General Bending (2.6) | Lecture Note - Edubirdie

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

2023
Views:

464

Pages:

2
Author:

Sabrina S.

2.6 General Bending Combined Loading σ = σA + σB = P/A - My/Iz m = fd Unsymmetric Loading My = Mn * sinθ σ = -Myz/Iy + Mzy/Iz Transverse Loading v = P N = Rx = Vx H = |Rx| = VQ/It T = VQ/5M τx = -Vx/2y ΔH = ΨQ/LΔx Transverse Loading Transverse Assumptions shear stress has no influence on normal stress free surfaces have zero shear stress shear stress only varies in the y direction & is constant in z Q = A'ȳ' τ > VA'ȳ'/It careful signd boards bonded together boards bonded together e.g. combined loading eccentric axial load a) find ση & σε b) where is the neutral axis? consider internal forces at section A-B in beam P = 20KN h = 40mm b = 10mm E = 200GPa c = 10mm a = 100mm L = 150mm -N + P = 0 N = P -Mz - Pe = 0 Me = -Pe consider stress distributions in the beam at the cut location σ = P/A σm = 12Pey/bh³ combine these to get σ(y) σ(y) = P/bh + 12Pey/bh³ at A: y = 0 σA = P/bh = 20x10³/10x40x10⁻⁶ = 50x10⁶ Pa = 50 MPa at B: y = h/2 σB = P/bh + 12Pe(h/2)/bh³ = 50x10⁶ + 6Pe/bh² = 125 MPa b) neutral axis is where σ = 0 0 = P/bh + 12Pey/bh³ y = -Pbh/12Pe = -h/12 = -40/12 = -3.33 mm

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