Chapter 9. Figure for Probs. 9-1 to Given, b = 50 mm, d = 50 mm, h = 5 mm, τ allow = 140 MPa.

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1 Chapter 9 Figure for Probs. 9-1 to Given, b 50 mm, d 50 mm, h 5 mm, ow 140 MPa. F hl ow 0.707(5)[(50)](140)(10 ) 49.5 kn Ans. 9- Given, b in, d in, h 5/16 in, ow 5 kpsi. F hl ow 0.707(5/16)[()](5).1 kip Ans. 9- Given, b 50 mm, d 0 mm, h 5 mm, ow 140 MPa. F hl ow 0.707(5)[(50)](140)(10 ) 49.5 kn Ans. 9-4 Given, b 4 in, d in, h 5/16 in, ow 5 kpsi. F hl ow 0.707(5/16)[(4)](5) 44. kip Ans. 9-5 Prob. 9-1 with E7010 Electrode. Table 9-6: f h kip/in [5 mm/(5.4 mm/in)].9 kip/in.9(4.45/5.4) 0.51 kn/mm F f l 0.51[(50)] 51. kn Ans. 9-6 Prob. 9- with E6010 Electrode. Table 9-6: f h kip/in 14.85(5/16) 4.64 kip/in Shigle s MED, 10 th edition Chapter 9 Solutions, Page 1/5

2 F f l 4.64[()] 18.6 kip Ans. 9-7 Prob. 9- with E7010 Electrode. Table 9-6: f h kip/in [5 mm/(5.4 mm/in)].9 kip/in.9(4.45/5.4) 0.51 kn/mm F f l 0.51[(50)] 51. kn Ans. 9-8 Prob. 9-4 with E6010 Electrode. Table 9-6: f h kip/in 14.85(5/16) 4.64 kip/in F f l 4.64[(4)] 7.1 kip Ans. 9-9 Table A-0: 1018 CD: S ut 440 MPa, S 70 MPa 1018 HR: S ut 400 MPa, S 0 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: min(0.0 Sut, 0.40 S ) min[0.0(400), 0.40(0)] min(10, 88) 88 MPa for both materials. Eq. (9-): F 0.707hl 0.707(5)[(50)](88)(10 ) 1.1 kn Ans Table A-0: 100 CD: S ut 68 kpsi, S 57 kpsi 100 HR: S ut 55 kpsi, S 0 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: min(0.0 Sut, 0.40 S ) min[0.0(55), 0.40(0)] min(16.5, 1.0) 1.0 kpsi for both materials. Eq. (9-): F 0.707hl 0.707(5/16)[()](1.0) 10.6 kip Ans. Shigle s MED, 10 th edition Chapter 9 Solutions, Page /5

3 9-11 Table A-0: 105 HR: S ut 500 MPa, S 70 MPa 105 CD: S ut 550 MPa, S 460 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: min(0.0 Sut, 0.40 S ) min[0.0(500), 0.40(70)] min(150, 108) 108 MPa for both materials. Eq. (9-): F 0.707hl 0.707(5)[(50)](108)(10 ) 8. kn Ans. 9-1 Table A-0: 105 HR: S ut 7 kpsi, S 9.5 kpsi 100 CD: S ut 68 kpsi, S 57 kpsi, 100 HR: S ut 55 kpsi, S 0 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: min(0.0 Sut, 0.40 S ) min[0.0(55), 0.40(0)] min(16.5, 1.0) 1.0 kpsi for both materials. Eq. (9-): F 0.707hl 0.707(5/16)[(4)](1.0) 1. kip Ans. 9-1 F ( 100)( 10 ) Eq. (9-): 141 MPa Ans. hl 5 ( ) 9-14 ( 40) ( 5 /16) ( + ) F Eq. (9-):.6 kpsi Ans. hl F Eq. (9-): 177 MPa Ans. hl 5 ( ) ( 40) ( 5 /16) ( 4 + ) F 9-16 Eq. (9-): 15.1 kpsi Ans. hl Shigle s MED, 10 th edition Chapter 9 Solutions, Page /5

4 9-17 b d 50 mm, c 150 mm, h 5 mm, and ow 140 MPa. (a) Primar shear, Table 9-1, Case (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kn and in MPa): ( 10 ) V F.89F A Secondar shear, Table 9-1: J u ( ) d b + d ( 10 ) mm 6 6 J h J u 0.707(5)(8.)(10 ) 94.6(10 ) mm 4 Mr 175F F J F F (1) ow 140 F 6.06 kn Ans..1.1 (b) For E7010 from Table 9-6, ow 1 kpsi 1(6.89) 145 MPa 100 HR bar: S ut 80 MPa, S 10 MPa 1015 HR support: S ut 40 MPa, S 190 MPa Table 9-, E7010 Electrode: S ut 48 MPa, S 9 MPa The support controls the design. Table 9-4: ow min(0.0s ut, 0.40S ) min[0.0(40), 0.40(190) min(10, 76) 76 MPa The owable load, from Eq. (1) is ow 76 F.9 kn Ans b d in, c 6 in, h 5/16 in, and ow 5 kpsi. Shigle s MED, 10 th edition Chapter 9 Solutions, Page 4/5

