1.571 Structural Analysis and Control Prof. Connor Section 1: Straight Members with Planar Loading. b Displacements (u, vβ),

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 Y v, X u, F M V Internal Forces 1.1.2 Deformation ­ Displacement Relations β a’ Y θ a v a’ b _{Displacements( , ,} u v β) a Assume βis small Longitudinal strain at location y: ε( )y = ∂∂ xu y( ) For small β ( ) ≈u 0 u y ( )– yβ v y( ) ≈ v 0( ) Then ( ) = u,_x – yβ,_x = ε + ε_b ε y _a ε = u,  = stretching strain

Shear Strain

β

a’ b’

a _θ

b

γ = decrease in angle between lines a and b

γ = θ β– dv θ ≈ 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 = ∫ σd A M = ∫ -yσd A V = ∫ τd A Consider initial strain for longitudinalactions ε_σ+ ε_o = ε_a + ε_b = ε_T where

= strain due to stress

ε_σ ε_o = initial strain = total strain = ε + ε_b ε_{T a} Then ε = ε – ε = --1-σ

σ = E(ε_T+ ε ) _o = E(ε_a+ ε_b – ε )_o = E u,( _x – yβ,_x – ε )_o ∴ F = ∫ σdA ( F = E u,∫ _x – yβ,_x – ε )_o dA F = u,_x∫ E A d + β,_x ∫-yE A d + -∫ ε_oE Ad Also M = ∫ -yσdA ( M = ∫ -yE u,_x – yβ,_x – ε )_o dA M = u,_x ∫ -yE A d + β,_x∫y 2 E A d + y∫ ε_oE Ad If one locates the  X­axis such that yE A = 0d ∫ the equations uncouple to give: F = u,_x∫ E A d + -∫ ε_oE Ad d + ∫ yε_oE A M = β,_x∫ y 2 E A d Define = E A = stretching rigidity d D_S ∫ 2 = y E A = bending rigidity d D_B ∫ F_o = -∫ ε_oE Ad M_o = ∫ yε_oE Ad Then F = D_Su,_x+ F_o M = D_Bβ,_x+ M_o Consider no inital shear strain ( τ = Gγ = G v,_x – β) V = G∫ γdA = G∫ (v,_x – β)dA

1.4  then V = D_T(v, _x – β) Consider V + ∆V M + ∆M C F + ∆F b_x∆x V b_y∆x  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 -F+F+∆F +b ∆F+ b_x∆x = 0 ∆F --- + b = 0_{∆ x} x -V+V+∆V +b ∆V+ b_y∆x = 0 ∆V --- + b = 0_y ∆x ∆x = 0 x ∆x = 0 y + – -M + M + ∆M m∆x b_y ∆x 2 ∆M+ m∆x+ V∆x b ---– _y 2 ∆x 2 --- + V∆x = 0 2 = 0 ∆M ∆x --- + m+ V–b --- = 0_y 2 ∆x Let ∆x→ 0 (i.e. ∆x → dx) ∂x ∂F b_x + = 0 ∂x ∂V b_y + = 0 ∂ ∂M V m + + = 0

1.5 Summary of Formulation Equations “uncouple” into 2 sets of equations; one setfor “axial” loading and the  other set for “transverse” loading. Axial (Stretching) F, + b = 0_{x x} F F + D Su, = _{o x} Boundary Condition F or u prescribed at each end Transverse (Bending) V, + b_{x y} = 0 M, +_x V+m = 0 M = D_Bβ, + M_{x o} V = D_T(v, – _x β) Boundary Conditions M orβ prescribed at each end and V or v prescribed at each end Note: These equations uncouple for two reasons 1. The location of the X­axis was selected to eliminate the coupling term yEdA ∫ 2. The longitudinal axis is straight and the rotation of the cross­sections is considered to be small.  This simplification does not apply when: i ­ the X­axis is curved (see Section 2) ii ­ the rotation, β, can not be considered small, creating  geometric non­linearity (see Section 4)

1.6 Fundamental Solution ­ Stretching Problem B A b_x F_B L x Governing Equations: ∂F ∂x + b = 0 x (i) ∂u F = F + D _o (ii) S∂_x Boundary Conditions F _B = FB u_A = u_A From (i) F x( ) = - b dx + C∫ _{x 1} F x( ) = -(∫ b_xdx) L + C1 = FB L ∴ C₁ = F_B+ (∫ b_xdx) L Then F x( ) = -∫ b_xdx + (∫ b_xdx) L + FB which can be written as L F x( ) = ∫ b_xdx + F_B x Note: you could also obtain this result by inspection: b_x F x( ) _FB B x L L F x( ) = ∫ b_xdx + F_B x

