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# Lecture 13 torsion in solid and hollow shafts 1

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### Lecture 13 torsion in solid and hollow shafts 1

1. 1. Unit 2- Stresses in BeamsTopics Covered  Lecture -1 – Review of shear force and bending moment diagram  Lecture -2 – Bending stresses in beams  Lecture -3 – Shear stresses in beams  Lecture -4- Deflection in beams  Lecture -5 – Torsion in solid and hollow shafts.
2. 2. TORSIONAL DEFORMATION OF A CIRCULAR SHAFT  Torsion is a moment that twists/deforms a member about its longitudinal axis  By observation, if angle of rotation is small, length of shaft and its radius remain unchanged3
3. 3. Torsional Deformation of Circular Bars  Assumptions   Plane sections remain plane and perpendicular to the torsional axis   Material of the shaft is uniform   Twist along the shaft is uniform.   Axis remains straight and inextensible 4
4. 4. Torsional Deformation L = angle of twist B F F’ F R F’ = shear strain φ is the shear strain, also remember that tanφ = φ,thus : FF Rθ φ= = L L Note that shear strain does not only change with the amount of twist, but also, it varies along the radial direction such that it is zero at the center and increases linearly towards the outer periphery (see next slide) 5€
5. 5. Torsional Deformation τ Cθ q = = R L rShear stress at any point in the shaft is proportional to the distance of the pointfrom the axis of the shaft. €
6. 6. Torque transmitted by shaft(solid) total turning moment due to turning force = total force on the ring x Distance of the ring from the axis r τ = × 2πr 3 dr R Total turning moment (or total torque) is obtained by integratingR the above equation between the limits O and R R R τ T = ∫ 0 dT = ∫ 0 × 2πr 3 dr R τ R 3 τ ⎡ r 4 ⎤ R = × 2π ∫ 0 r dr = × 2π ⎢ ⎥ R R ⎣ 4 ⎦ 0 π =τ × × R3 2 π = τD3 16 €
7. 7. Torque transmitted by shaft(hollow) total turning moment due to turning force = total force on the ring x Distance of the ring from the axis τ = × 2πr 3 dr R0 r Total turning moment (or total torque) is obtained by integratingR the above equation between the limits O and R Ro R0 τ T= ∫ Ri dT = ∫ Ri R0 × 2πr 3 dr τ R0 3 τ ⎡ r 4 ⎤ R 0 = × 2π R ∫ Ri r dr = R0 × 2π ⎢ ⎥ ⎣ 4 ⎦ R i π ⎡ R 0 4 − R i 4 ⎤ = τ × × ⎢ ⎥ 2 ⎣ R 0 ⎦ π ⎡ D0 4 − Di 4 ⎤ = τ ⎢ ⎥ 16 ⎣ D0 ⎦ €
8. 8. Power transmitted by shaft Power transmitted by the shafts N = r.p.m of the shaft T = Mean torque transmitted ω = Angular speed of shaft 2πNT * Power = 60 =ω × T€
9. 9. Torque in terms of polar moment of inertia Moment dT on the circular ring τ τ dT = × 2πr 3 dr = × r 2 × 2πrdr ⇒ (dA = 2πrdr) R R τ r = × r 2 × dA R RR Total Torque = ∫ 0 dT R R τ T= ∫ 0 dT = ∫ 0 R × r 2 dA τ R 2 = ∫ r dA R 0 r 2dA = moment of elemnetary ring about an axis perpendicular to the plane and passing though the center of the circle R 2 ∫ 0 r dA = moment of the circle about an axis perpendicular to the plane and passing though the center of the circle π = Polar moment of inertia = × D4 32 €
10. 10. Torque in terms of polar moment of inertia τ T = ×J R r T τR = J R τ Cθ = R L T τ Cθ C = Modulus of rigidity = = J R L θ = Angle of twist L = Length of the shaft €
11. 11. Polar ModulusPolar modulus is defined as ration of polar moment of inertia to the radiusof the shaft. J Zp = R π 4 For solid shaft => J = D 32 π 4 D π Z p = 32 = D3 D /2 16 π For hollow shaft => J = [ D0 4 − Di 4 ] 32 π [D04 − Di4 ] π 4 4 Z p = 32 = [D0 − Di ] D0 /2 16D0
12. 12. Torsional rigidityTorsional rigidity is also called strength of the shaft. It is defined as product ofmodulus of rigidity (C) and polar moment of inertia =C*J €
13. 13. Shaft in combined bending and Torsion stressesShear stress at any point due to torque Tq T T×r = ⇒q=r J J DShear Stress at a point on the surface of the shaft r = 2 T×r T D 16Tτc = = × = J π 4 2 πD 3 D 32Bending stress at any point due to bending momentM σ M×y = ⇒σ =I y I DBending Stress at a point on the surface of the shaft r = 2 M×y M D 32Mσb = = × = I π 4 2 πD 3 D 64 16T 2τ c 2×tanθ = = πD3 = T σb 32M M 3 πD
14. 14. Shaft in combined bending and Torsion stresses Major principal Stress σb ⎛ σ b ⎞ 2 = + ⎜ ⎟ + τ c 2 2 ⎝ 2 ⎠ 32M ⎛ 32M ⎞ 2 ⎛ 16T ⎞ 2 = 3 + ⎜ ⎟ + ⎜ ⎟ 2 × πD ⎝ 2 × πD 3 ⎠ ⎝ πD 3 ⎠ 16 = πD ( 3 M + M2 + T2 )SOLID SHAFT Minor principal Stress 16 = πD 3 (M − M2 + T2 ) Max shear Stress Max principal Stress - Min principal Stress = 2 16 = πD 3( M2 + T2 )
15. 15. Shaft in combined bending and Torsion stresses Major principal Stress 16D0 = [4 π D0 − Di 4 ] ( M + M2 + T2 ) Minor principal Stress 16D0 ]( )HOLLOW SHAFT = M − M2 + T2 [4 π D0 − Di 4 Max shear Stress 16D0 ]( ) = M2 + T2 π [ D0 − Di 4 4 €
16. 16. Application to a Bar Normal Force: Fn Fn Bending Moment: Mt Mt Shear Force: Ft Ft Torque or Twisting Moment: Mn Mn
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