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# Lesson 07, torsion

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Lesson 07, torsion

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### Lesson 07, torsion

1. 1. Lecturer; Dr. Dawood S. Atrushi January 2015 Torsion
2. 2. Chapter 3 Torsion of a structural member, y couples that produce gitudinal axis T2 = P2 d2 Introduction Torsion: twisting of a structural member, when it is loaded by couples that produce rotation about its longitudinal axis. January, 2015Torsion - DAT2 T1 = P1 d1 T2 = P2 d2
3. 3. The couples T1, T2 are called; ¢  Torques, ¢  Twisting couples or ¢  Twisting moments. Unit of T is: N-m, lb-ft January, 2015Torsion - DAT3 3.1 Introduction Torsion : twisting of a structural member, when it is loaded by couples that produce otation about its longitudinal axis T1 = P1 d1 T2 = P2 d2 the couples T1, T2 are called orques, twisting couples or twisting moments unit of T : N-m, lb-ft in this chapter, we will develop formulas or the stresses and deformations produced n circular bars subjected to torsion, such as drive shafts, thin-walled members
4. 4. Torsional Deformation of a Circular Bar Consider a bar or shaft of circular cross section twisted by a couple T, assume the left-hand end is fixed and the right- hand end will rotate a small angle Φ, called angle of twist. January, 2015Torsion - DAT4
5. 5. If every cross section has the same radius and subjected to the same torque, the angle Φ(x) will vary linearly between ends. January, 2015Torsion - DAT5 It is assumed 1.  plane section remains plane 2.  radii remaining straight and the cross sections remaining plane and circular 3.  if Φ is small, neither the length L nor its radius will change
6. 6. Consider an element of the bar dx, on its outer surface we choose an small element abcd, January, 2015Torsion - DAT6 3. if is small, neither the length L nor its radius will cha consider an element of the bar dx, on its outer surface we c ll element abcd,
7. 7. January, 2015Torsion - DAT7 During twisting the element rotate a small angle dΦ, the element is in a state of pure shear, and deformed into ab'c’d. during twisting the element rotate
8. 8. Its shear strain ϒmax is; January, 2015Torsion - DAT8 dΦ/dx represents the rate of change of the angle of twist Φ, denote θ= dΦ/dx as the angle of twist per unit length or the rate of twist, then ! ϒmax = r θ 2 during twisting the element rotate a small angle d a state of pure shear, and deformed into ab'c'd, its she b b' r d max = CC = CC a b dx
9. 9. January, 2015Torsion - DAT9 In general, Φ and θ are function of x, in the special case of pure torsion, θ is constant along the length (every cross section is subjected to the same torque) max = r in general, and are function of x, in the pure torsion, is constant along the length (every subjected to the same torque) r = C then max = CC L L and the shear strain inside the bar can be obtained = = C max r for a circular tube, it can be obtained max = r in general, and are function of x, in the pure torsion, is constant along the length (every subjected to the same torque) r = C then max = CC L L and the shear strain inside the bar can be obtained = = C max r for a circular tube, it can be obtained And the shear strain inside the bar can be obtained in general, and are functio pure torsion, is constant along th subjected to the same torque) = C then ma L and the shear strain inside the bar can = = C max r for a circular tube, it can be obtained
10. 10. January, 2015Torsion - DAT10 For a circular tube, it can be obtained: L side the bar can be obtained C max r an be obtained s are based only upon geometric concepts, they are and the shear strain inside the bar can be obta = = C max r for a circular tube, it can be obtained r1 min = C max r2 the above relationships are based only upon g
