Şafttaki normal alınarak enisbüyük kuvvet: Fig. ne the largest value of P for the axial forces E entire and soentire sectional area of the entire shaft is constant, region is 3P subjected to EC early normal-force diagram, Fig.; 1–28c. Since the crossarea of the iswe constant, region subjected to ss- shown on thethe A = sectional p10.03 m2 = EC maximumthe average normal stress.Applying Eq.2shaft 1–11, have P 3P maximum average normal stress.Applying Eq. 1–11, we have 6 2 2 bearing stress onentire themaximum collar does nots exceed an 55110 2have N>mm2 the average normal stress.Applying we ctional area of the region ECallowable is to 1–11, allow ;Eq. Asubjected = 3Pnormal p10.03 the maximum average stress.Applying Eq.=1–11, we6 have 2 to Pshaft is constant, P 3P SOLUTION 2in the shaft s SOLUTION EXAMPLE 55110 2 N>m 1.15 2 allow 1s 2 = 75 MPa and the average normal stress or each possible failure 3P ; A = A = p10.03 m2 b allow e maximum average normalPstress.Applying 1–11, we have 3P ; Eq.P p10.03 m2 = Ans. P = 51.8 kN 2= 6 2 2 6 failure2 ;1s 2To =sallow p10.03 m2 =55110 To solveAthe problem we will determine P for each possible 2 N>m solve the problem we will determine P for each possible failure s ; A = p10.03 m2 = 55110 2 N>m = 55 MPa. t exceed an allowable stress of allow 6 2 Ans. P = 51.8 kN alue. Why? allow 3P s 6# 2 P sallow # 2 t Bearing # # 55110 2 N>m Stress. As shown onthe thesmallest free-body diagram Fig.shaft 1–28d, 3P 55110 2Why? N>m inThe allow Then m2 we will choose the smallest value. Why? ; condition. = p10.03 = condition. Then we51.8 will choose value. shown in Fig. 1–28a is supported by th Ans. P = kN 6 2 P = load 51.8 of kNAs shown on the free-body diagram in Fig. 1–28d, ons, s the axial load within 55110 2 N>m C the collar at C P must resist which acts over a Ans. bearing allow Ans. = Bearing 51.8 kNtheStress. Ans. P =3P, 51.8 kN the attached to the shaft and located on the right si Normal Stress. Using the method of sections, the axial load within Yakadaki normal gerilmeler dikkate alınarak taşınabilecek en büyük kuvvet: Normal Stress. Using the method of sections, axial load within Stress. As shown on theshown free-body diagram Fig. 1–28d, 2 on the 2inresist -3 the collar at C must the load of2.3P, which3Pacts over a 3P bearing (d) Stress. As free-body diagram in2Fig. 1–28d, est axial force, 3P,Bearing occurs BkN 60Bearing mm area of Thus, A = [p10.04 m2 p10.03 m2 ] = 2.199110 m Ans. P = 51.8 A b 20 mm Bearing Stress. As shown on the free-body diagram in Fig. 1–28d, 3P Bearing Stress. As shown on the free-body diagram in Fig. 1–28d, ns. Determine the largest value of P for the axial 3P 2 2 -3 2 region FE of the shaft is 2P, whereas the largest axial force, 3P, occurs region FE of the shaft is 2P, whereas the largest axial force, 3P, occurs the collar C must resist load of 3P, acts a bearing Pat the P area ofwhich A [p10.04 m2 - p10.03 ] = 2.199110 2 mC. Thus, C collar at Cthe must resist the load 3P,over which acts over3Pm2 a bearing of the internal loading iscollar b = of mm P C on the collar doesCnot the at C within mustthe resist the load of 3P, which acts over ais bearing 2must P2Fig. earing Stress. As shown on the free-body diagram in 1–28d, 28d, P 3P 2 80 -3 2of3P collar at C resist the load 3P, which acts over a bearing that the bearing stress -3 2 2 2 -3 2 within region EC, Fig. 1–28b. The variation of the internal loading 3P region EC, Fig. 1–28b. The variation of the internal loading is of Ab = area [p10.04 m2 2.199110 . Thus, (d) 2.199110 Aof= Am2 ; p10.03m2 =[p10.04 -] 2=p10.03 m2=] 2-3 =m2.199110 (d) g. 1–28c. Since thearea cross26 2 P b m2 F at 22 m -3 23P 22 m . Thus, EAbload -32.199110 C 2ofdiagram, area ofthe = normal-force [p10.04 -Aacts p10.03 m2 ]bearing = 22.199110 m .2Thus, C 2the engcollar C must shown of 3P, which over s area [p10.04 m22.199110 - p10.03 m2Fig. ]N>m =1–28c. mof . # Thus, (d) 1sb2allow =(d)75 MPa and the average no 75110 allow AFig. =a1–28c. ;Since 