5 (a) Primar shear, Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and in kpsi): V F 1.1F A /16 Secondar shear, Table 9-1: d ( b + d ) ( ) + Ju 5. in 6 6 J h J u 0.707(5/16)(5.) in 4 Mr 7F F J F F (1) ow 5 F.71 kip Ans (b) For E7010 from Table 9-6, ow 1 kpsi 100 HR bar: S ut 55 kpsi, S 0 kpsi 1015 HR support: S ut 50 kpsi, S 7.5 kpsi Table 9-, E7010 Electrode: S ut 70 kpsi, S 57 kpsi The support controls the design. Table 9-4: ow min(0.0s ut, 0.40S ) min[0.0(50), 0.40(7.5) min(15, 11) 11 kpsi The owable load, from Eq. (1) is ow 11 F 1.19 kip Ans b 50 mm, c 150 mm, d 0 mm, h 5 mm, and ow 140 MPa. (a) Primar shear, Table 9-1, Case (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kn and in MPa): Shigle s MED, 10 th edition Chapter 9 Solutions, Page 5/5

6 ( 10 ) V F.89F A Secondar shear, Table 9-1: d ( b + d ) 50 ( 0 ) + 50 Ju 4.( 10 ) mm 6 6 J h J u 0.707(5)(4.)(10 ) 15.(10 ) mm 4 Mr 175F F J Mr 175F F J F F (1) ow 140 F.91 kn Ans (b) For E7010 from Table 9-6, ow 1 kpsi 1(6.89) 145 MPa 100 HR bar: S ut 80 MPa, S 10 MPa 1015 HR support: S ut 40 MPa, S 190 MPa Table 9-, E7010 Electrode: S ut 48 MPa, S 9 MPa The support controls the design. Table 9-4: ow min(0.0s ut, 0.40S ) min[0.0(40), 0.40(190) min(10, 76) 76 MPa The owable load, from Eq. (1) is ow 76 F.1 kn Ans b 4 in, c 6 in, d in, h 5/16 in, and ow 5 kpsi. Shigle s MED, 10 th edition Chapter 9 Solutions, Page 6/5

7 (a) Primar shear, Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and in kpsi): V F F A /16 4 Secondar shear, Table 9-1: J u ( ) d b + d in 6 6 J h J u 0.707(5/16)(18.67) 4.15 in 4 Mr 8F F J 4.15 Mr 8F.879F J F F (1) ow 5 F 5.15 kip Ans (b) For E7010 from Table 9-6, ow 1 kpsi 100 HR bar: S ut 55 kpsi, S 0 kpsi 1015 HR support: S ut 50 kpsi, S 7.5 kpsi Table 9-, E7010 Electrode: S ut 70 kpsi, S 57 kpsi The support controls the design. Table 9-4: ow min(0.0s ut, 0.40S ) min[0.0(50), 0.40(7.5) min(15, 11) 11 kpsi The owable load, from Eq. (1) is ow 11 F.7 kip Ans Shigle s MED, 10 th edition Chapter 9 Solutions, Page 7/5

8 9-1 Given, b 50 mm, c 150 mm, d 50 mm, h 5 mm, ow 140 MPa. Primar shear (F in kn, in MPa, A in mm ): V F ( 10 ) 1.414F A 1.414( 5)( ) Secondar shear: Table 9-1: Maimum shear: ( b + d ) ( ) Ju mm 6 6 J h J u 0.707(5)166.7(10 ) 589.(10 ) mm 4 Mr 175F 10 (5) 7.45F J F F ow 140 F 1.1 kn Ans Given, b in, c 6 in, d in, h 5/16 in, ow 5 kpsi. Primar shear: Secondar shear: Table 9-1: Maimum shear: V F F A /16 ( + ) ( b + d ) ( + ) Ju in 6 6 J h J u 0.707(5/16) in 4 Mr 7 F(1).970F J F F ow 5 F 5.41 kip Ans Given, b 50 mm, c 150 mm, d 0 mm, h 5 mm, ow 140 MPa. Shigle s MED, 10 th edition Chapter 9 Solutions, Page 8/5