11. 11. Circular Bars of Linearly Elastic Materials Shear stress τ in the bar of a linear elastic material is January, 2015Torsion - DAT11 nships are based only upon geometric concepts, they are ar of any material, elastic or inelastic, linear or nonlinear f Linearly Elastic Materials in the bar of a l is s of elasticity valid for a circular bar of any material, elastic o 3.3 Circular Bars of Linearly Elastic Materia shear stress in the bar of a linear elastic material is = G G : shear modulus of elasticity G : shear modulus of elasticity
12. 12. January, 2015Torsion - DAT12 With the geometric relation of the shear strain, it is obtainedwith the geometric relation of the shear strain, it is obtain max = G r = G = C max r and in circular bar vary linear with the radial dist the center, the maximum values max and max occur at the out the shear stress acting on the plane of the cross section are accompanied by shear stresses of the same magnitude acting on on of the shear strain, it is obtained C max r with the geometric relation of the shear strain, it is o max = G r = G = C max r and in circular bar vary linear with the radial the center, the maximum values max and max occur at the the shear stress acting on the plane of the with the geometric relation of the shear strain, it is max = G r = G = C max r and in circular bar vary linear with the radia the center, the maximum values max and max occur at th the shear stress acting on the plane of the
13. 13. January, 2015Torsion - DAT13 ¢  τ and ϒ in circular bar vary linear with the radial distance ρ from the center, the maximum values τmax and ϒmax occur at the outer surface. ¢  The shear stress acting on the plane of the cross section are accompanied by shear stresses of the same magnitude acting on longitudinal plane of the bar = G = C max r and in circular bar vary linear with the radial distance fr ter, the maximum values max and max occur at the outer surface shear stress acting on the plane of the section are accompanied by shear s of the same magnitude acting on dinal plane of the bar he material is weaker in shear on dinal plane than on cross-sectional nes, as in the case of a circular bar made of wood, the first crack
14. 14. January, 2015Torsion - DAT14 ¢  If the material is weaker in shear on longitudinal plane than on cross- sectional planes, as in the case of a circular bar made of wood, the first crack due to twisting will appear on the surface in longitudinal direction a rectangular element with sides at 45 o to the axis of the shaft will be subjected to tensile and compressive stresses. ccompanied by shear magnitude acting on the bar s weaker in shear on an on cross-sectional ase of a circular bar made of wood, the first crack due r on the surface in longitudinal direction ment with sides at 45 o to t will be subjected to ve stresses
15. 15. January, 2015Torsion - DAT Consider a bar subjected to pure torsion, the shear force acting on an element dA is τdA, the moment of this force about the axis of barisτρ dA: The Torsion Formula planes, as in the case of a circular bar made of wood, the first crack due wisting will appear on the surface in longitudinal direction a rectangular element with sides at 45 o to axis of the shaft will be subjected to sile and compressive stresses e Torsion Formula consider a bar subjected to pure torsion, shear force acting on an element dA dA, the moment of this force about axis of bar is dA dM = dA dM = τρ dA