2 m2Since =stress on the crossb = Cthe shown the normal-force diagram, cross# resist # clearly #-3 on (b) 3P 6 2 egion EC isclearly subjected to 2 P 2P -3 2 3P 2 2 m . Thus, sallow 75110 2 exceed N>m an allowable stress of 1st2allow -3 2 ea of Ab = sectional [p10.04 -of p10.03 m2 2.199110 ]shaft = ;2.199110 (d) does not P 3P A m2 =area ; 2 m = the entire is constant, region EC is subjected to P 3P 2.199110 A = 2 m = (d) sectional area of the entire shaft constant, region EC is subjected to -3 6 2 = 55.0 P-3 2 6 kN 2 Eq. 1–11, we have A =(a) s 2.199110 ; 2; m2 =75110 26N>m sallow 2.199110 A stress.Applying = 2have 2 N>m 2m = 61–11, 3P Axial P P we =2 55.0 kN the maximumallow Eq. 1–11, we75110 saverage the maximum average normal stress.Applying Eq. have 75110 2 N>m s -3 2 normal allow 75110 2 N>m allow the largest load that can be applied to the shaft is 2.199110 2 m By = ; = comparison, Force 6 2 P =3P 55.0 kN P P = 55.0 kN P 3P s2allow 75110 2kN, N>m Pany By comparison, the largest load be applied toAthe shaft is 2 than 60 mm B = 55.0 P loadkN larger willkN causethat thecan allowable N>m ; A = # p10.03 m22 ; = since = 655.0 A == 51.8 = P this #that 6 p10.03 2 m2 2 is P comparison, the largest load can be applied to the shaft sBy P = 51.8 kN, since any load larger than this will cause the allowable s 55110 2 N>m By comparison, the largest load that can be applied to the shaft is 55110 2 N>m allow allow stress in thethat shaft tolargest be P normal = the 55.0By kN 2 P shaft is By comparison, largest load can be exceeded. applied to the shaft is comparison, the load that can be exceeded. applied to the 3P P Ans. = 51.8 kN,Psince anykN, load larger than will cause the allowable normal stress in the shaft to be = 51.8 since any loadthis larger than this will cause the allowableAns. F Ans. P = 51.8 kN E P = 51.8 kN 2P P = 51.8 kN, since any load larger than this will cause the allowable Buna göre taşınabilecek en büyük kuvvet # dir. NOTE: Here we have not considered a possible shear failure of the ySyis comparison, the largest load that can be applied to the shaft is P 51.8 kN, since any load larger than this will cause the allowable in Fig. normal 1–28d, stressnormal 3P =toin Sdiagram T RTERSESS S in the shaft beSexceeded. shaft to be exceeded. Position 52 CAs H Ain P Tthe E Rstress 1 Swill Tto R Ethe S exceeded. NOTE: Here we have not considered a possible shear failure of the Bearing Stress. shown on the free-body diagram in Fig. 1–28d, 3P Bearing Stress. As shown on the free-body diagram in Fig. 1–28d, 3P normal stress shaft be collar as in Example 1.14. = 51.8 kN, since any load larger than this cause the allowable normal stress C in the shaft to be exceeded. hich acts over a bearing ble (a) NOTE: Here we have not considered a considered possible shear failure of the (c) asacts in Example 1.14. NOTE: Here weofhave not aof shear failure theAxial C -3 the collar attoCbe must resist the load 3P,collar which over apossible bearing C the collar at must resist the load 3P, which acts over of a bearing ormal stress in2.the shaft exceeded. Thus, .199110 2m (d)C NOTE: Here we2 have not considered a possible shear failure of the NOTE: Here we have not considered a possible shear failure of the 21.14. -3 2 2 2 -3 2 as[p10.04 in Example Fig. 1–28 collar as-1.14. inp10.03 Example area ofcollar Thus, A = m2 m2 ] = 2.199110 2 m . (d) Force area of Thus, A = [p10.04 m2 p10.03 m2 ] = 2.199110 2 m . (d) b b as shear collar in Example 1.14. OTE: Here we have notas considered a collar possible failure1.14. of the P in Example he EXAMPLE 1.16 3P 3P 1 # P 2in Example -3P 2 llar as 1.14. -3 2 2ON N>m 2.199110 A = ; 2.199110 A = 2 m; = 6 supported 2 2 mby=a steel rod 6 AC 3P sallow The rigid bar AB shown in Fig. 1–29a is AC2 s 75110 2 N>m The rigid bar AB shown in Fig. 1–29a is supported by a steel rod 75110 2 N>m allow 2 AC AB in Fig. 1–29a is supported byrod aeach steel rodolan ACfailure enthe problem we willrijit determine P çapı