9 Primar shear (F in kn, in MPa, A in mm ): V F ( 10 ) 1.768F A Secondar shear: Table 9-1: ( b + d ) ( ) Ju mm 6 6 J h J u 0.707(5)85.(10 ) 01.6(10 ) mm 4 Mr 175F 10 (15) 8.704F J Maimum shear: Mr 175F 10 (5) 14.51F J F F ow 140 F 7.58 kn Ans Given, b 4 in, c 6 in, d in, h 5/16 in, ow 5 kpsi. Primar shear: Secondar shear: Table 9-1: V F 0.77F A /16 4 ( + ) ( b + d ) ( 4 + ) Ju 6 in 6 6 J h J u 0.707(5/16) in 4 Mr 8 F(1) 1.006F J Maimum shear: Mr 8 F().01F J F F Shigle s MED, 10 th edition Chapter 9 Solutions, Page 9/5

10 ow 5 F 9.65 kip Ans Given, b 50 mm, d 50 mm, h 5 mm, E6010 electrode. A 0.707(5)( ) 50. mm Member endurance limit: From Table A-0 for AISI 1010 HR, S ut 0 MPa. Eq and Table 6-, pp. 95 and 96, respectivel: k a 7(0) k b 1 (uniform shear), k c 0.59 (torsion, shear), k d 1 Eqs. (6-8) and (6-18): S e 0.875(1)(0.59)(1)(0.5)(0) 8.6 MPa Electrode endurance: E6010, Table 9-, S ut 47 MPa Eq and Table 6-, pp. 95 and 96, respectivel: k a 7(47) As before, k b 1 (direct shear), k c 0.59 (torsion, shear), k d 1 S e 0.657(1)(0.59)(1)(0.5)(47) 8.8 MPa The members and electrode are basic of equal strength. We will use S e 8.6 MPa. For a factor of safet of 1, and with K fs.7 (Table 9-5) ow A 8.6( 50.) F 16.( 10 ) N 16. kn Ans. K fs Given, b in, d in, h 5/16 in, E6010 electrode. A 0.707(5/16)( + + ) 1.6 in Member endurance limit: From Table A-0 for AISI 1010 HR, S ut 47 kpsi. Eq and Table 6-, pp. 95 and 96, respectivel: k a 9.9(47) k b 1 (uniform shear), k c 0.59 (torsion, shear), k d 1 Eqs. (6-8) and (6-18): S e 0.865(1)(0.59)(1)(0.5)(47) 1.0 kpsi Electrode endurance: E6010, Table 9-, S ut 6 kpsi Eq and Table 6-, pp. 95 and 96, respectivel: k a 9.9(6) Shigle s MED, 10 th edition Chapter 9 Solutions, Page 10/5

11 As before, k b 1 (uniform shear), k c 0.59 (torsion, shear), k d 1 S e 0.657(1)(0.59)(1)(0.5)(6) 1.0 kpsi Thus the members and electrode are of equal strength. For a factor of safet of 1, and with K fs.7 (Table 9-5) ow A 1.0( 1.6) F 5.89 kip Ans. K fs Given, b 50 mm, d 0 mm, h 5 mm, E7010 electrode. A 0.707(5)( ) mm Member endurance limit: From Table A-0 for AISI 1010 HR, S ut 0 MPa. Eq and Table 6-, pp. 95 and 96, respectivel: k a 7(0) k b 1 (direct shear), k c 0.59 (torsion, shear), k d 1 Eqs. (6-8) and (6-18): S e 0.875(1)(0.59)(1)(0.5)(0) 8.6 MPa Electrode endurance: E6010, Table 9-, S ut 48 MPa Eq and Table 6-, pp. 95 and 96, respectivel: k a 7(48) As before, k b 1 (direct shear), k c 0.59 (torsion, shear), k d 1 S e 0.58(1)(0.59)(1)(0.5)(48) 8.7 MPa The members and electrode are basic of equal strength. We will use S e 8.6 MPa. For a factor of safet of 1, and with K fs.7 (Table 9-5) ow A 8.6( 459.6) F 14.1( 10 ) N 14.1 kn Ans. K fs Given, b 4 in, d in, h 5/16 in, E7010 electrode. A 0.707(5/16)( ).09 in Member endurance limit: From Table A-0 for AISI 1010 HR, S ut 47 kpsi. Eq and Table 6-, pp. 95 and 96, respectivel: k a 9.9(47) k b 1 (direct shear), k c 0.59 (torsion, shear), k d 1 Shigle s MED, 10 th edition Chapter 9 Solutions, Page 11/5

