Review of Mechanical Engineering Design | ME 3614, Study notes of Mechanical Systems Design

Download Review of Mechanical Engineering Design | ME 3614 and more Study notes Mechanical Systems Design in PDF only on Docsity! ME 3614 - Mechanical Design I Page 1 Review of Mechanical Engineering Designevie of echanical ngineering esign Dr. W. Tong Department of Mechanical Engineering Virginia Tech ME 3614 - Mechanical Design I Page 2 Contentsontents • Load and Stress Analysis - Axial - Bending - Torsion - Combination • Failure Theories - Maximum-shear-stress theory - Distortion-energy theory - Coulomb-Mohr theory • Fatigue Failure - Crack development process - Fatigue-life method - Stress-life method - Strain-life method - S-N diagram • Buckling - Euler & Johnson • Stress Concentration - Static & fatigue Overload and Factor of Safety • Design of shaft - Critical speed • Design of Nonpermanent Joints - Fastener stiffness - Member stiffness - Bolt strength - Preload • Welding – Permanent Joints - Butt and fillet welds - Stresses of welded joints ME 3614 - Mechanical Design I Page 5 ExampleExa ple F = 8,000 N M = 55,000 N-mm   2 22 2.314 4 20 4 mm D A   MPa A F axial 5.252.314 8000    4 44 8.853,7 64 20 64 mm D I z     MPa I Mc z bend 0.708.7853 10000,55 max,  MPabendaxial 5.95max,max   -70 MPa +70 MPa +25.5 MPa - 44.5 MPa +95.5 MPa ME 3614 - Mechanical Design I Page 6 Pure ShearPure Shear ME 3614 - Mechanical Design I Page 7 Static Stress ConcentrationStatic Stress oncentration For normal stress 0 max   tK Nominal normal stress (free of stress raiser) True stress in immediate neighborhood of stress raiser For shear stress 0 max   tsK Nominal shear stress ME 3614 - Mechanical Design I Page 10 Maximum Shear Stress Theory (MSS) for Ductile Materials axi u Shear Stress Theory ( SS) for Ductile aterials MSS predicts the yielding of ductile materials when ysSmax 2 y sy S S    3131max 2/ 2/      yyys SSSn Case 3 Sy Sy -Sy -Sy Sy Sy -Sy -Sy Case 2 Case 1 B A A B Inside – safe Outside – failure Online – n = 1 n > 1 n < 1 Factor of Safety ME 3614 - Mechanical Design I Page 11 Distortion Energy Theory (DE) – von Mises Theory Distortion Energy Theory (DE) – von ises Theory von Mises stress s’ is defined as:   a)     In terms of principal stresses       b)     In terms of applied stresses     Failure occurs as ,       2 2 13 2 32 2 21    2 321 2 1         2 )(6 222222 zxyzxyxzzyyx     222 3 xyyyxx   yS 3-D 2-D 3-D 2-D    ySnFactor of Safety ME 3614 - Mechanical Design I Page 12 Ductile Coulomb-Mohr Theory (DCM) Ductile Coulo b- ohr Theory (DC ) Coulomb-Mohr failure line -Sc St Tensile strengthCompressive strength 12 St St -Sc -Sc A B ME 3614 - Mechanical Design I Page 15 Selection of Failure Criteria Selection of Failure Criteria ME 3614 - Mechanical Design I Page 16 BucklingBuckling Inflection Point ME 3614 - Mechanical Design I Page 17 Euler-Johnson CurveEuler-Johnson Curve yS CE S 2 1  actual cr buckling P P n  ME 3614 - Mechanical Design I Page 20 S-N DiagramS- iagra “Knee” S’e S’103 Endurance Limit Low Cycle Fatigue Strength ME 3614 - Mechanical Design I Page 21 Estimating S’e for Ductile