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

Size: px
Start display at page:

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

Transcription

1 Capter 9 Figure for Probs. 9-1 to Given, b = 50 mm, d = 50 mm, = 5 mm, allow = 140 MPa. F = l allow = 0.707(5)[(50)](140)(10 ) = 49.5 kn Ans. 9- Given, b = in, d = in, = 5/16 in, allow = 5 kpsi. F = l allow = 0.707(5/16)[()](5) =.1 kip Ans. 9- Given, b = 50 mm, d = 0 mm, = 5 mm, allow = 140 MPa. F = l allow = 0.707(5)[(50)](140)(10 ) = 49.5 kn Ans. 9-4 Given, b = 4 in, d = in, = 5/16 in, allow = 5 kpsi. F = l allow = 0.707(5/16)[(4)](5) = 44. kip Ans. 9-5 Prob. 9-1 wit E7010 Electrode. Table 9-6: f = 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- wit E6010 Electrode. Table 9-6: f = kip/in = 14.85(5/16) = 4.64 kip/in Capter 9, Page 1/6

2 F = f l = 4.64[()] = 18.6 kip Ans. 9-7 Prob. 9- wit E7010 Electrode. Table 9-6: f = 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 wit E6010 Electrode. Table 9-6: f = 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 ot-rolled properties in te neigborood of te weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(400), 0.40(0)] min(10, 88) 88 MPa for bot materials. Eq. (9-): F = 0.707l all = 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 ot-rolled properties in te neigborood of te weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(55), 0.40(0)] min(16.5, 1.0) 1.0 kpsi for bot materials. Eq. (9-): F = 0.707l all = 0.707(5/16)[()](1.0) = 10.6 kip Ans. Capter 9, Page /6

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 ot-rolled properties in te neigborood of te weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(500), 0.40(70)] min(150, 108) 108 MPa for bot materials. Eq. (9-): F = 0.707l all = 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 ot-rolled properties in te neigborood of te weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(55), 0.40(0)] min(16.5, 1.0) 1.0 kpsi for bot materials. Eq. (9-): F = 0.707l all = 0.707(5/16)[(4)](1.0) = 1. kip Ans. 9-1 F Eq. (9-): 141 MPa Ans. l F 40 Eq. (9-):.6 kpsi Ans. l 5/16 F Eq. (9-): 177 MPa Ans. l /164 F 9-16 Eq. (9-): 15.1 kpsi Ans. l Capter 9, Page /6

4 9-17 b = d =50 mm, c = 150 mm, = 5 mm, and allow = 140 MPa. (a) Primar sear, Table 9-1, Case (Note: b and d are intercanged between problem figure and table figure. Note, also, F in kn and in MPa): V F 10.89F A Secondar sear, Table 9-1: J u d b d mm 6 6 J = J u = 0.707(5)(8.)(10 ) = 94.6(10 ) mm 4 Mr 175F 10 5 x 14.85F J x F F (1) allow 140 F 6.06 kn Ans..1.1 (b) For E7010 from Table 9-6, allow = 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 Te support controls te design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(40), 0.40(190) = min(10, 76) = 76 MPa Te allowable load, from Eq. (1) is allow 76 F.9 kn Ans b = d = in, c = 6 in, = 5/16 in, and allow = 5 kpsi. Capter 9, Page 4/6

5 (a) Primar sear, Table 9-1(Note: b and d are intercanged between problem figure and table figure. Note, also, F in kip and in kpsi): V F 1.1F A /16 Secondar sear, Table 9-1: db d Ju 5. in 6 6 J = J u = 0.707(5/16)(5.) = in 4 Mr 7F 1 x 5.94F J x F F (1) allow 5 F.71 kip Ans (b) For E7010 from Table 9-6, allow = 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 Te support controls te design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(50), 0.40(7.5) = min(15, 11) = 11 kpsi Te allowable load, from Eq. (1) is allow 11 F 1.19 kip Ans b =50 mm, c = 150 mm, d = 0 mm, = 5 mm, and allow = 140 MPa. (a) Primar sear, Table 9-1, Case (Note: b and d are intercanged between problem figure and table figure. Note, also, F in kn and in MPa): Capter 9, Page 5/6

6 V F 10.89F A Secondar sear, Table 9-1: db d Ju 4.10 mm 6 6 J = J u = 0.707(5)(4.)(10 ) = 15.(10 ) mm 4 Mr 175F x 17.1F J Mr 175F 10 5 x 8.55F J x F F (1) F allow kn Ans (b) For E7010 from Table 9-6, allow = 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 Te support controls te design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(40), 0.40(190) = min(10, 76) = 76 MPa Te allowable load, from Eq. (1) is allow 76 F.1 kn Ans b = 4 in, c = 6 in, d = in, = 5/16 in, and allow = 5 kpsi. Capter 9, Page 6/6

7 (a) Primar sear, Table 9-1(Note: b and d are intercanged between problem figure and table figure. Note, also, F in kip and in kpsi): V F F A /164 Secondar sear, Table 9-1: J u d b d in 6 6 J = J u = 0.707(5/16)(18.67) = 4.15 in 4 Mr 8F 1 x 1.99F J 4.15 Mr 8F x.879f J 4.15 x F F (1) allow 5 F 5.15 kip Ans (b) For E7010 from Table 9-6, allow = 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 Te support controls te design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(50), 0.40(7.5) = min(15, 11) = 11 kpsi Te allowable load, from Eq. (1) is allow 11 F.7 kip Ans Capter 9, Page 7/6

8 9-1 Given, b = 50 mm, c = 150 mm, d = 50 mm, = 5 mm, allow = 140 MPa. Primar sear (F in kn, in MPa, A in mm ): V F F A Secondar sear: Table 9-1: Maximum sear: bd Ju mm 6 6 J = J u = 0.707(5)166.7(10 ) = 589.(10 ) mm 4 175F 10 (5) Mr x 7.45F J x F F allow 140 F 1.1 kn Ans Given, b = in, c = 6 in, d = in, = 5/16 in, allow = 5 kpsi. Primar sear: Secondar sear: Table 9-1: Maximum sear: V F F A /16 bd Ju in 6 6 J = J u = 0.707(5/16)10.67 =.57 in 4 Mr 7 F(1) x.970f J.57 x F F allow 5 F 5.41 kip Ans Given, b = 50 mm, c = 150 mm, d = 0 mm, = 5 mm, allow = 140 MPa. Capter 9, Page 8/6

9 Primar sear (F in kn, in MPa, A in mm ): V F F A Secondar sear: Table 9-1: bd 50 0 Ju mm 6 6 J = J u = 0.707(5)85.(10 ) = 01.6(10 ) mm 4 175F 10 (15) Mr x 8.704F J Maximum sear: 175F 10 (5) Mrx 14.51F J x F F allow 140 F 7.58 kn Ans Given, b = 4 in, c = 6 in, d = in, = 5/16 in, allow = 5 kpsi. Primar sear: Secondar sear: Table 9-1: V F 0.77F A /16 4 bd 4 Ju 6 in 6 6 J = J u = 0.707(5/16)6 = in 4 Mr 8 F(1) x 1.006F J Maximum sear: Mrx 8 F().01F J x F F Capter 9, Page 9/6

10 allow 5 F 9.65kip Ans Given, b = 50 mm, d = 50 mm, = 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. 87, 88: k a = 7(0) = k b = 1 (uniform sear), k c = 0.59 (torsion, sear), 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. 87, 88: k a = 7(47) = As before, k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 S e = 0.657(1)(0.59)(1)(0.5)(47) = 8.8 MPa Te members and electrode are basicall of equal strengt. We will use S e = 8.6 MPa. For a factor of safet of 1, and wit K fs =.7 (Table 9-5) allow A F N 16. kn Ans. K.7 fs 9-6 Given, b = in, d = in, = 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. 87, 88: k a = 9.9(47) = k b = 1 (uniform sear), k c = 0.59 (torsion, sear), 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. 87, 88: k a = 9.9(6) = Capter 9, Page 10/6