12 Eqs. (6-8) and (6-18): S e 0.865(1)(0.59)(1)(0.5)(47) 1.0 kpsi Electrode endurance: E7010, Table 9-, S ut 70 kpsi Eq and Table 6-, pp. 95 and 96, respectivel: k a 9.9(70) As before, k b 1 (direct shear), k c 0.59 (torsion, shear), k d 1 S e 0.58(1)(0.59)(1)(0.5)(70) 1.0 kpsi Thus the members and electrode are of equal strength. For a factor of safet of 1, and with K fs.7 (Table 9-5) ow A 1.0(.09) F 9.8 kip Ans. K.7 fs 9-9 Primar shear: 0 (wh?) Secondar shear: Table 9-1: J u π r π (1.5) 1.1 in welds: J h J u 0.707(1/4)(1.1).749 in 4 F Mr F J.749 ow 1.600F 0 F 1.5 kip Ans. 9-0 l in in 10 ( 4) ( 0) 1.6 in 10 M FR F(10 1) 9 F r r in, in 1 r in Shigle s MED, 10 th edition Chapter 9 Solutions, Page 1/5

13 1 4 J G / in 1 1 JG J G / in 1 ( i i G i ) J J + A r i / / / in 1.6 α tan o r in Primar shear ( in kpsi, F in kip) : V F 0.456F A /16 10 Secondar shear: Mr 9F (.4).19F J 9.0 o o ( F ) ( F F ).19 sin cos F ow.74 F 5 F 6.71 kip Ans. Shigle s MED, 10 th edition Chapter 9 Solutions, Page 1/5

14 9-1 l mm mm 10 0( 50) + 50( 5) + 50( 0) 1.15 mm 10 M FR F( ) F (M in N m, F in kn) r r mm, mm 1 r mm 1 4 J G mm 1 1 JG J G mm 1 ( i i G i ) J J + A r i mm 1.15 α tan o r mm Primar shear ( in MPa, F in kn) : V F ( 10 ).176F A Secondar shear: Mr 186.9F ( 10 )( 4.55) 5.88F J Shigle s MED, 10 th edition Chapter 9 Solutions, Page 14/5

15 o o ( F ) ( F F ) 5.88 sin cos F ow 7.79 F 140 F 5.04 kn Ans. 9- Weld Pattern Figure of merit Rank 1. Ju a /1 a a fom 0.08 lh ah 1h h ( ) a a + a a a fom 0. 6 a h h h 4 a 6a a 5a a fom a + a ah 4h h 4 1 8a + 6a + a a a fom ah 1 a + a h a 1 8a a fom 0. 6h 4a 4ah h a a a π / 6. fom 0.5 π ah 4ah h Shigle s MED, 10 th edition Chapter 9 Solutions, Page 15/5

16 9- Weld Pattern Figure of merit Rank I ( a /1) u a 1. fom lh ah h.. 4.* 5. & & ( a / 6) a fom 0.08 ah h ( aa / ) a fom 0.5 ah h a /1 6a + a 7a a fom ah 6h h a, a a a + a a a a a Iu a + ( a + a) I ( a / ) 1 u a a fom lh ah 9 h h a / 6 a + a 1 a a fom ah 6 h h a / a a π fom 0.15 π ah 8h h *Note. Because this section is not smmetric with the vertical ais, out-of-plane deflection ma occur unless special precautions are taken. See the topic of shear center in books with more advanced treatments of mechanics of materials. 9-4 Attachment and member (1018 HR), S 0 MPa and S ut 400 MPa. The member and attachment are weak compared to the properties of the lowest electrode. Decision Specif the E6010 electrode Controlling propert, Table 9-4: min[0.(400), 0.4(0)] min(10, 88) 88 MPa For a static load, the parel and transverse fillets are the same. Let the length of a bead be l 75 mm, and n be the number of beads. Shigle s MED, 10 th edition Chapter 9 Solutions, Page 16/5

17 F n hl ( 0.707) F nh l where h is in millimeters. Make a table Number of beads, n Leg size, h (mm) mm Decision Specif h 6 mm on four sides. Weldment specification: Pattern: All-around square, four beads each side, 75 mm long Electrode: E6010 Leg size: h 6 mm 9-5 Decision: Choose a parel fillet weldment pattern. B so-doing, we ve chosen an optimal pattern (see Prob. 9-) and have thus reduced a snthesis problem to an analsis problem: Table 9-1, case, rotated 90 : A 1.414hd 1.414(h)(75) h mm Primar shear V A h h Secondar shear: d( b + d ) Ju 6 75[(75 ) + 75 ] mm 6 J 0.707( h)(81.) h mm With α 45, 4 Shigle s MED, 10 th edition Chapter 9 Solutions, Page 17/5