SteelsEsti ating S’e for uctile Steels S’e = 0.50 Sut (ksi or MPa) Sut  200 ksi (1,400 MPa) 100 ksi (700 MPa) Sut > 200 ksi (1,400 MPa) ME 3614 - Mechanical Design I Page 22 Fatigue Strength ModificationFatigue Strength odification Se = ka kb kc kd ke kf Se’ Our part? Moore Specimen • ka – Surface Factor • kb – Size/Shape Factor • kc – Loading Factor • kd – Operating Temperature Factor • ke – Reliability Factor • kf – Miscellaneous-Effects Factor Table 7-9 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Modified Geodman and Langer Failure Criteria Intersecting Equations Intersection Coordinates Load liner = = -* etee Sr = ee Load line r = = oe ghee! 6 Os e+$el So = S)— Spe Feit = Sof Sim Fatigue factor of safety n= ] & om "Sr {0 VirginiaTech Invent the Future ME 3614 - Mechanical Design I Page 25  Table 7-10 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria Intersecting Equations eve ta mM eres Ler Sy ($0)? _ eS 25," st(g) =| = FS [=r 1+(=) So _ 5a load liner = <= Sn = = So, Sm _ rs 3ts7! an lood liner = = 4 Saf Su\? _ _ Si 28,\* 5, Sy] | s-€[-(-@y0-9 Sa, Sn 5757! So = Sy — Sie fern = Saf Sm Fatigue factor of safety {0 VirginiaTech Invent the Future ME 3614 - Mechanical Design I ME 3614 - Mechanical Design I Page 27 Modified Goodman Theory r = a/m 1 y m y a S S S S mS aS m a ME 3614 - Mechanical Design I Page 30 Example: Spring ClipExa ple: Spring Clip 4 in 0.1094 in r = 1/8 in KT = 1.7 3/4 in A cantilever beam is serving as a spring for a latching mechanism. When assembled the free end is deflected 0.075 in, which corresponds to a force of 8.65 lb. When the latch operates the end deflects an additional 0.15 in Should we expect to see fatigue failure? Material: high carbon steel – HB = 490, machined finish, Sy = 225 ksi Reliability = 90% Static stress concentration factor KT at 1/8 in fillet radius is 1.70 ME 3614 - Mechanical Design I Page 31 Solutions: Eq. 3-17 Sut = 0.500 HB = 0.5(490) = 245 ksi > 212 ksi Eq. 7-8 Se’ = 100 ksi Critical section of spring clip is in bend at the wall. 1. Compute endurance limit at the critical section Se = ka kb kc kd ke kf Se’ Eq. 7-18 Table 7-4 a = 2.70 ksi b = - 0.265  ka = 0.6284 kb – non-circular, non-rotating cross-section Need to find de first. Table 7-50.0766de2 = 0.05 hb  de = 0.2314 in Since 0.11  de  2 in, kb = 0.879 de-0.107 = 1.028 Example: Spring Clip (Cont’d)Exa ple: Spring Clip (Cont’d) b uta aSk  A95% Max Stress Our Part = A95% Max Stress Equiv. Round-Rotating Section ME 3614 - Mechanical Design I Page 32 Example: Spring Clip (Cont’d)Exa ple: Spring Clip (Cont’d) kc = 1 (bending) kd = 1 (not specified) Table 7-7 ke = 0.897 kf = 1 (not specified) Thus, Se = ka kb kc kd ke kf Se’ = (0.6284)(1.028)(1)(1)(0.897)(1)(100) = 57.94 ksi 2. Compute the mean and alternating loads, Fm, Fa Fmin = Fi = 8.65 lbf on installation Fmax = kx max k = Fi/xi = 8.65/0.075 = 115.33 lbf/in Fmax = kxmax = k(xmin+ ) = 115.33(0.075 + 0.15) = 25.95 lbf Fa = (Fmax - Fmin) / 2 = (25.95 – 8.65)/2 = 8.65 lbf Fm = (Fmax + Fmin) / 2 = (25.95 + 8.65)/2 = 17.3 lbf 3. Compute the mean and alternating stresses, m, a a,norminal = Mac/I = 23.13 psi