11 As before, k b = 1 (uniform sear), k c = 0.59 (torsion, sear), k d = 1 S e = 0.657(1)(0.59)(1)(0.5)(6) = 1.0 kpsi Tus te members and electrode are of equal strengt. For a factor of safet of 1, and wit K fs =.7 (Table 9-5) allow A F 5.89 kip Ans. K.7 fs 9-7 Given, b = 50 mm, d = 0 mm, = 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. 87, 88: k a = 7(0) = k b = 1 (direct sear), k c = 0.59 (torsion, sear), 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. 87, 88: k a = 7(48) = 0.58 As before, k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 S e = 0.58(1)(0.59)(1)(0.5)(48) = 8.7 MPa Te members and electrode are basicall of equal strengt. We will use S e =8.6 MPa. For a factor of safet of 1, and wit K fs =.7 (Table 9-5) allow A F N 14.1 kn Ans. K fs Given, b = 4 in, d = in, = 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. 87, 88: k a = 9.9(47) = k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 Capter 9, Page 11/6

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. 87, 88: k a = 9.9(70) = 0.58 As before, k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 S e = 0.58(1)(0.59)(1)(0.5)(70) = 1.0 kpsi Tus te members and electrode are of equal strengt. For a factor of safet of 1, and wit K fs =.7 (Table 9-5) allow A F 9.8 kip Ans. K.7 fs 9-9 Primar sear: = 0 (w?) Secondar sear: Table 9-1: J u = r = (1.5) = 1.1 in J = J u = 0.707(1/4)(1.1) =.749 in 4 welds: Mr 8F F J.749 allow 1.600F 0 F 1.5 kip Ans. 9-0 l = = 10 in x in 1.6 in M = FR = F(10 1) = 9 F r r in, in 1 r in Capter 9, Page 1/6

13 1 4 JG / in 1 1 JG J G / in 1 J J Ar i1 i i G i / / / in 1.6 tan o r in Primar sear ( in kpsi, F in kip) : V F 0.456F A /16 10 Secondar sear: Mr 9F.4.19F J 9.0 o o F F F.19 sin cos F = allow.74 F = 5 F = 6.71 kip Ans. Capter 9, Page 1/6

14 9-1 l = = 10 mm x mm 1.15 mm M = FR = F( ) = F (M in Nm, F in kn) r 1 r mm, mm r mm mm 4 JG 1 1 JG J G mm 1 J J Ar i1 i i G i mm tan o r mm Primar sear ( in MPa, F in kn) : V F F A Secondar sear: Mr 186.9F F J Capter 9, Page 14/6

15 o o F F F 5.88 sin cos F = allow 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 l a a a a a a fom a a a a a a fom 0.08 a a a a 6a a a a fom a 1 a a a 1 8a a fom a 4a 1 a a a / fom 0.5 a 4a Capter 9, Page 15/6

16 9- Weld Pattern Figure of merit Rank I a /1 u a 1. fom l a.. 4.* 5. & & a /6 a fom 0.08 a 6 aa / a fom 0.5 a 1 a /1 6a a 7a a fom a 6 x a a, a a a a a a a Iu a aa I a / 1 u a a fom l a 9 a /6 a a 1 a a fom a 6 a / a a fom 0.15 a *Note. Because tis section is not smmetric wit te vertical axis, out-of-plane deflection ma occur unless special precautions are taken. See te topic of sear center in books wit more advanced treatments of mecanics of materials. 9-4 Attacment and member (1018 HR), S = 0 MPa and S ut = 400 MPa. Te member and attacment are weak compared to te properties of te lowest electrode. Decision Specif te E6010 electrode Controlling propert, Table 9-4: all = min[0.(400), 0.4(0)] = min(10, 88) = 88 MPa For a static load, te parallel and transverse fillets are te same. Let te lengt of a bead be l = 75 mm, and n be te number of beads. Capter 9, Page 16/6

17 F n l all all F n l were is in millimeters. Make a table Number of beads, n Leg size, (mm) mm Decision Specif = 6 mm on all four sides. Weldment specification: Pattern: All-around square, four beads eac side, 75 mm long Electrode: E6010 Leg size: = 6 mm 9-5 Decision: Coose a parallel fillet weldment pattern. B so-doing, we ve cosen an optimal pattern (see Prob. 9-) and ave tus reduced a sntesis problem to an analsis problem: Table 9-1, case, rotated 90: A = 1.414d = 1.414()(75) = mm Primar sear V A Secondar sear: d( b d ) Ju 6 75[(75 ) 75 ] mm 6 J 0.707( )(81.) mm Wit = 45, 4 Capter 9, Page 17/6

18 o cos 45 Mr 1 10 (187.5)(7.5) 44.4 Mr x J J x 44.4 ( ) Attacment and member (1018 HR): S = 0 MPa, S ut = 400 MPa Decision: Use E60XX electrode wic is stronger all min[0.(400), 0.4(0)] all 88 MPa mm MPa Decision: Specif 8 mm leg size Weldment Specifications: Pattern: Parallel orizontal fillet welds Electrode: E6010 Tpe: Fillet Lengt of eac bead: 75 mm Leg size: 8 mm 9-6 Problem 9-5 solves te problem using parallel orizontal fillet welds, eac 75 mm long obtaining a leg size rounded up to 8 mm. For tis problem, since te widt of te plate is fixed and te lengt as not been determined, we will explore reducing te leg size b using two vertical beads 75 mm long and two orizontal beads suc tat te beads ave a leg size of 6 mm. Decision: Use a rectangular weld bead pattern wit a leg size of 6 mm (case 5 of Table 9-1 wit b unknown and d = 75 mm). Materials: Attacment and member (1018 HR): S = 0 MPa, S ut = 400 MPa From Table 9-4, AISC welding code, all = 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 te unknown. Tis made solving for relativel eas. In tis problem, te terms will not be linear in b, and so we will use an iterative solution wit a spreadseet. Troat area and oter properties from Table 9-1: A = 1.414(6)(b + 75) = 8.484(b + 75) (1) Capter 9, Page 18/6

19 J u b 75, J = (6) J u = 0.707(b +75) () 6 Primar sear ( in MPa, in mm): V A 1 10 A () Secondar sear (See Prob. 9-5 solution for te definition of ) : Mr J Mr Mr b / (7.5) x cos cos (4) J J 0.707b 75 Mr Mr b / ( b / ) x sin sin (5) J J b 75 x Enter Eqs. (1) to (6) into a spreadseet and iterate for various values of b. A portion of te spreadseet is sown below. (6) b (mm) A (mm ) J (mm 4 ) ' (Mpa) " (Mpa) " x (Mpa) (Mpa) < 88 Mpa We see tat b 4 mm meets te strengt goal. Weldment Specifications: Pattern: Horizontal parallel weld tracks 4 mm long, vertical parallel weld tracks 75 mm long Electrode: E6010 Leg size: 6 mm 9-7 Materials: Member and attacment (1018 HR): S kpsi, Sut 58 kpsi Table 9-4: all min[0.(58), 0.4()] 1.8 kpsi Capter 9, Page 19/6

20 Decision: Use E6010 electrode. From Table 9-: S 50 kpsi, S 6 kpsi, all min[0.(6), 0.4(50)] 0 kpsi Decision: Since 1018 HR is weaker tan te E6010 electrode, use all 1.8 kpsi Decision: Use an all-around square weld bead track. Primar sear l 1 = 6 + a = = 1.5 in Troat area and oter properties from Table 9-1: A b ( d) 1.414( )(6 6) V F psi A A Secondar sear ( bd) (66) Ju 88 in J (88) 0.6 in Mr 0 10 (6.5 )() 76 psi x J x ( ) 76 ( ) psi Relate stress to strengt 476 all in Decision: Specif/8in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010 Tpe of weld: Fillet Weld bead lengt: 4 in Leg size: /8in Attacment lengt: 1.5 in ut Capter 9, Page 0/6