18 o Mr cos 45 Mr 1 10 (187.5)(7.5) 44.4 J J h h ( + ) ( ) h h Attachment and member (1018 HR): S 0 MPa, S ut 400 MPa Decision: Use E60XX electrode which is stronger h min[0.(400), 0.4(0)] MPa h mm MPa Decision: Specif 8 mm leg size Weldment Specifications: Pattern: Parel horizontal fillet welds Electrode: E6010 Tpe: Fillet Length of each bead: 75 mm Leg size: 8 mm 9-6 Problem 9-5 solves the problem using parel horizontal fillet welds, each 75 mm long obtaining a leg size rounded up to 8 mm. For this problem, since the width of the plate is fied and the length has not been determined, we will eplore reducing the leg size b using two vertical beads 75 mm long and two horizontal beads such that the beads have a leg size of 6 mm. Decision: Use a rectangular weld bead pattern with a leg size of 6 mm (case 5 of Table 9-1 with b unknown and d 75 mm). Materials: Attachment and member (1018 HR): S 0 MPa, S ut 400 MPa From Table 9-4, AISC welding code, min[0.(400), 0.4(0)] min(10, 88) 88 MPa Select a stronger electrode material from Table 9-. Decision: Specif E6010 Solving for b: In Prob. 9-5, ever term was linear in the unknown h. This made solving for h relativel eas. In this problem, the terms will not be linear in b, and so we will use an iterative solution with a spreadsheet. Throat area and other properties from Table 9-1: A 1.414(6)(b + 75) 8.484(b + 75) (1) Shigle s MED, 10 th edition Chapter 9 Solutions, Page 18/5

19 J u ( b + 75), J (6) J u 0.707(b +75) () 6 Primar shear ( in MPa, h in mm): V A 1( 10 ) A () Secondar shear (See Prob. 9-5 solution for the definition of α) : Mr J Mr Mr 1( 10 )( b / )(7.5) cosα cos α (4) J J b + 75 ( )( + b ) ( + 75) Mr Mr / ( b / ) sinα sin α (5) J J b + + Enter Eqs. (1) to (6) into a spreadsheet and iterate for various values of b. A portion of the spreadsheet is shown below. (6) b (mm) A (mm ) J (mm 4 ) ' (Mpa) " (Mpa) " (Mpa) (Mpa) < 88 Mpa We see that b 4 mm meets the strength goal. Weldment Specifications: Pattern: Horizontal parel weld tracks 4 mm long, vertical parel weld tracks 75 mm long Electrode: E6010 Leg size: 6 mm 9-7 Materials: Member and attachment (1018 HR): S kpsi, S 58 kpsi Table 9-4: min[0.(58), 0.4()] 1.8 kpsi ut Shigle s MED, 10 th edition Chapter 9 Solutions, Page 19/5

20 Decision: Use E6010 electrode. From Table 9-: S 50 kpsi, S 6 kpsi, min[0.(6), 0.4(50)] 0 kpsi Decision: Since 1018 HR is weaker than the E6010 electrode, use 1.8 kpsi Decision: Use an -around square weld bead track. l a in Throat area and other properties from Table 9-1: ut Primar shear A h( b + d) 1.414( h)(6 + 6) 16.97h V F psi A A 16.97h h Secondar shear ( b + d) (6 + 6) Ju 88 in J h(88) 0.6 h in Mr 0 10 (6.5 + )() 76 psi J 0.6h h ( + ) 76 + ( ) psi h h Relate stress to strength Decision: ( 10 ) h in h Specif/ 8in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010 Tpe of weld: Fillet Weld bead length: 4 in Leg size: / 8in Attachment length: 1.5 in Shigle s MED, 10 th edition Chapter 9 Solutions, Page 0/5

21 9-8 This is a good analsis task to test a student s understanding. (1) Solicit information related to a priori decisions. () Solicit design variables b and d. () Find h and round and output parameters on a single screen. Allow return to Step 1 or Step. (4) When the iteration is complete, the final displa can be the bulk of our adequac assessment. Such a program can teach too. 9-9 The objective of this design task is to have the students teach themselves that the weld patterns of Table 9- can be added or subtracted to obtain the properties of a contemplated weld pattern. The instructor can control the level of complication. We have left the presentation of the drawing to ou. Here is one possibilit. Stud the problem s opportunities, and then present this (or our sketch) with the problem assignment. Use b 1 as the design variable. Epress properties as a function of b 1. From Table 9-, case : A h( b b ) bd b d ( b b ) d Iu I 0.707hI u V F A h( b b1 ) Mc Fa( d / ) I 0.707hI u Parametric stud a 10 in, b 8 in, d 8 in, b in, 1.8 kpsi, l (8 ) 1 in Let 1 Shigle s MED, 10 th edition Chapter 9 Solutions, Page 1/5