From (ii) F– F_o --- = u, _x D_S F– F_o ( ) = ∫ ---dx + C₂ D_S F F u x – _o ⎛ ⎞ = ---dx + C₂ u_{A ⎝}∫ _⎠ D_{S 0} F– F ⎛ ⎞ = – ---dx C₂ u_A ∫ D_S o 0 ⎝ ⎠ x_{F– F} o ( ) = ∫ ---dx + u_A 0 DS u x x_F B ( ) = u_A+ ---dx + u_p( ) u x ∫ x 0 DS F_Bx ( ) = u_A+ --- + u_p( ) u x x D_S

where u_p ( )x = particular solution due to b_x and F_o. F_BL = + --- + u_{B o} u_B u_{A ,} D_S where = u L u_{B o, p}( )

1.7 Fundamental Solution: Bending Problem B A M_B , β_B L x V_B ,v_B V_A M_A V x( ) V_B,v_B M_B, β_B M x( ) B L– x Internal Forces V x( ) = V_B M x( ) = M_B+ V_B(L–x) Governing Equations for Displacement M– M_o M = D_Bβ, _x+ M_o → β, _x = ---­-D_B V V = D_T(v, _x – β) →v, _x = β+ ---­ D_T Integration leads to: M_Bx V_{B ⎛} 2 _⎞ ( ) = + --- + --- Lx – ----x - + β ( ) β x _{⎝ o} x 2 ⎠ β_A D_B D_B M_BL V_BL2 ( ) = = β_A+ --- + --- --- + β _, β L β_B 2 B o D_B D_B 2 3 MBx 2 V_{B ⎛} x _⎞ V_B + x ( ) = v_A+ β_AL + --- + --- L----x - – --- + ---x v ( ) v x _o 2 D_B⎝ 2 6 ⎠ D_T D_B M_BL2 V_BL3 V_B ( ) = v_B = v_A+ β_AL + --- + --- --- + ---L v _, v L D_B 2 D_B 3 D_T + B o

1.8 Particular Solutions Set β_{i o,} = end rotation at i due to span load v_{i o,} = end displacementat i due to span load Then β_i = β_{i e,} + β_{i o,} v_i = v_{i e,} + v_{i o,} where β_{i e,} = end rotation at i due to end actions v_{i e,} = end displacement atidue to end actions ConcentratedMoment B A L M* b a β_{B o,} = ---­-M* a D_B M* a 2 M* a v_{B o,} = --- + ---( L– a) 2 D_B 2 D_B ConcentratedForce B A L P* b a P* a 2 = ---­ , β_{B o} 2 D_B P* a P* a 3 P* a 2 v_{B o,} = --- + --- + ---( L– a) 3 D_B 2 D_B D_T

Distributed Loading B A L dx bdx

Replace P* with bdx and integrate from x = 0 to x = L L 2 = ---b xd , β_{B o ∫} 02 x D_B L _x L 3 L _x 2 v_{B o,} = ---b x + d ∫ ---b x + x d ---b xd ( L x– ) 03 DB ∫ 02 DB ∫ ₀ D_T

for b constant(ie uniformly distributed loading)

3 bL = ---­ , β_{B o} 6 D_B 2 4 bL bL = --- + ---­ , v_{B o} 2 D_T 8 D_B

1.9 Summary F_BL = _, + --- + u_A u_B u_{B o} D_S MBL2 VBL3 VB v_B = v_{B o,} + --- --- + --- --- + ---L+ v_A+ β_AL 2 D_B 3 D_B D_T M_BL V_BL2 + --- + --- + β_A β_B = _, 2 β_{B o} D_B D_B These equations can be written as u_B = v_B β_B u_{B o,} v_{B o,} β_{B o,} + L --- 0 0 D_S F_{B 1 0 0} u_A 3 2 L L L ₊ 0 --- + --- ---­- _{V 0 1 L vA} B 3 D_B D_T 2 D_B 0 0 1 β A 2 MB L 1 0 --- ---­-2 D_B D_B Rigid body transformation from A to B Also F_A = F_{A o,} – F_B VA = VA o, – VB M_A = M_{A o,} – M_B – LV_B FA VA MA = FA o, VA o, MA o, 1 0 0 – 0 1 0 0 L 1 FB VB MB

