NATIONAL TECHNICAL UNIVERSITY OF ATHENS NONUNIFORM TORSION, UNIFORM SHEAR AND TIMOSHENKO THEORY OF ELASTIC HOMOGENEOUS ISOTROPIC PRISMATIC BARS

Transcript

1 NATIONAL TECHNICAL UNIVERSITY OF ATHENS SCHOOL OF CIVIL ENGINEERING INTER-DEARTMENTAL OSTGRADUATE COURSES ROGRAMMES «ΔΟΜΟΣΤΑΤΙΚΟΣ ΣΧΕΔΙΑΣΜΟΣ ΚΑΙ ΑΝΑΛΥΣΗ ΚΑΤΑΣΚΕΥΩΝ» ANALYSIS AND DESIGN OF EARTHQUAKE RESISTANT STRUCTURES NONUNIFORM TORSION, UNIFORM SHEAR AND TIMOSHENKO THEORY OF ELASTIC HOMOGENEOUS ISOTROIC RISMATIC BARS Lecturer : Ε. J. Sapountzakis rofessor NTUA COURSE : ALIED STRUCTURAL ANALYSIS OF FRAMED AND SHELL STRUCTURES (A1) (1)

2 Eccentric Transverse Shear Loading Q Shear Loading Q Direct Torsional Loading Μ t =Q e) (Direct: Equilibrium Torsion, Indirect: Compatibility Torsion) Nonuniform shear stress flow due to Q Nonuniform shear stress flow due to M t Warping due to transverse shear Warping due to torsion ()

3 Due to Torsion WARING OF A HOLLOW SQUARE CROSS SECTION Due to Shear (3)

4 CROSS SECTIONS EXHIBITING SMALL AND SIGNIFICANT WARING SMALL WARING (Closed shaped cross sections) INTENSE WARING INTENSE WARING (Open shaped cross sections) (4)

5 COMARISON OF TORSIONAL DEFORMATIONS OF THIN WALLED TUBES HAVING CLOSED AND OEN SHAED CROSS SECTIONS θ 0 θ 0 θ 1 r m r I m = 100mm t = 1mm Close t = 30000I Open t θ 1 w r m t ο t t Distribution of primary shear stresses Distribution of primary shear stresses (5)

6 CLASSIFICATION OF TORSION AS A STRESS STATE Direct Torsion (Equilibrium Torsion) Indirect Torsion (Compatibility Torsion) Bridge deck of box shaped cross section curved in plan (ermanent) torsional loading due to self-weight Cracking due to creep and shrinkage effects Significant reduction of torsional rigidity (6)

7 Classification of shear & torsion according to longitudinal variation ation of warping (UNIFORM - NONUNIFORM SHEAR AND TORSION) Transverse Load, Twisting Moment: Constant Warping (Q, M t ): Free (Not Restrained) Uniform Shear Torsion Shear Stresses Exclusively (Saint Venant, 1855) Transverse Load, Twisting Moment: Variable Warping (Q, Mt): Restrained Nonuniform Shear Torsion (Wagner, 199) Shear Stresses (rimary (St. Venant)) Stresses due to Warping (Normal stresses & Secondary shear stresses) (7)

8 SHEAR STRESS DISTRIBUTION (NONUNIFORM TORSION) rimary Shear Stresses (torsional loading) Secondary (Warping) Shear stresses (torsional loading) E, G l h b M t Closed Bredt stress flow, 1896 Complex distribution in thick walled cross sections (thin-walled: Vlasov, 1963) (8)

9 RANDTL S S MEMBRANE ANALOGY Saint Venant (uniform) torsion has been depicted by randtl (1903) through the membrane analogy: Uniform torsion and membrane problems are described from analogous boundary value problems. φ 3 =0 Membrane Membrane φ 1 max Rectangular Cross section φ 3 φ 3 φ (φ 1 >φ ) φ 1 max T-shaped cross section The deformed membrane offers the following information: Contours correspond to the directions of the trajectories of shear stresses The slopes of the deformed membrane correspond to the values of shear stresses The volume of the deformed membrane corresponds to St. Venant s torsion constant (9)