22 A h(8 ) 8.48h in I (8 )(8 / ) 19 in u 4 I 0.707( h)(19) 15.7 h in psi 8.48h h (10)(8 / ) 948 psi 15.7h h psi h + h from which h 0.48 in. Do not round off the leg size something to learn. 19 fom ' I u 64.5 in hl 0.48(1) A 8.48(0.48).10 in 4 I 15.7(0.48).65 in h 0.48 vol l in I.65 eff 91. in vol psi psi psi 0.48 Now consider the case of uninterrupted welds, b 1 0 A 1.414( h)(8 0) 11.1h I (8 0)(8 / ) 56 in u 4 I 0.707(56) h 181 h in h h (10)(8 / ) h h h h 80 h in Do not round off h. Shigle s MED, 10 th edition Chapter 9 Solutions, Page /5

23 A 11.1(0.186).10 in 4 I 181(0.186).67 in psi, vol in psi I 56 fom ' u 86.0 in hl 0.186(16).67 eff I 11.7 in ( h / ) l (0.186 / )16 Conclusions: To meet owable stress limitations, I and A do not change, nor do and σ. To meet the shortened bead length, h is increased proportionatel. However, volume of bead laid down increases as h. The uninterrupted bead is superior. In this eample, we did not round h and as a result we learned something. Our measures of merit are also sensitive to rounding. When the design decision is made, rounding to the net larger standard weld fillet size will decrease the merit. Had the weld bead gone around the corners, the situation would change. Here is a follow up task analzing an alternative weld pattern From Table 9- For the bo A h( b + d) Subtracting b 1 from b and d 1 from d A 1.414h b b + d d 1 1 d d1 b1 d 1 1 Iu ( b + d) b b d + d d ( 1) ( 1 ) Length of bead l ( b b1 + d d1) fom Iu / hl Shigle s MED, 10 th edition Chapter 9 Solutions, Page /5

24 9-41 Computer programs will var kpsi. Use Fig. 9-17(a) for general geometr, but emplo beads and then beads. Horizontal parel weld bead pattern b in, d 6 in Table 9-: A 1.414hb 1.414( h)() 4.4 h in bd (6) I u 54 in I 0.707hI 0.707( h)(54) 8. h in u kpsi 4.4h h Mc 10(10)(6 / ) 7.85 kpsi I 8.h h kpsi h h + + Equate the imum and owable shear stresses h from which h 0.68 in. It follows that 4 I 8.(0.68) 6.1 in The volume of the weld metal is The effectiveness, (eff) H, is h l (0.68) ( + ) vol 1.40 in (eff) H I in Ans. vol 1.4 Shigle s MED, 10 th edition Chapter 9 Solutions, Page 4/5

25 Vertical parel weld beads b in d 6 in From Table 9-, case A 1.414hd 1.414( h)(6) 8.48 h in d 6 I u 7 in 6 6 I 0.707hI 0.707( h)(7) 50.9h u psi 8.48h h Mc 10(10)(6 / ) psi I 50.9h h kpsi h h Equating to gives h in. It follows that 4 I 50.9(0.501) 5.5 in h l vol (6 + 6) 1.51 in I 5.5 (eff ) V 16.7 in Ans. vol F 0, T 15 kip in. Table 9-1: J u π r π (1) 6.8 in, J 0.707(1/4) in 4 Tr kpsi Ans. J F kip, T 0. Table 9-: A π h r π (1/4)(1) in I u π r π (1).14 in, I 0.707(1/4) in 4 Shigle s MED, 10 th edition Chapter 9 Solutions, Page 5/5

26 V 1.80 kpsi A Mr kpsi I ( + ) 1/ ( ) 1/ 1.7 kpsi Ans F kip, T 15 kip in. Bending: Table 9-: A π h r π (1/4)(1) in Torsion: I u π r π (1).14 in, I 0.707(1/4) in 4 V 1.80 kpsi A Mr kpsi M I Table 9-1: J u π r π (1) 6.8 in, J 0.707(1/4) in 4 Tr kpsi T J kpsi Ans. M T 9-46 F kip, T 15 kip in. Bending: Table 9-: A π h r π h (1) 4.44h in I u π r π (1).14 in, I h (.14).1h in 4 V kpsi A 4.44h h Shigle s MED, 10 th edition Chapter 9 Solutions, Page 6/5

