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This 14 page Class Notes was uploaded by Jada Daniel on Wednesday September 23, 2015. The Class Notes belongs to MEM431 at Drexel University taught by GongyaoZhou in Fall. Since its upload, it has received 35 views. For similar materials see /class/212381/mem431-drexel-university in Mechanical Engineering at Drexel University.
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Date Created: 09/23/15
Machine Component Design New Chapter Fundamentals of Dimension Determination A Combined Chapters of 4I 6I 8 and 17 Mainly use the Teaching Notes Cha ter 6 eference book Fundamentals of Dimension Determination Linkages of machine parts need to have adequate strength and rigidity to support the forces acting on the parts There are several considerations that apply in the design of machine elements 61 Strength and Failure Strength It means to make part strong enough so that it will not fail in service Failure It not only means breaking of a part but also related with wear noise plastic and elastic deformation and creep of materials The designer in considering all the possible modes of failure must consider calculations of the followings l Allowable Stress 2 Rigidity 3 Stability 4 Wear Usually the calculations will give the minimum permissible dimensions In practical usage these need to be modified 62 Service Life and Reliability This will consider how long a mechanism or machine can run This is a statistical problem 63 Factor of Safety If life is to be infinite f3 W 1 normal service load If life is to be finite fs M 2 desired service life If stress is main concern from 1 by dividing area dama in stress f s A 3 design stress fs is always larger than one 64 Factor of Safety with Static Loading Static Loading the load varies relatively few times and slowly enough so that neither fatigue nor impact needs to be considered There are two kinds of materials ductile and brittle materials Ductile materials can sustain over 5 percent elongation before rupture Ductile materials steels aluminums brasses Brittle materials cast iron For ductile material f s 5 because it has a yielding stage the stress concentration is not considered in the factor of safety Where Sy yield strength S design stress A part made from a brittle material would give no warning in the form of yielding before breaking nor would there be any local yielding in the regions of high stresses due to stress concentrations Theoretical StressConcentration actor 5a g actual stress K t P oagt on no min al stress Ktgt 1 For brittle material f s S K515 Where Su ultimate strength and Kst Static notch factor K is purely theoretical whereas Kst refers to the effect of the stress concentration on failure of a real part 65 Factor of Safety with Fatigue Loading Fatigue loading based upon a complete reversal of stress during each cycle 9 Time Fig 62 Stresstime diagram for reversed stress loading Endurance Strength the stress can be carried for only a specified finite number of cycles before breaking Endurance Limit Se the stress can be carried on infinite number of cycles without breaking Stress The factor of safety for both ductile and brittle materials f s ES f Where Se endurance limit and Kf fatigue factor For steel Se z 045 to 065 Su S6 05 Su used in design Above is about normal reversed stress tensile or compressive For Shear loading Shear yield SSy 12 Sy Shear endurance limit Sse 12 Se In a general case the variation Sr is above and below an average value Sav 399 max Stress 0 V S min T Time Fig 64 Stresstime diagram for repeated stress loading Repeated Stress Loading Soderberg Triangle First find Sy and Se A B then divide by a chosen fs cf then if E is the design point for a ductile mateIial OD is Sav DE is KfSr Read TuVinall 3rd Edition pp263 6122 DE 0B or OC OD 0A and KS L f 7Si fs Se Sy S S nally f s y gtcompare f s y S 5 S K S S S V Se f 7 SW JKfs i 0 3 612 0C 0D 0A 3e 3 Selts E KfTsr t D C A 0 3 3y syfs Fig 66 Soderberg triangle for ductile materials 66 Application 0fS0derberg s Equations The above equations are useful in analyzing an existing part Here we discuss the design problem of the dimensions for ductile material Axial Loading S g lt gt P P load and A crosssection area 1 P S y ASy PWAK1PY 0139 A s 5 MI A B 3 MC Simple Bending S T f F2 Where M bending moment C distance from neutral axis to outer most fiber 1 moment of inertia of the area 1 s fS7 SyC I MWKMY quot7 Sy and E S 392 MW KfM x See Juvinnal Fourth Edition book pp834 properties for A I J and Z MI quot11 Z Section Modulus Simple torsion of circular shafts TC 1 T Torque C radius to outer most