Chapter 9. τ all = min(0.30s ut,0.40s y ) = min[0.30(58), 0.40(32)] = min(17.4, 12.8) = 12.8 kpsi 2(32) (5/16)(4)(2) 2F hl. = 18.1 kpsi Ans. 1.

Transcription

1 budynas_sm_c09.qxd 01/9/007 18:5 Page 39 Capter Eq. (9-3: F 0.707lτ 0.707(5/1(4( kip 9- Table 9-: τ all 1.0 kpsi f kip/in 14.85(5/1 4.4 kip/in F fl 4.4( kip 9-3 Table A-0: 1018 HR: S ut 58 kpsi, S y 3 kpsi 1018 CR: S ut 4 kpsi, S y 54 kpsi Cold-rolled properties degrade to ot-rolled properties in te neigborood of te weld. Table 9-4: for bot materials. Eq. (9-3: τ all min(0.30s ut,0.40s y min[0.30(58, 0.40(3] min(17.4, kpsi F 0.707lτ all F 0.707(5/1(4( kip 9-4 Eq. (9-3 τ F l (3 (5/1(4( 18.1 kpsi 9-5 b d in F " (a Primary sear Table 9-1 τ y V A F 1.414(5/1( 1.13F kpsi

2 budynas_sm_c09.qxd 01/9/007 18:5 Page Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design Secondary sear Table 9-1 J u d(3b + d [(3( + ] in 3 J 0.707J u 0.707(5/1( in 4 τ x τ y Mr y J 7F( F kpsi Maximum sear τ max τ x + (τ y + τ y F ( F kpsi F τ all kip 9. (1 (b For E7010 from Table 9-, τ all 1 kpsi Table A-0: HR 100 Bar: S ut 55 kpsi, S y 30 kpsi HR 1015 Support: S ut 50 kpsi, S y 7.5 kpsi Table 9-5, E7010 Electrode: S ut 70 kpsi, S y 57 kpsi Te support controls te design. Table 9-4: τ all min[0.30(50, 0.40(7.5] min[15, 11] 11 kpsi Te allowable load from Eq. (1 is F τ all kip b d in F Primary sear Secondary sear τ y V A 7" F 1.414(5/1( + 0.5F Table 9-1: J u (b + d3 ( in 3 J 0.707J u 0.707(5/1( in 4 τ x τ y Mr y J (7F(1.3.97F

3 budynas_sm_c09.qxd 01/9/007 18:5 Page 41 Capter 9 41 Maximum sear τ max τ x + (τ y + τ y F.97 + ( F kpsi F τ all 4.1 wic is twice τ max /9. of Prob Weldment, subjected to alternating fatigue, as troat area of A 0.707(( mm Members endurance limit: AS 1010 steel S ut 30 MPa, S e 0.5(30 10 MPa k a 7( k b 1 k c 0.59 k d 1 Electrode s endurance: 010 (direct sear (sear k f 1 1 K fs S se 0.875(1(0.59(0.37( MPa S ut ( MPa S e 0.5( MPa k a 7( k b 1 k c 0.59 k d 1 (direct sear (sear k f 1/K fs 1/ S se 0.57(1(0.59(0.37( MPa Tus, te members and te electrode are of equal strengt. For a factor of safety of 1, F a τ a A 30.(71(10 3.1kN

4 budynas_sm_c09.qxd 01/9/007 18:5 Page 4 4 Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design 9-8 Primary sear τ 0 (wy? Secondary sear Table 9-1: J u πr 3 π( cm 3 J 0.707J u 0.707(0.5(40 14 cm 4 M 00F N m (F in kn τ Mr J (00F(4.817F ( welds (14 F τ all τ kN fom J ( u l a3 /1 a a a fom a(3a + a ( a a (a fom (a4 a a ( 1(a + aa 5a a fom 1 [ 8a 3 + a 3 + a 3 ] a4 11 3a 1 a + a 3 fom (a3 fom π(a/3 πa 1 4a 8a3 4a a a3 4a a a ( a ( a ( a Rank Tese rankings apply to fillet weld patterns in torsion tat ave a square area a a in wic to place weld metal. Te object is to place as muc metal as possible to te border. f your area is rectangular, your goal is te same but te rankings may cange. Students will be surprised tat te circular weld bead does not rank first fom u l 1 a fom u l 1 a fom u l 1 a ( a 3 ( 1 1 ( a ( a ( a 3 ( a ( a 1 ( a ( a

