Baixe Soluções do Capítulo 9 de Shigley's Mechanical Engineering Design, 10ª Edição e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity!
Chapter 9
Figure for Probs.9-1 to 9-
9-1 Given, b = 50 mm, d = 50 mm, h = 5 mm, τallow = 140 MPa.
______________________________________________________________________________^ F^ = 0.707^ hl^ τallow^ = 0.707(5)2(50)(10−^3 ) = 49.5 kN^ Ans. 9-2 Given, b = 2 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi. F = 0.707 hl τallow = 0.707(5/16)2(2) = 22.1 kip Ans.
9-3 Given, b = 50 mm, d = 30 mm, h = 5 mm, τallow = 140 MPa. F = 0.707 hl τallow = 0.707(5)2(50)(10−^3 ) = 49.5 kN Ans.
9-4 Given, b = 4 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi. F = 0.707 hl τallow = 0.707(5/16)2(4) = 44.2 kip Ans.
9-5 Prob. 9-1 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm
______________________________________________________________________________^ F^ =^ f l^ = 0.512[2(50)] = 51.2 kN^ Ans. 9-6 Prob. 9-2 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in
F = f l = 4.64[2(2)] = 18.6 kip Ans.
9-7 Prob. 9-3 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm F = f l = 0.512[2(50)] = 51.2 kN Ans.
9-8 Prob. 9-4 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64[2(4)] = 37.1 kip Ans.
9-9 Table A-20:1018 CD: S 1018 HR: Sutut^ = 440 MPa,= 400 MPa,^ SSyy^ = 370 MPa= 220 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.Table 9-4: all min(0.30^ , 0.40^ ) min[0.30(400), 0.40(220)] min(120, 88) 88 MPa
τ = S ut Sy
= = for both materials. Eq. (9-3): F = 0_._ 707 hl τall = 0_._ 707(5)2(50)(10−^3 ) = 31.1 kN Ans.
9-10 Table A-20:1020 CD: S 1020 HR: Sutut^ = 68 kpsi,= 55 kpsi,^ SSyy^ = 57 kpsi= 30 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.Table 9-4: all min(0.30^ , 0.40^ ) min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi
τ = S ut Sy
= = for both materials. Eq. (9-3): F = 0_._ 707 hl τall = 0_._ 707(5 / 16)[2(2)](12_._ 0) = 10.6 kip Ans.
9-17 b = d = 50 mm, c = 150 mm, h = 5 mm, and τallow = 140 MPa. (a ) Primary shear , Table 9-1, Case 2 (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kN and τ in MPa):
τ y ′ = V A^ = 1.414 5^ F 50 =2.829 F
Secondary shear , Table 9-1:
( 3 2 2 ) 50 3 50(^2 )^502 83.33 10( 3 )mm 3
u 6 6 J =^ d^ b^ +^ d = ^ + = J = 0.707 h Ju = 0.707(5)(83.33)(10^3 ) = 294.6(10^3 ) mm^4
3 3
x y Mr^ y^ F F τ ′′^ = τ′′= (^) J = =
τ max = τ x ′′^2 + ( τ y ′^ + τ y ′′ ) 2 = F 14.85^2 + ( 2.829 + 14.85) 2 = 23.1 F (1)
F = τ 23.1^ allow^ = (^) 23.1^140 =6.06 kN Ans. ( b ) For E7010 from Table 9-6, τallow = 21 kpsi = 21(6.89) = 145 MPa 1020 HR bar: Sut = 380 MPa, Sy = 210 MPa 1015 HR support: Sut = 340 MPa, Sy = 190 MPa Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa The support controls the design. Table 9-4: τallow = min(0.30 Sut , 0.40 Sy ) =min[0.30(340), 0.40(190) = min(102, 76) = 76 MPa The allowable load, from Eq. (1) is F = τ 23.1^ allow^ = (^) 23.1^76 =3.29 kN Ans. ______ ________________________________________________________________________ 9-18 b = d = 2 in, c = 6 in, h = 5/16 in, and τallow = 25 kpsi.