10 SHEAR FORCE TWISTING MOMENT Warping due to shear Warping due to torsion Warping due to shear << Warping due to torsion (Significant in Open Shaped Cross Sections) Stress State (Stress field): Uniform Shear Strain/Deformation State: Shear Deformation Coefficients Indirect account of warping deformation (Timoshenko, 19) Stress State & Strain State: Nonuniform Torsion Seven degrees of freedom (14x14 [K]) Additional dof.: Twisting curvature Additional stress resultant: Warping moment (10)

11 roblem description of Nonuniform Torsion & Uniform Shear Technical Beam Theory Limited set of cross sections (of simple geometry) Warping restraints are ignored Compatibility equations are not employed Stress computations are performed studying equilibrium of a finite segment of a bar and not equilibrium of an infinitesimal material point (3d elasticity) Thin Walled Beam theory (Vlasov theory, 1964) Generalized Beam Theory (Schardt, 1966) Valid for thin walled cross sections (Midline employed) Warping restraints are taken into account Reliability: Depends on thickness of shell elements comprising the beam Valid for arbitrarily shaped cross sections (Thick or Thin walled) Warping restraints are taken into account BVs formulated employing theory of 3D D elasticity Numerical solution of BVs (11)

12 Analysis of Bars and Bar Assemblages Direct Stiffness Method lane Steel Frame lane Steel Grid Spatial Steel Frame Everyday Engineering ractice: Application of 1x1 Stiffness Matrix (6 dofs per node) Approximate Computation of Torsion Constant Approximate Computation of Shear Deformation Coefficients Approximate Computation of stresses due to shear and torsion Inaccuracies Non conservative Design (sometimes) (1)

13 ASSUMTIONS OF ELASTIC THEORY OF TORSION The bar is straight. The bar is prismatic. The bar s s longitudinal axis is subjected to twisting exclusively. Distortional deformations of the cross section are not allowed (cross( sectional shape is not altered during deformation (γ 3 =0, distortion neglected). Twisting rotation is considered small: Circular arc displacements s are approximated with the corresponding displacements along the chords. The material of the bar is homogeneous, isotropic, continuous (no o cracking) and linearly elastic: Constitutive relations of linear elasticity are valid. The distribution of stresses at the bar ends is such so that all the aforementioned assumptions are valid. Especially for (unrestrained) uniform (Saint Venant) torsion the following assumption is also valid: Longitudinal displacements (warping) are not restrained and do not n depend on the longitudinal coordinate (every cross section exhibits the same warping w deformations). (13)

14 a=0 x 3 L x b=l Displacement Field ELASTIC THEORY OF TORSION Ε, G t Γ t ( ) m= m x x 1 = 1 K + 1 Γ j = 1 j t s x 3 M x3 Ο x 3 M n M x u x,x,x x x,x ω 1 Ρ θ(x ) ( ) =+ θ ( ) ϕ ( ) M 3 u ω x x u 3 Ρ (Ω) M: C.of T.=S.C. (ν=0) t 3 = 0 Γ Κ+1 Γ Κ t = 0 Γ 1 Γ u ( x,x,x ) = ( )sin ω = (M ) θ (x )sinω = x θ (x ) u ( x,x,x ) = ( )cos ω = (M ) θ (x )cosω = x θ (x ) (14)

15 ELASTIC THEORY OF TORSION Components of the Infinitesimal Strain Tensor u ε θ ϕ ( x ) ( x,x ) 1 11 = = 1 1 M 3 x 1 ε 11 = 0 St.V. ε u = = x 0 ε 33 u3 = = x 3 0 γ 3 u u3 = + = x x 3 0 γ u1 u ϕ 1 = + = θ1 ( x 1 ) Μ x 3 x x 1 x γ u u1 3 ϕ 13 = + = θ1 ( x 1 ) Μ + x x 3 x 1 x 3 (15)

16 ELASTIC THEORY OF TORSION Components of the Cauchy Stress Tensor (ν=0) E τ = ν ε + ν ε + ϕ ( )( ) ( 1 ) ( ε 3 ) E 1 ( x ) M ( x,x3 1 ν 1 ν = θ + ) τ 11 = 0 St.V. E τ = 1 ν ε ν ε11 ε33 0 ν ν + + = ( )( ) ( ) ( ) 1+ 1 E τ33 = 1 v ε33 + v ε11 + ε = 0 ( )( ) ( ) ( ) 1+ v 1 v τ = G γ = ϕ τ M 1 = G γ1 = G θ ( x ) x x ϕ τ M 31 = G γ31 = G θ ( x ) + x 1 1 x 3 (16)