21 9-8 Tis 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 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 displa can be te bulk of our adequac assessment. Suc a program can teac too. 9-9 Te objective of tis design task is to ave te students teac temselves tat te weld patterns of Table 9- can be added or subtracted to obtain te properties of a contemplated weld pattern. Te instructor can control te level of complication. We ave left te presentation of te drawing to ou. Here is one possibilit. Stud te problem s opportunities, and ten present tis (or our sketc) wit te problem assignment. Use b as te design variable. Express properties as a function of b. 1 1 From Table 9-, case : A1.414 ( bb) bd bd ( b b) d Iu I 0.707I u V F A b ( b1 ) Mc Fa( d /) I 0.707I u Parametric stud a10 in, b8 in, d 8 in, b in, 1.8 kpsi, l (8 ) 1 in Let 1 all Capter 9, Page 1/6

22 A1.414 (8 ) 8.48in I (8 )(8 / ) 19 in u 4 I 0.707( )(19) 15.7 in psi (10)(8 / ) 948 psi psi from wic 0.48 in. Do not round off te leg size someting to learn. I 19 fom ' u 64.5 in l 0.48(1) A 8.48(0.48).10 in 4 I 15.7(0.48).65 in 0.48 vol l in I.65 eff 91. in vol psi psi psi 0.48 Now consider te case of uninterrupted welds, b1 0 A 1.414( )(8 0) 11.1 Iu (8 0)(8 / ) 56 in 4 I 0.707(56) 181 in (10)(8 / ) all in all Do not round off. Capter 9, Page /6

23 A 11.1(0.186).10 in I 181(0.186).67 in psi, vol in psi I 56 fom ' u 86.0 in l 0.186(16).67 eff I 11.7 in ( / ) l (0.186 / )16 4 Conclusions: To meet allowable stress limitations, I and A do not cange, nor do τ and σ. To meet te sortened bead lengt, is increased proportionatel. However, volume of bead laid down increases as. Te uninterrupted bead is superior. In 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. Had te weld bead gone around te corners, te situation would cange. Here is a follow up task analzing an alternative weld pattern From Table 9- For te box A b ( d) Subtracting b1 from b and d1 from d A 1.414bbd d 1 1 d d1 bd ( ) 1 1 Iu bd bb d d d Lengt of bead l ( bb1d d1) fom Iu / l Capter 9, Page /6

24 9-41 Computer programs will var. 9-4 Note to te Instructor. In te first printing of te nint edition, te loading was stated incorrectl. In te fourt line, bending moment of 100 kip in in sould read, 10 kip bending load 10 in from. Tis will be corrected in te printings tat follow. We apologize if tis as caused an inconvenience. all = 1 kpsi. Use Fig. 9-17(a) for general geometr, but emplo beads. beads and ten Horizontal parallel weld bead pattern b = in, d = 6 in Table 9-: A1.414b1.414( )() 4.4 in bd (6) Iu 54 in I 0.707I 0.707( )(54) 8. in u kpsi 4.4 Mc 10(10)(6 / ) 7.85 kpsi I kpsi Equate te imum and allowable sear stresses all 1 from wic 0.68 in. It follows tat I 8.(0.68) 6.1 in Te volume of te weld metal is l (0.68) ( ) vol 1.40 in Te effectiveness, (eff) H, is 4 Capter 9, Page 4/6

25 I 6.1 (eff) H 18.6 in vol (fom ') I u H 1. in l 0.68( ) Vertical parallel weld beads b in d 6 in From Table 9-, case A1.414d 1.414( )(6) 8.48 in d 6 Iu 7 in 6 6 I 0.707Iu 0.707( )(7) psi 8.48 Mc 10(10)(6 / ) psi I kpsi Equating to all gives in. It follows tat 4 I 50.9(0.501) 5.5 in l vol (6 6) 1.51 in I 5.5 (eff ) V 16.7 in vol (fom') I u V 1.0 in l 0.501(6 6) Te ratio of (eff ) V / (eff ) His 16.7 / Te ratio (fom') V / (fom') H is 1.0 / Tis is not surprising since I I 0.707Iu Iu eff fom' vol ( /) l ( /) l l Te ratios (e ff ) V / (eff ) H and (fom') V / (fom') H give te same information. Capter 9, Page 5/6

26 9-4 F = 0, T = 15 kipin. Table 9-1: J u = r = (1) = 6.8 in, J = 0.707(1/4) 6.8 = in 4 Tr kpsi Ans. J F = kip, T = 0. Table 9-: A = r = (1/4)(1) = in I u = r = (1) =.14 in, I = 0.707(1/4).14 = in 4 V 1.80 kpsi A Mr kpsi I = ( + ) 1/ = ( ) 1/ = 1.7 kpsi Ans F = kip, T = 15 kipin. Bending: Table 9-: A = r = (1/4)(1) = in Torsion: I u = r = (1) =.14 in, I = 0.707(1/4).14 = in 4 V 1.80 kpsi A Mr 1.6 kpsi M I Table 9-1: J u = r = (1) = 6.8 in, J = 0.707(1/4) 6.8 = in Tr 1.5 kpsi T J Capter 9, Page 6/6

27 kpsi Ans. M T 9-46 F = kip, T = 15 kipin. Bending: Table 9-: A = r = (1) = 4.44 in I u = r = (1) =.14 in, I = (.14) =.1 in 4 V kpsi A 4.44 Mr M kpsi I.1 Torsion: Table 9-1: J u = r = (1) = 6.8 in, J = (6.8) = 4.44 in 4 Tr T kpsi J M T kpsi 6.87 all in Ans. Sould specif a in weld. Ans mm, d 00 mm, b5mm From Table 9-, case : A = 1.414(9)(00) =.545(10 ) mm d 00 6 Iu 1.10 mm 6 6 I = I u = 0.707(9)(1.)(10 6 ) = 8.484(10 6 ) mm 4 Capter 9, Page 7/6

28 F MPa A.545(10 ) M = 5(150) = 750 Nm Mc I 8.484(10 ) 750(100) MPa MPa Ans Note to te Instructor. In te first printing of te nint edition, te vertical dimension of 5 in sould be to te top of te top plate. Tis will be corrected in te printings tat follow. We apologize if tis as caused an inconvenience. = 0.5 in, b =.5 in, d = 5 in. Table 9-, case 5: A = (b +d) = 0.707(0.5)[.5 + (5)] =.09 in d 5 in bd.5 5 d Iu d bd in I = I u = 0.707(1/4)(.) = in 4 Primar sear: F kpsi A.09 Secondar sear (te critical location is at te bottom of te bracket): = 5 = in M kpsi I kpsi all 18 n.48 Ans Capter 9, Page 8/6

29 9-49 Te largest possible weld size is 1/16 in. Tis is a small weld and tus difficult to accomplis. Te bracket s load-carring capabilit is not known. Tere are geometr problems associated wit seet metal folding, load-placement and location of te center of twist. Tis is not available to us. We will identif te strongest possible weldment. Use a rectangular, weld-all-around pattern Table 9-, case 6: A ( b d) 1.414(1 / 16)(1 7.5) in x b / 0.5 in d / 7.5 /.75 in d 7.5 Iu ( b d) [(1) 7.5] in 6 6 I 0.707Iu 0.707(1 / 16)(98.44) 4.50 in M ( ) W 4.5W V W 1.1W A Mc 4.5 W (7.5 / ).664W I 4.50 W W 4 Material properties: Te allowable stress given is low. Let s demonstrate tat. For te 100 CD bracket, use HR properties of S = 0 kpsi and S ut = 55. Te 100 HR support, S = 7.5 kpsi and S ut = 68. Te E6010 electrode as strengts of S = 50 and S ut = 6 kpsi. Allowable stresses: 100 HR: all = min[0.(55), 0.4(0)] = min(16.5, 1) = 1 kpsi 100 HR: all = min[0.(68), 0.4(7.5)] = min(0.4, 15) = 15 kpsi E6010: all = min[0.(6), 0.4(50)] = min(18.6, 0) = 18.6 kpsi Since Table 9-6 gives 18.0 kpsi as te allowable sear stress, use tis lower value. Terefore, te allowable sear stress is all = min(14.4, 1, 18.0) = 1 kpsi However, te allowable stress in te problem statement is 1.5 kpsi wic is low from te weldment perspective. Te load associated wit tis strengt is all.90w W 85 lbf.90 Capter 9, Page 9/6