(a ) Primary shear , Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and τ in kpsi):
y 1.414 5 /16 ( ) ( 2 ) 1.
V F F
τ ′ = (^) A = = Secondary shear , Table 9-1:
( 3 2 2 ) 2 3 2(^2 )^22 5.333 in 3
u 6 6 J =^ d^ b^ +^ d = ^ + = J = 0.707 h Ju = 0.707(5/16)(5.333) = 1.178 in^4
x y^ Mr^ y^ F^ F τ ′′^ = τ′′= (^) J = =
τ max = τ x ′′^2 + ( τ y ′ + τ y ′′ ) 2 = F 5.942^2 + (1.132 + 5.942 )^2 = 9.24 F (1)
allow (^25) 2.71 kip. F (^) 9.24 9.24 Ans = τ = = ( b ) For E7010 from Table 9-6, τallow = 21 kpsi 1020 HR bar: Sut = 55 kpsi, Sy = 30 kpsi 1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: τallow = min(0.30 Sut , 0.40 Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11) = 11 kpsi The allowable load, from Eq. (1) is F = τ 9.24^ allow^ = (^) 9.24^11 =1.19 kip Ans. ______ ________________________________________________________________________ 9-19 b = (a ) 50 mm, Primary shear c = 150 mm,, Table 9-1, Case 2 (Note: d = 30 mm, h = 5 mm, and b and d τare interchanged between problemallow = 140 MPa. figure and table figure. Note, also, F in kN and τ in MPa):
(a ) Primary shear , Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and τ in kpsi):
y 1.414 5 /16 ( ) ( 4 ) 0.
V F F
τ ′ = (^) A = = Secondary shear , Table 9-1:
( 3 2 2 ) 4 3 2(^2 )^42 18.67 in 3
u 6 6 J =^ d^ b^ +^ d = ^ + = J = 0.707 h Ju = 0.707(5/16)(18.67) = 4.125 in^4
x^ Mr^ y^ F^ F τ ′′ = (^) J = =
y Mr^ x^ F F τ ′′ = (^) J = =
τ max = τ x ′′^2 + ( τ y ′^ + τ y ′′ ) 2 = F 1.939^2 + ( 0.5658 + 3.879 )^2 = 4.85 F (1)
allow (^25) 5.15 kip. F (^) 4.85 4.85 Ans = τ = = ( b ) For E7010 from Table 9-6, τallow = 21 kpsi 1020 HR bar: Sut = 55 kpsi, Sy = 30 kpsi 1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: τallow = min(0.30 Sut , 0.40 Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11) = 11 kpsi The allowable load, from Eq. (1) is F = τ 4.85^ allow^ = (^) 4.85^11 =2.27 kip Ans. ______ ________________________________________________________________________
9-21 Given, b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, τallow = 140 MPa. Primary shear ( F in kN, τ in MPa, A in mm^2 ): ( ) ( )( )
τ (^) y ′ = V A^ = (^) 1.414 5^ F 50 + 50 =1.414 F Secondary shear : Table 9-1: (^ )^ (^ )^ ( )
J (^) u =^ b^ + 6 d = + 6 =166.7 10 mm J = 0.707 h Ju = 0.707(5)166.7(10^3 ) = 589.2(10^3 ) mm^4 ( ) ( )
3 3
x y Mr^ y^ F F τ ′′ = τ′′= (^) J = = Maximum shear: τ (^) max = τ (^) x ′′^2 + (^) ( τ (^) y ′^ + τ′′ y ) 2 = F 7.425^2 + (^) (1.414 + 7.425 (^) )^2 =11.54 F (^140) 12.1 kN. 11.54 11. F = τ^ allow = = Ans ______ ________________________________________________________________________ 9-22 Given, b = 2 in, c = 6 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi. Primary shear : y 1.414 5 /16 ( )( (^2 2) ) 0.