5 budynas_sm_c09.qxd 01/9/007 18:5 Page 43 Capter 9 43 fom u l 1 [(a] ( a (3a + a 1 ( a ( a 0.17 x b a, ȳ d b + d a u d3 3 d ( a 3 fom u l a3 /3 3a 1 9 u πr 3 πa3 8 3a a 3 ( a ( + (b + d a3 9 3 a3 a 3 + 3a a3 9 3 ( a ( a ( a fom u l πa3 /8 πa a Te CEE-section pattern was not ranked because te deflection of te beam is out-of-plane. f you ave a square area in wic to place a fillet weldment pattern under bending, your objective is to place as muc material as possible away from te x-axis. f your area is rectangular, your goal is te same, but te rankings may cange Materials: Attacment (1018 HR S y 3 kpsi, S ut 58 kpsi Member (A3 S y 3 kpsi, S ut ranges from 58 to 80 kpsi, use 58. Te member and attacment are weak compared to te E0XX electrode. Decision Specify E010 electrode Controlling property: τ all min[0.3(58, 0.4(3] min(1., kpsi For a static load te parallel and transverse fillets are te same. f n is te number of beads, Make a table. τ n Decision: Specify 1/4" leg size Decision: Weld all-around F n(0.707l τ all F 0.707lτ all Number of beads n (3( Leg size /" /1" /4" 4

6 budynas_sm_c09.qxd 01/9/007 18:5 Page Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design Weldment Specifications: Pattern: All-around square Electrode: E010 Type: Two parallel fillets Two transverse fillets Lengt of bead: 1 in Leg: 1/4 in For a figure of merit of, in terms of weldbead volume, is tis design optimal? 9-1 Decision: Coose a parallel fillet weldment pattern. By so-doing, we ve cosen an optimal pattern (see Prob. 9-9 and ave tus reduced a syntesis problem to an analysis problem: Table 9-1: A 1.414d 1.414((3 4.4 in 3 Primary sear Secondary sear Table 9-1: J u d(3b + d τ y V A [3(3 + 3 ] J 0.707(( in 4 τ x Mr y J τ max 3000(7.5( τ x + (τ y + τ y 1 18 in 3 57 τ y 57 + ( Attacment (1018 HR: S y 3 kpsi, S ut 58 kpsi Member (A3: S y 3 kpsi Te attacment is weaker Decision: Use E0XX electrode τ all min[0.3(58, 0.4(3] 1.8 kpsi τ max τ all psi in Decision: Specify 3/8" leg size Weldment Specifications: Pattern: Parallel fillet welds Electrode: E010 Type: Fillet Lengt of bead: in Leg size: 3/8 in

7 budynas_sm_c09.qxd 01/9/007 18:5 Page 45 Capter An optimal square space (3" 3" weldment pattern is or or. n Prob. 9-1, tere was roundup of leg size to 3/8 in. Consider te member material to be structural A3 steel. Decision: Use a parallel orizontal weld bead pattern for welding optimization and convenience. Materials: Attacment (1018 HR: S y 3 kpsi, S ut 58 kpsi Member (A3: S y 3 kpsi, S ut kpsi; use 58 kpsi From Table 9-4 ASC welding code, τ all min[0.3(58, 0.4(3] min(1., kpsi Select a stronger electrode material from Table 9-3. Decision: Specify E010 Troat area and oter properties: A 1.414d 1.414((3 4.4 in x b/ 3/ 1.5in ȳ d/ 3/ 1.5in Primary sear: J u d(3b + d 3[3(3 + 3 ] 18 in 3 J 0.707J u 0.707(( in 4 τ x V A x y r r x x r y y x Secondary sear: τ Mr J τ x τ cos 45 Mr J cos 45 Mr x J τ x 3000( + 1.5( τ y τ x 51 51