30 If te welding can be accomplised (1/16 leg size is a small weld), te weld strengt is psi and te load associated wit tis strengt is W = 1 000/.90 = 077 lbf. Can te bracket carr suc a load? Tere are geometr problems associated wit seet metal folding. Load placement is important and te center of twist as not been identified. Also, te load-carring capabilit of te top bend is unknown. Tese uncertainties ma 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 emplos screw fasteners F F F F B x B B 100 lbf, all kpsi 100(16 / ) 5. lbf 5.cos lbf 5.cos0 46 lbf x It follows tat RA 56 lbf and RA 66.7 lbf, RA = 6 lbf M = 100(16) = 1600 lbf in Te OD of te tubes is 1 in. From Table 9-1, case 6: A 1.414( r) (1.414)( )(1 / ) 4.44 in Ju r (1 / ) in J (0.707) J 1.414(0.7854) in u 4 Capter 9, Page 0/6

31 V A 4.44 Tc Mc 1600(0.5) 70.1 J J Te sear stresses, and, are additive algebraicall ( ) psi 860 all / 16 in 000 Decision: Use 5/16 in fillet welds Ans For te pattern in bending sown, find te centroid G of te weld group. 6mm x I I Ax G 6mm 5 mm mm I 9mm mm I = I 6 mm + I 9 mm = ( )(10 6 ) = 5.69(10 6 ) mm 4 Te critical location is at B. Wit in MPa, and F in kn Capter 9, Page 1/6

32 V F F A Mc 00F F 6 I F F Materials: 1015 HR (Table A-0): S = 190 MPa, E6010 Electrode(Table 9-): S = 45 MPa Eq. (5-1), p. 5 all = 0.577(190) = MPa / / all n F 61. kn Ans In te textbook, Fig. Problem 9-5b is a free-bod diagram of te bracket. Forces and moments tat act on te welds are equal, but of opposite sense. (a) M = 100(0.66) = 49 lbf in Ans. (b) F = 100 sin 0 = 600 lbf Ans. (c) F x = 100 cos 0 = 109 lbf Ans. (d) From Table 9-, case 6: A 1.414(0.5)(0.5.5) 0.97 in d.5 Iu ( b d) [(0.5).5].9 in 6 6 Te second area moment about an axis troug G and parallel to z is I 0.707I (0.5)(.9) in Ans. u (e) Refer to Fig. Problem 9-5b. Te sear stress due to F is F A psi Te sear stress along te troat due to F x is F x psi A 0.97 Te resultant of 1 and is in te troat plane Capter 9, Page /6

33 psi Te bending of te troat gives Mc 49(1.5) I psi Te imum sear 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 A ns (g) Bending in te attacment near te base. Te cross-sectional area is approximatel equal to b. A1 b 0.5(.5) 0.65 in Fx 109 x 166 psi A I bd 0.5(.5) 0.60 in c 6 6 At location A, F M A1 I / c psi Te von Mises stress is x 648 (166) 91 psi Tus, te factor of safet is, S n 8.18 Ans..91 Te clip on te mooring line bears against te side of te 1/-in ole. If te clip fills te ole Capter 9, Page /6

34 F psi td 0.5(0.50) S (10 ) n. Ans Furter investigation of tis situation requires more detail tan is included in te task statement. () In sear fatigue, te weakest constituent of te weld melt is 1018 HR wit S ut = 58 kpsi, Eq. (6-8), p. 8, gives S e 0.504S 0.504(58) 9. kpsi ut Eq. (6-19), p. 87: k a = 14.4(58) = For te size factor estimate, we first emplo Eq. (6-5), p. 89, for te equivalent diameter d b (.5)(0.5) 0.57 in e Eq. (6-0), p. 88, is used next to find k b k b de Eq.(6-6), p. 90: k c = 0.59 From Eq. (6-18), p. 87, te endurance strengt in sear is S se = 0.780(0.940)(0.59)(9.) = 1.6 kpsi From Table 9-5, te sear stress-concentration factor is K f s =.7. Te loading is repeatedl-applied 1.57 a m K f s.7.07 kpsi Table 6-7, p. 07: Gerber factor of safet n f, adjusted for sear, wit S su = 0.67S ut 1S su a m S se n f 1 1 m Sse Ssua (58).07 (.07)(1.6) Ans (58)(.07) Attacment metal sould be cecked for bending fatigue. 9-5 (a) Use b = d = 4 in. Since = 5/8 in, te primar sear is Capter 9, Page 4/6

35 F 0.89F 1.414(5 / 8)(4) Te secondar sear calculations, for a moment arm of 14 in give J u J x 4[(4) 4] 4.67 in Ju 0.707(5 / 8) in Mr 14 F() 1.485F J Tus, te imum sear and allowable load are: F ( ).09F all 5 F 10.8 kip Ans Te load for part (a) as increased b a factor of 10.8/.71 =.99 Ans. (b) From Prob. 9-18b, all = 11 kpsi F all all 4.76 kip Te allowable load in part (b) as increased b a factor of 4.76/1.19 = 4 Ans Purcase te ook aving te design sown in Fig. Problem 9-54b. Referring to text Fig. 9-9a, tis design reduces peel stresses (a) l / 1 l/ Pcos( x) l/ A1 dx A / 1 cos( x) dx sin( x) l l 4bsin( l / ) l/ l / A1 A1 [sin( l / ) sin( l / )] [sin( l / ) ( sin( l / ))] A1 sin( l / ) P P [sin( l / )] Ans. 4bl sin( l / ) bl P cos( l/ ) P (b) (/) l Ans. 4bsin( l/ ) 4btan( l/ ) Capter 9, Page 5/6

36 (c) K ( l / ) P bl l / 4btan( l / ) P tan( l / ) Ans. For computer programming, it can be useful to express te perbolic tangent in terms of exponentials: l exp( l / ) exp( l / ) K Ans. exp( l / ) exp( l / ) 9-56 Tis is a computer programming exercise. All programs will var. Capter 9, Page 6/6

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.

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. budynas_sm_c09.qxd 01/9/007 18:5 Page 39 Capter 9 9-1 Eq. (9-3: F 0.707lτ 0.707(5/1(4(0 17.7 kip 9- Table 9-: τ all 1.0 kpsi f 14.85 kip/in 14.85(5/1 4.4 kip/in F fl 4.4(4 18.5 kip 9-3 Table A-0: 1018

More information

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

Chapter 9. Figure for Probs. 9-1 to Given, b = 50 mm, d = 50 mm, h = 5 mm, τ allow = 140 MPa. Chapter 9 Figure for Probs. 9-1 to 9-4 9-1 Given, b 50 mm, d 50 mm, h 5 mm, ow 140 MPa. F 0.707 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 0.707 hl ow 0.707(5/16)[()](5).1

More information

min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi

min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi 9-10 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: all min(0.0 Sut,

More information

6. Non-uniform bending

6. Non-uniform bending . Non-uniform bending Introduction Definition A non-uniform bending is te case were te cross-section is not only bent but also seared. It is known from te statics tat in suc a case, te bending moment in

More information

3. Using your answers to the two previous questions, evaluate the Mratio

3. Using your answers to the two previous questions, evaluate the Mratio MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 0219 2.002 MECHANICS AND MATERIALS II HOMEWORK NO. 4 Distributed: Friday, April 2, 2004 Due: Friday,

More information

June : 2016 (CBCS) Body. Load

June : 2016 (CBCS) Body. Load Engineering Mecanics st Semester : Common to all rances Note : Max. marks : 6 (i) ttempt an five questions (ii) ll questions carr equal marks. (iii) nswer sould be precise and to te point onl (iv) ssume

More information

Exam 1 Review Solutions

Exam 1 Review Solutions Exam Review Solutions Please also review te old quizzes, and be sure tat you understand te omework problems. General notes: () Always give an algebraic reason for your answer (graps are not sufficient),

More information

A = h w (1) Error Analysis Physics 141

A = h w (1) Error Analysis Physics 141 Introduction In all brances of pysical science and engineering one deals constantly wit numbers wic results more or less directly from experimental observations. Experimental observations always ave inaccuracies.

