1.571 Structural Analysis and Control Prof. Connor Section 2: Linear Formulation for a General Planar Member. x X

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1 .57 Structural nalysis and Control Prof. Connor Section 2: Linear Formulation for a General Planar Member 2. Geometric elations: Plane Curve Y t 2 t j y r i X = ( θ) y = y ( θ) θ = Parameter to define the curve r = position v ector for a point on the curve r = i + yj 2.. rc lenth Define Then Differential arc lenth = ( 2 + dy 2 ) 2 = ds 2 dy dθ ds = dθ dθ 2 dy dθ dθ α = ds = αdθ Prof Connor Pae of 24

2 2..2 Tanent and Normal Vectors r ( +, y + y) ( y, ) r r 2 r = r 2 r = r(θ + θ) r ( θ) r = (( + )i + (y + y)j) (i + yj) r = ( i + yj) 2 2 r = ( 2 + y ) = s Then for continuous and well-behaved functions dr = (i + dyj) dr ---- dr dy = = t = -----i j ds dr ds ds t = un it tanent vector dy t = i j αdθ αdθ Define t 2 su ch that and enforce ives t t 2 = k (u nit vector in z -direction) t t 2 ( perpendicular) dy t 2 = i j αdθ αdθ Prof Connor Pae 2 of 24

3 2..3 Differentiation formulae for the tanent and normal vectors t and t 2 2 t t t = = 2 t 2 t 2 = t 2 = d dt dt dt ---- ( t t ) = t + t = 2 t ds ds d s d s dt Therefore, must be to t, and thus proportional to t d s 2 The result is written as dt = -- t d s 2 dt 2 d s where = -- t 2 2 dy dy -- = α 3 dθ 2 dθ dθ 2 dθ Prof Connor Pae 3 of 24

4 2..4 Eample: Curvilinear Formulation Take the anle θ as the parameter definin the curve Y θ t 2 θ t y X = sin θ y = ( cos θ) 2 2 d 2 d sin θ + θ) d θ d θ α = ( cos α = {cos 2 θ + sin 2 θ} 2 = t = -- { cos θi + sin θj } = cos θi + sin θj t 2 = sin θi + cos θj dt dt dt = -- = -- = -- ( sin θi + cos θj ) = -- -t d s αds ds 2 dt 2 dt 2 dt 2 = -- = -- = -- ( cos θi sin θj ) = -- -t d s αds ds This illustrates the equations on the previous pae. Prof Connor Pae 4 of 24

5 2.2 Force Equilibrium Formulation: Curvilinear Formulation t 2 V Y t F j i t 3 = t t 2 X F = Ft + Vt 2 M = Mt 3 Loadin per unit arc lenth b = b t + b 2 t 2 m = mt Vector Equations F + F M + M M F s Then F + F F+ b s M + M M+ m s st ( F) df b ds dm m+ t ds F Prof Connor Pae 5 of 24

6 2.2.2 Scalar Equations F = Ft + Vt df V -- + b ds dv F b 2 ds For the moment equation df d ( Ft + Vt ds d s 2 ) F dt ds - t df - V dt 2 dv = = t ds ds 2 ds df F df V dv = -- t 2 + t t + t ds ds ds df F df V dv b = -- t 2 + t t + t b t + ds ds ds Separatin into components of t a nd t 2 leads to dm m+ t (Ft + Vt 2 ) ds Epandin and writin in scalar form dm m+ V ds b 2 t 2 Summary df V -- + b ds dv F b 2 ds dm m+ V ds Note: F and V are coupled Prof Connor Pae 6 of 24

7 2.3 Force Displacement elations : Curvilinear Coordinates ( u β)t vt 2 ut β 2.3. Strains : Displacement components defined wrt the local frame (t, t 2 ) θ l ( ) l ( ) = ( ) θ s s = θ θ = s l ( ) = ( )----- l ( ) = -- s s u = ut + vt 2 βt ε = ε γ Prof Connor Pae 7 of 24

8 du ε = dl Separatin t a nd t 2 terms dl = -- ds du ε = ds du -- u du v dv = t t -- t t ds ds ds 2 β -- t dβ t ds Define du v dβ ε = ds d s -- du v ε a = ds ε = ε dv u γ = -- β β + ds -- γ = γ -- For most members «then -- 0 so dv u γ β+ -- o ds lso if the member is thin -- a o dβ ds Then, the strain distribution reduces to a linear distribution over the cross-section. Prof Connor Pae 8 of 24

9 2.3.2 Stress Strain elations ε = -- σ+ εo σ = E (ε ε o) E γ = ---τ τ = G γ G Force-Deformation elations F = σ d M = σ d V = τ d Substitutin for the strains leads to E d β F = E (ε ε o ) d = ε a E εo d d s -- Similarly E du E v E d β F = d d d E εo d d s d s E E F = ε d β, s d E ε a o d E E 2 M = ε d + β, s d + E ε a o d G V = γo d -- Prof Connor Pae 9 of 24