16. 16. January, 2015Torsion - DAT16 Equation of moment equilibriumequation of moment equilibrium T = dM = dA = G 2 dA = G A A A = G Ip [ = G ] in which Ip = 2 dA is the polar moment of iner A r4 d4 Ip = CC = CC for circular cross section 2 32 the above relation can be written T = CC A = G 2 dA = G 2 dA A A G ] s the polar moment of inertia for circular cross section equation of moment equilibrium T = dM = dA = G 2 dA = G A A A A = G Ip [ = G ] in which Ip = 2 dA is the polar moment of inertia A r4 d4 Ip = CC = CC for circular cross section 2 32 the above relation can be written T = CC equation of moment equilibrium T = dM = dA = G 2 dA = G A A A A = G Ip [ = G ] in which Ip = 2 dA is the polar moment of inertia A r4 d4 Ip = CC = CC for circular cross section 2 32 the above relation can be written T equation of moment equilibrium T = dM = dA = G 2 dA = G 2 A A A A = G Ip [ = G ] in which Ip = 2 dA is the polar moment of inertia A r4 d4 Ip = CC = CC for circular cross section 2 32 the above relation can be written T
17. 17. January, 2015Torsion - DAT17 The above relation can be written GIp: torsional rigidity the above relation can be written T = CC G Ip G Ip : torsional rigidity the angle of twist can be expressed as T L = L = CC is measured in r G Ip L torsional flexibility f = CC The angle of twist Φ can be expressed as T = CC G Ip G Ip : torsional rigidity the angle of twist can be expressed as T L = L = CC is measured in radians G Ip L torsional flexibility f = CC G Ip G Ip torsional stiffness k = CC L and the shear stress is Torsional flexibility; G Ip G Ip : torsional rigidity the angle of twist can be expressed as T L = L = CC is measured G Ip L torsional flexibility f = CC G Ip G Ip torsional stiffness k = CC L and the shear stress is Torsional stiffness; G Ip G Ip : torsional rigidity the angle of twist can be expressed as T L = L = CC is measur G Ip L torsional flexibility f = CC G Ip G Ip torsional stiffness k = CC L and the shear stress is
18. 18. January, 2015Torsion - DAT18 The shear stress is; The maximum shear stress τmax at ρ = r is; 5 torsional stiffness k = CC L and the shear stress is T T = G = G CC = CC G Ip Ip the maximum shear stress max at = r is T r 16 T max = CC = CC Ip d3 for a circular tube
19. 19. T r 16 T = CC = CC Ip d3 ular tube = (r2 4 - r1 4 ) / 2 = (d2 4 - d1 4 ) / 32 ow tube is very thin (r2 2 + r1 2 ) (r2 + r1) (r2 - r1) / 2 (2r2 ) (2r) (t) = 2 r3 t = d3 t / 4 January, 2015Torsion - DAT19 For a circular tube If the hollow tube is very thin for a circular tube Ip = (r2 4 - r1 4 ) / 2 = (d2 4 - d1 4 ) / 32 if the hollow tube is very thin Ip (r2 2 + r1 2 ) (r2 + r1) (r2 - r1) / 2 = (2r2 ) (2r) (t) = 2 r3 t = d3 t limitations 1. bar have circular cross section (either solid or holl p for a circular tube Ip = (r2 4 - r1 4 ) / 2 = (d2 4 - d1 4 ) / 32 if the hollow tube is very thin Ip (r2 2 + r1 2 ) (r2 + r1) (r2 - r1) / 2 = (2r2 ) (2r) (t) = 2 r3 t = d3 t / 4 limitations 1. bar have circular cross section (either solid or hollow) 2. material is linear elastic for a circular tube Ip = (r2 4 - r1 4 ) / 2 = (d2 4 - d1 4 ) / 32 if the hollow tube is very thin Ip (r2 2 + r1 2 ) (r2 + r1) (r2 - r1) / 2 = (2r2 ) (2r) (t) = 2 r3 t = d3 t / 4 limitations 1. bar have circular cross section (either solid or hollow 2. material is linear elastic = CC = CC Ip d3 r tube (r2 4 - r1 4 ) / 2 = (d2 4 - d1 4 ) / 32 w tube is very thin (r2 2 + r1 2 ) (r2 + r1) (r2 - r1) / 2 (2r2 ) (2r) (t) = 2 r3 t = d3 t / 4 circular cross section (either solid or hollow) 16 T CC d3 = (d2 4 - d1 4 ) / 32 thin (r2 + r1) (r2 - r1) / 2 ) = 2 r3 t = d3 t / 4
20. 20. January, 2015Torsion - DAT20 Limitations 1.  bar have circular cross section (either solid or hollow) 2.  material is linear elastic 5 and the shear stress is T T = G = G CC = CC G Ip Ip the maximum shear stress max at = r is The shear stress; = G = the maximum shear stres Note that the above equations cannot be used for bars of noncircular shapes, because their cross sections do not remain plane and their maximum stresses are not located at the farthest distances from the midpoint.
