6.Deflection

6.Deflection - Deflection of beams Theory and Design of...

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Theory and Design of Structures I Deflection Deflection of beams Assumptions 1. The beam is prismatic, initially straight and has an axial plane of symmetry which is taken as the x-y plane. 2. The material is homogeneous and obeys Hooke’s Law. 3. Plane sections before bending remain plane after bending. As a result of deformation, fibres on the concave side are shortened slightly (compression), while those on the convex side are elongated slightly (tension). Somewhere in between, the section remains unchanged in length, i.e. neutral surface. its intersection with any cross-section is the neutral axis dx d θ M M R y Beam segment under pure bending z y dA σ Compression Tension Beam section Bending stress Beam under pure bending Consider two adjacent cross-sections at a small distance dx apart. After bending they intersect at o and the angle between them is denoted by d θ . Let R = radius of curvature of the neutral surface Then d θ = dx / R dx d θ M M R y Beam segment under pure bending z y dA σ Compression Tension Beam section Bending stress Beam under pure bending At a depth y from the neutral axis, the strain is ε = (y d θ ) / dx = y / R the stress is σ = E ε = E y/ R the force on a small area dA is σ dA = (E/R) y dA dx d θ M M R y Beam segment under pure bending z y dA σ Compression Tension Beam section Bending stress Beam under pure bending dx d θ M M R y Beam segment under pure bending z y dA σ Compression Tension Beam section Bending stress Beam under pure bending Since there must be no resultant normal force on the section under pure bending, σ dA = (E/R) y dA = 0 A = 0 Therefore must be equal to 0. In other words, the neutral axis of a section in pure bending passes through the centroid. y y
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dx d θ M M R y Beam segment under pure bending z y dA σ Compression Tension Beam section Bending stress Beam under pure bending The moment of the elemental force about the neutral axis of the section is dM = ( σ dA) y The total moment on the section is therefore M = σ y dA = (E/R) y 2 dA = (E/R) y 2 dA ∫∫ dx d θ M M R y Beam segment under pure bending z y dA σ Compression Tension Beam section Bending stress Beam under pure bending The integral y 2 dA is the second moment of area of the section and is denoted by I . Therefore M = (E/R) I or 1/R = M / (EI) = d θ/ dx Hence σ = (M y) / I For a symmetrical prismatic beam bent by transverse loads acting in a plane of symmetry, it can be shown that the curvature of the neutral surface is given by 1 / R = M / EI at each cross-section. Thus, the curvature 1/R is seen to vary along the beam as the bending moment varies. dx d θ M M R y Beam segment under pure bending z y dA σ Compression Tension Beam section Bending stress Beam under pure bending Consider any segment ds of the deflection curve (or elastic line). One has ds =R d θ or If the deflection is relatively small, ds dx R 1 ds d θ = dx dy θ 2 2 dx y d 1 R Taking the sign convention and the direction of the coordinate axes into consideration, we get the
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6.Deflection - Deflection of beams Theory and Design of...

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