8 budynas_sm_c09.qxd 01/9/007 18:5 Page 4 4 Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design τ max 479 psi Relate stress and strengt: τ max τ all Weldment Specifications: (τ x + τ x + τ y 1 ( in 3/8in Pattern: Horizontal parallel weld tracks Electrode: E010 Type of weld: Two parallel fillet welds Lengt of bead: in Leg size: 3/8 in Additional tougts: Since te round-up in leg size was substantial, wy not investigate a backward C weld pattern. One migt ten expect sorter orizontal weld beads wic will ave te advantage of allowing a sorter member (assuming te member as not yet been designed. Tis will sow te inter-relationsip between attacment design and supporting members Materials: Member (A3: S y 3 kpsi, S ut 58 to 80 kpsi; use S ut 58 kpsi Attacment (1018 HR: S y 3 kpsi, S ut 58 kpsi τ all min[0.3(58, 0.4(3] 1.8 kpsi Decision: Use E010 electrode. From Table 9-3: S y 50 kpsi, S ut kpsi, τ all min[0.3(, 0.4(50] 0 kpsi Decision: Since A3 and 1018 HR are weld metals to an unknown extent, use τ all 1.8 kpsi Decision: Use te most efficient weld pattern square, weld-all-around. Coose " " size. Attacment lengt: l 1 + a in Troat area and oter properties: A 1.414(b + d 1.414(( x b 3in, ȳ d 3in

9 budynas_sm_c09.qxd 01/9/007 18:5 Page 47 Capter 9 47 Primary sear Secondary sear τ y V A F A psi J u (b + d3 ( + 3 J 0.707( in 4 τ x τ y Mr y J τ max 88 in (.5 + 3(3 7 psi 03. τ x + (τ y + τ y ( psi Relate stress to strengt τ max τ all in Decision: Specify 3/8 in leg size Specifications: Pattern: All-around square weld bead track Electrode: E010 Type of weld: Fillet Weld bead lengt: 4 in Leg size: 3/8 in Attacment lengt: 1.5 in 9-15 Tis is a good analysis task to test te students understanding (1 Solicit information related to a priori decisions. ( Solicit design variables b and d. (3 Find and round and output all parameters on a single screen. Allow return to Step 1 or Step. (4 Wen te iteration is complete, te final display can be te bulk of your adequacy assessment. Suc a program can teac too. 9-1 Te objective of tis design task is to ave te students teac temselves tat te weld patterns of Table 9-3 can be added or subtracted to obtain te properties of a comtemplated weld pattern. Te instructor can control te level of complication. ave left te

10 budynas_sm_c09.qxd 01/9/007 18:5 Page Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design presentation of te drawing to you. Here is one possibility. Study te problem s opportunities, ten present tis (or your sketc wit te problem assignment. A Section AA b 1 8" 1" d 1018 HR A3 a A lbf Use b 1 as te design variable. Express properties as a function of b 1. From Table 9-3, category 3: A 1.414(b b 1 x b/, ȳ d/ u bd b 1d u τ V A τ Mc Body welds not sown (b b 1d F 1.414(b b 1 Fa(d/ u τ max τ + τ Parametric study Let a 10 in, b 8 in, d 8 in, b 1 in, τ all 1.8 kpsi, l (8 1 in A 1.414( in u (8 (8 / 19 in (( in 4 τ psi τ (10(8/ 948 psi τ max from wic 0.48 in. Do not round off te leg size someting to learn. fom u l (1 4.5 A 8.48( in 135.7( in 4 8" b Attacment weld pattern considered

11 budynas_sm_c09.qxd 01/9/007 18:5 Page 49 Capter 9 49 vol l in 3 vol eff 0.39 τ psi 0.48 τ psi 0.48 τ max psi Now consider te case of uninterrupted welds, b 1 0 A 1.414(( u (8 0(8 / 5 in 3 Do not round off (5 181 in 4 τ τ (10(8/ τ max τ all τ max in τ all A 11.31( in 181( τ psi, vol in τ psi 0.18 fom u l (1 8.0 eff ( /l 33.7 (0.18 / Conclusions: To meet allowable stress limitations, and A do not cange, nor do τ and σ. To meet te sortened bead lengt, is increased proportionately. However, volume of bead laid down increases as. Te uninterrupted bead is superior. n tis example, we did not round and as a result we learned someting. Our measures of merit are also sensitive to rounding. Wen te design decision is made, rounding to te next larger standard weld fillet size will decrease te merit.