17 ELASTIC THEORY OF TORSION Differential Equilibrium Equations of 3D Elasticity (Body forces neglected) ϕm G θ1 ( x 1) x 3 = 0 x ϕ G θ M 1 ( x 1) + x = 0 x 3 Not Satisfied! Inconsistency of Theory of Nonuniform Torsion: Overall equilibrium of the bar is satisfied (energy principle). However, only the longitudinal equilibrium equation (along x 1 ) is satisfied locally (St.V. Identical satisfaction of all diff. equil. eqns) ϕm ϕm G θ1 ( x 1) x 3 + G θ1 ( x 1) + x + Ε θ1 ( x1) ϕm = 0 x x x3 x3 x 1 ε = θ ϕ ϕm ϕm x x 3 ( ) ϕ : ϕ x,x,x M M 1 3 ( x ) ( x,x ) M 3 ( x ) ( x ) Ε θ1 1 + = ϕ G θ Inconsistency 1 1 M DECOMOSITION OF SHEAR STRESSES (17)

18 hysical Meaning of Decomposing Shear Stresses τ σ x (A) τ S 1 = τ S = τ S Cauchy rinciple M t τ S 1 τ S σ x +dσ x M t hase Ι rimary Twisting : Moment Angle of Twist er Unit Length (Torsional Curvature): Constant St.V. M w : hase ΙΙ Warping Moment S M t : Secondary Twisting Moment hase ΙΙΙ (18)

19 rimary (St. Venant) Shear Stresses τ τ ϕm 1 = G θ1 ( x 1) x 3 x ϕm 13 = G θ1 ( x 1) + x x 3 Secondary (Warping) Shear Stresses τ τ S 1 S 13 Normal Stresses ϕ = G x = G ϕ x S M S M 3 ELASTIC THEORY OF TORSION ( ) ( ) w 11 = E 1 x1 M x,x3 τ θ ϕ S S w τ11 τ 13 τ1 τ13 τ = 0 x x x x x τ x 1 S 1 τ + x 13 3 S 13 τ τ τ + + x x ϕ M = 0 = 0 w 11 3 x 1 E θ ( x ) 0 S 1 1 ϕm ϕm = G = Reliability of the shear stress decomposition ( ) ϕ ( ) S 1 1 = θ 1 1 M 3 u u x x,x (19)

20 ( ) ( ) n = cos x,n = cosα n = sin x,n = sinα 3 3 Γ Ω τ τ dx 3 τ 1t t ds dx ELASTIC THEORY OF TORSION Boundary Conditions τ 13 α τ 1n ϕm 1n G θ1 ( x1 ) x3 n 3 S 1n S M S M = 0 n τ 1 M Shear stresses along the normal direction n on the boundary VANISH: 1n S 1n S 1 S 13 3 τ = τ n + τ n = 0 τ = τ n + τ n = 0 ϕ = + x n = 0 = x 3 n x n n n ϕ = G = 0 n ϕ n ϕ M S 1t 1 ( 1 ) 3 3 1t ϕ τ = G θ x + x n + x n τ = G t t S M 3 (0)

21 ELASTIC THEORY OF TORSION Boundary Value roblems rimary Warping Function Laplace differential eqn with Neumann type boundary conditions ϕm ϕ ϕ M M x x 3 M = x 3 n x n 3, = + = 0, Ω ϕ n Secondary Warping Function oisson differential eqn with Neumann type boundary conditions Γ ( x ) S S S ϕm ϕm Ε θ1 1 M M x G x 3 S ϕm ϕ = + = ϕ, Ω n = 0, Γ (DEs) (1)

22 Twisting Moment: rimary Twisting Moment: ELASTIC THEORY OF TORSION Stress Resultants S 1 = M M M M x x d M G I x ϕ M ϕ M 1 = τ1 3 + τ13 + Ω 1 = t θ1 1 x x Ω 3 I M M t x x 3 x ϕ x ϕ = x d Ω Torsion constant 3 x Ω (Saint-Venant) Secondary Twisting Moment: M S S ϕm S ϕ M S 1 = τ1 τ13 d Ω M 1 = E C M θ1 x 1 x x Ω 3 ( ) ( ) C M = Ω ( ϕ ) M dω Warping Constant (Wagner) ()