21. 21. ions do not remain plane and their maximum rthest distances from the midpoint section G = 80 GPa = ? 5o , T = ? 16 x 340 N-M Example 1 A solid bar of circular cross section d = 40 mm, L = 1.3 m, and G = 80 GPa. a)  for T = 340 N-m, find τmax and Φ. b)  for τallowable = 42 MPa and Φallowable = 2.5o, find T. January, 2015Torsion - DAT21 1.3 m 40 mm
22. 22. Solution January, 2015Torsion - DAT22 6 a solid bar of circular cross section d = 40 mm, L = 1.3 m, G = 80 GPa (a) T = 340 N-m, max, = ? (b) all = 42 MPa, all = 2.5o , T = ? (a) 16 T 16 x 340 N-M max = CC = CCCCCCC = 27.1 d3 (0.04 m)3 Ip = d4 / 32 = 2.51 x 10-7 m4 T L 340 N-m x 1.3 m = CC = CCCCCCCCCC = 0. G Ip 80 GPa x 2.51 x 10-7 m4 a) The maximum shear stress τmax ; 6 r cross section 3 m, G = 80 GPa max, = ? all = 2.5o , T = ? 16 x 340 N-M = CCCCCCC = 27.1 MPa (0.04 m)3 2 = 2.51 x 10-7 m4 340 N-m x 1.3 m CCCCCCCCCC = 0.02198 rad = 1.26o 80 GPa x 2.51 x 10-7 m4 The angle of twist Φ; 6 Example 3-1 a solid bar of circular cross section d = 40 mm, L = 1.3 m, G = 80 GPa (a) T = 340 N-m, max, = ? (b) all = 42 MPa, all = 2.5o , T = ? (a) 16 T 16 x 340 N-M max = CC = CCCCCCC = 27.1 d3 (0.04 m)3 Ip = d4 / 32 = 2.51 x 10-7 m4 T L 340 N-m x 1.3 m = CC = CCCCCCCCCC = 0 G Ip 80 GPa x 2.51 x 10-7 m4 arthest distances from the midpoint section G = 80 GPa = ? 5o , T = ? 16 x 340 N-M CCCCCCC = 27.1 MPa (0.04 m)3 2.51 x 10-7 m4 340 N-m x 1.3 m CCCCCCCC = 0.02198 rad = 1.26o GPa x 2.51 x 10-7 m4
23. 23. January, 2015Torsion - DAT23 b) Due to τallowable = 42 MPa (b) due to all = 42 MPa T1 = d3 all / 16 = (0.04 m)3 x 42 MPa / 16 = 528 due to all = 2.5o = 2.5 x rad / 180o = 0.0 T2 = G Ip all / L = 80 GPa x 2.51 x 10-7 m4 x = 674 N-m thus Tall = min [T1, T2] = 528 N-m Example 3-2 a steel shaft of either solid bar or circular tube b) Due to Φallowable = 2.5o = 2.5 x PI/180 = 0.04363 rad; (b) due to all = 42 MPa T1 = d3 all / 16 = (0.04 m)3 x 42 MPa / 16 = 528 N-m due to all = 2.5o = 2.5 x rad / 180o = 0.0436 T2 = G Ip all / L = 80 GPa x 2.51 x 10-7 m4 x 0.0 = 674 N-m thus Tall = min [T1, T2] = 528 N-m Example 3-2 a steel shaft of either solid bar or circular tube e to all = 42 MPa T1 = d3 all / 16 = (0.04 m)3 x 42 MPa / 16 = 528 N-m ue to all = 2.5o = 2.5 x rad / 180o = 0.04363 rad T2 = G Ip all / L = 80 GPa x 2.51 x 10-7 m4 x 0.04363 / 1.3 m = 674 N-m hus Tall = min [T1, T2] = 528 N-m 3-2 (b) due to all = 42 MPa T1 = d3 all / 16 = (0.04 m)3 x 42 MPa / 16 = 528 N-m due to all = 2.5o = 2.5 x rad / 180o = 0.0436 T2 = G Ip all / L = 80 GPa x 2.51 x 10-7 m4 x 0.0 = 674 N-m thus Tall = min [T1, T2] = 528 N-m Example 3-2 m)3 x 42 MPa / 16 = 528 N-m x rad / 180o = 0.04363 rad GPa x 2.51 x 10-7 m4 x 0.04363 / 1.3 m = 528 N-mThus;