12 budynas_sm_c09.qxd 01/9/007 18:5 Page Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design Had te weld bead gone around te corners, te situation would cange. Here is a followup task analyzing an alternative weld pattern. b 1 d 1 d b 9-17 From Table 9- For te box A 1.414(b + d Subtracting b 1 from b and d 1 from d A (b b 1 + d d 1 u d (3b + d d3 1 b 1d 1 (b b 1d + 1 ( d 3 d 3 1 lengt of bead l (b b 1 + d d Computer programs will vary. fom u /l 9-19 τ all psi. Use Fig. 9-17(a for general geometry, but employ beads and ten beads. Horizontal parallel weld bead pattern " 8" b in d 8in From Table 9-, category 3 A b 1.414(( 8.48 in x b/ / 3in, u bd (8 19 in 3 ȳ d/ 8/ 4in u 0.707(( in 4 τ psi

13 budynas_sm_c09.qxd 01/9/007 18:5 Page 51 Capter 9 51 τ Mc (10(8/ τ max τ + τ 1 ( / 3175 Equate te maximum and allowable sear stresses. τ max τ all from wic 0.48 in. t follows tat 135.7( in 4 Te volume of te weld metal is Te effectiveness, (eff H,is vol l 0.48 ( + psi 0.39 in 3 (eff H vol in (fom H u l ( + 4.5in Vertical parallel weld beads psi 8" " b in d 8in From Table 9-, category A 1.414d 1.414(( in x b/ / 3in, u d in3 ȳ d/ 8/ 4in u 0.707(( τ τ Mc psi (10(8/ τ max τ + τ 1 ( / 9 psi psi

14 budynas_sm_c09.qxd 01/9/007 18:5 Page 5 5 Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design Equating τ max to τ all gives 0.53 in. t follows tat 0.3( in 4 vol l 0.53 ( in 3 (eff V vol in (fom V u l ( in Te ratio of (eff V /(eff H is 14.4/ Te ratio (fom V /(fom H is 10./ Tis is not surprising since eff vol ( /l u ( /l u l fom Te ratios (eff V /(eff H and (fom V /(fom H give te same information. 9-0 Because te loading is pure torsion, tere is no primary sear. From Table 9-1, category : J u πr 3 π(1 3.8 in 3 J J u 0.707(0.5( in 4 τ Tr J 0( kpsi in, d 8in, b 1in From Table 9-, category : A 1.414(0.375(8 4.4 in u d in u 0.707(0.375(85.3.in 4 τ F A kpsi 4.4 M 5( 30 kip in c ( / 4in τ Mc 30( kpsi τ max τ + τ kpsi

15 budynas_sm_c09.qxd 01/9/007 18:5 Page 53 Capter cm, b cm, d 1 cm. Table 9-3, category 5: B A G A 0.707(b + d 0.707(0.[ + (1] 1.7cm ȳ d b + d 1 + (1 4.8cm u d3 3 d ȳ + (b + dȳ (13 3 (1 (4.8 + [ + (1] cm u 0.707(0.(41 19 cm 4 τ F A 7.5( ( MPa M 7.5( N m c A 7.cm, c B 4.8cm Te critical location is at A. τ A Mc A 900( MPa τ max τ + τ ( / 33.MPa n τ all τ max Te largest possible weld size is 1/1 in. Tis is a small weld and tus difficult to accomplis. Te bracket s load-carrying capability is not known. Tere are geometry problems associated wit seet metal folding, load-placement and location of te center of twist. Tis is not available to us. We will identify te strongest possible weldment. Use a rectangular, weld-all-around pattern Table 9-, category : 7.5" 1" A (b + d 1.414(1/1( in x b/ 0.5in ȳ d in