Ma = 17.3(4) = 69.2 in-lbf c = h/2 = 0.0547 in I = bh3/12 = 8.1833 x 10-5 in4 ME 3614 - Mechanical Design I Page 35 Example: Spring Clip - OverloadExa ple: Spring Clip - Overload Determine the factors of safety in fracture and yielding assuming overload occurs by an increase in Fmax 6. Compute Fm* and Fa* 7. Compute m* and a* = 34,691.8 n + 11,563.9 psi lbf325.4975.12 2 65.895.25 2 * min * max*      n nFF Fa 53 * * min, * 10x1833.8 )0547.0)(4)(325.4975.12( 12/ )2/)(4(    n bh hF I cM mm alnormm  lbf325.4975.12 2 65.895.25 2 * min * max*      n nFF Fm lbf65.8min * min  iFFF lbf)95.25(max * max nnFF  ME 3614 - Mechanical Design I Page 36 Example: Spring Clip - OverloadExa ple: Spring Clip - Overload Fracture line Yield line 53 * * min, 10x1833.8 )0547.0)(4)(325.4975.12( 12/ )2/)(4(    n bh hF I cM aa alnora psi9.563,118.691,34  n psi6.658,191.976,58)9.563,118.691,34(7.1* min, *  nnK alnorafa  1 ut m e a S S S S Sa = a* Sm = m* 1 000,225 9.563,118.691,34 940,57 6.658,191.976,58     nn 10.1fn 1 y m y a S S S S Sa = a* Sm = m* 1 000,225 9.563,118.691,34 000,225 6.658,191.976,58     nn 49.2yieldn ME 3614 - Mechanical Design I Page 37 Example: Spring Clip - OverloadExa ple: Spring Clip - Overload 20 20 40 60 80 100 120 140 160 180 200 220 240 260 280 40 60 80 100 120 160 180 140 200 240 a mSut=245sy Sy = 225 Se = 57.945 Yield line Modified Goodman line e = 39.32 m = 46.25 Sm Sa Graphic Check of Solution From the graph, it reads that Sm = 53 ksi & Sa = 45 ksi Similarly for yield It reads that Sm = 118 ksi and Sa = 99 ksi 14.1 32.39 45  a a f S n  15.1 25.46 53  m m f S n  52.2 32.39 99  a a yield S n  55.2 25.46 118  m m yield S n  Stress von Mises (WCS) (N 7 m%*2) Deformed Scale 5.8767E+02 Loadset:LoadSet! .B27et+a7 -B24etar -O2letar »O15eta7 -O15e+a7 -O12et+a7 -809e+47 -805etar -803e+a7 SPNWAUANAYW @VirginiaTech ME 3614 - Mechanical Design I Page 40 Invent the Future ME 3614 - Mechanical Design I Page 41 Shaft Critical SpeedShaft Critical Speed Shaft critical speed is defined as the speed at which the shaft becomes unstable, with deflection increasing without upper bound. A shaft critical speed can be calculated as    A gEI l 2 1        A – cross section area E – Young’s modulus of elasticity g – acceleration due to gravity  – specific weight ME 3614 - Mechanical Design I Page 42 ExampleExa ple A 1.0 inch diameter steel shaft is supported with bearings at each end. The distance along the shaft between bearings is 24 inches. Your team has been asked to increase the critical speed of the shaft by 200%. E – Young’s modulus of elasticity = 30E6 psi g – acceleration due to gravity = 386 in/s2  – specific weight = 0.282 lb/in3 a) What is the original design’s critical speed? b) What will be the new design’s shaft diameter?            