University:
McMaster University
Course:
MATLS 2H04 | Introduction to Materials Engineering
Academic year:

2023
Views:

131

Pages:

3
Author:

Sabrina S.

3.5 Torsion 3.5.1 Torsion of Circular Shafts Torsion loading is important for transmitting power or torque in rotary machinery Typically occurs using circular shafts Assumptions made in this deformation model: The cross-sections of a shaft twisted by an applied torsional moment only rotate without any in-plane or out of plane distortion (i.e. each cross-sectional disk along the shaft maintains its circular shape during torsional loading) φ(x) = angle of twist (angle from the centre line of the shaft between the undeformed & deformed slope of the shaft) varies w z, & is max @ L θ = φ(L) / L = Specific angle of twist γθ = Rθ = shear strain γL = Rθ = K * (φ(L) / L) = shear strain @ outer surface Shear Strain Normal strain describes the change in length of a line element in a body as it deforms ε = lim (L' - L₀) / L₀ L₀→0 Shear strain describes the change in shape of the body as it deforms, measured by the change of angle of 2 line elements which are originally 90° apart Shear Stress Bending ("normal") stress σ = Eε Shear stress τ = Gγ G = shear modulus Normal strain: ε = lim (L' - L₀) / L₀ L₀→0 Shear strain: γ = π/2 - lim θ' L,c→0 dT = τ dA r T = ∫∫_A r τ dA = Gθ ∫∫_A r² dA Polar Moment of Area Ip = πR⁴ / 2 = 2I₂ / 1 T = Gθ ∫₀^R ∫₀^(2π) r³ dα dr T = Gθ 2π ∫₀^R r³ dr = Gθ 2π R⁴ / 4 T = Gθ Ip / 2 τ = T / Ip r θ = T / GIp φ(z) = Tz / GIp Extensions to the fundamental model of a circular shaft under torsional loading Stress Concentrations Inelastic torsion Non-circular cross-sections 3.5.2 Torsion of Thin-Walled Closed Members Thickness, ti, may vary along circumferential dimensions s Smaller segment in ds Take average shear stress & moment Constant thickness Shear Stress dT = hdF = h(τ_avg tds) T = ∫ hτ_avg tds T = τ_avg t ∫ hds T = 2τ_avg t Am τ_avg = T / (2tAm) q = T / (2Am) 4 Stress State Analysis 4.3 Stress State 4.3.1 General Stress States Failure modes brittle - fail due to a critical level of normal stress which cause material planes to separate, which results in cracks to initiate & propagate until the part fails ductile - fail due to the movements of dislocations in the crystal lattice, which are initiated & controlled by shear stresses creates permanent plastic deformations Assume out of plane thickness is a unit length of 1 ΣFx = 0 = σx da1 - σx da1 + τxy da1 - τxy da1 = 0 ΣFy = 0 ΣMz = 0 = τxy da1 da - τyx da1 da τxy = τyx Shear Stress Reciprocity τxy = τyx τxz = τzx τyz = τzy ∴ only have 6 components to worry about σx, σy, σz, τxy, τxz, τyz