tailieunhanh - Lecture Mechanics of materials (Third edition) - Chapter 6: Shearing stresses in beams and thinwalled members

The following will be discussed in this chapter: Introduction, shear on the horizontal face of a beam element, determination of the shearing stress, longitudinal shear on a beam element of arbitrary shape, shearing stresses in thin-walled members, unsymmetric loading of thin-walled members. | Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University Shearing Stresses in Beams and ThinWalled Members © 2002 The McGraw-Hill Companies, Inc. All rights reserved. Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shearing Stresses in Beams and Thin-Walled Members Introduction Shear on the Horizontal Face of a Beam Element Example Determination of the Shearing Stress in a Beam Shearing Stresses τxy in Common Types of Beams Further Discussion of the Distribution of Stresses in a . Sample Problem Longitudinal Shear on a Beam Element of Arbitrary Shape Example Shearing Stresses in Thin-Walled Members Plastic Deformations Sample Problem Unsymmetric Loading of Thin-Walled Members Example Example © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 6-2 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Introduction • Transverse loading applied to a beam results in normal and shearing stresses in transverse sections. • Distribution of normal and shearing stresses satisfies Fx = ∫ σ x dA = 0 Fy = ∫ τ xy dA = −V Fz = ∫ τ xz dA = 0 ( ) M x = ∫ y τ xz − z τ xy dA = 0 M y = ∫ z σ x dA = 0 M z = ∫ (− y σ x ) = 0 • When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces • Longitudinal shearing stresses must exist in any member subjected to transverse loading. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 6-3 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Shear on the Horizontal Face of a Beam Element • Consider prismatic beam • For equilibrium of beam element ∑ Fx = 0 = ∆H + ∫ (σ D − σ D )dA A ∆H = M D − MC ∫ y dA I A • Note, Q = ∫ y dA A M D − MC = dM ∆x = V ∆x dx • Substituting, VQ ∆x I ∆H VQ q= = = shear flow ∆x I ∆H = © 2002 The McGraw-Hill Companies, Inc. All rights .

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