164/ 64/ 2 2 2 422 1 gEd ld dgE lA gEI l                         rpm E 1.868 282.016 1630386 24 22         a) b) Increasing critical speed by 200% rpm2.736,1%200x1.8681     16 22 1 gEd l       From It follows that     in EgE l d 0.2 282.016 630386 24 2.1736 16 22 1                    ME 3614 - Mechanical Design I Page 45 Self-LockingSelf-Locking   )(2 fld ldfFd T m mm L      Required torque for lowering the load is TL > 0 Self-Locking, i.e., 0 ldf m md l f   tanfThe condition for self-locking is For car jacks, self-locking is extremely critical for safety! ME 3614 - Mechanical Design I Page 46 Power Screw EfficiencyPower Screw Efficiency Required torque for rising the load is If f = 0, the torque requires only to raise the load The power screw efficiency is When the screw is loaded axially, a thrust or collar bearing must be employed between the rotating and stationary members in order to carry the axial load. If fc is the coefficient of collar friction, the torque required is Therefore, the total torque for rising the load is TR,t = TR + Tc and the total torque for lowering the load is TL,t = TL + Tc   )(2 fld dflFd T m mm R      20 Fl T  )(22 ,, 0 cRtRtR TT Fl T Fl T T e    2 cc c dFf T  ME 3614 - Mechanical Design I Page 47 Exam 8-7 (p.444)Exa 8-7 (p.444) A C clamp shown in the figure has a handle with diameter 3/16 in made of cold-drawn AISI 1006 steel. The overall length is 3 in. The screw is 7/16 in-14 UNC and is 5-3/4 in long, overall. Distance A = 2 in. The clamp will accommodate parts up to 4-3/16 in high. a) What screw torque will cause the handle to bend permanently? b) What clamping force will the answer to part (a) cause if the collar friction is neglected and if the thread friction is 0.075? c) What clamping force will cause the screw to buckle? d) Are there any other stresses or possible failure to be checked? Solutions a) The force F is perpendicular to the paper in 406.2 32 7 4 1 8 1 3 L FT 406.2 ¼ in ME 3614 - Mechanical Design I Page 50 Exam 8-7 (Cont’d)Exa 8-7 (Cont’d)                     F F fld dflFd T m mm R 02845.0155.1 0714.0 075.03911.02 155.1 3911.0 075.00714.03911.0 )sec(2 sec            lbf030,1 02845.0 2.29 02845.0  T Fclamp c) The column has one end fixed and the other end rounded (pivoted), C = 1.2. Using the mean diameter dm = 0.3911 in, L = 4.1875 in, and E = 30 x 106 psi, it follows that in0978.0 4 3911.0 4  m d k       7.131 000,41 10 30 2.122 6 1   yS CE S 8.42 0978.0 1875.4  k L S Since S < S1, go to Johnson column 2 2 1         SS CE S A P y y cr                                      222 2 8.4241000 6302.1 1 000,41 4 3911.0 2 1    e SS CE SAPF yycrclamp lbf663,4 ME 3614 - Mechanical Design I Page 51 Bolted Connectionolted onnection (Grip of connection) 21 111 kkk  21 21 kk kk k   ld lt t t t l EA k  d d d l EA k  dttd td b lAlA EAA k  Bolt Stiffness ME 3614 - Mechanical Design I Page 52 Summarization of Simple Bolted JointSu arization of Si ple Bolted Joint Member Bolt Preload Fm = - Fi Compression Fb = Fi Tension External Load Fm = Pm - Fi = (1 – C)P - Fi Fm < 0 (Eq. 8-25) Fb = Pb + Fi = CP + Fi Fm < 0 (Eq. 8-24) Separation: Fm = 0 Fi = Pm = (1 – C)PIndicating that Pm – Fi = 0 Introduce the factor of safety against joint separation nsep Fi = P(1 – C)nsep Thus,  CP F n isep   1 Eq. 8-29 ME 3614 - Mechanical Design I Page 55 Stiffness of Members in Clamped ZoneStiffness of e bers in Cla ped Zone In this book, use  = 30° unless specified l = grip length dw = D = 1.5d For the joints with n