# 재료역학(130527)

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재료역학(130527)

Dep. Of Materials Science and Engineering Definition and applications of beams Beam? : Long and slender structural elements that are called on to support transverse as well as axial loads. Definition and applications of beams Beam vs. Truss : A truss is special beam element that can resist axial deformation only Shear Force & Bending Moment 1. Free-body diagram 2. Static (equilibrium) equations: F ?, F M ? x y ? continued. V(x) x M(x) x continued. 3. Geometric considerations V(x) x M(x) x continued. Example : Statics analysis P1 A P2 C m n B Distributed loads Free body diagram Accumulated load up to a certain point is of your interest. The shear force and moment at the point will be counterbalanced with the accumulated load. An equivalent point load to the distributed load can be considered for your better understanding. Equivalent point load Q The distributed load q(x) is statically equivalent to a concentrated load of magnitude Q placed at the centroid of the area under q(x) diagram. Successive integration method Example 4.1 Example Find RA and MA at the left end A Shear Force Diagram(S.F.D.) & Bending Moment Diagram(B.M.D.) - 1 concentrated load S.F.D. B.M.D. Shear Force Diagram(S.F.D.) & Bending Moment Diagram(B.M.D.) - uniformly distributed load Shear Force Diagram(S.F.D.) & Bending Moment Diagram(B.M.D.) - 2 concentrated load P1 A P2 C D B Normal stresses in beams - Stress distributions in bending beams Understanding of the stresses induced in beams by bending loads took many years to develop. Galileo worked on this problem, but the theory as we use it today is usually credited principally to the great mathematician Leonard Euler (1707~1783). Beams develop normal stresses in the lengthwise direction that vary from a maximum in tension at one surface, to zero at the beam's midplane, to a maximum in compression at the opposite surface. Shear stresses are also induced, although these are often negligible in comparision with the normal stresses when the length-to-height ratio of the beam is large. continued. Pure bending state Uniform bending moment V = dM / dx = 0 M = constant continued. Assumptions in Pure Bending Theory The material of the beam is homogeneous and isotropic. The value of Young's Modulus of Elasticity is same in tension and compression. The transverse sections which were plane before bending, remain plane after bending also. The beam is initially straight and all longitudinal filaments bend into circular arcs with a common centre of curvature. The radius of curvature is large as compared to the dimensions of the cross-section. Each layer of the beam is free to expand or contract, independently of the layer, above or below it. continued. Neutral axis Compression Tension Neutral surface continued. Shear strain, vB vA dy v v Δv vB v A v A dx v A dx x x δ v tanγ γ dx x Calculation procedures of normal stresses Relating the magnitude of these axial normal stresses to the bending moment. 1. Geometric statement: u v tanθ ~ y x u yv,x 2. Kinematic equation: du ε xx yv,xx dx Spatial rate of change of the slope of v,xx the beam deflection curve, the “slope of the slope” “curvature” of the beam. 3. Constitutive equation: d 2v 2 dx σ xx Eε xx yEv,xx 4. Calculating bending moment: Moment of inertia with respect to the centroidal axis, I. - For a rectangular cross section of height h and width b, I is: • The stress varies linearly from zero at the neutral axis to a maximum at the outer surface. • The stress varies inversely with the moment of inertia of the cross section, and it is independent of the material's properties.