1 571 Structural Analysis and Control Prof Connor Section 1 Straight Members with Planar Loading Governing Equations for Linear Behavior 1 1 Notation Y v V a M F X u Internal Forces a 1 1 2 Deformation Displacement Relations a Y a v a b a Assume is small Displacements u v Longitudinal strain at location y For small Then y u y x u y u 0 y v y v 0 y u x y x a b a u x stretching strain b y x bending strain 1 571 Structural Analysis and Control Prof Connor Section 1 Page 1 of 17 Shear Strain a a b b decrease in angle between lines a and b dv v x dx v x 1 3 Force Deformation Relations E stress strain relations for linear elastic material G V M F X F dA M y dA V dA Consider initial strain for longitudinal actions where o a b T strain due to stress o initial strain Then T total strain a b 1 T o E 1 571 Structural Analysis and Control Prof Connor Section 1 Page 2 of 17 E T o E a b o E u x y x o F dA F E u x y x o dA F u x E dA x yE dA o E dA Also M y dA M yE u x y x o dA 2 M u x yE dA x y E dA y o E dA If one locates the X axis such that yE dA 0 the equations uncouple to give F u x E dA o E dA 2 Define M x y E dA y o E dA D S E dA stretching rigidity DB y 2 E dA bending rigidity F o o E dA Mo Then y o E dA F D S u x Fo M D B x M o Consider no inital shear strain G G v x V G dA G v x dA V v x GdA Define DT G dA 1 571 Structural Analysis and Control Prof Connor transverse shear rigidity Section 1 Page 3 of 17 then V D T v x 1 4 Force Equilibrium Equations V V M M M C F F F b x x m x V b y x Consider the rate of change of the internal force quantities over an interval x F x F F F b x x 0 F b x x 0 F b x 0 x Fy V V V b y x 0 V b y x 0 V b y 0 x 2 x M c M M M m x b y 2 V x 0 2 x M m x V x b y 0 2 M x m V b y 0 x 2 Let x 0 i e x dx F bx 0 x V by 0 x M V m 0 x 1 571 Structural Analysis and Control Prof Connor Section 1 Page 4 of 17 1 5 Summary of Formulation Equations uncouple into 2 sets of equations one set for axial loading and the other set for transverse loading Axial Stretching F x b x 0 F Fo D S u x Boundary Condition F or u prescribed at each end Transverse Bending V x b y 0 M x V m 0 M D B x M o V D T v x Boundary Conditions M or prescribed at each end and V or v prescribed at each end Note These equations uncouple for two reasons 1 The location of the X axis was selected to eliminate the coupling term yEdA 2 The longitudinal axis is straight and the rotation of the cross sections is considered to be small This simplification does not apply when i the X axis is curved see Section 2 ii the rotation can not be considered small creating geometric non linearity see Section 4 1 571 Structural Analysis and Control Prof Connor Section 1 Page 5 of 17 1 6 Fundamental Solution Stretching Problem bx FB B A L x Governing Equations F bx 0 x F Fo D S Boundary Conditions F B FB u A uA From i i u x ii F x b x dx C 1 F x Then L b x dx C 1 F B L C 1 F B b x dx L F x b x dx b x dx FB L which can be written as F x L x b x dx FB Note you could also obtain this result by inspection bx F x FB B x F x L L x b x dx FB 1 571 Structural Analysis and Control Prof Connor Section 1 Page 6 of 17 From ii F Fo u x DS u x F Fo dx C 2 DS F Fo u A dx C 2 DS 0 F Fo C 2 u A dx DS 0 u x F Fo dx u A 0 DS x xF u x u A B dx up x 0 DS FB x u x u A u p x DS where up x particular solution due to b x and F o where FB L u B u A u B o DS u B o u p L 1 571 Structural Analysis and Control Prof Connor Section 1 Page 7 of 17 1 7 Fundamental Solution Bending Problem V B v B VA M B B MA B A L x V x V B v B M B B M x B L x Internal Forces V x VB M x M B VB L x Governing Equations for Displacement M Mo M D B x Mo x DB V V D T v x v x DT Integration leads to 2 M B x VB x x A Lx o x DB DB 2 2 MB L VB L L B A B o DB DB 2 VB M B x 2 V B x2 x 3 v x v A A L L x v o x DB 2 DB 2 6 DT M B L 2 VB L3 V B v L v B v A A L L v B o DB 2 DB 3 DT 1 571 Structural Analysis and Control Prof Connor Section 1 Page 8 of 17 1 8 Particular Solutions Set i o end rotation at i due to span load v i o end displacement at i due to span load Then i i e i o v i v i e v i o where i e end rotation at i due to end actions v i e end displacement at i due to end actions Concentrated Moment b a B A M L M a B o DB 2 Concentrated Force M a M a v B o L a 2 DB 2 DB P b a B A L 2 P a B o 2 DB 3 v B o 2 P a P a P a L a 3 DB 2 DB DT 1 571 …