23 Warping Moment: ELASTIC THEORY OF TORSION Stress Resultants w W = ϕmτ11 Ω W = Mθ 1 Ω Schardt, 1966: Higher order Stress M d M EC Warping Moment as External Loading Normal Stresses with Nonuniform Distribution Bending Moments applied in planes parallel to the longitudinal bar axis located at distance from the center of twist Concentrated Axial Forces: K Mw = ϕ j M j= 1 ( ) ( ) e.g. Z-shaped cross section with equal length flanges ϕm 0 at centroid j Resultant Difficult visualization/depiction Self-equilibrated stress distribution h M f M f σ xx M w = M f h (3)

24 Warping Moment due to Axial loading z (Roik, 1978) z y z y y /4 /4 Axial loading (Ν) z /4 /4 Bending Loading (Μ z ) z /4 /4 /4 y /4 y /4 /4 /4 /4 Bending loading (Μ y ) /4 /4 /4 /4 Warping Moment Loading (Μ w ) (4)

25 M M 3 ELASTIC THEORY OF TORSION Center of Twist (Μ) x,x : oint with respect to which the cross sections rotate (no transverse 1 13 t displacements) (or point where rotation causes no axial and bending stress resultants τ, τ,i : Independent of the center of twist (St. Venant could not calculate the S S w M position of the center of twist!) u, τ, τ, τ,c : Dependent of the center of twist ( ) M M ϕm x,x 3 = φo(x,x 3 ) xx3 + x3x + c ϕo ϕo = 0, Ω = x3 n x n 3, Γ n Method of equilibrium: Under any coordinate system due to warping normal stresses Energy Method: Minimization of Strain Energy due to warping normal stresses N = M = M3 = 0 CM CM CM = = = 0 x x c 3 (5)

26 where: ELASTIC THEORY OF TORSION Center of Twist (Μ) Μ Μ = S Μ Μ = Μ Μ = 3 S x S x Ac R I x I x S c R I x I x S c R A= dω S = x dω S = x dω 3 3 Ω Ω Ω = 3 33 = 3 = 3 Ω Ω Ω S = O = 3 O 3 = O Ω Ω Ω I x dω I x dω I x x dω R ϕ dω R x ϕ dω R x ϕ dω (6)

27 ELASTIC THEORY OF TORSION Global Equilibrium Equation & Boundary conditions Method of Equilibrium or Energy Method x 1 mt dx 1 M M dx 1 x 1 dx 1 x 3 x M 1 TOTAL OTENTIAL ENERGY 1 1 Π θ θ θ L oλ = 0 GI t 1 + EC Μ 1 mt 1 dx 1 F d F d F 0 θ θ + θ = 1 dx 1 1 dx 1 1 F (Euler Lagrange eqns) Torsional Damping Coefficient mt = G It θ1 + E CM θ1 aθ + a M = a βθ + β M = β W 3 GIt 15 ε = L : ECM < 15 Inside the bar interval Uniform Torsion Nonuniform Torsion At the bar ends (Ramm & Hofmann 1995) (7)

28 ELASTIC THEORY OF TORSION Alternative Solution of the Uniform Torsion roblem Conjugate function ψ of function ϕ M ψ ψ ψ = + = x x3 1 ( ψ = x ) + x3 + C ψ randtl Stress function F(x,y) 0 ψ ψ τ1 = Gθ x3 τ13 = Gθ + x x3 x ψ ψ It = x + x3 x x3 d Ω Ω x x3 F F = + = x x3 F = C F F 1 F F τ = Gθ τ = Gθ 1 13 x3 x F F It = x + x3 dω Ω x x3 Constants C ψ,c F are unknown and must be determined at each boundary of a multiply connected region (occupied by the cross section) Complex roblem. (8)