10 Geometric elations Now Makin But d Finally, one can write where = = = d = d d = = d = -- d d d = d + -- d d i f i s measured from the centroid of the section I ' d = d d = I ' = d -- We can use this epression to epand the force-deformation relations. Prof Connor Pae 0 of 24

11 Summarizin F = D S ε a + D SB β, s + Fo M = D SB ε a + D B β, s + M o where V = D T γ o E D S = d -- E D SB = d -- G D T = d -- 2 E D B = d -- For the homoeneous case E and G ar e constant, and the above simplify to I' D S = E + E I' D SB = E-- D B = EI' I' D T = G + G dditionally, for a thin curved member, one can nelect wrt, obtainin the simplified form EI D S = E EI D SB = D B = EI GI D T = G Prof Connor Pae of 24

12 2.3.4 Eample for a ectanular Section C L b d 2 I' = d -- d = bd I' = d 2 b d d 2 -- d d n and for -- <, -- 3 d 2 3 d 4 I' = I bd I = «for n > Summary of Equations F = D du v + D dβ S SB ds + F ds o du v dβ M = D SB D B M ds ds o dv u V = D T β + -- ds Prof Connor Pae 2 of 24

13 2.3.5 Fleibility Matri - Circular Member X Y Note: k into pae θ B B V B, v B F B,, β B u B}End ctions and displacements referred to the local frame at B M B ssumptions. Thin linear elastic member 2. Nelect stretchin and transvers shear deformations. This is valid for nonsha llow cases only u B = u B = u B v B F B = β B f B F B F B V B M B sin θ B + -- sin θ B cos θ B 2 cos θ B sin 2 θ B (θ B sin θ B ) f = B 2 2 cos θ -- B sin 2 θ B (θ B sin θ B cos θ B ) ( cosθ B ) 2 (θ B sin θ B ) ( cosθ B ) θ B Prof Connor Pae 3 of 24

14 2.4 Cartesian Formulation - Governin Equations Consider the case where the centroidal ais is defined by y = y ( ). The i ndependent variable is, the position coordinate on the -ais. In the anular formulation, we worked with displacement components and loads re ferred to the local curvilinear frame defined by the unit vectors t and t 2. In this section, we are oin to work with quantities referred to the lobal cartesian reference frame. This approach was oriinally suested by Maruerre. Yv, t 2 v j y ψ t u i ds α = ds = α dy dy si n ψ = = ds α cos ψ = = -- ds α s X u, V N y F ψ M N b y m b The above fiure defines the cartesian notation. The distributed loadin ( b, b,m ) y is defined with respect the the and y directions and the loadin per unit projected lenth ( ). Similarly, the displacement components ( u, v ) are referred to the cartesian lobal reference directions. Prof Connor Pae 4 of 24

15 2.4. Force Equilibrium Equations b y F + F M + M M o b dy F then Knowin we et Separatin F = F + F + ( bi + b y j) F F + bi+ byj df bi+ byj F = (Fcos ψ V sin ψ)i + ( Fsin ψ + Vcos ψ)j d -----(( Fcos ψ Vsin ψ)i + (F sin ψ + Vcos ψ)j) + bi+ b y j d -----(F cos ψ Vsin ψ) + b d -----(F sin ψ + Vcos ψ) + b y Similarly M o = ( i + yj) ( F) + (M + M M+ m ) --( i + yj) (b i + b j) 2 y Second order term oes to 0 in the limit 0 But ( i + yj) (( Fcos ψ Vsin ψ)i + (F sin ψ + Vcos ψ)j) + ( M+ m )k ( ( Fsinψ + V cos ψ) y( Fcos ψ Vsin ψ) + M+ m )k y y M F sinψ + V cos ψ F cosψ V sinψ m dy dy dm F sinψ + V cos ψ -----Fcos ψ -----Vsin ψ m dy = α sin ψ cos ψ = -- α Prof Connor Pae 5 of 24

16 Then dm F sinψ + V cos ψ α sin ψf--- α sin ψv sin ψ m α dm V cosψ + αsin 2 ψv m Multiply both sides by -- ᾱ Therefore dm -- - V cos ψ + αsin 2 ψv m α dm Vcos 2 ψ + Vsin 2 ψ m α α dm V m α α d -----(F cos ψ Vsin ψ) + b d -----(F sin ψ + Vcos ψ) + b y N N y = F cosψ V sin ψ = Fsin ψ + Vcos ψ Prof Connor Pae 6 of 24