for possible 2P alanı by Theçelik rigid çubuğu bar AB shown in Fig. 1–29a iskesit supported a steel AB çubuğu, 20mm AC ve B noktasındaki 1800 mmrod N inshown Fig. 1–29a is supported by a steel AC having a diameter of 20 mm and an aluminum block having a crosshaving a diameter of 20 mm and an aluminum block having a crosseter of 20 mm and an aluminum block having a crossP = 55.0 kN n. Then we will choose the smallest value. Why? P = 55.0 kN having a diameter of 20 mm and an aluminum block having a C having a cross2 blok mm andtoan aluminum block olan isbir aluminyum tarafından taşınmaktadır. ACCare veare C noktalarında bulunan 18mmcross- Position e0 applied shaft 1800 mm . The sectional area 18-mm-diameter pinsatatAAand and mm .2A sectional area ofof1800 The 18-mm-diameter pins 2 the 2 1800 mm . of The 18-mm-diameter pins at and C are 2 1800 mm . sectional area of Theto18-mm-diameter pins at A(c) and C are By subjected comparison, the largest that can be applied tothat the shaft P and Stress. Using the method of sections, the axial load within By comparison, the largest load can beisapplied the shaft is Steel mm .cause The 18-mm-diameter pins at A C are will the allowable çapındaki perçinler tek tesirlidir. Çelik ve aluminyum için göçme normal gerilmeleri subjected single shear. If the failure stress forthe the steel and aluminum toto single shear. Ifload the failure stress for steel and aluminum ngle shear. If the failure stress for the steel and aluminum subjected to single shear. If the failure stress for the steel and aluminum P stress = 51.8 kN, any larger than this will cause thethan allowable thefailure shaft is whereas the largest axial force, 3P, occurs Pload = and 51.8 kN, any load larger this will cause the allowable r.EIfofthe for steel and aluminum 1s 2the = 680 2since = 70 MPa, is2P, respectively, and the =since 680 70 MPa, is52 respectively, and the C H A MPa P TMPa E R 1and S T1s R1s Eal S2 Sfail fail al fail=the st2st fail 80 MPa 1s 21s =The 70 MPa, and respectively, and # ve # , ayrıca perçinler için2 göçme kayma gerilmesi 1s 2 = 680 MPa 1s = 70 MPa, is and respectively, and the al fail normal stress in the shaft to be exceeded. st fail egion EC, Fig. 1–28b. variation of the internal loading is normal stress in the shaft to be exceeded. 1s 2 = 70 MPa, and respectively, and the B fail failure Bal tfail 900MPa, MPa,determine failure shear stressfor foreach eachpin pin determinethe thelargest largest al fail ==900 shear isistfail A stress sible shear failure of the B SOLUTION t = 900 MPa, ess for each pin is determine the largest t = 900 MPa, failure shear stress for each pin is determine the largest failP kuvvetini hesaplayınız. hown normal-force diagram, Fig. Since cross# fail olduğuna göre çubuğa uygulanabilecek büyükfailure tfailNOTE: =load 900 ach pinon is the determine the1–28c. largest Here we have not considered possible shear failure of the=en NOTE: Here weathe have not considered possible of the F.S. =2.2.shear load P thatcan can beapplied applied tothe the bar.Apply factor safety minum PMPa, that be to bar.Apply a afactor ofofsafety ofaofF.S. num To solve the problem we will determine P for F.S. = 2. be applied to the bar.Apply a factor of safety of F.S. = 2. load P that can be applied to the bar.Apply a factor of safety of Aluminum l area ofbar.Apply thecollar entire is of constant, EC is subjected 0.75 m 2.Example ed to the ashaft factor safety ofregion as in Example 1.14. collar 1.14.to Emniyet katsayısı # F.S.as=in alınacaktır. EXAMPLE 1.16 condition. Then we will choose the smallest valu SOLUTION 1 normal SOLUTION mum average stress.Applying2 m Eq. 1–11, we have SOLUTION UsingEqs. Eqs. 1–9and and 1–10,the theallowable allowablestresses stressesare are Using 1–9 1–10, Normal Stress. Using the method of section Pand 1–10, the allowable 3P stresses are Using 1–10, stresses are The rigidEqs. bar 1–9 AB and shown in the Fig.allowable 1–29a is supported by a steel rod AC (a) ; allowable