1.571 Structural Analysis and Control Prof. Connor Section 5: Non-linear Analysis of Members. π -- γ. v 2. v 1. dx du. v, x. u, x.

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1 5. Deformation nalysis Y.57 Structural nalysis and Control rof. Connor Section 5: Non-linear nalysis of Members dy ( + ε y )dy π -- γ y ( + ε ) X 5.. Deformation of a fiber initially aligned with. u u u u v v + u u + du + u u u du ( + u, ) v v v ( + ε ) v, dv dv v, ( + u, ) ( + ε ) ( + u, ) + v, + ε + ε + u, + u, + v, ε ( + ε ) u, ( + u, ) + v, ε u, ε v, u, For small strain ( ε «) u, v, ε u, Structural nalysis and Control Section 5 rof Connor age of 7

2 If the member does not eperience large relative rotations, then the non-linear term can be ignored. u, Then for small relative rotations ε u, + --v, 5.. Deformation of a fiber dy initially aligned with y. roceed as done for to obtain: For small strain v, ε y v, y y + --u, y For small relative rotations ε y v, y + --u, y 5..3 Shear deformation ~ u, y For small rotations ~ v, γ u, y + v,.57 Structural nalysis and Control Section 5 rof Connor age of 7

3 5. Straight eams - Non-inear nalysis for lane Members Y y 5.. ial Strain For small rotations X u u ysinβ v v y( cosβ ) v β «sinβ β cosβ ε ε ( u ) ( v) ( u ysinβ) -- v ε u, yβ, + -- ( v, ) ε ε yβ, ε o yχ 5.. Shear Strain γ γ u v y β + v, ( u y ysinβ) ( v ).57 Structural nalysis and Control Section 5 rof Connor age 3 of 7

4 5..3 Summarizing Define ε ( y) ε ' γ γ' ε ' ε yβ, ε u, + --v, γ' v, β 5..4 pply the principle of virtual displacements - Take a deformable body resisting some eternal loads - roduce a perturbation (a virtual displacement) - If body is in state of equilibrium, the first order work done by the stresses is equal to the first order work done by the eternally applied forces i.e. ( σ ε ) d Vol Forces Virtual Displacements For a beam: Internal virtual work is caused by stresses σ, τ y {( σ 'δε ' + τ y 'δγ y ) d } d Forces Virtual Displacements Then ε ' ε yβ, ε yχ ε u, + --v, u v ( ε + δε ) ( u + δu) ( v δv) ( ε + δε ) u δu v δv ( ε + δε ) u δu v v δv δv δε δu v δv δv δu, + v, δv, β δβ y( χ + δχ) y ( + ) y β y δβ yδχ y δβ y β,.57 Structural nalysis and Control Section 5 rof Connor age 4 of 7

5 Therefore δε ' δε yδβ, Right hand side of the virtual work equation Giving v γ' v, β β γ ( v + δv) γ' + δγ' β δβ γ+ δγ δv δγ' δβ δv, δβ δγ {( σ 'δε ' + τ y 'δγ y ) d} σ 'δε ' ( σ 'δε σ yδβ, ) d σ 'δε ' ( Fδε + Mδβ, ) τ y 'δγ' y d Vδγ RHS ( Fδε + Mδβ, + Vδγ) Integrating by parts uv' uv u'v RHS ( Fδu + Fv, δv + Mδβ + Vδv) δuf, δv ( Fv, ) V, + + δβ( M, + V) eft hand side of the virtual work equation HS b y δv + b δu + mδβ + δp Differentiating both RHS and HS wrt, recognizing that the first RHS term is a constant and that both the second RHS term and the HS are integrated over the same interval, we get ( F, + b )δu + ( V, + ( Fv, ), + b y )δv+ ( M, + V+ m)δβ.57 Structural nalysis and Control Section 5 rof Connor age 5 of 7

6 Since δu, δv, and δβ are independent, each term becomes and independent epression F, + b V, + ( Fv, ), + b y M, + V+ m These are the governing equations of equilibrium for a non-linear member. Note: in V, + ( Fv, ), + b y, ( Fv, ), source of -δ effect. 5.3 Compatibility Equations F σ ' d Eε ' d E u, + --v, yβ, d F u, Ed+ --v, E d β, yed If y is measured from the mechanical centroid ye d Then where where F D S u, + --v, M Ed yσ ' d u, + --v, Eyd + β, Ey d M D β, D Ey d M τ y d Gγ' d Gv, ( β) d V ( v, β) Gd.57 Structural nalysis and Control Section 5 rof Connor age 6 of 7

7 Summarizing F, + b V, + ( Fv, ), + b y M, + V+ m F D S u, + --v, M D β, V ( v, β) oundary conditions u or F β or M v or V+ Fv, y Note y effective shear 5.3. Eamples of boundary conditions a k r k u v β F M k r β( ) ; M without spring V v, k e v ( ) u v β F β V v,.57 Structural nalysis and Control Section 5 rof Connor age 7 of 7