V F F
τ ′ = (^) A = (^) + = Secondary shear : Table 9-1: (^ )^ (^ )
J (^) u =^ b^ + 6 d = + 6 =10.67 in J = 0.707 h Ju = 0.707(5/16)10.67 = 2.357 in^4 7 (1) (^) 2.
x y^ Mr^ y^ F^ F Maximum shear:^ τ^ ′′^ =^ τ′′=^^ J =^ = τ (^) max = τ (^) x ′′^2 + (^) ( τ (^) y ′^ + τ′′ y ) 2 = F 2.970 2 + (^) ( 0.566 + 2.970 (^) )^2 =4.618 F (^25) 5.41 kip. F (^) 4.618^ allow 4.618 Ans = τ = = ______ ________________________________________________________________________ 9-23 Given, b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, τallow = 140 MPa.
(^25) 9.65 kip. F (^) 2.592^ allow 2.592 Ans = τ = = ______ ________________________________________________________________________ 9-25 Given, b = 50 mm, d = 50 mm, h = 5 mm, E6010 electrode. A = 0.707(5)(50 +50 + 50) = 530.3 mm^2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa. Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 272(320)−0.995^ = 0. kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa Electrode endurance: E6010, Table 9-3, Sut = 427 MPa Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 272(427)−0.995^ = 0. As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.657(1)(0.59)(1)(0.5)(427) = 82.8 MPa The members and electrode are basically of equal strength. We will useFor a factor of safety of 1, and with K Se = 82.6 MPa. fs = 2.7 (Table 9-5) allow 82.6 530.3^ (^ )^ 16.2 10( 3 )N 16.2 kN. fs^ 2. F A Ans K = τ = = = ______ ________________________________________________________________________ 9-26 Given, b = 2 in, d = 2 in, h = 5/16 in, E6010 electrode. A = 0.707(5/16)(2 +2 + 2) = 1.326 in^2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi. Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 39.9(47)−0.995^ = 0. kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi Electrode endurance: E6010, Table 9-3, Sut = 62 kpsi Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 39.9(62)−0.995^ = 0.
As before, kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.657(1)(0.59)(1)(0.5)(62) = 12.0 kpsi Thus the members and electrode are of equal strength. For a factor of safety of 1, andwith K fs = 2.7 (Table 9-5) allow 12.0 1.326^ (^ )^ 5.89 kip. fs^ 2. F A Ans K = τ = = ______ ________________________________________________________________________ 9-27 Given, b = 50 mm, d = 30 mm, h = 5 mm, E7010 electrode. A = 0.707(5)(50 +50 + 30) = 459.6 mm^2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa. Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 272(320)−0.995^ = 0. kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa Electrode endurance: E6010, Table 9-3, Sut = 482 MPa Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 272(482)−0.995^ = 0. As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.582(1)(0.59)(1)(0.5)(482) = 82.7 MPa The members and electrode are basically of equal strength. We will useFor a factor of safety of 1, and with K Se =82.6 MPa. fs = 2.7 (Table 9-5) allow 82.6 459.6^ (^ )^ 14.1 10( 3 )N 14.1 kN. fs^ 2. F A Ans K = τ = = = ______ ________________________________________________________________________ 9-28 Given, b = 4 in, d = 2 in, h = 5/16 in, E7010 electrode. A = 0.707(5/16)(4 +4 + 2) = 2.209 in^2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi. Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 39.9(47)−0.995^ = 0. kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