29 Example r=1cm Steel rofile UE 100 t f =0.75cm Bar ends simply supported Loading: mt = 1kNm m Length: l=1.0m ε = 366. < 15 h=10cm M C e MC t w =0.45cm Ε=.1Ε8, ν=0.3 b=5.5cm FΕΜ (Kraus, 005) ΒΕΜ maxux ( cm ) = ( x = 0010.,. m) ( σ w ) S 4 maxux ( cm ) = ( x = 0010.,. m) φm φ m S x maxu = maxu x ΒΕΜ Warping along the thickness direction IS NOT CONSTANT Thin walled beam theory not valid (9)

30 Example My (knm): Καμπτική Ροπή (min/max: 500) Mw (knm): Διρροπή (min: , max: ) z 0.45 Β m =100kN l/=0m ε = 3.3> m Α y 3.45 (E=3.0Ε7, G=1.5E7) Bar of Box Shaped Cross Section Clamped at Both Ends Bar Length (m) φ M rimary Warping σb: Κάμψη (θλ: , εφ: ) σw: Διρροπή (θλ: , εφ: ) σtot: Κάμψη & Διρροπή (θλ: , εφ: ) NORMAL STRESSES σb: Κάμψη (θλ: , εφ: ) σw: Διρροπή (θλ: , εφ:+.61) σtot: Κάμψη & Διρροπή (θλ: , εφ: ) -40 oint Α Discrepancy 30% oint Β Bar Length (m) Bar Length (m) (30)

31 Example 3.65m =100kN l/=0m Mtp: Πρωτογενής Στρεπτική Ροπή (min/max: 180.7) Mts: Δευτερογενής Στρεπτική Ροπή (min/max: 18.50) Mt: Συνολική Στρεπτική Ροπή (min/max: 18.50) ε = 3.3>15-50 Bar of Box Shaped Cross Section Clamped at Both Ends Bar length (m) SHEAR STRESSES Discrepancy 30% (31)

32 Computation of Shear Stresses Computation of Shear Center osition α Β Γ Δ Q z z Β 1 Β 3 (C: Centroid) C Δ 1 Δ 3 b(y) α UNIFORM SHEAR BEAM THEORY Computation of Shear Deformation Coefficients (required for Timoshenko beam theory) Displacement Field GBT BV Shear Stresses TWBT ΤBT y Q z : τ xz τ = xy QS z I = yy 0 Stress Field cut y b oisson ratio ν taken into account Cross sections possessing at least one axis of symmetry Simply connected cross section oisson ratio ν neglected τ xz : constant along the width b Β 1 Β 3 & Δ 1 Δ 3 τ xz : vanishing Β Δ point Γ τ xz : Discontinuity (3)

33 ASSUMTIONS OF ELASTIC THEORY OF UNIFORM SHEAR The bar is straight. The bar is prismatic. Distortional deformations of the cross section are not allowed (γ3=0,( distortion neglected). The material of the bar is homogeneous, isotropic, continuous (no o cracking) and linearly elastic: Constitutive relations of linear elasticity are e valid. The distribution of stresses at the bar ends is such so that all the aforementioned assumptions are valid. Deflections and bending rotations are considered to be small (geometrically linear theory). Longitudinal displacements (warping) are not restrained and do not depend on the longitudinal coordinate (every cross section exhibits the same warping deformations). (33)

34 THEORY OF UNIFORM SHEAR DISLACEMENT FIELD x 3 x 1 x l-x 1 Q C Displacement field ( ) = θ ( ) θ ( ) + ϕ ( x,x ) u x,x,x x x x x ( ) = ( ) u x,x,x u x 1 3 S M ( ) = ( ) u x,x,x u x l Q M 3 C S θ Q 3 Q 3 S x 1 c 3 By ignoring Shear Strains: u ( x ) θ ( x ) u 3 1 = 3 1 = x1 x1 (34)

35 THEORY OF UNIFORM SHEAR DISLACEMENT FIELD Components of the Infinitesimal Strain Tensor u θ θ u u ε = = x x ε = = 0 ε = = x1 x1 x1 x x3 ε 1 u u 1 u ϕ 1 ϕ 1 c c 1 = + = θ3 + + = x x1 x1 x x ε 3 1 u u3 = + = x3 x 0 ε 1 u u 1 ϕ u 1 ϕ 1 3 c 3 c 13 = + = θ + + = x3 x1 x3 x1 x3 (35)