8 Eample b y b m D,, constant F, + b F ( ) constant D S u, + --v, u, v, u ( ) u ( ) v, M D β, M, D β, M, + V+ m D β, + V V D β, V v, D β T v, V β β D β, V, + ( Fv, ), + b y D β, v, + b y D D β, β, β, D + b y T D β, β, + b y Define β, β, D µ D b y D Structural nalysis and Control Section 5 rof Connor age 8 of 7

9 Solving for β and then v, starting with β, + µ β, µ -----b y β C + µ ( C 4 sin µ + C 5 cosµ ) + β part v β D β, v C + C 3 + C 4 cosµ + C 5 sin µ + v part For b y M D β, ( C 4 cosµ C 5 sinµ ) + D ( β part ), V D β, V µ( C 4 sinµ + C 5 cosµ ) D ( β part v C 3 + C 4 M C 4 C 4 C v C + C 5 sinµ M C 5 sinµ C For a non-trivial solution sinµ (ie µ nπ, n,, ) For So n µ π µ π π D cr π D π D Structural nalysis and Control Section 5 rof Connor age 9 of 7

10 5.4 Member Relations Geometrically Non-inear Case Y, v u β β v v v v u V M X, u V F v v F ( µ) λ D D EI D ( cosµ) µ sinµ D µ ( sin µ µ cosµ ) D µ ( µ sinµ) + ssumption: Neglect transverse shear deformation wrt bending deformation u u v u u v β β F F V F F V M M.57 Structural nalysis and Control Section 5 rof Connor age of 7

11 5.4. Member Equations u u v, Set e r -- --v, u u e r F ( u u + e r ) (member in compression) M M D ( β + β ) D ( β + β ) V V Write member equations as If F V M ikewise D D D V ( β + β ) D D u v β Forces due to span loads D D u F k T ( r) () i u + k u + F + F Note: This formulation is wrt local frame and must be transformed to use global frame 3 D D v β e r ( r) F k u k u F () i F F V M ( r) F k u + k u + F + D D D D u v β () i F D D u D D v β e r F k T ( r) u k u F.57 Structural nalysis and Control Section 5 rof Connor age of 7

12 5.5 pplications. Frame W W H I H I C I C Consider I H W v oundary Conditions β β v v ---- v v W V H End actions at Write y D V D v W W v H D M v V H kv For W k D W D If we apply 6 k D W cr π D µ π π k D W π for W 6 cr cr.57 Structural nalysis and Control Section 5 rof Connor age of 7

13 Cantilever W H, v oundary Conditions β v M V H W Solution v M β ---- V D H β v ---- v ---- k H v Note, from elementary mechanics for a cantilevered column D k H Then If W cr π D ( ) π D If W t the critical load 3D k H 3 inear case k H µ D ( D ) π -- π D -- π D -- π D 3 -- k H.57 Structural nalysis and Control Section 5 rof Connor age 3 of 7

14 t -- cr µ π cosµ.445 sinµ k H D π D So, about a 4% reduction in stiffness due to aial loading. 3 Simply Supported eam-column M * v oundary Conditions Solution M v M M * v M β + β If β ---- β M D β k β k 3 D (inear case).57 Structural nalysis and Control Section 5 rof Connor age 4 of 7

15 4 Multi Span eam-column Hv, C C Segment - (from eample 3) M, β M k β Segment -C k β Hv, C C β C v C oundary Conditions M C v M k β V C H M, β (be careful of sign convention) H Solution i) ii) M M C D v C β + β C k D v C β C + β β Use i) and ii) to solve β and β C in terms of v C D D D v C β k β C β + β C ---- v 3 C Then substitute in epression for H H D β C β v C v C Structural nalysis and Control Section 5 rof Connor age 5 of 7

16 5.5 Incremental Formulation 5.5. Some simplifications. v v Take v, Then e r v, e r ( ) ssume s are constant during incremental motion. This results in k, k, and k constant. (ie dk ) 3. Consider small increment, ie work with first order change. ( r) df k u + k u + df + () i df df T ( r) u + k u + df + k () i df ( r) df ( r) df ( r) df df ( r) e r ( ) F F r D S F F D S ( u u ) + e r operating on df ( r) df r) F d + df d v d v D df S ( u u ) v v Structural nalysis and Control Section 5 rof Connor age 6 of 7

17 Then df ( r) F ( v v ) ( v v ) D S + ( u u ) ( v v ) df ( r) K R ( u u ) 5.5. Summarizing K R F D S i df ( k + K R ) u + ( k K R ) u + df () df ( k + K R ) T i u + ( k + K R ) u + df () These two epressions define the tangent stiffness for a member. One usually retains the non-linear terms only when the aial force is compressive since the stiffness is reduced. tensile aial force increases the stiffness inearized Stability nalysis. Take in K R.. Find aial forces with linear analysis (ie ignore K R.) Evaluate s. 3. Test determinant of tangent stiffness matri..57 Structural nalysis and Control Section 5 rof Connor age 7 of 7

4.5 The framework element stiffness matrix

4.5 The framework element stiffness matrix 45 The framework element stiffness matri Consider a 1 degree-of-freedom element that is straight prismatic and symmetric about both principal cross-sectional aes For such a section the shear center coincides