24. 24. Example 2 A steel shaft of either solid bar or circular tube is exposed to T = 1200 N-m. τall = 40 MPa, θ=0.75o/m, G= 78 Gpa. January, 2015Torsion - DAT24 p all / L = 80 GPa x 2.51 x 10-7 m4 x 0.04363 / 1.3 m N-m = min [T1, T2] = 528 N-m ther solid bar or circular tube m, all = 40 MPa m G = 78 GPa 0 of the solid bar a)  Determine do of the solid bar b)  For the hollow shaft t=d2/10, determine d2 c)  Determine d2/do, Whollow/Wsolid
25. 25. Solution January, 2015Torsion - DAT25 a) For the solid shaft, due to τmax = 40 MPa; (a) determine d0 of the solid bar (b) for the hollow shaft, t = d2 / 10, determine d2 (c) determine d2 / d0, Whollow / Wsolid (a) for the solid shaft, due to all = 40 MPa d0 3 = 16 T / all = 16 x 1200 / 40 = 152.8 d0 = 0.0535 m = 53.5 mm due to all = 0.75o / m = 0.75 x rad / 180o / m = 0 Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 117 d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 1197 d0 = 0.0588 m = 58.8 mm thus, we choose d0 = 58.8 mm [in practical design, d olid bar d2 / 10, determine d2 hollow / Wsolid o all = 40 MPa l = 16 x 1200 / 40 = 152.8 x 10-6 m3 = 53.5 mm m = 0.75 x rad / 180o / m = 0.01309 rad / m 1200 / 78 x 109 x 0.01309 = 117.5 x 10-8 m4 = 32 x 117.5 x 10-8 / = 1197 x 10-8 m4 = 58.8 mm 58.8 mm [in practical design, d0 = 60 mm] all (a) determine d0 of the solid bar (b) for the hollow shaft, t = d2 / 10, determine d2 (c) determine d2 / d0, Whollow / Wsolid (a) for the solid shaft, due to all = 40 MPa d0 3 = 16 T / all = 16 x 1200 / 40 = 152. d0 = 0.0535 m = 53.5 mm due to all = 0.75o / m = 0.75 x rad / 180o / m = 0 Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 11 d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 119 d0 = 0.0588 m = 58.8 mm thus, we choose d0 = 58.8 mm [in practical design, all = 0.75o / m G = 78 GPa (a) determine d0 of the solid bar (b) for the hollow shaft, t = d2 / 10, determine d2 (c) determine d2 / d0, Whollow / Wsolid (a) for the solid shaft, due to all = 40 MPa d0 3 = 16 T / all = 16 x 1200 / 40 = 152.8 d0 = 0.0535 m = 53.5 mm due to all = 0.75o / m = 0.75 x rad / 180o / m = 0 Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 11 d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 119 d0 = 0.0588 m = 58.8 mm thus, we choose d0 = 58.8 mm [in practical design, 40 MPa = 78 GPa olid bar d2 / 10, determine d2 hollow / Wsolid o all = 40 MPa l = 16 x 1200 / 40 = 152.8 x 10-6 m3 = 53.5 mm m = 0.75 x rad / 180o / m = 0.01309 rad / m 1200 / 78 x 109 x 0.01309 = 117.5 x 10-8 m4 = 32 x 117.5 x 10-8 / = 1197 x 10-8 m4 = 58.8 mm T = 1200 N-m, all = 40 MPa all = 0.75o / m G = 78 GPa (a) determine d0 of the solid bar (b) for the hollow shaft, t = d2 / 10, determine d2 (c) determine d2 / d0, Whollow / Wsolid (a) for the solid shaft, due to all = 40 MPa d0 3 = 16 T / all = 16 x 1200 / 40 = 15 d0 = 0.0535 m = 53.5 mm due to all = 0.75o / m = 