p10.03 m22are = 0, the stresses 2 region FE of the shaft is 2P, whereas the largest 1s 2 6 21s 680 MPa st fail 680 MPa st fail 1sst2fail 55110 2 N>m low 680 CMPa having=a=340 diameter and an aluminum 680 MPa block having a crossst2fail 1s 2340 = 340MPa MPa of 20 mm1s 1s 2stallow == 1sst2fail allow= 680 MPa st= within region EC, Fig. 1–28b. The variation of 1s 2 = = MPa 1s 2 = = = 340 MPa 2 st allow F.S. F.S. 2 2Ans. st allow = = F.S. mm . The sectional area of 1800 18-mm-diameter pins at A and C are P ==Steel 51.8MPa kN 2340 P F.S. 2 clearly shown on the normal-force diagram, Fig. F.S. 2 1s 2 1sal2alfailfail 7070MPa MPa subjected to single shear. stress for the steel and aluminum 1sal 2fail on Stress. As shown free-body diagram in=Fig. 1–28d, 1sfailure 3P If the 70the MPa 70 MPa al2fail sectional 1s 2 = = = 35 MPa 1s 2 = = 35 MPa 1s 2 area of the entire shaft is al allow 70 MPa al allow al fail = = resist the = =load MPa acts 35 MPa and the constant, reg MPa and= 1salC2fail == 70 MPa, = respectively, F.S. F.S. 2is2 1sst2fail = 6801s al2allow ars C must of = 3P,35which over a bearing =alat2allow 35 MPa F.S. 2 F.S. 2 the maximum average normal stress.Applying E A -3 B900 2 MPa MPa, determine the largest failure shear stress for each pin tfail MPa tfail Ab = F.S. [p10.04 m222-900 p10.03 m22] = 2.199110 2 m900 . Thus, (d)t is tfail = 900 P t MPa 3P 900 MPa fail t = = = 450 MPa fail t = = = 450 MPa 2 tfail = 900 MPa allow allow = 2. of safety of F.S. m2 Aluminum =m450 MPa = tallow to = the bar.Apply = A = a factor =; 450 MPa p10.03 F.S. 2 2 load P that can be applied P= tallow 3P F.S. = F.S. = -3 = 245020.75 MPa s F.S. 2 5511062 N> allow 2.199110 ;F.S. m = 2 The 2free-body 6 2the 2m diagram of the bar is shown in Fig. 1–29b. There are The free-body diagram of bar is shown in Fig. 1–29b. There are SOLUTION 75110 2 N>m low dy of isthe isinshown in Fig.There 1–29b. There are The free-body diagram of the bar is shown in Fig. 1–29b. There are51.8 kN P = # bar unknowns. m diagram of the bar shown Fig. 1–29b. are three Here wewill will applythe themoment moment equations ofequilibrium equilibrium three unknowns. Here we apply equations Using Eqs. 1–9 of and 1–10, the allowable stresses are (a) kN s.e Here we will apply the moment equations of equilibrium three unknowns. Here we will apply the moment equations of equilibrium P = 55.0 Bearing Stress. As shown on the free-body will apply the moment equations equilibrium F orderto toexpress express andF termsofofthe theapplied appliedload loadP.We P.Wehave have Pof FF in and ininterms 1sstF2fail AC AC BB Çözüm: 680of MPa Flargest ess F and ininorder terms of the applied P.We have FAC=and in is order to express the at applied load P.Wethe have B of B in terms Fthat must resist load of 3P, whic parison, canP.We beload applied to the shaft FAC and terms theload applied have AC load 1sst2allow =the collar =C 340 MPa B in the d + ©M = 0; P11.25 m2 F 12 m2 = 0 (1) d + ©M = 0; P11.25 m2 F 12 m2 = 0 (1) 2 F.S. 2 AC B B 12 m2 = 0 AC P11.25 m2 F (1) area of A = [p10.04 m2 p10.03 d + ©M = 0; P11.25 m2 F 12 m2 = 0 (1) m22] = 2.1 8 kN, since any load larger than this will cause the allowable AC b B AC P11.25 m2 - FAC12 m2 = 0 (1) 1s 2 70 MPa al fail B B m2 P10.75 m2==0 0 (2) stress in the shaft be exceeded. d +d +to ©M ==0;0; FF m2 (2) P = 35 3P B12 AA B12 A B- -P10.75 2 m2 -©M P10.75 (2) m2 d + ©MA = 1s 0; al2allow = FB12 m2 =-AP10.75 0 2.199110-32 m(2) = 2 m2 ; = MPa = FB12 m2F-B12 P10.75 m2 = 0 m2 = 0 (2) F.S. 6 s We will each value thatcreates createsthe theallowable allowable 1.25 meach Here we have not considered adetermine possible shear failure the 0.75 mdetermine We will now value ofofof PPthat determine each value ofcreates Pnow that creates the allowable We will now determine each 900 value of allow P that creates the allowable 75110 2 N Feach BFB t MPa ine value of P that the allowable fail F stress therod, rod,block, block,and andpins, pins, respectively. inblock, Example 