6 5. Draw the shear and moment diagrams for the beam.

Transcription

1 6 2. Draw the shear and moment diagrams for the shaft. The bearings at and D eert only vertical reaction on the shaft.the loading is applied to the pulleys at and C and E. 14 in. 2 in. 15 in. 12 in. C D E 8 lb 11 lb 35 lb V (lb) M (lb in)

2 6 5. Draw the shear and moment diagrams for the beam. 1 kn 8 kn 15 kn m 2 m 3 m V (kn) 18 8 M (kn m)

3 6 18. Draw the shear and moment diagrams for the beam, and determine the shear and moment throughout the beam as functions of. 2 kip/ft 1 kip 8 kip 4 kip ft Support Reactions: s shown on FD. Shear and Moment Function: For 6 6 ft: + c F y = ; V = 6 ft 4 ft V = {3. - 2} kip a+ M N = ; M a 2 b - 3. = For 6 ft 6 1 ft: M = { } kip # ft + c F y = ; V - 8 = V = 8. kip a+ M N = ; -M - 8(1 - ) - 4 = M = { } kip # ft For 6 6 ft: V = {3. - 2} kip, M = { } kip # ft, For 6 ft 6 1 ft: V = 8. kip, M = { } kip # ft V (kip) M (kip ft) (ft) (ft)

4 6 25. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of, where 4 ft < < 1 ft. 2 lb ft 15 lb/ft 2 lb ft 4 ft 6 ft 4 ft + c F y = ; -15( - 4) - V + 45 = V = a+ M = ; ( - 4) -2-15( - 4) - M + 45( - 4) = 2 M = V = M = V (lb) M (lb ft)

5 6 33. The shaft is supported by a smooth thrust bearing at and smooth journal bearing at. Draw the shear and moment diagrams for the shaft. 4 N m 1 m 1 m 1 m 9 N Equations of Equilibrium: Referring to the free-body diagram of the shaft shown in Fig. a, a+ M = ; + c F y = ; y (2) + 4-9(1) = y = 25 N y = y = 65 N Shear and Moment Diagram: s shown in Figs. b and c. V (N) (m) 25 M (N m) (m) 493

6 6 35. Draw the shear and moment diagrams for the beam and determine the shear and moment as functions of. 2 N/m 4 N/m 3 m 3 m Support Reactions: s shown on FD. Shear and Moment Functions: For 6 3 m: a+ M N = ; M - 2 = For 3 m 6 6 m: + c F y = ; 2-2( - 3) c 2 ( - 3) d( - 3) - V = 3 1 V = e f N Set V =, = m a + c F y = ; 2 - V = V = 2 N + M N = ; M c 2 3 M = 52 6 N # m - 3 ( - 3) d( - 3)a b 3 + 2( - 3)a - 3 b - 2 = 2 1 M = e f N # m Substitute = 3.87 m, M = 691 N # m For 6 3 m: V = 2 N, M = (2) N # m, For 3 m 6 6 m: V = e f N, 1 M = e f N # m V (N) (m) 7 M (N m) (m) 495

7 6 39. Draw the shear and moment diagrams for the double overhanging beam. 4 lb 2 lb/ft 4 lb 3 ft 6 ft 3 ft Equations of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, a+ M = ; + c F y = ; y (6) + 4(3) - 2(6)(3) - 4(9) = y = 1 lb y (6) - 4 = y = 1 lb Shear and Moment Diagram: s shown in Figs. b and c. V (lb) (ft) M (lb ft) (ft)

8 6 49. Determine the maimum tensile and compressive bending stress in the beam if it is subjected to a moment of M = 4 kip # ft..5 in..5 in. 3 in. 3 in. C.5 in. 1 in. M D.5 in. Section Properties: y = y~ =.25(4)(.5) + 2[2(3)(.5)] + 5.5(1)(.5) 4(.5) + 2[(3)(.5)] + 1(.5) = 3.4 in. N = 1 12 (4)(.53 ) + 4(.5)( ) 2 = in 4 + 2c 1 12 (.5)(33 ) +.5(3)(3.4-2) 2 d (.5)(13 ) +.5(1)( ) 2 Maimum ending Stress: pplying the fleure formula s ma = Mc (s t ) ma = 4(13 )(12)( ) (s c ) ma = 4(13 )(12)(3.4) = psi = 3.72 ksi = psi = 1.78 ksi (s t ) ma = 3.72 ksi, (s c ) ma = 1.78 ksi 51