More information

Section 15.6 Directional Derivatives and the Gradient Vector

Section 15.6 Directional Derivatives and the Gradient Vector Section 15.6 Directional Derivatives and te Gradient Vector Finding rates of cange in different directions Recall tat wen we first started considering derivatives of functions of more tan one variable,

More information

Chapter 1 Functions and Graphs. Section 1.5 = = = 4. Check Point Exercises The slope of the line y = 3x+ 1 is 3.

Chapter 1 Functions and Graphs. Section 1.5 = = = 4. Check Point Exercises The slope of the line y = 3x+ 1 is 3. Capter Functions and Graps Section. Ceck Point Exercises. Te slope of te line y x+ is. y y m( x x y ( x ( y ( x+ point-slope y x+ 6 y x+ slope-intercept. a. Write te equation in slope-intercept form: x+

More information

The Derivative as a Function

The Derivative as a Function Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka Te Derivative as a Function DEFINITION: Te derivative of a function f at a number a, denoted by f (a), is if tis limit exists. f (a) f(a + )

More information

Sample Problems for Exam II

Sample Problems for Exam II Sample Problems for Exam 1. Te saft below as lengt L, Torsional stiffness GJ and torque T is applied at point C, wic is at a distance of 0.6L from te left (point ). Use Castigliano teorem to Calculate

More information

232 Calculus and Structures

232 Calculus and Structures 3 Calculus and Structures CHAPTER 17 JUSTIFICATION OF THE AREA AND SLOPE METHODS FOR EVALUATING BEAMS Calculus and Structures 33 Copyrigt Capter 17 JUSTIFICATION OF THE AREA AND SLOPE METHODS 17.1 THE

More information

Chapter 2 GEOMETRIC ASPECT OF THE STATE OF SOLICITATION

Chapter 2 GEOMETRIC ASPECT OF THE STATE OF SOLICITATION Capter GEOMETRC SPECT OF THE STTE OF SOLCTTON. THE DEFORMTON ROUND PONT.. Te relative displacement Due to te influence of external forces, temperature variation, magnetic and electric fields, te construction

More information

MAT 145. Type of Calculator Used TI-89 Titanium 100 points Score 100 possible points

MAT 145. Type of Calculator Used TI-89 Titanium 100 points Score 100 possible points MAT 15 Test #2 Name Solution Guide Type of Calculator Used TI-89 Titanium 100 points Score 100 possible points Use te grap of a function sown ere as you respond to questions 1 to 8. 1. lim f (x) 0 2. lim

More information

Lines, Conics, Tangents, Limits and the Derivative

Lines, Conics, Tangents, Limits and the Derivative Lines, Conics, Tangents, Limits and te Derivative Te Straigt Line An two points on te (,) plane wen joined form a line segment. If te line segment is etended beond te two points ten it is called a straigt

More information

MVT and Rolle s Theorem

MVT and Rolle s Theorem AP Calculus CHAPTER 4 WORKSHEET APPLICATIONS OF DIFFERENTIATION MVT and Rolle s Teorem Name Seat # Date UNLESS INDICATED, DO NOT USE YOUR CALCULATOR FOR ANY OF THESE QUESTIONS In problems 1 and, state

More information

LIMITATIONS OF EULER S METHOD FOR NUMERICAL INTEGRATION

LIMITATIONS OF EULER S METHOD FOR NUMERICAL INTEGRATION LIMITATIONS OF EULER S METHOD FOR NUMERICAL INTEGRATION LAURA EVANS.. Introduction Not all differential equations can be explicitly solved for y. Tis can be problematic if we need to know te value of y

More information

Moments and Product of Inertia

Moments and Product of Inertia Moments and Product of nertia Contents ntroduction( 绪论 ) Moments of nertia of an Area( 平面图形的惯性矩 ) Moments of nertia of an Area b ntegration( 积分法求惯性矩 ) Polar Moments of nertia( 极惯性矩 ) Radius of Gration

More information

MTH-112 Quiz 1 Name: # :

MTH-112 Quiz 1 Name: # : MTH- Quiz Name: # : Please write our name in te provided space. Simplif our answers. Sow our work.. Determine weter te given relation is a function. Give te domain and range of te relation.. Does te equation

More information

2.8 The Derivative as a Function

2.8 The Derivative as a Function .8 Te Derivative as a Function Typically, we can find te derivative of a function f at many points of its domain: Definition. Suppose tat f is a function wic is differentiable at every point of an open

More information

Continuity and Differentiability Worksheet

Continuity and Differentiability Worksheet Continuity and Differentiability Workseet (Be sure tat you can also do te grapical eercises from te tet- Tese were not included below! Typical problems are like problems -3, p. 6; -3, p. 7; 33-34, p. 7;

More information

Preface. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

Preface. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed. Preface Here are my online notes for my course tat I teac ere at Lamar University. Despite te fact tat tese are my class notes, tey sould be accessible to anyone wanting to learn or needing a refreser

More information

Section 2.7 Derivatives and Rates of Change Part II Section 2.8 The Derivative as a Function. at the point a, to be. = at time t = a is

Section 2.7 Derivatives and Rates of Change Part II Section 2.8 The Derivative as a Function. at the point a, to be. = at time t = a is Mat 180 www.timetodare.com Section.7 Derivatives and Rates of Cange Part II Section.8 Te Derivative as a Function Derivatives ( ) In te previous section we defined te slope of te tangent to a curve wit

More information

158 Calculus and Structures

158 Calculus and Structures 58 Calculus and Structures CHAPTER PROPERTIES OF DERIVATIVES AND DIFFERENTIATION BY THE EASY WAY. Calculus and Structures 59 Copyrigt Capter PROPERTIES OF DERIVATIVES. INTRODUCTION In te last capter you

More information

Math 1241 Calculus Test 1

Math 1241 Calculus Test 1 February 4, 2004 Name Te first nine problems count 6 points eac and te final seven count as marked. Tere are 120 points available on tis test. Multiple coice section. Circle te correct coice(s). You do

More information

Teaching Differentiation: A Rare Case for the Problem of the Slope of the Tangent Line

Teaching Differentiation: A Rare Case for the Problem of the Slope of the Tangent Line Teacing Differentiation: A Rare Case for te Problem of te Slope of te Tangent Line arxiv:1805.00343v1 [mat.ho] 29 Apr 2018 Roman Kvasov Department of Matematics University of Puerto Rico at Aguadilla Aguadilla,

More information

11-19 PROGRESSION. A level Mathematics. Pure Mathematics

11-19 PROGRESSION. A level Mathematics. Pure Mathematics SSaa m m pplle e UCa ni p t ter DD iff if erfe enren tiatia tiotio nn - 9 RGRSSIN decel Slevel andmatematics level Matematics ure Matematics NW FR 07 Year/S Year decel S and level Matematics Sample material

More information

. Compute the following limits.