1.14. stress ininthe Brespectively. = and pins, = respectively. = 450 MPa d, pins, respectively. P = 55.0 kN stress in the rod,tallow block, and pins,and respectively. F.S. 2 (b) AC . This requires Rod AC . This requires Rod s requires comparison, largest load AC. Thisdiagram requires the barBy Rod es The free-body is shown in Fig. the 1–29b. There are that can be a 2 6 62 N>m 2 2] = 106.8of 1s 2Fig. 1A 340110 [p10.01m2 m2 kN FF 1A 2 22=] ==340110 2kN N>m2 [p10.01 ] = 106.8 kN 6 = =1s 6= 51.8 2kN, since any 2 load larger than this wi AC allow AC P 1–29 AC st22st[p10.01 allow AC 2allow 1A 2 = 340110 2 N>m m2 106.8 6 2 2 F = 1s 2 1A 2 = 340110 2 N>m [p10.01 m2 ] = 106.8 kN AC three unknowns. will apply the moment equations of equilibrium AC stHere allow weAC ] = 106.8 kN #2be 0 exceeded. C2 = 340110 2 N>m [p10.01 m2 normal stress in the shaft to UsingEq. Eq.1,1, P Using in order to express FAC and FB in terms of the applied load P.We have FAC 1106.8kN212 kN212m2 m2Using Eq. 1, 1106.8 NOTE: Here have not considered a possib 1106.8 kN212 m2 1106.8 =171 171 P = P = 1106.8 kN212 m2 d +=©M =kN 0; P11.25 m2 -kN212 FAC12m2 m2= =171 0 we (1) B kN = 171 kN P = kN P = 1.25 m 1.25 m collar as in Example 1.14. = 1.25 =m 171 kN 1.25 m bjectedarea to single shear. failure stress for the steel and aluminum 2 at 1800 mmIf .the ctional ofhaving The 18-mm-diameter pins A and C are of 20 mm and an aluminum block having a the cross1s 1800 mm sectional area pinsaand at Athe C are 1s 2allow =stress =and 340 MPa 70 =MPa al2steel fail diameter ofsubjected 20of mm and an. The block having crossstfailure Pving a diameter to shear. If for the 1sst2aand =1s680 MPa 1s 2single =aluminum 7018-mm-diameter MPa, isMPa and respectively, fail2the alrespectively, fail F.S. 2=and aluminum 1sst2failarea =single 680 2 = 70 MPa, and the 1s 2 = = 35 MPa 2 al fail al allow bjected to shear. If failure stress for the steel and aluminum 1800 mm . ctional of The 18-mm-diameter pins at A and C are subjected to single shear. If the failure stress for the steel and aluminum 1800 mm . sectional area of The 18-mm-diameter pins at A and C are B F.S. 2 1sstMPa, 2pin 680=MPa 1s iseach andlargest respectively, and the 900 the MPa, failure shear stress for determine thealMPa, fail is=t al2fail = 70 fail 1s 2largest t = 900 ilure stress for each pin is determine 70 MPa fail fail 1s = 680 MPa 1s 2 = 70 MPa, and respectively, and the A st2shear bjected to single shear. If the failure stress for the steel and aluminum fail al2fail 1s =failure 680Ifto MPa 1s =the 70 of MPa, istocan and respectively, and 900 the MPa subjected single shear. the stress for steel aluminum B st al2fail 1s 22. ==of =the 35largest MPa tfailand 900 shear stress for pin is F.S.MPa, =t=fail 2.determine load P that befailapplied thefailure bar.Apply aeach factor safety num allow F.S. =althe ad that=can beB applied topin the aMPa, factor ofpin safety of t2fail =stress 900 lure shear stress for each isalbar.Apply determine the largest F.S. tallow = the =2 = 450 MPa 1sP 680 MPa 1s 70 MPa, and respectively, and t = 900 MPa, failure shear for each is determine the largest st2 fail 1s 2 = 680 MPa 1s 2 = 70 MPa, is and respectively, and fail st fail al fail load P that can be applied the bar.Apply Aluminum The rigid AB shown in Fig. 1–29a is supported by atosteel rod AC a factor F.S.of safety2of F.S. = 2. 0.75can m bebar F.S.largest = a2.factor ofthe ad PAluminum that applied to the aapplied factor oftosafety ofMPa, SOLUTION tsafety 900 MPa tfailfor =each 900 MPa, lure shear stress for each pin the B F.S. = 2. load Pisbar.Apply that can be bar.Apply of faillargest = block 900 failure shear stress pin is tdetermine determine OLUTION fail the having a 2diameter of 20 mm and an aluminum having a crosst = = MPa The free-body is 450 shown in Fig. 1–29b. There are m allow diagram of the bar = SOLUTION Using Eqs. 1–9 and 1–10, the allowable stresses are F.S. = 2. ad P that can be applied to the bar.Apply a