36 THEORY OF UNIFORM SHEAR DISLACEMENT FIELD 1 3 Components of the Cauchy Stress Tensor (ν=0) 3 11 E 11 E θ x θ 3 x τ = ε = τ = 0 τ33 = 0 τ3 = 0 x1 x1 ϕc ϕc τ1 = G ε1 = G τ13 = G ε13 = G x x 3 Differential Equilibrium Equations of 3D Elasticity (Body forces neglected) τ11 τ1 τ31 ϕc ϕc + + = 0 E( θ x3 θ3 x) + G + G = 0 x x x x x τ τ τ τ = 0 1 = 0 x1 x x3 x 1 τ13 τ3 τ33 τ13 : + + = 0 = 0 x1 x x3 x1 Identical Satisfaction 3 (36)

37 THEORY OF UNIFORM SHEAR DISLACEMENT FIELD Determination of: E ( θ x θ x ) 3 3 M = τ11x3 dω = E θ x3 dω θ3 xx3dω = E( θ I θ3 I 3) Ω Ω Ω M3 = τ11x dω = E θ xx3dω θ3 x dω = E( θ3 I33 θ I 3) Ω Ω Ω Ι, Ι 33, Ι 3 : Moments of inertia M Q3 = = E( θ I θ3 I3) x Equilibrium of 1 bending moments M 3 Q = = E( θ I3 θ3 I33) x1 ( θ θ ) ( ) + ( ) QI QI x QI QI x = II33 I3 E x x (37)

38 THEORY OF UNIFORM SHEAR DISLACEMENT FIELD oisson artial Differential Equation ϕc ϕc 3 3 x x3 ( ) ( ) ϕc x, x = + = f x, x, Ω 1 g( x, x ) = ( Q I Q I ) x ( Q I Q I ) x D Γ Ω dx 3 τ 1t t ds dx τ 13 α f 1 G ( x, x ) = g ( x, x ) 3 3 Boundary Condition τ 1n n τ 1 D = II33 I3 τ1n = τ1n + τ13n3 = 0 ϕc ϕc ϕc G = G n + G n3 = 0 n x x3 ϕc = 0, Γ (Neumann) n (38)

39 Beam theory: 3 THEORY OF UNIFORM SHEAR STRESS FIELD ΜΙ +Μ Ι ΜΙ +Μ Ι τ & τ11 = x + x = τ33 = τ3 = 0 M M 3 Q 3 =,Q = (statically determinate beam x1 x1 M,M are computed through the global equilibrium equation s) Analysis: Q ΙΙ33 -I3 ΙΙ33 -I3 Differential Equilibrium Equations of 3D Elasticity (Body forces neglected) τ11 τ1 τ 31 τ 1 τ 31 Q3 + + = 0 + = xi x x x x x I I I 1 3 (Q correspondingly and subsequent superposition of results) τ1 τ τ 3 τ1 + + = 0 = 0 x1 x x3 x 1 : τ13 τ3 τ33 τ = 0 = 0 x1 x x3 x ( xi ) Shear stresses depend on x & x 3, exclusively, that is they are the same at each cross section of the bar (39)

40 THEORY OF UNIFORM SHEAR STRESS FIELD τ ε 11 = E ε Components of the infinitesimal strain tensor 11 τ G 1 1 = = ε1 3 ν ε = ε = τ = νε E ( x,x ) ε τ = = ε13 3 G Compatibility Equations ( x,x ) Strain field Satisfies 4 compatibility equations identically τ13 τ1 νqi 3 33 = x x x 1+ I I I ( )( ν ) τ13 τ1 νqi 3 3 = x x x 1+ I I I ( )( ν ) (ANALYSIS: Q 3 ) ( v 0) ε 3 = 0 (40)

41 THEORY OF UNIFORM SHEAR STRESS FIELD Stress Function (Analysis: Q 3 ) Shear stresses: They satisfy Compatibility Equations Identically ( ) 3 τ Φ x,x : Q d= d i + d i = Φ Q Φ = d τ13 = d3 B x B x3 Stress function with continuous partial derivatives up to nd order 3 3 ν I x x I I I x x x x3 x x i + ν i3 ( )( ν ) 33 3 B= 1+ I I I (41)