0.75 x rad / 180o / m = Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 1 d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 11 d0 = 0.0588 m = 58.8 mm = 40 MPa = 78 GPa e solid bar = d2 / 10, determine d2 Whollow / Wsolid to all = 40 MPa all = 16 x 1200 / 40 = 152.8 x 10-6 m3 m = 53.5 mm / m = 0.75 x rad / 180o / m = 0.01309 rad / m = 1200 / 78 x 109 x 0.01309 = 117.5 x 10-8 m4 = 32 x 117.5 x 10-8 / = 1197 x 10-8 m4
26. 26. January, 2015Torsion - DAT26 7 due to all = 0.75o / m = 0.75 x rad / 180o / m = 0 Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 117 d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 1197 d0 = 0.0588 m = 58.8 mm thus, we choose d0 = 58.8 mm [in practical design, d (b) for the hollow shaft d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8 d (a) for the solid shaft, due to all = 40 MPa d0 3 = 16 T / all = 16 x 1200 / 40 = 152.8 x d0 = 0.0535 m = 53.5 mm due to all = 0.75o / m = 0.75 x rad / 180o / m = 0.0 Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 117.5 d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 1197 x d0 = 0.0588 m = 58.8 mm thus, we choose d0 = 58.8 mm [in practical design, d0 (b) for the hollow shaft d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8 d2 7 d0 = 0.0535 m = 53.5 mm due to all = 0.75o / m = 0.75 x rad / 180o / m = 0 Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 117 d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 1197 d0 = 0.0588 m = 58.8 mm thus, we choose d0 = 58.8 mm [in practical design, (b) for the hollow shaft d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8 7 d0 = 0.0535 m = 53.5 mm due to all = 0.75o / m = 0.75 x rad / 180o / m = Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 11 d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 119 d0 = 0.0588 m = 58.8 mm thus, we choose d0 = 58.8 mm [in practical design (b) for the hollow shaft d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8 7 / all = 16 x 1200 / 40 = 152.8 x 10 m 535 m = 53.5 mm .75o / m = 0.75 x rad / 180o / m = 0.01309 rad / m all = 1200 / 78 x 109 x 0.01309 = 117.5 x 10-8 m4 p / = 32 x 117.5 x 10-8 / = 1197 x 10-8 m4 588 m = 58.8 mm d0 = 58.8 mm [in practical design, d0 = 60 mm] aft - 2 t = d2 - 0.2 d2 = 0.8 d2 b) For the hollow shaft 7 d0 3 = 16 T / all = 16 x 1200 / 40 = 152.8 x 10 d0 = 0.0535 m = 53.5 mm due to all = 0.75o / m = 0.75 x rad / 180o / m = 0.0130 Ip = T / G all = 1200 / 78 x 109 x 0.01309 = 117.5 x d0 4 = 32 Ip / = 32 x 117.5 x 10-8 / = 1197 x 10 d0 = 0.0588 m = 58.8 mm thus, we choose d0 = 58.8 mm [in practical design, d0 = 6 (b) for the hollow shaft d1 = d2 - 2 t = d2 - 0.2 d2 = 0.8 d2 Ip = (d2 4 - d1 4 ) / 32 = [d2 4 - (0.8d2)4 ] / 32 = 0 due to all = 40 MPa Ip = 0.05796 d2 4 = T r / all = 1200 (d2/2) / 4 4 - (0.8d2)4 ] / 32 = 0.05796 d2 4