ber J Polar moment J S and SS S Tm ySejK T syfSi Where Kfs torsionfatigue notch factor For the usual case the shaft is made of steel f s 2 1 T SSEJK Tr Combined Bending and torsion of Circular Shafts D m D a u D N Usually a machine element has this kind of arrangement in which the shaft will transmit torque and has a bending load At this situation we need consider combination of stresses There are many theories of failure but not one is perfect or can fit every situation Here we introduce a simple and commonly used maximumshearstress theory for ductile material For the shaft there are normal and shear stress acting on it the maximum shear stress is S s SW s 1 Where S is designed normal stress and SS is shear stress For static loads the safety factor S 2 1 2 1 S 2 For cychc loads 3 15 ZSm Kfs 3 A d 2 SW 2 n S Sm SSE K ss 4 Substitute 3 4 into 2 For Hollow Shaft d outer diameter 61 inner diameter 2d 2df i C d 2 Where C Substitute 7 into 6 The most of rotating shafts cany a steady torque and the loads remain fixed in space in both direction and manitude For this case Mav 0 Mr M Tav T and Once the right hand values of 9 or 10 are calculated the dimension d can be solved Hollow shaft is used in the situation of considering weight and reliability Because center part does not have much contribution to strength 6 7 Stress Concentration Generally stress concentration occurs whenever there are changes in shape such as shoulders holes notches or keyways P P P a Poor 1 Good 0 Pveferred d Pvefened Flg 61 Reduction of sums caneenlration in tension members Stress owlines The example of reduction of stress concentration in shaft can be seen in figure 8 9 pp114N115 II 7A WIIA y a S II a D 37106 108 I b C Fig H Reduclinn of stress wmnlration in arm tted assemblies 68 Stress Concentration Factors It has been known the stressconcentration factor actualstrem Kr no min alstrem P 7 P Amm 7 C 7 2agtltd The actual stress can be calculated from the Kirsch solution for an elliptical hole in an infinitive late The nominal stress S The theoretical StressConcentration factor gt u 1 gtquotu 393me s 0 D 0 II b 11 P P P 411 2 2 Z b 1 sm5u sm25 sm3s a b c Fig 5I0 Stress wncentralion due to elliptical holes in at plates Since almost all design problems are concerned with dynamic loads Kf is more important The fatigue notch factor is determined from experimental tests as K endurancelim itwithoutstress concentration I endurance lirn it with stress concentration To determine the Kfvalue it needs a lot of experiments and is timeconsuming But to find Kt is much easier It is convenient ifwe can find relation OfoWlth K The sensitivity of the specimen to the presence of a notch is known as the notch sensitivity factor Where Kfs torsionfatigue Notch Factor Km torsional stress concentration factor 9 1 qZK lm quot011 11 Based on duh for hrlt3 Notch radius 139V in Flg 6 Average notch sensitivity Mms Handbookquot 1954 Supplzmznl p 102 MelalPragr July 15 1954 Fig 612 shows the variation of q with notch radius for quenches and tempered steel normalized or annealed steel and wrought aluminum alloys So K 1qK 71 andKf 1qK 71 The K and Kt values can be find in Figures 615 pp 122 621 Usually q is very close to 1 We have waKtand Kfszm When key is used to transmit toque or connect two components the fatigue notch needs to be considered For Qled runner kevwav Kf 16 for nrofiled kev an K5 20 u Sled runner b Profiled Fig 6H Kcyways Sled runner Kf16 pro ledKf20 12 439 1 3 ledrunner i keywa S Y 5 d l Example 51 quot 1 Dd 12 r 01d F 550 lb T 21001bin Material hot rolled 1020 steel Find d D and r for safety factor 3 The load is static we have equation r Plane of Dell clces Known fs 3 From appendix C4a pp786 Text book 43911 Ed Sy 48 ksi Su 65 ksi C 0 From Fig63 pp104 for S 65000 Se 26000 psi machined T 210001bin There are two points need to consider the end of the keyway and the shoulder fillet Since the notch isensitivity factor q is dependent on the dimensions of r d and D and no q for keyway It can be replace by Kw Kt So for keyway KtM 16 x 550 x 2625 2310 T ed runner keyway For shoulder fillet KtM 162 x 550 x 3 2670 where K 162 from Fig617 pp124 for Dd 12 and rd 01 Therefore considerin the shoulder fillet as the critical section Round up the d to a closed fraction value 1 113911 15625171 Then r 0d 056in D 12d 1875 in Now that r is known it is possible to deteimine bey use of the q from Fig612 pp119 q 093 19 qK71 1 0931627 1 158 K sz Homework Do the example 61 and 613 616 617 618 in handout Text book 4th Ed HW text book sample problem 171D and Problem 1717 1714 for 3rel Edition 3rd ed P328 fig 824 p145 fig 435 439