17 4.2 Deformation - Displacement elations e 2 v e = s tretchin deformation s β o e u e 2 = tr ansverse shear deformation = co s ψ ds = ds cos ψ u = (u i+ vj) du du = cos ψ ds du du dv = i j du du dv = cos ψ i j ds du du dv e = t = cos ψ i j ( cos ψi + si n ψj) ds du dv e = cos 2 ψ si n ψ co s ψ du du dv e 2 = t 2 β = co s ψ i j ( si n ψi + cos ψj) β ds du dv e 2 = si n ψ cos ψ cos 2 ψ β dβ k = bendin deformation = = co s ψ dβ ds note: e and e 2 a re referred to the centroidal aes (s) Prof Connor Pae 7 of 24

18 2.4.3 Force Deformation elations Thin slihtly curved member: use relations developed for the prismatic case ssume ε = e γ = e 2 F = ( E e e, o ) d D S = Ed V = Ge 2 d D T = Gd M = D B = E2 d Summarize as follows F e = = e 2 D T = (e e, o )D S o o E ( e e, o ) d = Ek ( k ) 2 d = ( k k ) E 2 d M = (k k 0 )D B e, o V e 2 = D T M k = k o D B D S Thick Case ( 0 wrt ) : chane in l as a function of must be considered o Use equations developed in section with e a = e γ o = e 2 dβ k = ds Prof Connor Pae 8 of 24

19 4.4 pproimate Formulation - Shallow Car tesian - Sha llow assumption ψ 2 «Then cos ψ dy si n ψ ta n ψ = y, α = (ds ) - M aruerre nelects V terms in the equilibrium equation N = F N y = V+ Fy, So, resultin equations (shallow and Maruerre) Equilibrium df b dv d dy F b y dm + V+ m Deformation - Displacement du dydv e = = ε a dv e 2 = β = γ o dβ k = Prof Connor Pae 9 of 24

20 For the Force-Displacement elations, thin and thick should be considered separately - Thi n, Shallow, Marquerre du dydv F = D S D S e, o D S = Ed dv V = D T β dβ M = D B D B k o D T = Gd - Thi ck, Shallow, Maruerre D B = E2 d du dydv dβ F = D S D SB ds ds du dydv dβ M = D SB D B ds dv V = D T β E D S = d ( ) E D SB = d ( ) G D T = d ( ) 2 E D B = d ( ) Boundary Conditions in eneral, the B.C. s are u or N v or Ny pre scribed at each end β or M for the Maruerre formulation u or F v or V + y, F pres cribed at each end β or M Prof Connor Pae 20 of 24

21 Eample Y h X L 2 y = --a 2 y, = a y, = a 2 H a = L H y, = al-- = L L L oes from 0 to ; then 2 H L provides a measure of y,. The maimum value of y, is 2 H L. usin θ = Suppose H = L 0 then α dy α = dθ dθ 2 α = { + (y, ) 2 } H α ma = L ma =. 02. Thus, it is reasonable to assume α is constant and equal to. In en eral, when (y, ) 2 is small wrt, one can simplify these equations. This is what we call the shallow cartesian formulation. The r esultin approimate equations are: α -- y, t i+ y, j t 2 y, i + j cos ψ si n ψ y, Prof Connor Pae 2 of 24

22 2.4.5 Shallow Cartesian Solution - Parabolic Geometry b Y V B, v B M B, β B B F B, u B y = -- 2 a2 Use Maruerre formulation, assume thin and shallow b, b = constant = b, m y y, = a at u = v = β (fi ed support) at = L N = N B N y = N By M = M B Governin equations Equilibrium F, V, + (Fa), + b M, + V Force Deformation (set u = u ; v = v ) F = u, + av, D S V = v, β D T M = β, DB Boundary Conditions at = L at F = N B u = v = β V = N By alf M = M B X Prof Connor Pae 22 of 24

23 Solution for the internal forces F, F = C = N B V, + (N B a), + b V = b ( a )N B + C 2 N By alf = b ( a)n B + C 2 C 2 = N By + bl V = b ( a )N B + N By + bl V = N By + b( L ) ( a)n B Solution for displacements M = N By + b 2 2 L a---- N B + C L L M B = N By L+ b----- a---- N B + C L L C 3 = M B N By L b a----- N 2 2 B b an B 2 2 M = --(L ) (L ) + (L )N By + M B = D B β, b an B 2 3 D B β = --(L ) L L ---- N By + M B + C 4 b 3 C 4 = --L 6 V v, = β D T 2 2 v = N By + bl ---- a---- N D 2 2 B T an 4 B bl---- L L L N M B + C D B By C 5 Prof Connor Pae 23 of 24

24 F u, = av, D S F u, = V ( a) β D S D T a N B a al 2 a u = N By N B + b D 2 3 S D T a ab L ---- L a N B L an ---- By L M B D + C 6 B C 6 Fleibility matri - cartesian formulation Set u B = u B v B F B = β B N B N By M B u B = u Bo, + T B u + f B F B T B = 0 h 0 L 00 N = N, o N B N y = N y, o N By M = M, o M B LN By + hn B Prof Connor Pae 24 of 24