16 budynas_sm_c09.qxd 01/9/007 18:5 Page Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design u d 7.5 (3b + d [3( ] 98.4in u 0.707(1/1( in 4 M ( W 4.5W τ V A τ Mc W W 4.5W(7.5/ W τ max τ + τ W W Material properties: Te allowable stress given is low. Let s demonstrate tat. For te A3 structural steel member, S y 3 kpsi and S ut 58 kpsi. For te 100 CD attacment, use HR properties of S y 30 kpsi and S ut 55. Te E010 electrode as strengts of S y 50 and S ut kpsi. Allowable stresses: A3: τ all min[0.3(58, 0.4(3] min(17.4, kpsi 100: τ all min[0.3(55, 0.4(30] τ all min(1.5, 1 1 kpsi E010: τ all min[0.3(, 0.4(50] min(18., kpsi Since Table 9- gives 18.0 kpsi as te allowable sear stress, use tis lower value. Terefore, te allowable sear stress is τ all min(14.4, 1, kpsi However, te allowable stress in te problem statement is 0.9 kpsi wic is low from te weldment perspective. Te load associated wit tis strengt is τ max τ all 3.90W 900 W lbf 3.90 f te welding can be accomplised (1/1 leg size is a small weld, te weld strengt is psi and te load W 3047 lbf. Can te bracket carry suc a load? Tere are geometry problems associated wit seet metal folding. Load placement is important and te center of twist as not been identified. Also, te load-carrying capability of te top bend is unknown. Tese uncertainties may require te use of a different weld pattern. Our solution provides te best weldment and tus insigt for comparing a welded joint to one wic employs screw fasteners.

17 budynas_sm_c09.qxd 01/9/007 18:5 Page 55 Capter y F x A R x A F B 0 R y A B F 100 lbf, τ all 3 kpsi F B 100(1/ lbf FB x cos 0.7 lbf F y B cos 30 4 lbf t follows tat R y A 5 lbf and Rx A.7 lbf, R A lbf M 100(1 100 lbf in Te OD of te tubes is 1 in. From Table 9-1, category : A 1.414(πr( (1.414(π(1/ 4.44 in J u πr 3 π(1/ in 3 J (0.707J u 1.414( in 4 τ V A τ Tc J Mc J 100( Te sear stresses, τ and τ, are additive algebraically Decision: Use 5/1 in fillet welds τ max 1 81 ( psi τ max τ all /1 "

18 budynas_sm_c09.qxd 01/9/007 18:5 Page 5 5 Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design 9-5 y 1" 4 1" 4 B g g 9" G 7" g g 3" 8 3" 8 x For te pattern in bending sown, find te centroid G of te weld group. (0.707(1/4(3 + (0.707(3/8(13 x (0.707(1/4 + (0.707(3/8 9in 1/4 ( G + A x [ 0.707(1/4( in 4 [ 0.707(3/8( 3 3/ in 4 Te critical location is at B. From Eq. (9-3, τ ] (1/4(( ] (3/8((4 1/4 + 3/ in 4 τ Mc F [(0.707(3/8 + 1/4] 0.189F (8F( F τ max τ + τ F F Materials: A3 Member: S y 3 kpsi 1015 HR Attacment: S y 7.5 kpsi E010 Electrode: S y 50 kpsi τ all min(3, 7.5, kpsi F τ all/n / 14.8 kip Figure P9-b is a free-body diagram of te bracket. Forces and moments tat act on te welds are equal, but of opposite sense. (a M 100( lbf in (b F y 100 sin lbf (c F x 100 cos lbf

