"CESE 656 – Advanced Mechanicsfor Structural EngineersCivil Structural EngineeringUniversity of Alabama at BirminghamBending Non-symmetric BendingBending when loads are not applied in a plane of symmetry2 Bending CESE 656 Non-symmetric Bending? Try applying the bending formula and examine M y? ? = - ? ?? ? = 0? ? ? ? From the bending formula, ? = - ???? ? ? ? ?? ? ? ? = - ? - ? ?? = ? ? ?? = 0? ? ? ??? ? ? ? Equilibrium is only satisfied if ? ? ?? = 0? x3 Bending CESE 656 Principal Axes for an Area? ?? = ? , where I is called the Product of Inertia? xy ? Symmetric sections? I = 0 if x- or y-axis is an axis of symmetry xy ? Other sections? there is a particular pair of axes for which I = 0xy ? These are called the Principal Axes, and the bending formulacan be used for moments applied about these axes4 Bending CESE 656?? ??Locating the Principal Axes? At the Principal Axes I = 0 and I and I take are thexy x y largest and smallest they will be for any axis orentation? Moments of Inertia transform in a manner similar tostresses and strains? The orientation of the principal axis is found using- 2 ? ? ? ? tan(2 ? ) =? ? - ? ? ? ? The Principal Moments of Inertia are2 ? + ? ? - ? ? ? ? ? 2 ? ? = ± + ?? ?? , ? ?? 2 2 5 Bending CESE 656??Using the Bending Formula for PrincipalAxes for Non-Symmetric Bending? For bending about the principal axes (with thecoordinate axes shown)? ? ? ? ? ? ? ? = - +? ? ? ? ? The neutral axis is located where? ? ? ? ? ? ? ? = 0 = - +? ? ? ? ? ? ? si n ? ? ? ? ? ? ? ? ? = ? = ? = tan ? ?? ? ? cos ? ? ? ? ? ? ? ? The slope of the line of the NA w.r.t to the z- axis is ? ? tan ? =? ? ? ? tan ? = tan ?(+a is CW or z toward y)? ? 6 Bending CESE 656 Example – Load not applied along an axis ofsymmetry? A rectangular cross-section has a moment of M = 50 lb·ft applied as shown.a. Determine the stress at each corner of the cross-sectionb. Determine the orientation of the NA w.r.t to the z-axis? The section is symmetric, so y and z are principal axes4? I = 32 in y 4? I = 72 in z ? M = M sin(30°) = 25 lb·ft = 300 lb·iny ? M = M cos(30°) = 43.3 lb·ft = 520 lb·in z 7 Bending CESE 656 a) Stress at each corner? ? ? ? ? ? ? ? = - +? ? ? ? ? Corner A ? y = 3 in., and z = -2 in.520 3 ?? 300 - 2 ? ? ? ? = - +? 4 4 72 ? ? 32 ? ? ? ? = -40.4? ? ?? ? Corner B ? y = 3 in., and z = 2 in.520 3 ?? 300 2 ? ? ? ? = - +? 4 4 72 ? ? 32 ? ? ? ? = -2.92? ? ?? ? Corner C ? y = -3 in., and z = -2 in.520 - 3 ? ? 300 - 2 ? ? ? ? = - +? 4 4 72 ? ? 32 ? ? ? ? = 2.92? ? ?? ? Corner D ? y = -3 in., and z = 2 in.520 3 ?? 300 - 2? ? ? ? = - +? 4 4 72 ? ? 32 ? ? ? ? = 40.4? ? ?? 8 Bending CESE 656 b) Orientation of Neutral Axis? ? ? tan ? = tan ?? ? 4 72 ? ? ? tan ? = tan 30°4 32 ? ? ? ? = 52.4°a9 Bending CESE 656 Example – Non-symmetric cross-sectionusing the bending formula for Principal Axes? The Z-section shown is subjected to a bending moment of M = 20 kN·m directedalong the axis of the horizontal piece of the cross-sectiona. Determine the principal moments of inertia, I and I , and the orientation of thez' y' Principal Axes, ? Pb. Determine the stress at point Pc. Determine the orientation of the NA w.r.t. the z-axisQSR10 Bending CESE 656 "