. Compute the following limits. Today: Tangent Lines and te Derivative at a Point Warmup:. Let f(x) =x. Compute te following limits. f( + ) f() (a) lim f( +) f( ) (b) lim. Let g(x) = x. Compute te following limits. g(3 + ) g(3) (a) lim

More information

2.11 That s So Derivative

2.11 That s So Derivative 2.11 Tat s So Derivative Introduction to Differential Calculus Just as one defines instantaneous velocity in terms of average velocity, we now define te instantaneous rate of cange of a function at a point

More information

2.3 Product and Quotient Rules

2.3 Product and Quotient Rules .3. PRODUCT AND QUOTIENT RULES 75.3 Product and Quotient Rules.3.1 Product rule Suppose tat f and g are two di erentiable functions. Ten ( g (x)) 0 = f 0 (x) g (x) + g 0 (x) See.3.5 on page 77 for a proof.

More information

= 0 and states ''hence there is a stationary point'' All aspects of the proof dx must be correct (c)

= 0 and states ''hence there is a stationary point'' All aspects of the proof dx must be correct (c) Paper 1: Pure Matematics 1 Mark Sceme 1(a) (i) (ii) d d y 3 1x 4x x M1 A1 d y dx 1.1b 1.1b 36x 48x A1ft 1.1b Substitutes x = into teir dx (3) 3 1 4 Sows d y 0 and states ''ence tere is a stationary point''

More information

Chapter 10 1" 2 4" 1 = MPa mm. 201 = kpsi (0.105) S sy = 0.45(278.7) = kpsi D = = in. C = D d = = 10.

Chapter 10 1 2 4 1 = MPa mm. 201 = kpsi (0.105) S sy = 0.45(278.7) = kpsi D = = in. C = D d = = 10. budynas_sm_ch10.qxd 01/29/2007 18:26 Page 261 Chapter 10 10-1 1" 4" 1" 2 1" 4" 1" 2 10-2 A = Sd m dim( A uscu ) = dim(s) dim(d m ) = kpsi in m dim( A SI ) = dim(s 1 ) dim ( d m ) 1 = MPa mm m A SI = MPa

More information

f a h f a h h lim lim

f a h f a h h lim lim Te Derivative Te derivative of a function f at a (denoted f a) is f a if tis it exists. An alternative way of defining f a is f a x a fa fa fx fa x a Note tat te tangent line to te grap of f at te point

More information

1 The concept of limits (p.217 p.229, p.242 p.249, p.255 p.256) 1.1 Limits Consider the function determined by the formula 3. x since at this point

1 The concept of limits (p.217 p.229, p.242 p.249, p.255 p.256) 1.1 Limits Consider the function determined by the formula 3. x since at this point MA00 Capter 6 Calculus and Basic Linear Algebra I Limits, Continuity and Differentiability Te concept of its (p.7 p.9, p.4 p.49, p.55 p.56). Limits Consider te function determined by te formula f Note

More information

Path to static failure of machine components

Path to static failure of machine components Pat to static failure of macine components Load Stress Discussed last week (w) Ductile material Yield Strain Brittle material Fracture Fracture Dr. P. Buyung Kosasi,Spring 008 Name some of ductile and

More information

Pre-Calculus Review Preemptive Strike

Pre-Calculus Review Preemptive Strike Pre-Calculus Review Preemptive Strike Attaced are some notes and one assignment wit tree parts. Tese are due on te day tat we start te pre-calculus review. I strongly suggest reading troug te notes torougly

More information

UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics

UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics UNIVERSITY O SASKATCHEWAN Department of Pysics and Engineering Pysics Pysics 117.3 MIDTERM EXAM Regular Sitting NAME: (Last) Please Print (Given) Time: 90 minutes STUDENT NO.: LECTURE SECTION (please ceck):

More information

Lecture 21. Numerical differentiation. f ( x+h) f ( x) h h

Lecture 21. Numerical differentiation. f ( x+h) f ( x) h h Lecture Numerical differentiation Introduction We can analytically calculate te derivative of any elementary function, so tere migt seem to be no motivation for calculating derivatives numerically. However

More information

Numerical Differentiation

Numerical Differentiation Numerical Differentiation Finite Difference Formulas for te first derivative (Using Taylor Expansion tecnique) (section 8.3.) Suppose tat f() = g() is a function of te variable, and tat as 0 te function

More information

1 1. Rationalize the denominator and fully simplify the radical expression 3 3. Solution: = 1 = 3 3 = 2

1 1. Rationalize the denominator and fully simplify the radical expression 3 3. Solution: = 1 = 3 3 = 2 MTH - Spring 04 Exam Review (Solutions) Exam : February 5t 6:00-7:0 Tis exam review contains questions similar to tose you sould expect to see on Exam. Te questions included in tis review, owever, are

More information

The total error in numerical differentiation

The total error in numerical differentiation AMS 147 Computational Metods and Applications Lecture 08 Copyrigt by Hongyun Wang, UCSC Recap: Loss of accuracy due to numerical cancellation A B 3, 3 ~10 16 In calculating te difference between A and

More information

Exercises for numerical differentiation. Øyvind Ryan

Exercises for numerical differentiation. Øyvind Ryan Exercises for numerical differentiation Øyvind Ryan February 25, 2013 1. Mark eac of te following statements as true or false. a. Wen we use te approximation f (a) (f (a +) f (a))/ on a computer, we can

More information

Introduction to Derivatives

Introduction to Derivatives Introduction to Derivatives 5-Minute Review: Instantaneous Rates and Tangent Slope Recall te analogy tat we developed earlier First we saw tat te secant slope of te line troug te two points (a, f (a))

More information

Mathematics 5 Worksheet 11 Geometry, Tangency, and the Derivative

Mathematics 5 Worksheet 11 Geometry, Tangency, and the Derivative Matematics 5 Workseet 11 Geometry, Tangency, and te Derivative Problem 1. Find te equation of a line wit slope m tat intersects te point (3, 9). Solution. Te equation for a line passing troug a point (x

More information

2.3. Applying Newton s Laws of Motion. Objects in Equilibrium

2.3. Applying Newton s Laws of Motion. Objects in Equilibrium Appling Newton s Laws of Motion As ou read in Section 2.2, Newton s laws of motion describe ow objects move as a result of different forces. In tis section, ou will appl Newton s laws to objects subjected

More information

Section 3: The Derivative Definition of the Derivative

Section 3: The Derivative Definition of the Derivative Capter 2 Te Derivative Business Calculus 85 Section 3: Te Derivative Definition of te Derivative Returning to te tangent slope problem from te first section, let's look at te problem of finding te slope

More information

Polynomial Interpolation

Polynomial Interpolation Capter 4 Polynomial Interpolation In tis capter, we consider te important problem of approximatinga function fx, wose values at a set of distinct points x, x, x,, x n are known, by a polynomial P x suc

More information

Numerical Analysis MTH603. dy dt = = (0) , y n+1. We obtain yn. Therefore. and. Copyright Virtual University of Pakistan 1

Numerical Analysis MTH603. dy dt = = (0) , y n+1. We obtain yn. Therefore. and. Copyright Virtual University of Pakistan 1 Numerical Analysis MTH60 PREDICTOR CORRECTOR METHOD Te metods presented so far are called single-step metods, were we ave seen tat te computation of y at t n+ tat is y n+ requires te knowledge of y n only.