19 budynas_sm_c09.qxd 01/9/007 18:5 Page 57 Capter 9 57 (d From Table 9-, category : A 1.414(0.5( in u d.5 (3b + d [3( ] 3.39 in3 Te second area moment about an axis troug G and parallel to z is u 0.707(0.5( in 4 (e Refer to Fig. P.9-b. Te sear stress due to F y is τ 1 F y A psi 0.97 Te sear stress along te troat due to F x is τ F x A psi 0.97 Te resultant of τ 1 and τ is in te troat plane Te bending of te troat gives τ ( τ 1 + τ 1/ ( / 134 psi τ Mc Te maximum sear stress is 439( psi τ max (τ + τ 1/ ( / 1537 psi (f Materials: 1018 HR Member: S y 3 kpsi, S ut 58 kpsi (Table A-0 E010 Electrode: S y 50 kpsi (Table 9-3 n S sy 0.577S y 0.577(3 1.0 τ max τ max (g Bending in te attacment near te base. Te cross-sectional area is approximately equal to b. A 1. b 0.5( in τ xy F x psi A c bd 0.5( in 3 At location A σ y F y A 1 + M /c σ y psi 0.0

20 budynas_sm_c09.qxd 01/9/007 18:5 Page Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design Te von Mises stress σ is σ ( σ y + 3τ xy 1/ [48 + 3(1 ] 1/ 391 psi Tus, te factor of safety is, n S y σ Te clip on te mooring line bears against te side of te 1/-in ole. f te clip fills te ole σ F td ( psi n S y σ 3( Furter investigation of tis situation requires more detail tan is included in te task statement. ( n sear fatigue, te weakest constituent of te weld melt is 1018 wit S ut 58 kpsi Table 7-4: S e 0.5S ut 0.5(58 9 kpsi k a 14.4( For te size factor estimate, we first employ Eq. (7-4 for te equivalent diameter. d e b (.5( in Eq. (7-19 is used next to find k b ( de k b 0.30 ( Te load factor for sear k c, is Te endurance strengt in sear is k c 0.59 S se 0.780(0.940(0.59(9 1.5 kpsi From Table 9-5, te sear stress-concentration factor is K fs.7. Te loading is repeatedly-applied. τ max τ a τ m K fs kpsi Table 7-10: Gerber factor of safety n f, adjusted for sear, wit S su 0.7S ut n f 1 [ ] 0.7(58 (.07 [ ] (.07( (58( Attacment metal sould be cecked for bending fatigue.

21 budynas_sm_c09.qxd 01/9/007 18:5 Page 59 Capter Use b d 4 in. Since 5/8 in, te primary sear is τ F 1.414(5/8(4 0.83F Te secondary sear calculations, for a moment arm of 14 in give J u 4[3(4 + 4 ] 4.7 in 3 J 0.707J u 0.707(5/ in 4 τ x τ y Mr y J 14F( F Tus, te maximum sear and allowable load are: τ max F ( F F τ all kip.30 From Prob. 9-5b, τ all 11 kpsi F all τ all kip.30 Te allowable load as tus increased by a factor of Purcase te ook aving te design sown in Fig. P9-8b. Referring to text Fig. 9-3a, tis design reduces peel stresses. 9-9 (a τ 1 l l/ l/ Pω cos(ωx 4b sin(ωl/ dx l/ A 1 cos(ωx dx l/ A 1 ω sin(ωx l/ l/ A 1 [sin(ωl/ sin( ωl/] ω A 1 [sin(ωl/ ( sin(ωl/] ω A 1 sin(ωl/ ω Pω [ sin(ωl/] 4bl sin(ωl/ τ P bl

22 budynas_sm_c09.qxd 01/9/007 18:5 Page 0 0 Solutions Manual nstructor s Solution Manual to Accompany Mecanical Engineering Design (b τ(l/ Pω cos(ωl/ 4b sin(ωl/ Pω 4b tan(ωl/ (c K τ(l/ τ Pω 4b sin(ωl/ ( bl P K ωl/ tan(ωl/ For computer programming, it can be useful to express te yperbolic tangent in terms of exponentials: K ωl exp(ωl/ exp( ωl/ exp(ωl/ + exp( ωl/ 9-30 Tis is a computer programming exercise. All programs will vary.