材料力学第五章A.ppt

Chapter 5 Stresses in Bending Contents: 2. Normal stresses in bending 3. Shearing stresses in bending 4. Strength of beams 1. Geometrical properties of plane areas 5. Principal moment of inertia 7. Shear center 6. Unsymmetrical bending 5.1 Geometrical properties of plane areas 1. Static moment of an area Sz = ?A y dA , Sy = ?A z dA 2. Centroid yC = ?A y dA /A = Sz /A , Sz = yC A zC = ?A z dA /A = Sy /A , Sy = zC A 3. Calculations for the composite area Sz = ? Ai yi , Sy = ? Ai zi 1 2 ? n y z O yi zi ? Ci ? ? ? yC = , ? Ai yi ? Ai zC = ? Ai zi ? Ai the static moment of the area with respect to its centroid axis is zero. z1 ? ? z2 zC 1 2 ? ? ? y z yC y1 y2 yC = For example: A1 y1 + A2 y2 A1 + A2 ? ? yC zC C O y z dA y z A 4. Moment of inertia of the area Iz = ?A y2 dA , Iy = ?A z2 dA the moment of inertia of area to z and y. Ip = ?A ?2dA = ?A(y2 + z2)dA = Iz + Iy (1) For the rectangular section: C z y h b y dy Iz = ?A y2 dA = ?-h/2 y2bdy h/2 = bh3/12 Similarly, Iy = hb3/12 (2) For the circular section: From Iy = Iz , we get: Iy = Iz = Ip / 2 (3) For the tubular section: C z y d z y d D where ? = d / D . Iy = Iz = ?D4(1-? 4)/64 = ?d 4/64 O y z dA y z ? 5. Moment of inertia for the composite area (1) Parallel axis theorem: O y z dA yC zC ? a b C yC zC A Iy = Iyc + a2 A ( 5.9a) Iz = Izc + b2 A ( 5.9b) (2) For the composite area: z1 ? ? ? ? ? ? z2 z y1 y2 1 2 ? ? ? y Iz1 + y1 A1 2 Iz(1) = Iz(2) = Iz2 + y2 A2 2 Iz = Iz(1) + Iz(2) = Iz1 + y1 A1 2 + Iz2 + y2 A2 2 5.2 Normal stresses in bending 1. Geometrical relation M M Neutral layer dx a? a? ? d? y ? = M M a a dx (?+y)d? -?d? ? d? = y ? Hypothesis of plane section (a) 2. Physical relation ? = E ? = E y ? (b) 3. Statical relation ? ?max y z FN = ?A ? dA = 0 (c) Mz = ?A (? dA) y = M (d) From (b), (c), ?A y dA = Sz = 0 Neutral axis z must be centroidal axis. From (b),(d), E ? ?A y2dA = E ? Iz =