More information

Lab 6 Derivatives and Mutant Bacteria

Lab 6 Derivatives and Mutant Bacteria Lab 6 Derivatives and Mutant Bacteria Date: September 27, 20 Assignment Due Date: October 4, 20 Goal: In tis lab you will furter explore te concept of a derivative using R. You will use your knowledge

More information

Chapter 2 Limits and Continuity

Chapter 2 Limits and Continuity 4 Section. Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 6) Quick Review.. f () ( ) () 4 0. f () 4( ) 4. f () sin sin 0 4. f (). 4 4 4 6. c c c 7. 8. c d d c d d c d c 9. 8 ( )(

More information

Reinforced Concrete Structures/Design Aids

Reinforced Concrete Structures/Design Aids AREA OF BARS (mm 2 ) Reinforced Concrete Structures/Design Aids Size of s (mm) Number of s 1 2 3 4 5 6 7 8 8 50 101 151 201 251 302 352 402 10 79 157 236 314 393 471 550 628 12 113 226 339 452 566 679

More information

INTRODUCTION AND MATHEMATICAL CONCEPTS

INTRODUCTION AND MATHEMATICAL CONCEPTS INTODUCTION ND MTHEMTICL CONCEPTS PEVIEW Tis capter introduces you to te basic matematical tools for doing pysics. You will study units and converting between units, te trigonometric relationsips of sine,

More information

Section 3.1: Derivatives of Polynomials and Exponential Functions

Section 3.1: Derivatives of Polynomials and Exponential Functions Section 3.1: Derivatives of Polynomials and Exponential Functions In previous sections we developed te concept of te derivative and derivative function. Te only issue wit our definition owever is tat it

More information

Logarithmic functions

Logarithmic functions Roberto s Notes on Differential Calculus Capter 5: Derivatives of transcendental functions Section Derivatives of Logaritmic functions Wat ou need to know alread: Definition of derivative and all basic

More information

2.1 THE DEFINITION OF DERIVATIVE

2.1 THE DEFINITION OF DERIVATIVE 2.1 Te Derivative Contemporary Calculus 2.1 THE DEFINITION OF DERIVATIVE 1 Te grapical idea of a slope of a tangent line is very useful, but for some uses we need a more algebraic definition of te derivative

More information

11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR

11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR SECTION 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 633 wit speed v o along te same line from te opposite direction toward te source, ten te frequenc of te sound eard b te observer is were c is

More information

Math 102 TEST CHAPTERS 3 & 4 Solutions & Comments Fall 2006

Math 102 TEST CHAPTERS 3 & 4 Solutions & Comments Fall 2006 Mat 102 TEST CHAPTERS 3 & 4 Solutions & Comments Fall 2006 f(x+) f(x) 10 1. For f(x) = x 2 + 2x 5, find ))))))))) and simplify completely. NOTE: **f(x+) is NOT f(x)+! f(x+) f(x) (x+) 2 + 2(x+) 5 ( x 2

More information

Chapter 5 FINITE DIFFERENCE METHOD (FDM)

Chapter 5 FINITE DIFFERENCE METHOD (FDM) MEE7 Computer Modeling Tecniques in Engineering Capter 5 FINITE DIFFERENCE METHOD (FDM) 5. Introduction to FDM Te finite difference tecniques are based upon approximations wic permit replacing differential

More information

Chapters 19 & 20 Heat and the First Law of Thermodynamics

Chapters 19 & 20 Heat and the First Law of Thermodynamics Capters 19 & 20 Heat and te First Law of Termodynamics Te Zerot Law of Termodynamics Te First Law of Termodynamics Termal Processes Te Second Law of Termodynamics Heat Engines and te Carnot Cycle Refrigerators,

More information

Derivatives of Exponentials

Derivatives of Exponentials mat 0 more on derivatives: day 0 Derivatives of Eponentials Recall tat DEFINITION... An eponential function as te form f () =a, were te base is a real number a > 0. Te domain of an eponential function

More information

Printed Name: Section #: Instructor:

Printed Name: Section #: Instructor: Printed Name: Section #: Instructor: Please do not ask questions during tis exam. If you consider a question to be ambiguous, state your assumptions in te margin and do te best you can to provide te correct

More information

CHAPTER 2 Functions and Their Graphs

CHAPTER 2 Functions and Their Graphs CHAPTER Functions and Teir Graps Section. Linear Equations in Two Variables............ 9 Section. Functions......................... 0 Section. Analzing Graps of Functions............. Section. A Librar

More information

SFU UBC UNBC Uvic Calculus Challenge Examination June 5, 2008, 12:00 15:00

SFU UBC UNBC Uvic Calculus Challenge Examination June 5, 2008, 12:00 15:00 SFU UBC UNBC Uvic Calculus Callenge Eamination June 5, 008, :00 5:00 Host: SIMON FRASER UNIVERSITY First Name: Last Name: Scool: Student signature INSTRUCTIONS Sow all your work Full marks are given only

More information

Lesson 6: The Derivative

Lesson 6: The Derivative Lesson 6: Te Derivative Def. A difference quotient for a function as te form f(x + ) f(x) (x + ) x f(x + x) f(x) (x + x) x f(a + ) f(a) (a + ) a Notice tat a difference quotient always as te form of cange

More information

Notes on wavefunctions II: momentum wavefunctions

Notes on wavefunctions II: momentum wavefunctions Notes on wavefunctions II: momentum wavefunctions and uncertainty Te state of a particle at any time is described by a wavefunction ψ(x). Tese wavefunction must cange wit time, since we know tat particles

More information

MA455 Manifolds Solutions 1 May 2008

MA455 Manifolds Solutions 1 May 2008 MA455 Manifolds Solutions 1 May 2008 1. (i) Given real numbers a < b, find a diffeomorpism (a, b) R. Solution: For example first map (a, b) to (0, π/2) and ten map (0, π/2) diffeomorpically to R using

More information

MAT244 - Ordinary Di erential Equations - Summer 2016 Assignment 2 Due: July 20, 2016

MAT244 - Ordinary Di erential Equations - Summer 2016 Assignment 2 Due: July 20, 2016 MAT244 - Ordinary Di erential Equations - Summer 206 Assignment 2 Due: July 20, 206 Full Name: Student #: Last First Indicate wic Tutorial Section you attend by filling in te appropriate circle: Tut 0

More information

5.1. Cross-Section and the Strength of a Bar

5.1. Cross-Section and the Strength of a Bar TRENGTH OF MTERL Meanosüsteemide komponentide õppetool 5. Properties of ections 5. ross-ection and te trengt of a Bar 5. rea Properties of Plane apes 5. entroid of a ection 5.4 rea Moments of nertia 5.5

More information

A general articulation angle stability model for non-slewing articulated mobile cranes on slopes *

A general articulation angle stability model for non-slewing articulated mobile cranes on slopes * tecnical note 3 general articulation angle stability model for non-slewing articulated mobile cranes on slopes * J Wu, L uzzomi and M Hodkiewicz Scool of Mecanical and Cemical Engineering, University of

More information

Solutions Manual for Precalculus An Investigation of Functions

Solutions Manual for Precalculus An Investigation of Functions Solutions Manual for Precalculus An Investigation of Functions David Lippman, Melonie Rasmussen 1 st Edition Solutions created at Te Evergreen State College and Soreline Community College 1.1 Solutions

More information

Differentiation. Area of study Unit 2 Calculus

Differentiation. Area of study Unit 2 Calculus Differentiation 8VCE VCEco Area of stud Unit Calculus coverage In tis ca 8A 8B 8C 8D 8E 8F capter Introduction to limits Limits of discontinuous, rational and brid functions Differentiation using first

More information

Exponentials and Logarithms Review Part 2: Exponentials

Exponentials and Logarithms Review Part 2: Exponentials Eponentials and Logaritms Review Part : Eponentials Notice te difference etween te functions: g( ) and f ( ) In te function g( ), te variale is te ase and te eponent is a constant. Tis is called a power

More information

1. Consider the trigonometric function f(t) whose graph is shown below. Write down a possible formula for f(t).

1. Consider the trigonometric function f(t) whose graph is shown below. Write down a possible formula for f(t). . Consider te trigonometric function f(t) wose grap is sown below. Write down a possible formula for f(t). Tis function appears to be an odd, periodic function tat as been sifted upwards, so we will use

More information

Investigating Euler s Method and Differential Equations to Approximate π. Lindsay Crowl August 2, 2001

Investigating Euler s Method and Differential Equations to Approximate π. Lindsay Crowl August 2, 2001 Investigating Euler s Metod and Differential Equations to Approximate π Lindsa Crowl August 2, 2001 Tis researc paper focuses on finding a more efficient and accurate wa to approximate π. Suppose tat x