J G 1 = 121 ( 0.707) ( 5 /16 ) ( 23 )=0.1473 in^4
J G 2 = JG 3 = 121 ( 0.707 )( 5 / 16) ( 43 )=1.178 in^4
(^3 ) (^12 ) 2 4
1.178 0.707 5 /16 4 1.887 9.220 in
J = (^) i = J (^) i + A ri G i = + + +
tan 1 1.6 28.07o α (^4 ) = −^ ^ = −
r = 1.6 2 + ( 4 − 1 ) 2 =3.4 in
Primary shear ( τ in kpsi, F in kip) :
V F F
τ ′ = (^) A = = Secondary shear :
Mr F F τ ′′ = (^) J = =
max (^ 3.319^ sin 28.07^ o^ )^2 (^ 3.319^ cos 28.07^ o 0.4526 )^2
F F F
F
τ = + +
τmax = τallow ⇒ 3.724 F = 25 ⇒ F = 6.71 kip Ans. ______ ________________________________________________________________________
9-31 l = 30 + 50 + 50 = 130 mm
30 15 50 0 50 25 (^) 13.08 mm 130 30 50 50 25 50 0 (^) 21.15 mm 130
x y
= +^ + =
= +^ + =
M = FR = F (200 − 13.08)
= 186.92 F ( M in N⋅m, F in kN)
r 1 = ( 15 − 13.08 ) 2 + ( 50 − 21.15 ) 2 = 28.92 mm, r 2 = 13.08^2 + ( 25 − 21.15) 2 =13.63 mm
r 3 = ( 25 −13.08 )^2 + 21.15^2 =24.28 mm
J G 1 = 121 ( 0.707 )( ) 5 ( 303 ) =7.954 10( 3 )mm^4
J G 2 = JG 3 = 121 ( 0.707 )( ) 5 ( 50 3 ) =36.82 10( 3 )mm^4
( ) (^ )(^ )( ) ( ) ( ) (^ )^ ( )
(^3 ) (^13 2 3 ) 3 2 3 4
36.82 10 0.707 5 50 24.28 307.3 10 mm
J = (^) i = J (^) i + A ri G i = + + +
tan 1 21.15 29.81o α (^50) 13. = −^ ^ = −
r = 21.15^2 + ( 50 − 13.08 )^2 =42.55 mm
Primary shear ( τ in MPa, F in kN) :
τ ′ = V A^ = (^) 0.707 5^ F 130 =2.176 F Secondary shear :
3 3
Mr F F τ ′′ = (^) J = =
9-33 Weld Pattern Figure of merit Rank______
- (^ )
fom ′ = I lh^ u^ = aah = 0.0833^ ^ ah 6
- (^ )
fom ′ =^ a 2^ ah = 0.0833^ ^ ah 6
- (^ )
fom ′ =^ aa 2^ ah = 0.25^ ^ ah 1 4.* (^ )(^ )
fom ′ =^ a^ 3 ah a^^ +^ a^ = 36 ah^ = 0.1944 ^ ah 2
- & 7. x = 2^ a^ , y = (^) a + a^^22 a = 3 a ( )
I (^) u = 3^ a^ − 2 a (^) 3 a^ + a + 2 a ^3 a^ = a 3 fom (^3 / 3^ )^12 0.1111^2 u (^) 3 9 I a a a lh ah h h
′ = = = ^ ^ = ^
^5
- & 8. (^ )(^ )
fom ′ =^ a^ 4 aha^ +^ a^ = 6 ^ ah^ ^ =0.1667^ ah 3
- (^ )
/ 2^3 2
fom ′ = π π^ a^ ah = 8 ah^ = 0.125^ ^ ah 4 *Note. Because this section is not symmetric with the vertical axis, out-of-planedeflection may occur unless special precautions are taken. See the topic of “shear center”
______in books with more advanced treatments of mechanics of materials. ________________________________________________________________________ 9-34 Attachment and member (1018 HR), Sy = 220 MPa and Sut = 400 MPa. The member and attachment are weak compared to the properties of the lowest electrode. Decision Specify the E6010 electrode Controlling property, Table 9-4: τall = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa For a static load, the parallel and transverse fillets are the same. Let the length of a bead be l = 75 mm, and n be the number of beads.
( 0.707)^ all
F
τ= (^) n hl =τ ( ) ( )( )
3 all
nh F = (^) l τ = = where h is in millimeters. Make a table Number of beads, n Leg size, h (mm) 12 21.4310. 3 7. 4 5.36 → 6 mm Decision Specify h = 6 mm on all four sides. Weldment specification: Pattern: All-around square, four beads each side, 75 mm long Electrode: E
______ ________________________________________________________________________ Leg size:^ h^ = 6 mm 9-35 Decision : Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an optimal pattern (see Prob. 9-32) and have thus reduced a synthesis problem to an analysisproblem: Table 9-1, case 2, rotated 90°: A = 1_._ 414 hd = 1_._ 414( h )(75) = 106.05 h mm^2 Primary shear 12 10 ( 3 ) 113. y 106.