9 6 59. The aluminum machine part is subjected to a moment of M = 75 kn # m. Determine the maimum tensile and compressive bending stresses in the part. 1 mm 1 mm 2 mm 2 mm 1 mm 1 mm.5(.8)(.1) (.4)(.1)4 y =.8(.1) + 2(.4)(.1) = 1 12 (.8)(.13 ) +.8(.1)( ) =.175 m 1 mm 4 mm N C M 75 N m + 2 c 1 12 (.1)(.43 ) +.1(.4)( )d =.3633(1-5 ) m 4 (s ma ) t = Mc = 75( ).3633(1-6 ) = 6.71 MPa (s ma ) c = My = 75(.175) = 3.61 MPa.3633(1-6 ) (s ma ) t = 6.71 MPa, (s ma ) c = 3.61 MPa 52

10 6 61. The beam is subjected to a moment of 15 kip # ft. Determine the percentage of this moment that is resisted by the web D of the beam. 1 in. 5 in. M 15 kip ft 8 in. 1 in. 1 in. D y = y~ = 2.27 in 4 =.5(1)(5) + 5(8)(1) + 9.5(3)(1) 1(5) + 8(1) + 3(1) (3)(13 ) + 3(1)( ) 2 Using fleure formula s = My = in. = 1 12 (5)(13 ) + 5(1)( ) (1)(83 ) + 8(1)( ) 2 3 in. s = s = 15(12)( ) (12)( ) 2.27 = ksi = 4.17 ksi F C = 1 (3.896)(3.4375)(1) = kip 2 F T = 1 (4.17)(4.5625)(1) = kip 2 M = 5.312(2.2917) (3.417) = kip # in. = kip # ft % of moment carried by web = * 1 = 22.6 % 15 % of moment carried by web = 22.6 % 522

11 6 69. Two designs for a beam are to be considered. Determine which one will support a moment of M = 15 kn # m with the least amount of bending stress. What is that stress? 2 mm 15 mm 2 mm 3 mm 3 mm 3 mm 3 mm 15 mm 15 mm 3 mm Section Property: For section (a) For section (b) Maimum ending Stress: pplying the fleure formula For section (a) = 1 12 (.2) (.17)(.3)3 =.21645(1-3 ) m 4 = 1 12 (.2) (.185).33 =.36135(1-3 ) m 4 s ma = Mc (a) (b) s ma = 15(13 )(.165).21645(1-3 ) = MPa For section (b) s min = 15(13 )(.18).36135(1-3 ) = MPa = 74.7 MPa s min = 74.7 MPa 53

12 6 71. The boat has a weight of 23 lb and a center of gravity at G. f it rests on the trailer at the smooth contact and can be considered pinned at, determine the absolute maimum bending stress developed in the main strut of the trailer. Consider the strut to be a bo-beam having the dimensions shown and pinned at C. D 3 ft G 1 ft 1 ft 5 ft 4 ft C 1.75 in. 3 in in. oat: 1.5 in. : + F = ; a+ M = ; + c F y = ; = -N (9) + 23(5) = N = lb y = y = lb ssembly: a+ M C = ; + c F y = ; -N D (1) + 23(9) = N D = 27 lb C y = C y = 23 lb = 1 12 (1.75)(3) (1.5)(1.75)3 = in 4 s ma = Mc = (12)(1.5) = 21.1 ksi s ma = 21.1 ksi 532

13 6 81. f the beam is made of material having an allowable tensile and compressive stress of (s allow ) t = 125 MPa and (s allow ) c = 15 MPa, respectively, determine the maimum allowable internal moment M that can be applied to the beam. M 3 mm 3 mm C 15 mm 15 mm 3 mm Section Properties: The neutral ais passes through centroid C of the cross section as shown in Fig. a. The location of C is y = ~ y =.15(.3)(.3) +.18(.3)(.3).3(.3) +.3(.3) =.975 m Thus, the moment of inertia of the cross section about the neutral ais is = 1 12 (.3)(.33 ) +.3(.3)( ) (.3)(.33 ) +.3(.3)( ) 2 =.197(1-3 ) m 4 llowable ending Stress: The maimum compressive and tensile stress occurs at the top and bottom-most fibers of the cross section. For the top-most fiber, (s allow ) c = Mc 15(1 6 ) = M( ).197(1-3 ) M = N # m = 123 kn # m (controls) For the bottom-most fiber, (s allow ) t = My 125(1 6 ) = M(.975).197(1-3 ) M = N # m = 244 kn # m M = 123 kn # m 542

14 6 97. log that is 2 ft in diameter is to be cut into a rectangular section for use as a simply supported beam. f the allowable bending stress for the wood is s allow = 8 ksi, determine the largest load P that can be supported if the width of the beam is b = 8 in. h b 2 ft P 24 2 = h h = in. 8 ft 8 ft M ma = P (96) = 48 P 2 s allow = M ma c 8(1 3 ) = 48P( ) 1 12 (8)(22.63)3 P = 114 kip P = 114 kip 558