members, the total stiffness of the members is nm kkkkk 11111 321  The spring stiffness of a single frustum is                 dDdDt dDdDt Ed k 155.1 155.1 ln 5774.0  Eq. 8-20 ME 3614 - Mechanical Design I Page 56 Example 8-24Exa ple 8-24 The figure illustrates the connection of a cylinder head (steel) to a pressure vessel using 10 bolts with a diameter of ½ in and confined-gasket seal. The effective sealing diameter is 5.9 in. Other dimensions are: A = 3.94 in, B = 7.87 in, C = 11.81 in, D = ½ in, and E = 5/8 in. The cylinder is cast iron with a modulus of elasticity of 18 Mpsi. The ½ in SAE washer to be used under the nut has OD = 1.062 in and is 0.095 in thick. Find the stiffness of the bolt and the members and the joint constant C. Solutions Calculate bolt stiffness Nut thickness H = 7/16 in Grip in 22.1095.0 8 5 2 1  washerG tEDL in 4 1 1 4 1 2 1 2 4 1 2        dLTThreaded length The bolt total length L > LG + H = 1.22 + 7/16 = 1.66 in Take L = 1.75 in ME 3614 - Mechanical Design I Page 57 Example 8-24 (Cont’d)Exa ple 8-24 (Cont’d) The length of useful unthreaded portion The length of threaded portion within the grip The area of unthreaded portion The area of threaded portion The stiffness for the threaded portion in50.0 4 1 175.1  Td LLl in72.050.022.1  dG lLl t   in1963.0 4 5.0 4 22   d Ad (UNC)in 1419.0tA   Mlbf/in 9125.5 72.0 301419.0  t t t l EA k The stiffness for the unthreaded portion   Mlbf/in 778.11 50.0 301963.0  d d d l EA k The bolt stiffness dtb kkk 111  Mlbf/in 936.3   dt dt b kk kk k ME 3614 - Mechanical Design I Page 60 Example 8-26 (Cont’d)Exa ple 8-26 (Cont’d) Member stiffness for three frusta Top frustum: D = 0.75”, t = 0.4225”, d = 0.5”, E = 30 Mpsi  k1 = 36.14 Mlbf/in 2nd frustum: D = 1.018”, t = 0.1725”, d = 0.5”, E = 30 Mpsi  k2 = 134.6 Mlbf/in 3rd frustum: D = 0.75”, t = 0.25”, d = 0.5”, E = 14.5 Mpsi  k3 = 23.49 Mlbf/in   3 1321 11111 i im kkkkk Mlbf/in87.12 1 1 3 1         i i m k k 281.0 87.1204.5 04.5      mb b kk k C ME 3614 - Mechanical Design I Page 61 Statically Loaded Tension Joint with PreloadStatically Loaded Tension Joint with Preload The tensile stress in the bolt with preload is The limiting value of b is the proof strength Sp, i.e., . Sp  b, With the introduction of a load factor n, it can be derived as that This gives the upper bound on preload (Eq. 8-28 mod.). Define Fp = AtSp (Eq. 8-31) as the proof load, thus t i tt i t b b A F A CP A F A F  t i t p A F A CP S  CPnSAF pti      p p i F F F 90.0 75.0 CPnFF pi  for nonpermanent connections, reused fasteners for permanent connections ME 3614 - Mechanical Design I Page 62 Bolt Joint SummaryBolt Joint Su ary Stiffness Model a) Equivalent bolt stiffness kb b) Equivalent member stiffness km dttd td b lAlA EAA k   • ld = L – LT is the length of useful unthreaded portion. • lt = LG – ld is the length of threaded portion within grip LG. • At is the tensile stress area. • Ad is the major diameter area. Eq. 8-17 1 321 1111         n m kkkk k Eq. 8-18 • For each member • Use D = 1.5d if no information provided                 dDdDt dDdDt dE k i i i i 155.1 155.1 ln 5774.0  Eq. 8-20 If k1 << ki (i = 2, 3,…n)  km = k1 t t t l EA k  d d d l EA k  ME 3614 - Mechanical Design I Page 65 ExampleExa ple t1 = 6 mm t2 = 10 mm ld = 9 mm lt = 9 mm Consider the single bolt and clamped members of a bolted joint in tension. The bolt is M10 x 1.5 with the grade of 9.8. Let the preload Fi = 10,000 N and the external load P = 13,500 N. The member thickness is t1 = 6 mm and t2 = 10 mm, respectively. Em = EB = 207 GPa. ld = lt = 9mm. For the bolt (Table 8-1), the tensile stress area At = 58.0 mm2, the shank diameter dshank = dnonimal = 10 mm Sp = 650 MPa (Table 8-11)  shankbolt A d A  2 22 mm 54.78 4 10 4  Compute bolt stiffness kb   2 6 mm N 10 x 806.1 9 000,20754.78  d boltd d l EA k   2 6 mm N 10 x 334.1 9 000,2070.58  t boltt t l EA k ME 3614 - Mechanical Design I Page 66 Example (Cont’d)Exa ple (Cont’d) Compute stiffness of the members, km Both members are made from steel but with different thicknesses.                      mm N 10 x 019.6 101510156155.1 101510156155.1 ln 10207,0005774.0 155.1 155.1 ln 5774.0 6 1 1 1                     dDdDt dDdDt Ed k                      mm N 10 x 601.4 1015101510155.1 1015101510155.1 ln 10207,0005774.0 155.1 155.1 ln 5774.0 6 2 2 2                     dDdDt dDdDt Ed k ME 3614 - Mechanical Design I Page 67 Example (Cont’d)Exa ple (Cont’d) Compute stiffness ratio, C     2273.0 10 x 6078.210 x 7673.0 10 x 7673.0 66 6      mb b kk k C Analyze the Current Design a) Computer factor of safety against separation   PnCF sepi  1       9586.0 500,132273.01 000,10 1      PC F n isep Because nsep < 1, the joint fails. Thus, the bolt carries all the external load, Fb = P b) Computer load factor in bolt This shows that the joint fails but the bolt carries load without yielding.       03.9 500,132273.0 000,1058650      CP FAS n itp ME 3614 - Mechanical Design I Page 70 Fatigue Failure ModelsFatigue Failure odels 1 ut m e a S S S S Modified Goodman Gerber ASME-elliptic 1 2        ut m e a S S S S 1 22                p m e a S S S S Note: since Sp < Sy, Sy is replaced by Sp in the above equation. iam SS +   eut iute a SS SS S     + iam SS    eiutieeutut e a SSSSSSS S  24 2 1 22  + iam SS   eiiepp ep e a SSSSSS S S     222 22 (Eq. 8-40) (Eq. 8-42) (Eq. 8-43) ME 3614 - Mechanical Design I Page 71 Factor of Safety against FatigueFactor of Safety against Fatigue a a f S n  The factor of safety against fatigue is Where and use Modified Goodman criterion, t a A CP 2    eut iute a SS SS S        eut itute f SSCP FASS n    2 (Eq. 8-44) (Eq. 8-45) Without preload, C = 1 and Fi = 0  eut tute f SSP ASS n   2 0, (Eq. 8-46) ME 3614 - Mechanical Design I Page 72 Shear Joints – Centroid GShear Joints – Centroid G Rotational Pivot Point       n i n ii A xA AAAAA xAxAxAxAxA x 1 1 54321 5544332211       n i n ii A yA AAAAA yAyAyAyAyA y 1 1 54321 5544332211 Ai = cross-sectional area of the ith pin xi, yi = coordinates of the ith pin For 2D problems, the coordinates of the Centroid are: y1 x1 ME 3614 - Mechanical Design I Page 75 Example 8-50 (Cont’d)Exa ple 8-50 (Cont’d) Bolt shear   2 22 in 1963.0 4 5.0 4   d As Bearing on bolt 2in 1875.0)2/1)(8/3(  dtAb The resultant