factor of safety of 2applied F.S. 2 F.S. = 2. load P that can be to the bar.Apply a factor of safety of um sing Eqs. 1–9 and 1–10, the allowable stresses are OLUTION 2 m sectional area of 1800 mm . The 18-mm-diameter pins at A and C are SOLUTION three unknowns. Here we will apply the moment equations of equilibrium Using Eqs. 1–9 and 1–10, the allowable stresses are 1s 2 680free-body MPaand aluminum st are failfor 1sallowable The diagram of the bar isinshown inthe Fig.applied 1–29b.load There are have 680 MPa ing 1–9(a)and 1–10, Using the stresses st2P fail subjected to single shear. If the failure stress the steel OLUTION Eqs. 1–9 and 1–10, the allowable stresses areF 1s 2 = = = 340 MPa F in order to express and terms of P.We a) Eqs. st allow AC B F 1s 2 1sSOLUTION 2 = = = 340 MPa AC MPa st fail we 680 st allow F.S. 2 three unknowns. Here will apply the moment equations of equilibrium = 1–10, 680Eqs. MPa = allowable 70are MPa, is 1s1–9 and the 1sF.S. 2and sing Eqs. and the 2the 680 MPa st2fail al2stresses fail stallowable fail 1sststrespectively, 2stresses =680 = m2 340 MPa Using and1s 1–10, areMPa fail allow P 1–9 dto +=©M = F0; == P11.25 Fapplied = P.We 0 (1) BF.S. AC12 m2 1sstF2allow = is t1sst21s 340 MPa 2MPa 2fail = 340 F in order express and in terms of the load have 70 MPa al= = 900 MPa, failure shear stress= for each pin determine the largest 1s 2 allow AC B 1s 2 fail 70 MPa AC al fail 680 MPa st 1s fail 2 1s 2 F.S. 2 680 MPa st fail F.S. 2 = = = 35 MPa 1s 2fail = ==a35 MPa Abe = 1sstalcan 22allow ==allow 340 MPa B al allow = 70 2. MPa load P that applied bar.Apply of of340 1stoal2the =2 F.S. = = d2+safety =F.S. 0;MPa FB12 m2 - m2 P10.75 (2) dfactor +1s ©M =©M m2 12 = 0m2 = 0 (1) 1sF.S. B F.S. 70 MPa al2failst allow 1s =0;70AMPa P11.25 = = F 35AC MPa F.S. 2 al22allow fail al 1.25 m 0.75 m 1sal2allow = 1s 2 = 1sal2tallow = 35 MPa F.S. 2 = = = 35 MPa 900 MPa We will now determine value allowable tF.S. MPa A B MPa 70 SOLUTION al fail 900 2= 1sfail fail + ©M FB12 m2 -each P10.75 m2of= P0 that creates the (2) 70 MPa FB2450 F.S. 2 MPa900 tallow =35 A = =0;450 = =al # mand # d MPa 1s 2tallow == 1.251s = =fail MPa t35 MPa and pins, respectively. allow 2 = = = MPa fail stress in the rod, block, F.S. 2 al allow m Using Eqs.al0.75 1–9 1–10, the allowable stresses are tF.S. 2 2 F.S. 900 MPa F.S. (b) = MPa = 450 MPa fail tWe 2 900 allow now determine each=value of P that creates the allowable failtwill tallow = 1s 450=bar MPa F.S. 2MPa FB t= =RodThere =1–29b. 450 The = free-body diagram of the isMPa shown in There are AC .Fig. This requires allow 680 MPa tfailthe bar st2is 900 MPa The free-body diagram of shown in Fig. 1–29b. are F.S. 2fail t stress in the rod, block, and pins, respectively. 900 F.S. 340 MPa 2 fail 450 MPa 1sunknowns. == Here tallow =2(b) allow Fig. 6 2 The free-body of shown2 in Fig. 1–29b. There are m22] = 106.8 kN 1–29 tthe ==will= = = 450 MPa three we apply the=diagram moment equations ofisequilibrium allow ree unknowns. west will apply equations ofbar equilibrium F =the 1sbar 2allow 1A 340110 2 N>m [p10.01 F.S. 2Fig. F.S. AC stin AC = There The free-body Here diagram of the bar is2moment shown in 1–29b. There are F.S. 2 The free-body diagram of the is shown Fig. 1–29b. are AC . This requires Rod three unknowns. Here we will apply the moment equations of equilibrium F F in order to express and in terms of the applied load P.We have Bequations FAC Fapply order to express in terms of the applied have 1s 2moment 70 MPa B alAC fail ee unknowns. Here weand will the ofP.We equilibrium P Using Eq. 1,1–29b. 6 2 The free-body diagram of the bar is shown Fig. 1–29b. There are 1–29 three unknowns. Here we willload apply the moment equations of equilibrium The free-body diagram bar is shown Fig. There are F =MPa 1sin 2allow = the 340110 N>m [p10.01 m22] = 106.8 kN FAC F inP11.25 order toin express and terms applied load P.We 1sFig. = =of the = 35 AC st= AC2 of B0in1A F alB2allow d + ©M = 0; m2 F 12 m2 (1)2 have AC 1106.8 kN212 m2 have AC FACwe order and in of 12 the load P.We have +ree ©M 0; Here P11.25 m2terms -the Fwe m2 = 0 (1) unknowns. will moment equations of equilibrium F.S. 2 Bapply Fapplied F inForder to express and in terms of the applied load P.We Bto=express AC three unknowns. Here will apply the moment equations of equilibrium AC B = 171 Eq.P11.25 1, d of + ©M = 0;Using m2 - FACP 12 =m2 = 01.25 m (1)kN B F F©M order and in terms the applied P.We have B to express t-= 900 MPa P11.25 m2 12 m2 == 00- load (1) AC B0; FF F order to express and in terms of the applied load=P.We have d+ ©M = F 12 m2 P10.75 m2 = 0 (2) fail AC d + 0; P11.25 m2 F 12 m2 0 (1) AC B A B 1106.8 kN212 m2 +©M ©MBA == 0;0; in F 12 m2 P10.75 m2 (2) B AC tBallow = = = 450 MPa B m2 Block Bcreates .0m2 InP-this = 171 kN (2) =P10.75 + ©M == 0; FB=12 m2 = 0 ©MB = = 0; 0; B d + ©M P11.25 - P10.75 FdAC 12 m2 m2 0- FACof12 (1) F.S. 2m2 A =12 0; P11.25 We will now determine each value Pm2 that thecase, allowable # B ©M F m2 = 0 (2) 1.25(1) m2 We will now determine each value of P that creates the allowable A B d + ©M = 0; F (2) 2 FB m 1.25 m 0.75 6 A B12 m2 - P10.75 m2 = 0 Fdetermine =B.1s1–29b. A = are 35110 N>m [1800the mmallowable 110-62 m2>mm2] = 63.0 kN We will each value of P 2that creates free-body diagram the bar ispins, shown inB Fig. There stress in thepins, rod,of block, and respectively. al2allow B 1.25 m The In this case, ress in and respectively. ©M =now 0;rod, F m2 P10.75 m2 =now 0Block (2) Bwill F=12 We determine each value of P that creates the allowable A the B dblock, + ©M 0; F 12 m2 P10.75 m2 = 0 (2) B We will now determine each value of P that creates the allowable A B AC çubuğu için hesap: mm and veya r=0.01 m 2 stress thed=20 rod,equations block, respectively. three unknowns. Here we will apply theinmoment of pins, equilibrium F Rod AC . each This requires ess in the rod, block, and pins, respectively. Using 2, 6the FB creates = 1spins, 2allow A =creates 35110 2 N>m [1800 mm2 110-62 m2>mm2] = 63.0 kN . now This requires od (b)B determine WeAC will value of Pblock, that the allowable stress in theinrod, and respectively. alof B Eq. We will now determine each value P that allowable F F in order to express and terms of the applied load P.We have AC B6 Rod AC 6 2 2 ) FB . This requires 163.0 2 FAC = =1s 2allow 2 2=pins, 340110 2 N>m [p10.01 kN kN212 m2 ess in the1s rod, block, pins, respectively. st AC This requires dFAC stress inand rod, block, and respectively. 2allow 1A 2the 340110 2 N>m [p10.01 m2 ] = 2,106.8 kNm2 6] = 106.8 AC . 1A This requires AC .= Fig. P 2= = 168 kN Using Eq. 2 d + ©MstB1–29 = 0;# AC RodP11.25 m2 F 12 m2 = 0 (1) AC FAC 2 N>m 2 [p10.01 m2m ] = 106.8 kN 0.75 6 2 = 1sst2allow 2 1AAC62 = 340110 Using Eq. 1, 2 AC requires od 1–29 163.0 kN212 m2 FAC =. 1s 2 1A 2 = 340110 2 N>m [p10.01 m2 ] = 106.8 kN AC . This requires Rod sing Eq. 1,This st allow AC FAC = 1sst2allow 1AAC2 = 340110 m2 ] = 106.8 kN A2 N>m or C. [p10.01 PDue = to single shear, = 168 kN 6 Using 2Eq. 1,kN212 d += ©M = 0;1A 21106.8 FBkN212 12 m2 - 1106.8 P10.75 m26m2 = 2m2 0]Pin m2 2 (2) FACEq. 1sstA2allow = 340110 2PAC N>m [p10.01 =2=106.8 kN ing 1, AC FPAC = 1s 2 1A 2 = 340110 2 N>m [p10.01 m2 ] = 106.8 kNm 171 kN = Using Eq. 1, st allow 1106.8 kN212 m2 0.75 = 171 kN = 6 2 2 mAcreates FAC =allowable V =single tallowshear, A= =171 450110 now each value of1.25 P Pin that the 1106.81.25 kN212 m2 m or C . Due to 1106.8 kN212 m2 kN 2 N>m [p10.009 m2 ] = 114.5 kN P = # P determine sing Eq.We 1, willUsing Eq. 1, = 171 kN = 1.25 m kN Block .1106.8 Inand this case,respectively. = 171 P =kN212 m2 the rod, B block, pins, kN212 ock stress B. Ininthis case, 1106.8 1.25 m m2 1.25 m -6 From Eq. FAC = kN t1, A 2= 45011062 N>m2 [p10.009 m22] = 114.5 kN allow 6 2 =B 2 In 2== V 2 171 kN P = Block . this case, 171 P = B deki aluminyum blok için hesap: 6 2 -6 2 2 F 1sal2requires AB1.25 = 2 N>m mm ] 110 2m kN kN 12 m2 AC This Rod 114.5 allow ock . A In this