Bending of Beams with Unsmmetrical Sections Assume that CZ is a neutral ais. C = centroid of section Hence, if > 0, da has negative stress. From the diagram below, we have:
δ = α and s = αρ and δ ε = = s ρ E σ = = κ E ρ if is the onl load, we have: σ da = 0 κ E da = 0 or da = 0 hence the neutral ais passes through the centroid C. A similar result holds for and the Y ais. oment equilibrium about the Z ais: and about the Y ais we have: σ da = 2 κ E da = = κ E σ da = κ E da = = κ E and, if CZ is a neutral ais, we will have moments such that: = Similarl, if CY is a neutral ais, we will have moments such that: =
However, if CZ is a principal ais, = 0. Therefore, if CZ is also the neutral ais, we have = 0, i.e. bending takes place in the XY plane just as for smmetrical bending. Therefore the plane of the bending moment is perpendicular to the neutral surface onl if the Y and Z aes are principal aes. Hence, we can tackle bending of beams of non smmetric cross section b: (1) finding the principal aes of the section (2) resolving moment into components in the principal ais directions (3) calculating stresses and deflections in each direction (4) superimpose stresses and deflections to get the final result Let Y and Z be the principal aes and let be the bending moment vector. Resolving into components with respect to the principal aes we get: ' = sinθ ' = cosθ f, are the principal moments of inertia σ = ' ' sin σ = ' ' ' ' ( θ ) ' cos( θ ) ' ' '
For the neutral ais, σ = 0 b definition, hence as the point (, ) lies on the neutral ais in this case, we have the neutral ais at angle β with respect to the principal ais CZ and hence ' tan β = tanθ ' Note that β 0 in general. The above method is most useful when the principal aes are known or can be found easil b calculation or inspection. The method is also useful for finding deflections (see below). t is also possible to calculate stresses with respect to a set of non principal aes. σ = ( + ) ( + ) The neutral ais is at an angle φ given b: tan φ = + + 2 This method is useful if the principal aes are not easil found but the components, and of the inertia tensor can be readil determined. Deflections: Using the first method described above, deflections can be found easil b resolving the applied lateral forces into components parallel to the principal aes and separatel calculating the deflection components parallel to these aes. The total deflection at an point along the beam is then found b combining the components at that point into a resultant deflection vector. Note that the resulting deflection will be perpendicular to the neutral ais of the section at that point.
Rotation Transformations: φ f, are the coordinates of point P in the the sstem YZ shown above, then the coordinates of P in the sstem Y Z are: ' = cosφ sinφ ' = sinφ + cosφ This transformation is useful in finding the coordinates of points with respect to the principal aes of a section.
Problems on Unsmmetrical Beams 1. An angle section with equal legs is subject to a bending moment vector having its direction along the Z-Z direction as shown below. Calculate the maimum tensile stress σt and the maimum compressive stress σc if the angle is a L 663/4 steel section and = 20000 in.lb. ( σ t = 3450 psi : σ c = 3080 psi ). 2. An angle section with unequal legs is subjected to a bending oment having its direction along the Z-Z direction as shown below. Calculate the maimum tensile stress σt and the maimum compressive stress σc if the angle is a L 861 and = 25000 lb.in. (σt = 1840 psi : σ c = 1860 psi ). 3. Solve the previous problem for a L 741/2 section and = 15000 lb. in. (σ t = 2950 psi : σ c = 2930 psi ). See net page for section properties needed in these problems.
Section properties for structural steel angle sections. Weight Ais ZZ Ais YY Ais Y'Y' Designation per ft. Area ZZ r ZZ d YY r YY c r min tan α in. lb. in 2 in 4 in. in. in 4 in. in. in. L663/4 28.7 8.44 28.2 1.83 1.78 28.2 1.83 1.78 1.17 1 L861 44.2 13 80.8 2.49 2.65 38.8 1.73 1.65 1.28 0.543 L741/2 17.9 5.25 26.7 2.25 2.42 6.53 1.11 0.917 0.872 0.335 1. Aes ZZ and YY are centroidal aes parallel to the legs of the section. 2. Distances c and d are measured from the centroid to the outside surfaces of the legs. 3. Aes Y Y and Z Z are the principal centroidal aes. 4. The moment of inertia for ais Y Y is given b Y Y = Ar 2 min. 5. The moment of inertia for ais Z Z is given b Z Z = YY + ZZ Z Z. Y c Y Z Z Z C d Z Y α Y