Mechanics of Materials: Stress, Strain, and Torsion
Chapter 1-4 Summary
Chapter 1: Friction and Belts
FBD (Free Body Diagrams) are starters for all questions!
For friction, Fr = μ x N, where three surface conditions exist: smooth dry, rough dry, and lubricated surfaces.
For friction between coaxial discs, let the friction on an infinitesimal element of area be δF = (2π x δr)p x μ. The frictional torque on the element of area is δT = r x δF. For the whole disc:
T = 2πμ∫r1r2 r2 p dr, where p varies with r.
Case 1: If p is constant, then T = (2π/3) μp(r23 – r13).
Case 2: If constant wear of disc is assumed, then p x r = constant, and T = πμ(pr)(r22 – r12).
For flexible belts, force equilibrium is M + T1R = T2R, so T2 = M/R + T1, where T2 > T1.
Consider an infinitesimal element of the belt with unit thickness: Fx = 0, Fy = 0, δFr = μδN. Since δθ is small, cos(δθ/2) ≈ 1, sin(δθ/2) ≈ δθ/2, and δT x δθ/2 ≈ 0 when δθ ≈ 0.
dT = dF = μdN, dN = Tdθ, dT = μTdθ, dT/T = μdθ, ∫T1T2 dT/T = ∫0β μdθ, ln(T2/T1) = μβ or T2/T1 = eμβ.
Note: β (in radians) can be > 2π and T2 > T1.
Chapter 2: Stress and Strain
Normal stress: σ = P/A
Average shear stress: τ = P/A
(Units for both require conversion)
1 MPa = 1 N/mm2 = 1 MN/m2
1 GPa = 1 GN/m2 = 103 MPa
1 lb/in2 = 1 psi
1 ksi = 103 psi
1 ft = 12 in
Actual units: N/mm2 (MPa), N/m2 (Pa), lb/in2 (psi), lb/ft2
Normal strain: ε = ΔL/L0
Shear strain: γ = tan(Θ) = δs/h = Θ (radians)
Volume strain: Δ = ΔV/V0
ΔV = V – V0 = xyz – x0y0z0 = (x0 + Δx)(y0 + Δy)(z0 + Δz) – x0y0z0 (Only associated with normal strains, not shear strains; shear strains cause distortion of body, no volume change)
Elongation = ((lf – l0)/l0) x 100%
Reduction in area = ((A0 – Af)/A0) x 100%
Young’s Modulus: E = σ/ε = tan(θ)
Poisson’s ratio: v = -εy/εx
Example: εx = σx/E
Thermal Strain: Δlx = lx α ΔT, Δly = ly α ΔT, Δlz = lz α ΔT
Thermal strain = Δl/l = α ΔT
Volumetric strain with thermal strain is ΔT = εx + εy + εz = 3α ΔT
The strains in x, y, and z directions are expressed as:
εx = σx/E – v(σy/E + σz/E)
εy = σy/E – v(σz/E + σx/E)
εz = σz/E – v(σx/E + σy/E)
Volumetric strain Δ = εx + εy + εz = (1/E)(σx + σy + σz)(1-2v)
With thermal effect existing also:
εx = σx/E – v(σy/E + σz/E) + α ΔT
εy = σy/E – v(σz/E + σx/E) + α ΔT
εz = σz/E – v(σx/E + σy/E) + α ΔT
Plane stress condition: σz = 0 and εz ≠ 0 (thin plates), σz ≠ 0 and εz = 0 (thick plates)
Hydrostatic stress state: σx = σy = σz = -p
Bulk Modulus: K = -p/Δ = E/(3(1-2v))
Δ = (1/E)(-3p)(1-2v)
Shear Modulus: G = τ/γ
Relationship between shear modulus and Young’s Modulus: G = E/(2(1+v))
Chapter 3: Pressure Vessels and Rotating Rings
1. Thin-walled pressurized sphere for force equilibrium: p πR2 = σ 2πRt, σ = pR/2t (everywhere in the wall in each direction)
2. Close-ended thin-walled pressurized cylinder: pπR2 = σl 2πRt, σl = pR/2t (Longitudinal Stress). Consider force equilibrium in the vertical direction: p 2Rl = 2σθ tl, σθ = pR/t (hoop/circumferential stress)
3. Thin-rotating ring: Hoop stress is assumed to be uniform through wall thickness and radial stress is zero.
Fr = 2σθ A sin(δθ/2), A – Cross-sectional Area. The centrifugal force is Fr = (ρ A R δθ)R ω2, where ρ is material density. Since δθ is small, sin(δθ/2) ≈ δθ/2, so δθ/2 σθ = ρR2 ω2.
P.8 (LAST CASE)
Chapter 4: Torsion
Torsion of Thin-walled cylinder: shear stress is assumed to be uniform in wall thickness. T = τ 2πRt R, τ = T/(2πR2t) -> τ is uniform along circumference. Consider shear strain γ: θ/L = γ/r (Angle of twist). Since τ = G γ then T/(2πR3t) = τ/R = Gθ/L (Torsion of uniform solid circular shaft) -> Torque carried by elemental cylinder is dT = 2πr3 dr Gθ/L. Then the total torque carried by the entire cylinder is Gθ/L ∫0R r2 dA, with ∫r2 dA = J (Defined as Polar second moment of Area) -> T = (Gθ/L) J -> J = π/2 R4.
From eq. 4.5 following expression is obtained τ = Tr/J, τmax = TR/J.
For hollow circular shafts: J = π/2 (Ro4 – Ri4)
For thin-walled cylinders: J = 2πtR3.
FOR A SHAFT with two different diameters is subjected to a torque T
FOR Two concentric shafts jointed rigidly at ends are