27. 27. January, 2015Torsion - DAT27 Ip = (d2 4 - d1 4 ) / 32 = [d2 4 - (0.8d2)4 ] / 32 = 0.05796 d2 4 due to all = 40 MPa Ip = 0.05796 d2 4 = T r / all = 1200 (d2/2) / 40 d2 3 = 258.8 x 10-6 m3 d2 = 0.0637 m = 63.7 mm due to all = 0.75o / m = 0.01309 rad / m all = 0.01309 = T / G Ip = 1200 / 78 x 109 x 0.05796 d2 4 d2 4 = 2028 x 10-8 m4 d2 = 0.0671 m = 67.1 mm thus, we choose d0 = 67.1 mm [in practical design, d0 = 70 mm] (c) the ratios of hollow and solid bar are
28. 28. January, 2015Torsion - DAT28 c) The ratios of hollow and solid bar are thus, we choose d0 = 67.1 mm [in practical design, (c) the ratios of hollow and solid bar are d2 / d0 = 67.1 / 58.8 = 1.14 Whollow Ahollow (d2 2 - d1 2 )/4 CCC = CCC = CCCCCC = 0 Wsolid Asolid d0 2 /4 the hollow shaft has 14% greater in diameter but 53% le Example 3-3 a hollow shaft and a solid shaft has same material, same length, same outer radius R, and r = 0.6 R for the hollow shaft d2 = 0.0671 m = 67.1 mm thus, we choose d0 = 67.1 mm [in practical design, d (c) the ratios of hollow and solid bar are d2 / d0 = 67.1 / 58.8 = 1.14 Whollow Ahollow (d2 2 - d1 2 )/4 CCC = CCC = CCCCCC = 0.4 Wsolid Asolid d0 2 /4 the hollow shaft has 14% greater in diameter but 53% les Example 3-3 a hollow shaft and a solid shaft has same material, same length, same outer radius R, 10-8 m4 = 67.1 mm 67.1 mm [in practical design, d0 = 70 mm] solid bar are / 58.8 = 1.14 ollow (d2 2 - d1 2 )/4 CC = CCCCCC = 0.47 olid d0 2 /4 % greater in diameter but 53% less in weightThe hollow shaft has a 14% greater diameter but 53% less weight.
29. 29. max R / L x R / L T1 Non-uniform Torsion 1)  Constant torque through each segment January, 2015Torsion - DAT29 SH is 36% greater than SS 3.4 Nonuniform Torsion (1) constant torque through each segment TCD = - T1 - T2 + T3 TBC = - T1 - T2 TAB = - T1 n n Ti Li = i = CC i=1 i=1 Gi Ipi (2) constant torque with continuously varying cross section SH is 36% greater than S 3.4 Nonuniform Torsion (1) constant torque through each segm TCD = - T1 - T2 + T3 TBC = - T1 - T2 TAB = n n Ti Li = i = CC i=1 i=1 Gi Ipi (2) constant torque with continuousl varying cross section SS = TS / WS = 0.5 max R / L SH is 36% greater than SS 3.4 Nonuniform Torsion (1) constant torque through each segment TCD = - T1 - T2 + T3 TBC = - T1 - T2 TAB = - T1 n n Ti Li = i = CC i=1 i=1 Gi Ipi (2) constant torque with continuously varying cross section
30. 30. 2)  Constant torque with continuously varying cross section January, 2015Torsion - DAT30 9 2 AB 1 n Ti Li CC i=1 Gi Ipi ith continuously ion T dx d = CCC G Ip(x) L L T dx = d = CCC 0 0 G Ip(x) (3) continuously varying cross section and T dx d = CCC G Ip(x) L L T dx = d = CCC 0 0 G Ip(x) (3) continuously varying cross section and