case, sal2B =. =35110 2 Block N>m [1800 110 2[1800 mm>mm = 63.0 kN>mm ] = 263.0-6 mInmm B.35110 this case, allow BB 1.25 6 2 = 183 P 110 = 2 m2>mm2] = 63.0 F = 1s 2 A = 35110 2 N>m [1800 mm kNkN From Eq. 1, 6 2 2 6 2 2 -6 2 2 B al allow B 6 2 2 -6 2 2 ock B. F In this35110 case, 1s 2allow 1A 2 =case, 340110 2 N>m [p10.01 ] = kN 106.8 kN 1.25 m s 2Eq. [1800 mm 110 22mN>m >mm[1800 ] m2 = 63.0 B .N>m InAC this AC stF Using 2, al2allow B == Block = 1s 2 A = 35110 mm 110 2 m >mm ] = 63.0 kN 114.5 kN 12 m2 B al allow B sing Eq.A2, # 6 2 -6 22 By-6 comparison, 163.0 m2 6 2 110 2 2, 2 P = 2 as P reaches its smallest = 183 value kN (168 kN), the allowable Using Eq. sal2allow A Eq. = 35110 2 N>m mm 2kN212 m2>mm ] 110 = 63.0 kN Using 1, 163.0 kN212 m2 AB =[1800 35110 2=N>m [1800 mm 2m ] = 63.0 kN ing Eq. 2,FBB = 1salP2allow 1.25 m P = 168 kN>mm Using Eq. 2, 163.0 kN212 m2 = = 168 kN normal stress will first be developed in the aluminum block. Hence, 1106.8 kN212 m2 0.75163.0 m comparison, 163.0 kN212 m m2 smallest kN212 P = m2 as P reaches=its168 kN value (168 kN), the allowable = By 171 kN P sing Eq. 2, Using Eq. 2,. =0.75 P = = 168 kN A or C Due to single shear, Pin 0.75 m 1.25 m P =normal stress will first = 168 kN shear, n A or C. Due to single P = 168 kN Ans. 163.0 kN212 kN212 m2 0.75 m m2 163.0 m shear, be developed in the aluminum block. Hence, #P C.kN Due0.75 to2= single Pin= A=or168 = Block B . In this case, 6 2 P 168 kN 2= 450110 2 or =C.V Due toFsingle shear, nA = 450110 VA0.75 =ort6allow A 2m2 N>m kN C Due to 0.75 single shear, Pin m AC = FAC = tallow A 2 .2N>m [p10.009 ] =2 [p10.009 114.5 kNm26 ] = 114.5 m Ans. 6 2 -6 2 2 P = 168 kN2 F A = 35110 2 N>m [1800 mm 110 2 m >mm ] = 63.0 kN F = V = t A = 450110 2 N>m [p10.009 m2 A ve C deki perçinler için hesap: d=18 mm veya r=0.009 m ] = 114.5 kN A=or1s C Due to single shear, nBF al.2allow B 6 2 2 AC allow A or C . Due to single shear, Pin 6114.5 kN 2 2 = V = t A = 450110 2 N>m [p10.009 m2 ] = From Eq. 1, allow FAC = V = tallow A = 450110 2 N>m [p10.009 m2 ] = 114.5 kN omAC Eq. 1, 6 From 114.5 12 2m2 6 m2 Eq. 1,kN Using Eq. 2, 114.5 kN m2=2 [p10.009 F = V = t A = 450110 212 N>m ] 2=[p10.009 114.5 kNm22] = 114.5 kN AC allow F = V tallow A 2 N>m om Eq. 1, = 183 kNkN 12 m2 = =450110 1, P # PAC=From=Eq. 114.5 183 kN 163.0 kN212 m2 1.25 m kN 114.5 kN 12 m2 1.25 m 114.5 kN 12 m2 = 183 kN P = P = = 168 om Eq. 1, From Eq. 1, = 183 kN P = 114.5 By as0.75 P reaches its= smallest value (1681.25 mthe m =kN), 183 kNallowable P(168 By comparison, as P comparison, reaches its smallest kN), the allowable kN 114.5 kN 12 m2 1.25 m12 m2 value 1.25 m By comparison, as P=aluminum reaches itsblock. smallest value (168 kN), the allowable A or Cnormal .first Due to single shear, Pin will first be developed in the Hence, = 183 kNkN), =stress 183 kN P the = value ormal stress will be developed aluminum block. Hence, By comparison, as# PPreaches smallest (168 the allowable Byits comparison, asstress P1.25 reaches its smallest value (168 kN), the allowable 1.25 min m normal will first be developed in the aluminum block. Hence, 6 2 2 rmal stress be developed in the aluminum block. Hence, FACwill = first V = t A = 450110 2 N>m [p10.009 m2 ] = 114.5 kN P = 168 kN Ans. By comparison, as P reaches its smallest value (168 kN), the allowable normal stress will first be developed in the aluminum block. Hence, allow By comparison, P reaches its smallest value (168 kN), the allowable P =as168 kN Ans. P = 168 kN Ans. rmal stress willnormal first bestress developed in be the aluminum block. Hence,Ans. will in the göre en büyük kuvvet P taşınabilecek =first 168 kNdeveloped From Eq. 1, Buna P =aluminum 168 kN# block. Hence, Ans. P 114.5 = 168kN kN12 m2 Ans. P = = 183 168 kN kN Ans. P = 1.25 m By comparison, as P reaches its smallest value (168 kN), the allowable normal stress will first be developed in the aluminum block. Hence, P = 168 kN Ans. #21

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