More information

Key Concepts. Important Techniques. 1. Average rate of change slope of a secant line. You will need two points ( a, the formula: to find value

Key Concepts. Important Techniques. 1. Average rate of change slope of a secant line. You will need two points ( a, the formula: to find value AB Calculus Unit Review Key Concepts Average and Instantaneous Speed Definition of Limit Properties of Limits One-sided and Two-sided Limits Sandwic Teorem Limits as x ± End Beaviour Models Continuity

More information

5.1 We will begin this section with the definition of a rational expression. We

5.1 We will begin this section with the definition of a rational expression. We Basic Properties and Reducing to Lowest Terms 5.1 We will begin tis section wit te definition of a rational epression. We will ten state te two basic properties associated wit rational epressions and go

More information

1watt=1W=1kg m 2 /s 3

1watt=1W=1kg m 2 /s 3 Appendix A Matematics Appendix A.1 Units To measure a pysical quantity, you need a standard. Eac pysical quantity as certain units. A unit is just a standard we use to compare, e.g. a ruler. In tis laboratory

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019 ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 6. Differential Calculus 6.. Differentiation from First Principles. In tis capter, we will introduce

More information

Definition of the Derivative

Definition of the Derivative Te Limit Definition of te Derivative Tis Handout will: Define te limit grapically and algebraically Discuss, in detail, specific features of te definition of te derivative Provide a general strategy of

More information

Low Decibel Linear Way E

Low Decibel Linear Way E U.S. PATNTD Low Decibel Linear Way LWQLWTQLWSQ Low Decibel Linear Way series Sape lange type mounted from bottom Low Decibel Linear Way is a linear motion rolling guide for smoot and quiet motion. Its

More information

Combining functions: algebraic methods

Combining functions: algebraic methods Combining functions: algebraic metods Functions can be added, subtracted, multiplied, divided, and raised to a power, just like numbers or algebra expressions. If f(x) = x 2 and g(x) = x + 2, clearly f(x)

More information

5. (a) Find the slope of the tangent line to the parabola y = x + 2x

5. (a) Find the slope of the tangent line to the parabola y = x + 2x MATH 141 090 Homework Solutions Fall 00 Section.6: Pages 148 150 3. Consider te slope of te given curve at eac of te five points sown (see text for figure). List tese five slopes in decreasing order and

More information

MATH1151 Calculus Test S1 v2a

MATH1151 Calculus Test S1 v2a MATH5 Calculus Test 8 S va January 8, 5 Tese solutions were written and typed up by Brendan Trin Please be etical wit tis resource It is for te use of MatSOC members, so do not repost it on oter forums

More information

7.1 Using Antiderivatives to find Area

7.1 Using Antiderivatives to find Area 7.1 Using Antiderivatives to find Area Introduction finding te area under te grap of a nonnegative, continuous function f In tis section a formula is obtained for finding te area of te region bounded between

More information

In Leibniz notation, we write this rule as follows. DERIVATIVE OF A CONSTANT FUNCTION. For n 4 we find the derivative of f x x 4 as follows: lim

In Leibniz notation, we write this rule as follows. DERIVATIVE OF A CONSTANT FUNCTION. For n 4 we find the derivative of f x x 4 as follows: lim .1 DERIVATIVES OF POLYNOIALS AND EXPONENTIAL FUNCTIONS c =c slope=0 0 FIGURE 1 Te grap of ƒ=c is te line =c, so fª()=0. In tis section we learn ow to ifferentiate constant functions, power functions, polnomials,

More information

. If lim. x 2 x 1. f(x+h) f(x)

. If lim. x 2 x 1. f(x+h) f(x) Review of Differential Calculus Wen te value of one variable y is uniquely determined by te value of anoter variable x, ten te relationsip between x and y is described by a function f tat assigns a value

More information

3.4 Worksheet: Proof of the Chain Rule NAME

3.4 Worksheet: Proof of the Chain Rule NAME Mat 1170 3.4 Workseet: Proof of te Cain Rule NAME Te Cain Rule So far we are able to differentiate all types of functions. For example: polynomials, rational, root, and trigonometric functions. We are

More information

Higher Derivatives. Differentiable Functions

Higher Derivatives. Differentiable Functions Calculus 1 Lia Vas Higer Derivatives. Differentiable Functions Te second derivative. Te derivative itself can be considered as a function. Te instantaneous rate of cange of tis function is te second derivative.

More information

1. Questions (a) through (e) refer to the graph of the function f given below. (A) 0 (B) 1 (C) 2 (D) 4 (E) does not exist

1. Questions (a) through (e) refer to the graph of the function f given below. (A) 0 (B) 1 (C) 2 (D) 4 (E) does not exist Mat 1120 Calculus Test 2. October 18, 2001 Your name Te multiple coice problems count 4 points eac. In te multiple coice section, circle te correct coice (or coices). You must sow your work on te oter

More information

INTRODUCTION TO CALCULUS LIMITS

INTRODUCTION TO CALCULUS LIMITS Calculus can be divided into two ke areas: INTRODUCTION TO CALCULUS Differential Calculus dealing wit its, rates of cange, tangents and normals to curves, curve sketcing, and applications to maima and

More information

Quantization of electrical conductance

Quantization of electrical conductance 1 Introduction Quantization of electrical conductance Te resistance of a wire in te classical Drude model of metal conduction is given by RR = ρρρρ AA, were ρρ, AA and ll are te conductivity of te material,

More information

A.P. CALCULUS (AB) Outline Chapter 3 (Derivatives)

A.P. CALCULUS (AB) Outline Chapter 3 (Derivatives) A.P. CALCULUS (AB) Outline Capter 3 (Derivatives) NAME Date Previously in Capter 2 we determined te slope of a tangent line to a curve at a point as te limit of te slopes of secant lines using tat point

More information

Printed Name: Section #: Instructor:

Printed Name: Section #: Instructor: Printed Name: Section #: Instructor: Please do not ask questions during tis exam. If you consider a question to be ambiguous, state your assumptions in te margin and do te best you can to provide te correct

More information

Chapter 1D - Rational Expressions

Chapter 1D - Rational Expressions - Capter 1D Capter 1D - Rational Expressions Definition of a Rational Expression A rational expression is te quotient of two polynomials. (Recall: A function px is a polynomial in x of degree n, if tere

More information

Instructional Objectives:

Instructional Objectives: Instructiona Objectives: At te end of tis esson, te students soud be abe to understand: Ways in wic eccentric oads appear in a weded joint. Genera procedure of designing a weded joint for eccentric oading.

More information

Polynomial Interpolation

Polynomial Interpolation Capter 4 Polynomial Interpolation In tis capter, we consider te important problem of approximating a function f(x, wose values at a set of distinct points x, x, x 2,,x n are known, by a polynomial P (x

More information

The Verlet Algorithm for Molecular Dynamics Simulations

The Verlet Algorithm for Molecular Dynamics Simulations Cemistry 380.37 Fall 2015 Dr. Jean M. Standard November 9, 2015 Te Verlet Algoritm for Molecular Dynamics Simulations Equations of motion For a many-body system consisting of N particles, Newton's classical

More information

Lecture XVII. Abstract We introduce the concept of directional derivative of a scalar function and discuss its relation with the gradient operator.

Lecture XVII. Abstract We introduce the concept of directional derivative of a scalar function and discuss its relation with the gradient operator. Lecture XVII Abstract We introduce te concept of directional derivative of a scalar function and discuss its relation wit te gradient operator. Directional derivative and gradient Te directional derivative

More information

MATH 1020 Answer Key TEST 2 VERSION B Fall Printed Name: Section #: Instructor:

MATH 1020 Answer Key TEST 2 VERSION B Fall Printed Name: Section #: Instructor: Printed Name: Section #: Instructor: Please do not ask questions during tis exam. If you consider a question to be ambiguous, state your assumptions in te margin and do te best you can to provide te correct

More information