L12 - Chapter 5 Analysis and Design of Beams for Bending...

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Unformatted text preview: Chapter 5 Analysis and Design of Beams for Bending • Shear and Bending Moment Diagrams • Relations among Load, Shear, and Bending Moment • Design of Prismatic Beams for Bending So far: (in chapter 4) bending Beam - structural member that supports loads at various points along the member Concentrated load Distributed load beam Loads result in: a shear force (from the shear stress distribution) + a bending couple (from the normal stress distribution) Normal stress is often the critical design criteria My "x = ! I Mc M "m = = I S Requires determination of the location and magnitude of largest bending moment Classification of Beam Supports Shear and Bending Moment Diagrams Why do we need shear and bending moment diagrams? To determine maximum normal and shearing stresses and their respective locations– design safe beams Moment diagram M x Shear diagram V x To determine shear force and bending couple at a point: - Draw FBD passing a section through the beam - Do equilibrium analysis on the beam portions on either side of the section. • Sign conventions for shear forces V and V’ and bending couples M and M’ Shear Diagrams 0< x< 1 L 2 V= 1 L<x<L 2 1 P 2 1 V =! P 2 Equivalent load: -wx V == !wx Vwx ! Relations Among Load, Shear, and Bending Moment • Relationship between load and shear: # Fy = 0 : V " (V + !V )" w !x = 0 !V = " w !x dV = "w dx VD " VC = " ! w dx xC xD • Relationship between shear and bending moment: #x " MC ! = 0 : ( M + #M ) $ M $ V #x + w#x 2 = 0 2 #M = V #x $ 1 w (#x ) 2 dM =V dx M D ! MC = xD xC " Vdx Bending Moment Diagrams: 0< x< 1 L 2 V= 1 L<x<L 2 1 P 2 1 V =! P 2 0< x< 1 L 2 M= 1 L<x<L 2 1 Px 2 1 1 1 M = PL ! P ( x ! L) 4 2 2 V = !wx 12 M = ! wx 2 Design of Prismatic Beams for Bending • The largest normal stress is found at the surface where the maximum bending moment occurs. !m = M max c M max = I S • A safe design requires: • maximum normal stress be less than the allowable stress for the material used. • leads to the determination of the minimum acceptable section modulus. ! m " ! all M max S min = ! all • Pick the smallest weight per unit length or cross sectional area will be the least expensive and the best choice. Section modulus: S=194X10-6 m3 ...
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This note was uploaded on 12/14/2010 for the course ENGR 2530 taught by Professor Lindaschadlerfeist during the Fall '08 term at Rensselaer Polytechnic Institute.

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