Engineering science lesson 7
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The document discusses torsion in circular shafts. It covers the assumptions in torsion theory including the determination of shear stress, strain, and modulus. It also describes the distribution of shear stress and angle of twist in solid and hollow circular shafts. Key points include: - Shear stress is highest at the surface and decreases linearly towards the center of a shaft. - Angle of twist is proportional to both the applied torque and length of the shaft. - Cross sections remain plane during twisting for circular shafts but may become distorted for non-circular shafts.
Engineering science lesson 7
- 1. Chapter 1- Static engineering systems 1.3 Torsion in circular shafts 1.3.1 theory of torsion and its assumptions (eg determination of shear stress, shear strain, shear modulus) 1.3.2 distribution of shear stress and angle of twist in solid and hollow circular section shafts 1
- 2. Torsional loads on circular shafts • Interested in stresses and strains of circular shafts subjected to twisting couples or torques • Turbine exerts torque T on the shaft • Shaft transmits the torque to the generator 133 2
- 5. Shear strain • The value of shear stress τ is expressed by d Tr T Tr 16 T τ= τ= J = 24 = πd π d3 J 32 • The shear strain γ is given by where θ=angle of twist or degree of rotation L= length of the specimen
- 6. Torque twist diagram • The elastic properties in torsion can be obtained by using the torque at the proportional limit where the shear stress is calculated corresponding to the twisting moment • The torsional elastic limit or yield strength can be obtained from testing a tubular specimen since the stress gradient are practically eliminated.
- 7. Twisting moment • Consider a cylindrical bar subjected to a torsional moment at one end. • The twisting moment is resisted by shear stresses set up in the cross section of the bar. (zero at centre, max at surface) • Twisting moment is the sum of shear torques over the cross section • Since is the polar moment of inertia of the area with respect to the axis of the bar
- 8. • The maximum shear stress at the surface of the bar is • For a tubular specimen, the shear stress on the outer surface is Where D1=Outside diameter of tube D2=Inside diameter of tube • The modulus of elasticity in shear G or the modulus of rigidity is as follows:
- 9. Problem SOLUTION: • Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0 • Apply a kinematic analysis to relate the angular rotations of the gears • Find the maximum allowable torque Two solid steel shafts are connected on each shaft – choose the smallest by gears. Knowing that for each shaft G = 11.2 x 106 psi and that the • Find the corresponding angle of twist allowable shearing stress is 8 ksi, for each shaft and the net angular determine (a) the largest torque T0 rotation of end A that may be applied to the end of shaft AB, (b) the corresponding angle through which end A of shaft AB rotates. 156 9
- 10. SOLUTION: • Apply a static equilibrium analysis on • Apply a kinematic analysis to relate the two shafts to find a relationship the angular rotations of the gears between TCD and T0 rBφ B = rCφC ∑ M B = 0 = F ( 0.875 in.) − T0 rC 2.45 in. φB = φC = φC ∑ M C = 0 = F ( 2.45 in.) − TCD rB 0.875 in. TCD = 2.8 T0 φ B = 2.8φC 156 10
- 11. • Find the T0 for the maximum • Find the corresponding angle of twist for each allowable torque on each shaft – shaft and the net angular rotation of end A choose the smallest T L φ A / B = AB = ( 561lb ⋅ in.)( 24in.) ( J ABG π ( 0.375 in.) 4 11.2 × 106 psi 2 ) = 0.387 rad = 2.22o TAB c T ( 0.375 in.) τ max = 8000 psi = 0 π ( 0.375 in.) 4 T L 2.8 ( 561lb ⋅ in.)( 24in.) φC / D = CD = ( ) J AB 2 J CDG π ( 0.5 in.) 4 11.2 × 106 psi 2 T0 = 663 lb ⋅ in. TCD c 2.8 T0 ( 0.5 in.) = 0.514 rad = 2.95o ( ) τ max = 8000 psi = π ( 0.5 in.) 4 J CD 2 φ B = 2.8φC = 2.8 2.95o = 8.26o T0 = 561lb ⋅ in. T0 = 561lb ⋅ in φ A = φ B + φ A / B = 8.26o + 2.22o φ A = 10.48o 156 11
- 12. Design of Transmission Shafts • Principal transmission shaft • Determine torque applied to shaft at performance specifications are: specified power and speed, power P = Tω = 2πfT speed P P T= = ω 2πf • Designer must select shaft material and crosssection to • Find shaft crosssection which will not meet performance specifications exceed the maximum allowable without exceeding allowable shearing stress, shearing stress. Ta τ max = J J π 3 T = a = ( solid shafts ) a 2 τ max π J = a2 2 a2 ( a24 − a14 ) = τ T ( hollow shafts ) max 165 12
- 13. Stress concentrations • The derivation of the torsion formula, Ta τ max = J assumed a circular shaft with uniform crosssection loaded through rigid end plates. • The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and crosssection discontinuities can cause stress concentrations • Experimental or numerically determined concentration factors are applied as Ta τ max = K J 167 13
- 14. Stress distribution of a solid shaft T*r T τ= ; T (N-m) τ (N/m ) 2 J J = π d4 / 32 ; (m4) r=d/2 T’ Shear stress Circular shafts subjected to torsion G τmax = T * a / J τ max τa = T * r / J τ=G*γ r γ T
- 15. • Multiplying the previous equation by the shear modulus, r Gγ = Gγ max a From Hooke’s Law, τ = Gγ , so a r r τ = τ max a The shearing stress varies linearly with the J = 1 π a4 2 radial position in the section. a1 a2 r ( J = 1 π a2 − a14 2 4 )
- 16. Angle of twist • From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length. θ ∝T θ ∝L • When subjected to torsion, every crosssection of a circular shaft remains plane and θ undistorted. • Crosssections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric. • Crosssections of noncircular (non axisymmetric) shafts are distorted when subjected to torsion. 136 16