V
τ ′ = (^) A = (^) h = h
Secondary shear :
( ) ( ) ( )
2 2 (^2 23 ) 3 3 4
75[3(75 )^6 75 ]
6 281.3 10^ mm 0.707( )(281.3) 10 198.8 10 mm
J u^ d^ b^ d
J h h
=^ +
With α = 45°,
( 75 )^3 u 6 J =^ b + , J = 0.707 (6) Ju = 0.707( b +75)^3 (2) Primary shear ( τ in MPa, h in mm):
y^ V^ 12 10^ (^3 ) (3) τ ′ = (^) A = A Secondary shear (See Prob. 9-35 solution for the definition of α) :
( )( ) ( ) ( )(^ ) ( ) ( )
3 3 3 3 max^22
cos cos 12 10^150 / 2 (37.5) (4) 0.707 75 sin sin 12 10^150 / 2 (^ / 2) (5) 0.707 75 (6)
x^ y y x y x y
Mr J Mr Mr b J J (^) b Mr Mr b^ b J J (^) b
τ τ τ α α τ τ α α τ τ τ τ
= ′^ + ′′^ + ′′
Enter Eqs. (1) to (6) into a spreadsheet and iterate for various values of b. A portion of the spreadsheet is shown below. b (mm) A (mm^2 ) J (mm^4 ) τ 'y (Mpa) τ "y (Mpa) τ "x (Mpa) (Mpa)^ τmax 41 984.144 1103553.5 12.19334 69.5254 38.00722 90. 42 992.628 1132340.4 12.08912 67.9566 38.05569 88. 43 1001.112 1161623.6 11.98667 66.43718 38.09065 87.18485 < 88 Mpa 44 1009.596 1191407.4 11.88594 64.96518 38.11291 85. We see that b ≥ 43 mm meets the strength goal. Weldment Specifications: Pattern: Horizontal parallel weld tracks 43 mm long, vertical parallel weld tracks 75 mmlong Electrode: E6010Leg size: 6 mm ______ ________________________________________________________________________ 9-37 Materials: Member and attachment (1018 HR): S (^) y = 32 kpsi, Sut =58 kpsi Table 9-4: τ (^) all = min[0.3(58), 0.4(32)] =12.8 kpsi
Decision: Use E6010 electrode. From Table 9-3: S (^) y = 50 kpsi, Sut =62 kpsi, τ (^) all = min[0.3(62), 0.4(50)] =20 kpsi Decision: Since 1018 HR is weaker than the E6010 electrode, use τ (^) all =12.8 kpsi Decision : Use an all-around square weld bead track. l 1 = 6 + a = 6 + 6.25 = 12.25 in Throat area and other properties from Table 9-1: A = 1.414 ( h b + d ) = 1.414( )(6 h + 6) =16.97 h Primary shear
20 10 ( 3 ) 1179 psi
y 16.
V F
τ ′ = (^) A = (^) A = (^) h = h Secondary shear ( )^3 (6 6)^3 288 in 3 u 6 6 J =^ b^ +^ d = + = J = 0.707 (288) h =203.6 h in^4
20 10 ( 3 )(6.25 3)(3) 2726 psi
x y^ Mr^ y τ τ (^) J h h
max x^2 (^ y y )^2 1 27262 (1179^ 2726)^2 4762 psi τ = τ ′′^ + τ ′^ + τ′′ = (^) h + + = h Relate stress to strength
τ max = τall ⇒ 4762 h^ = 12.8 10 ( 3 ) ⇒ h = 12.8 10^4762 ( 3 )=0.372 in
Decision: Specify 3 / 8 in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010Type of weld: Fillet Weld bead length: 24 inLeg size: 3 / 8 in
______ ________________________________________________________________________ Attachment length: 12.25 in
