1.571 Structural Analysis and Control Prof. Connor Section 1: Straight Members with Planar Loading Governing Equations for Linear Behavior 1.1 Notation a a Yv, Xu, F M V Internal Forces 1.1.2 Deformation - Displacement Relations β a’Y θ a v a’ b ,,Displacements (uvβ) a Assume βis small Longitudinal strain at location y: ε y()= ∂∂ xuy() For small β ()≈ u 0uy ()– yβ vy () ≈ () v 0Then()= u,x – yβ,x = ε + εbε ya ε= u,x = stretching straina εb = –yβ,x = bending strain 1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 1 of 17Shear Strain β a’ b’ a θ b γ = decrease in angle between lines a and b γ θβ = – dvθ ≈x d = v,x γ = v,x – β 1.3 Force - Deformation Relations σ = Eε⎫ ⎬ stress strain relations for linear elastic material τ = Gγ ⎭ τ σ F M V X F = σ Ad ∫ M = -yσ Ad ∫ V = τ Ad ∫ Consider initial strain for longitudinal actions εσ+ ε = ε + εb = o a εT where = strain due to stress εσ ε = initial strain o = total strain = ε + εbεT a Then εσ = εT – ε = --1-σoE 1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 2 of 17σ = E(εT+ ε ) = E(ε + εb – ε )o a o = Eu,x – yβ,x – ε )( o ∴ F= σdA∫ (F = Eu,x – yβ,x – ε )dA∫ o F = u,x∫ EA β,x -yE A dd + d + -ε EA∫ ∫ oAlso M= -yσdA∫ (M= -yE u,x – yβ,x – ε )dA∫ o M= u,x -yE A β,xy 2 EA dd + d + yεoEA∫ ∫ ∫ If one locates the X-axis such that yE A = 0d ∫ the equations uncouple to give: F = u,x∫ EA dd + -ε EA∫ od + ∫ yεoEAM = β,x∫ y 2 EA d Define = EA= stretching rigidity dDS ∫ 2= y EA= bending rigidity dDB ∫ F= - ε EAdo ∫ oM= ∫ yεoEAdo Then F= DSu,x+ Fo M= DBβ,x+ Mo Consider no inital shear strain (τ = Gγ = Gv,x – β) V = GγdA = G(v,x – β)dA ∫ ∫ V= (v,x – β)∫ GdA Define = GA= transverse shear rigidity dDT ∫ 1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 3 of 171.4 then V DT(v, x – β)= ConsiderV + ∆V M + ∆M C F + ∆F bx∆x V b∆xy the rate of change of the internal force quantities over an interval ∆x Force Equilibrium Equations m∆x M F F = ∑ x F = ∑ y M= ∑ c -FF∆Fb++ + ∆Fb∆x = 0+ x∆F -------+ b = 0x∆x -VV∆Vb++ + ∆Vb∆x = 0+ y∆V -------+ b = 0y∆x ∆x = 0x∆x = 0y+ –-M + M+ ∆Mm∆xby ∆x 2 ∆Mm∆xV∆xb---------+ + – y2 ∆x 2 ---------+ V∆x = 02 = 0 ∆M ∆x ---------+ mV–b------= 0+ y2∆x Let ∆x → 0 (i.e. ∆x dx)→ ∂x ∂Fbx+ = 0 ∂x ∂Vby+ = 0 ∂x ∂MVm++ = 0 1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 4 of 171.5 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, + b = 0x x FF+ DSu, = o x Boundary Condition F or u prescribed at each end Transverse (Bending) V, + by = 0xM, ++Vm= 0xMDBβ, + M= x o VDT(v, – β)= xBoundary Conditions M or β 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, β, can not be considered small, creating geometric non-linearity (see Section 4) 1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 5 of 171.6 Fundamental Solution - Stretching Problem B A bx FB L x Governing Equations: ∂F ∂x + b= 0 (i)x ∂uF = F + D (ii) oS∂x Boundary Conditions F = FBB u = A uA From (i) Fx()= - b dx + C1∫ x Fx()= -(∫ bxdx)L + C1= FBL ∴ C1= FB+ (∫ bxdx)L Then Fx()= -∫ bxdx + (∫ bxdx)L + FB which can be written as L Fx()= ∫ bxdx + FB x Note: you could also obtain this result by inspection: bx Fx() FB Bx L L Fx()= ∫ bxdx + FB x 1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 6 of 17From (ii) FF– o---------------= u, xDS FF– o()= ∫ ---------------dx+ C2DS FFux– o ⎛ ⎞= ---------------dx + C2uA ⎝ ⎠ ∫ DS 0 FF– ⎛ ⎞= – ---------------dxC2 uA ∫ DS o 0⎝ ⎠ xFF– o()= ∫ ---------------dx + uA 0 DS uxxFB()= uA+ ------dx+ up()uxx∫ 0 DS FBx()= uA+ ---------+ up()ux xDS where u ()= particular solution due to b and Fxp x o. FBL = + ----------+ uBouB uA ,DS where = u LuBo p(), 1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 7 of 171.7 Fundamental Solution: Bending Problem BA MB βB, L x VB vB,VA MA Vx() VB, vB MB, βBMx() B Lx– Internal Forces Vx()= VB Mx –()= MB+ VB(Lx) Governing Equations for Displacement MM– oM = DBβ, x+ M → β, x = ----------------- -o DB VV= DT(v, x – β) → v, x = β + -------DT Integration leads to: MBx VB 2 ⎛ ⎞()= + -----------+ -------Lx – ----x -+ β ()β xxo ⎝ 2 ⎠ βA DB DB MBL VBL2 ()= = βA+ -----------+ ------------+ β, β L βB 2 BoDB DB 2 3MBx 2 VB xVB ⎛ ⎞ + x()= vA+ βAL + ------- -----+ -------L----x -– -----+ -------xv()vxo2 DB⎝ 6 ⎠2 DTDB MBL2 VBL3 VB()= vB = vA+ βAL + ------- -----+ ------------+ -------Lv, vLDB2 DB3 DT + Bo1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 8 of 171.8 Particular Solutions Set βio, = end rotation at i due to span load vio, = end displacement at i due to span load Then βi = βie, βio, + vi = vie, vio, + where βie, = end rotation at i due to end actions vie, = end displacement at i due to end actions Concentrated Moment B A L M* ba βBoM* a = -----------, DB M* a 2 M* a vBo= --------------+ -----------( La)– , 2 DB 2 DB Concentrated Force B A L P* ba P* a 2 = ------------, βBo2 DB P* a P* a 3 P* a 2 vBo= ----------+ ------------+ ------------( La)– , 3 DB 2 DBDT 1.571 Structural Analysis and Con trol Section 1 Prof Connor Page 9 of 17Distributed Loading B A Ldx bdx Replace P* with bdx and integrate fromx = 0 to x = …