load on bolt psi 600,9 1875.0 800,1  bA F  lbf500,11650150"'  AAA FFF lbf800,11650150"'  BBB FFF psi 170,9 1963.0 1800  s B A F  (Note: only one shear surface)   ksi05.4985577.0577.0  psp SS 35.517.9 05.49   spSn 85.8 6.9 85   pSn ME 3614 - Mechanical Design I Page 76 Example 8-49 (Cont’d)Exa ple 8-49 (Cont’d) Bearing on member hole 2in 1875.0)2/1)(8/3(  holet dtA psi 600,9 1875.0 800,1  tA F  Beading stress in plate     inlbf500,4114300 FLM 63.5 6.9 54   ySn 433 in 2461.0)5.02( 12 8/3 I   psi 285,18 2461.0 1500,4  I Mc  95.2 285.18 54   ySn ME 3614 - Mechanical Design I Page 77 Example 8-7 (p. 441)Exa ple 8-7 (p. 441) Shown in the figure below is a 15- by 200-mm rectangular steel bar cantilevered to a 250-mm steel channel using four tightly fitted bolts located at A, B, C, and D. For a F = 16 kN load, find (a) The resultant load on each bolt (b) The maximum shear stress in each bolt (c) The maximum bearing stress (d) The critical bending stress in the bar kN4 4 16 '  n V FPrimary shear force The moment about the centroid O     mFLM  N800,67550300161 mm 04.967560 22 r   kN7.17 964 6800 4 " 2 1  r rM F Since the secondary shear forces are equal, it follows that ME 3614 - Mechanical Design I Page 80 Butt Joint – Normal and Shear StressesButt Joint – Nor al and Shear Stresses hl F  hl F  The average normal stress The average shear stress h h ME 3614 - Mechanical Design I Page 81 Transverse Fillet WeldsTransverse Fillet elds Fs Fn F  Fs = Fsin Fn = Fcos At angle  the forces on each weldment consist of a normal force Fn and a shear force Fs The throat length t and surface A are  sincos   h t  sincos   hl tlA l – length of the weld     2sincossinsincossin  hl F hl F A Fs     sincoscossincoscos 2  hl F hl F A Fn The nominal stresses at the angle  in the weldment ME 3614 - Mechanical Design I Page 82 Transverse Fillet WeldsTransverse Fillet elds The von Mises stress at the angle  is     222222 cossinsin3cossincos3   hl F To find the maximum von Mises stress, differentiate the above equation with respect to  and let d’/d = 0 At  = 62.5 degree, the maximum von Mises stress is hl F 16.2'max  The maximum shear stress can be found by differentiating the nominal shear stress equation with respect to  and equating to zero. The stationary point occurs at  = 67.5 degree with a corresponding hl F 196.1 hl F 623.0 hl F 207.1max  hl F 5.0 ME 3614 - Mechanical Design I Page 85 Strength of Welded JointsStrength of elded Joints + Shear stress on on base metal should not exceed 0.40 Sy of base metal. First 2 (or 3) digits are ultimate tensile strength. ME 3614 - Mechanical Design I Page 86 Example 9-3 (p. 499)Exa ple 9-3 (p. 499) The members being joined in Prob. 9-1 are cold-rolled 1018 for the horizontal bar and hot-rolled 1018 for the vertical support. What load on the weldment is allowable because member is incorporated into the welds? From Table A-20 1018 HR: Sut = 58 ksi, Sy = 32 ksi 1018 CR: Sut = 64 ksi, Sy = 54 ksi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. From Table 9-4:     ksi8.128.12,4.17min40.0,30.0min  yutall SS for both materials.      kip3.118.122216/5707.0707.0  allhlF 