27 Mr M kpsi I.1h h Torsion: Table 9-1: J u π r π (1) 6.8 in, J h (6.8) 4.44 in 4 Tr T kpsi J 4.44h h M T + + kpsi h h h h h 0.19 in Ans. h Should specif a in weld. Ans h 9 mm, d 00 mm, b 5 mm From Table 9-, case : A 1.414(9)(00).545(10 ) mm d 00 I u 1.( 10 ) mm I 0.707h I u 0.707(9)(1.)(10 6 ) 8.484(10 6 ) mm 4 F MPa A.545(10 ) M 5(150) 750 N m Mc 750(100) 6 ( 10 ) 44.0 MPa I 8.484(10 ) MPa Ans h 0.5 in, b.5 in, d 5 in. Table 9-, case 5: A 0.707h (b +d) 0.707(0.5)[.5 + (5)].09 in Shigle s MED, 10 th edition Chapter 9 Solutions, Page 7/5

28 d 5 in b + d d Iu d + ( b + d ) 5 ( 5 ) ( 5) ( ). in I h I u 0.707(1/4)(.) in 4 Primar shear: F kpsi A.09 Secondar shear (the critical location is at the bottom of the bracket): 5 in M kpsi I kpsi 18 n.48 Ans The largest possible weld size is 1/16 in. This is a sm weld and thus difficult to accomplish. The bracket s load-carring capabilit is not known. There are geometr problems associated with sheet metal folding, load-placement and location of the center of twist. This is not available to us. We will identif the strongest possible weldment. Use a rectangular, weld--around pattern Table 9-, case 6: Shigle s MED, 10 th edition Chapter 9 Solutions, Page 8/5

29 A h( b + d) 1.414(1 / 16)( ) in b / 0.55 in d / 7.55 /.75 in d 7.5 Iu ( b + d ) [(1) + 7.5] in 6 6 I 0.707hIu 0.707(1 / 16)(98.44) 4.50 in M ( ) W 4.5W V W 1.1W A Mc 4.55 W (7.5 / ).664W I W W Material properties: The owable stress given is low. Let s demonstrate that. For the 100 CD bracket, use HR properties of S 0 kpsi and S ut 55. The 100 HR support, S 7.5 kpsi and S ut 68. The E6010 electrode has strengths of S 50 and S ut 6 kpsi. Allowable stresses: HR: 100 HR: E6010: min[0.(55), 0.4(0)] min(16.5, 1) 1 kpsi min[0.(68), 0.4(7.5)] min(0.4, 15) 15 kpsi min[0.(6), 0.4(50)] min(18.6, 0) 18.6 kpsi Since Table 9-6 gives 18.0 kpsi as the owable shear stress, use this lower value. Therefore, the owable shear stress is min(14.4, 1, 18.0) 1 kpsi However, the owable stress in the problem statement is 1.5 kpsi which is low from the weldment perspective. The load associated with this strength is W.90W lbf.90 If the welding can be accomplished (1/16 leg size is a sm weld), the weld strength is psi and the load associated with this strength is W 1 000/ lbf. Can the bracket carr such a load? There are geometr problems associated with sheet metal folding. Load placement is important and the center of twist has not been identified. Also, the load-carring capabilit of the top bend is unknown. Shigle s MED, 10 th edition Chapter 9 Solutions, Page 9/5

30 These uncertainties ma require the use of a different weld pattern. Our solution provides the best weldment and thus insight for comparing a welded joint to one which emplos screw fasteners F F F F B B B 100 lbf, kpsi 100(16 / ) 5. lbf o 5.cos lbf o 5.cos 0 46 lbf It follows that R A 56 lbf and R A 66.7 lbf, R A 6 lbf M 100(16) 1600 lbf in The OD of the tubes iss 1 in. From Table 9-1, case 6: A Ju J [ ] 1.414( π hr) (1.414)( π h)(1 / ) 4.44h π r π (1 / ) in ( (0.707) hj 1.414(0.7854) h h in u The weld onl carries the torsional load between the handle and tube A. Consequentl, the primar shear in the weld is zero, and the imum shear stresss is comprised entirel of the secondar shear. 4 in Shigle s MED, 10 th edition Chapter 9 Solutions, Page 0/5

31 Tc Mc 1600(0.5) 70.1 psi J J 1.111h h h 70.1 h / 4 in 000 Decision: Use 1/4 in fillet welds Ans For the pattern in bending shown, find the centroid G of the weld group. 6mm + ( 6)( 150) + ( 9)( 150) mm ( G ) I I + A I 9mm 6mm ( 6)( 150)( 5 75) 1.0( 10 ) mm ( 9)( 150)( ).67( 10 ) mm 1 I I 6 mm + I 9 mm ( )(10 6 ) 5.69(10 6 ) mm 4 The critical location is at B. With in MPa, and F in kn ( 10 ) V F 0.14F A 0.707( 6 + 9)( 150) Mc 00F F I Shigle s MED, 10 th edition Chapter 9 Solutions, Page 1/5