31. 31. 3)  Continuously varying cross section and continuously varying torque January, 2015Torsion - DAT31 ) L T dx = CCC 0 G Ip(x) ying cross section and ing torque L T(x) dx = CCC 0 G Ip(x) CDE, d = 30 mm T dx d = CCC G Ip(x) L L T dx = d = CCC 0 0 G Ip(x) (3) continuously varying cross section and continuously varying torque L L T(x) dx = d = CCC 0 0 G Ip(x) T dx d = CCC G Ip(x) L L T dx = d = CCC 0 0 G Ip(x) (3) continuously varying cross section and continuously varying torque L L T(x) dx = d = CCC 0 0 G Ip(x)
32. 32. Example 3 In a solid shaft ABCDE; d = 30 mm; T1 = 275 N-m T2 = 175 N-m T3 = 80 N-m G = 80 GPa L1 = 500 mm Determine τmax in each pat and ΦBD. January, 2015Torsion - DAT32 L T(x) dx d = CCC 0 G Ip(x) ft ABCDE, d = 30 mm m T2 = 450 N-m m G = 80 GPa m L2 = 400 mm x in each part and BD T2 - T1 = 175 N-m
33. 33. Solution January, 2015Torsion - DAT33 T3 = 175 N-m G = 80 GPa L1 = 500 mm L2 = 400 mm determine max in each part and BD TCD = T2 - T1 = 175 N-m TBC = - T1 = - 275 N-m 16 TBC 16 x 275 x 103 BC = CCC = CCCCCC = 51.9 MPa d3 303 16 TCD 16 x 175 x 103 CD = CCC = CCCCCC = 33 MPa d3 303 BD = BC + CD 4 4 T3 = 175 N-m G = 80 GPa L1 = 500 mm L2 = 400 mm determine max in each part and BD TCD = T2 - T1 = 175 N-m TBC = - T1 = - 275 N-m 16 TBC 16 x 275 x 103 BC = CCC = CCCCCC = 51.9 MPa d3 303 16 TCD 16 x 175 x 103 CD = CCC = CCCCCC = 33 MPa d3 303 BD = BC + CD d4 304 T3 = 175 N-m G = 80 GPa L1 = 500 mm L2 = 400 mm determine max in each part and BD TCD = T2 - T1 = 175 N-m TBC = - T1 = - 275 N-m 16 TBC 16 x 275 x 103 BC = CCC = CCCCCC = 51.9 MPa d3 303 16 TCD 16 x 175 x 103 CD = CCC = CCCCCC = 33 MPa d3 303 BD = BC + CD 4 4 T3 = 175 N-m G = 80 GPa L1 = 500 mm L2 = 400 mm determine max in each part and BD TCD = T2 - T1 = 175 N-m TBC = - T1 = - 275 N-m 16 TBC 16 x 275 x 103 BC = CCC = CCCCCC = 51.9 MPa d3 303 16 TCD 16 x 175 x 103 CD = CCC = CCCCCC = 33 MPa d3 303 BD = BC + CD d4 304
34. 34. January, 2015Torsion - DAT34 10 CD = CCC = CCCCCC = 33 MPa d3 303 BD = BC + CD d4 304 Ip = CC = CCC = 79,520 mm2 32 32 10 CD CD = CCC = CCCCCC = 33 MPa d3 303 BD = BC + CD d4 304 Ip = CC = CCC = 79,520 mm2 32 32 TBC L1 - 275 x 103 x 500 BC = CCC = CCCCCCCC = - 0.0216 rad G Ip 80 x 103 x 79,520 TCD L2 175 x 103 x 400 CD = CCC = CCCCCCCC = 0.011 rad G Ip 80 x 103 x 79,520 BD = BC + CD = - 0.0216 + 0.011 = - 0.0106 rad = - 0.61o ample 3-5 TBC L1 - 275 x 103 x 500 BC = CCC = CCCCCCCC = - 0.0216 rad G Ip 80 x 103 x 79,520 TCD L2 175 x 103 x 400 CD = CCC = CCCCCCCC = 0.011 rad G Ip 80 x 103 x 79,520 BD = BC + CD = - 0.0216 + 0.011 = - 0.0106 rad = - 0.61o ample 3-5 TBC L1 - 275 x 103 x 500 BC = CCC = CCCCCCCC = - 0.0216 rad G Ip 80 x 103 x 79,520 TCD L2 175 x 103 x 400 CD = CCC = CCCCCCCC = 0.011 rad G Ip 80 x 103 x 79,520 BD = BC + CD = - 0.0216 + 0.011 = - 0.0106 rad = - 0.61o ample 3-5