1.10 Matrix Formulation ­ Straight Members Define: u_B = u_B v_B β_B Displacement Matrix F_B = F_B V_B End Action Matrix M_B General Force Displacemnt Relation Express displacement at B as: u_B = u_{B o,} + f F_B + T_ABu_A B , : Due to applied loading u_{B o ⎫} ⎪ f BFB: Due to forces at B ⎬ Based on cantilever model ⎪ Effect of motion at A⎭ T_ABu_A: Interpret f  = Member flexibility matrix B = Rigid body transformation from A to B T_AB For the prismatic case L --- 0 0 D_S 3 2 L L L f = 0 --- + --- ---­ B _{3 D} B DT 2 DB 2 L 1 0 --- ---­-2 D_B D_B

Force Displacement Relations –1 Define k_B = f_B  = Member stiffness matrix Start with u_B = u_{B o,} + f BFB + TABuA Solve for F_B f_BF_B = u_B – T_{A B}u_A – u_{B o,} F_B = k_Bu_B – k_BT_ABu_A – k_Bu_{B o,} Define F_{B i,} = k– _Bu_{B o,} Then F_B = k_Bu B – kBTABuA + FB i, Next, determine F_A F_A = F_{A o,} – T_AB T F_B F_A = ( –T_AB T k_B )u_B + T( _AB T k_BT_AB)u_A + F_{A i,} where F_{A i,} = F_{A o,} – T_AB T F_{B i,}

Note F_{A i,}  and F_{B i,}  are the initial endactions with no end displacements

Finally, rewrite as

F_B = k_BBu + F_{B i} B + kBAuA , T

F_A = k_{B A}u_B + k_{A A}u_A +F_{A i,}

Matrices for Prismatic Case u_B v_B β_B u_A v_A β_A D_S D_S --- 0 0 –--- 0 0 F_B L L 12 D* _B ( )–6 D*_B (–12 )D* _B ( )–6 D* _B V_B 0 --- --- 0 --- ---­-3 2 3 2 L L L L –6 _B ( )D* (4 + a)D* 6 D* _B (2 + a)D* _B M_B 0 --- ---B- 0 --- ---­-L2 L L2 L D_S D_S –--- 0 0 --- 0 0 F_A L L (–12 )D* _B 6 D* _B 12 D* _B 6 D* _B V_A 0 --- --- 0 --- ---­-3 2 3 2 L L L L –6 _B ( )D* (2 + a)D* 6 D* _B (4 + a)D* _B M_A 0 --- ---B- 0 --- ---­-L2 L L2 L 12 D_B D_B a = --- D* _B = ---­-L2 D_T (1 + a)

1.11 Transformation Relations Rigid Body Displacement Transformation B A ωA u_A r_{AB ω} B u_B ωB = ωA u_B = u_A +ωA × rAB u_B = u_{A ⎫} ⎪ v_B = v_A +ω_AL _⎬ ⎪ ω_B = ω_{A ⎭} u_{B 1 0 0} u_A = 0 1 L v_B v_A 0 0 1 ω_B ω_A u_B = T_ABu_A Statically Equivalent Force Transformation in two dimensions T A B A m_A PA r_AB m_B P_B ranslate force system acting atB to point P_A = P_B m_A = m_B + r_AB × PB T F = F – T F

Coordinate Transformation x’ x y y’ z,z’ a = a_x a_y a_z a_x' a' = a_y' a_z' a' = Ra cos θ sin θ 0 R = –sin θ cos θ 0 0 0 1 The inverse is cos θ = cos (–θ)⎫ ⎪ T –1 sin θ = –sin ( )–θ _{⎬ R = R} –1 ⎪ R = R(–θ) _⎭ Take (x y z, , ) = Global frame (x', ,y' z') = Local frame l ( ) F g F( ) = R gl ( ) R( )gl = R l

Given k  in local frame ( k( ) ), transform to global frame l = k( )u l l ( ) ( ) F( ) l ( ) = k( )R gl u g ( ) = R lg l l ( ) ( ) F g ( )F( ) = R( ) ( )lg k R gl u g If ( ) = k g u( ) F g ( ) g Then ( ) _{( ) ( )}l _{( ) ( )} T ( )l _{( )}

1.12 StructuralStiffness Matrix assembly g F_B = kg_{B B}ug_B + kg_{B A}ug _A + F_{B i}g_, g g g g F_A = ( kg _BA)T g u_B + k_AAu_A + F_{A i,} F_ig = R FT l _i Use direct stiffness method to generate the system equations referred to the global  frame. Take B as the positive end and A as the negative end. B → n+ A → n-for member n Write system equation as = P_I + KU E Work with the partitioned form of system stiffness matrix K.  in n+,n+ k_BB  in n­,n­ k_AA T  in n+,n­ k_BA T  in n­,n+ k_BA with  in n+ of P_I F_{B i,}  in n­ of P_I F_{A i,}