32 + F Materials: 1015 HR (Table A-0): S 190 MPa, E6010 Electrode(Table 9-): S 45 MPa F Eq. (5-1), p (190) MPa / / n F 61. kn Ans In the tetbook, Fig. Problem 9-5b is a free-bod diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. (a) M 100(0.66) 49 lbf in Ans. (b) F 100 sin lbf Ans. (c) F 100 cos lbf Ans. (d) From Table 9-, case 6: A 1.414(0.5)( ) 0.97 in d.5 Iu ( b + d) [(0.5) +.5].9 in 6 6 The second area moment about an ais through G and parel to z is I 0.707hI (0.5)(.9) in Ans. u (e) Refer to Fig. Problem 9-5b. The shear stress due to F is F A psi The shear stress along the throat due to F is F psi A 0.97 The resultant of 1 and is in the throat plane psi The bending of the throat gives Mc 49(1.5) I psi Shigle s MED, 10 th edition Chapter 9 Solutions, Page /5

33 The imum shear stress is psi Ans. (f) Materials: 1018 HR Member: S kpsi, S ut 58 kpsi (Table A-0) E6010 Electrode: S 50 kpsi (Table 9-) n Ss 0.577S 0.577() 1.0 Ans (g) Bending in the attachment near the base. The cross-sectional area is approimatel equal to bh. A1 bh 0.5(.5) 0.65 in F psi A I bd 0.5(.5) 0.60 in c 6 6 At location A, F M σ + A1 I / c σ psi The von Mises stress σ is σ σ (166) 91 psi Thus, the factor of safet is, S n 8.18 Ans. σ.91 The clip on the mooring line bears against the side of the 1/-in hole. If the clip fills the hole F 100 σ 9600 psi td 0.5(0.50) S (10 ) n. Ans. σ 9600 Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 HR with S ut 58 kpsi, Eq. (6-8), p. 90, gives Shigle s MED, 10 th edition Chapter 9 Solutions, Page /5

34 S 0.5S 0.5(58) 9.0 kpsi e ut Eq. (6-19), p. 95: k a 14.4(58) For uniform shear stress on the throat, assume k b 1. Eq.(6-6), p. 98: k c 0.59 From Eq. (6-18), p. 95, the endurance strength in shear is S se 0.780(1)(0.59)(9.0) 1.5 kpsi From Table 9-5, the shear stress-concentration factor is K f s.7. The loading is repeatedl-applied 1.57 a m K f s.7.07 kpsi Table 6-7, p. 15: Gerber factor of safet n f, adjusted for shear, with S su 0.67S ut 1 S su a ms se n f m Sse Ssu a (58).07 (.07)(1.5) Ans (58)(.07) Attachment metal should be checked for bending fatigue. 9-5 (a) Use b d 4 in. Since h 5/8 in, the primar shear is F 1.414(5 / 8)(4) 0.89F The secondar shear calculations, for a moment arm of 14 in give 4[(4 ) + 4 ] Ju 4.67 in 6 J 0.707hJu 0.707(5 / 8) in Mr 14 F() 1.485F J Thus, the imum shear and owable load are: 4 Shigle s MED, 10 th edition Chapter 9 Solutions, Page 4/5

35 F ( ).09F 5 F 10.8 kip Ans The load for part (a) has increased b a factor of 10.8/ Ans. (b) From Prob. 9-18b, 11 kpsi F kip The owable load in part (b) has increased b a factor of 4.76/ Ans Purchase the hook having the design shown in Fig. Problem 9-54b. Referring to tet Fig. 9-9a, this design reduces peel stresses (a) l / l / Pω ω l / A1 d A / 1 cosh( ) d sinh( ) l 4 sinh( ω / ) l / ω l / 1 cosh l ω ω b l A1 A1 [sinh( ωl / ) sinh( ωl / )] [sinh( ωl / ) ( sinh( ωl / ))] ω ω A1 sinh( ωl / ) Pω P [sinh( ωl / )] Ans. ω 4bl sinh( ωl / ) bl Pω cosh( ωl / ) Pω (b) ( l / ) Ans. 4b sinh( ωl / ) 4b tanh( ωl / ) (c) K ( l / ) Pω bl ωl / 4b tanh( ωl / ) P tanh( ωl / ) Ans. For computer programming, it can be useful to epress the hperbolic tangent in terms of eponentials: ωl ep( ωl / ) ep( ωl / ) K Ans. ep( ωl / ) + ep( ωl / ) 9-56 This is a computer programming eercise. All programs will var. Shigle s MED, 10 th edition Chapter 9 Solutions, Page 5/5

Chapter 9. Figure for Probs. 9-1 to Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.

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