42 THEORY OF UNIFORM SHEAR STRESS FIELD oisson artial Differential Equation Φ Φ Φ, = + =, Ω Γ Ω dx 3 ( x x ) ( x I x I ) τ 1t x x3 t ds dx τ 13 Boundary Condition α τ 1n n τ 1 (Analysis: Q 3 ) τ1n = τ1n + τ13n3 = 0 Φ = dn + dn 3 3 = n d, Γ n (Neumann) (4)

43 THEORY OF UNIFORM SHEAR STRESS FIELD (Analysis: Q ) oisson artial Differential Equation ( I x I x ) Θ=, Ω 3 3 Boundary Condition Θ = en + en 3 3 = n e, Γ n (Neumann) e= e i + e i = 3 3 ν I I x x I I x x x x3 x x3 3 3 i + ν i3 (43)

44 THEORY OF UNIFORM SHEAR Shear Center (S) oint where Internal Shear Stress Resultant is subjected oisson ratio ν = 0 S.C. (S) coincides with C. of T. (Μ) (Weber, 194), (Trefftz, 1935) Determination: With respect to an arbitrary point Μ t ext = Μ t int Μ t ext : Twisting Moment at S.C. arising from externally applied forces Μ t int : Twisting Moment arising from shear stresses due to transverse shear K1 + j 1 Γ = = Γ j s Γ x 3 t n Γ 1 s Γ K+1 Γ K S C (G,v) (Ω) x Ext Q x Q x = S S 3 3 Int τ x τ x dω Ω ( ) (44)

45 Q THEORY OF UNIFORM SHEAR SHEAR CENTER DISLACEMENT FIELD Q = 0& Q3 = 1: = 1&Q = 0 : ϕ ϕ S cx cx = Ω 3 x3 x x G x x d ϕ ϕ S cx cx 3 3 = Ω 3 x3 x 3 3 x G x x d SHEAR CENTER STRESS FIELD S 1 Φ Φ Q = 0&Q3 = 1: x = x x3 xd3+ x3d dω B Ω x3 x 1 Θ Θ Q x = x x x e + xe3 dω S = 1&Q3 = 0: B Ω x x3 Ω Ω (45)

46 s3 3 a3 THEORY OF UNIFORM SHEAR τ γ = ( x,x) ( x,x) : Actual shear strains ( ) G γ Q3= Gκ 3Aγ 13 = GA s 3γ13 1 A = A= κ A: Effective Shear Area Ω ( x,x) γ 13 x,x 1 3 = : Average shear strains A γ13 : Shear strains of Timoshenko beam theory They need correction since they have unsatisfactory (constant) distribution Shear Deformation Coefficient α 3 (>1) Q 3 γ13 = a3γ13 Shear Correction Factor κ 3 (<1) 1 γ13 = γ13 = κ3γ13 a 3 (46)

47 THEORIES OF SHEAR DEFORMATION COEFFICIENTS 1) Timoshenko Theory (191, 19): ) Cowper Theory (1966): κ 3 = Average value of shear stresses Actual shear stress at centroid (if centroid does not lie in the cross section?) Global equilibrium equations formulated by integrating the 3d elasticity differential equilibrium equations 3) Energy Approach (Bach & Baumann, 194): The formulas of the approximate shear strain energy per unit length and the exact one are equated α 3 must depend on the ratio of the sides (b/h( b/h) that is then κ = 1 0 so that 1 γ13 = γ13 = κ3γ13 0 a h bh 3 a3 If α3 is independent of b/h then GA s EI 3 Unacceptably large values!!! 3 Unrealistic results FEM: Shear Locking (47)

48 Shear Deformation Coefficients Exact formula of shear strain energy per unit length = Approximate formula of shear strain energy per unit length U exact U appr τ1 + τ13 αq α3q3 α3qq3 dω = + + G ΑG ΑG ΑG Ω rincipal Shear System 3 tan S a ϕ = a a 3 rincipal Bending System 3 tan B I ϕ = I I 33 Bending Deflections: COULED ϕ = ϕ Axis of symmetry S B (48)

49 THEORY OF UNIFORM SHEAR SHEAR DEFORMATION COEFFICIENTS DISLACEMENT FIELD Q 0, Q = 0 : { } 3 AG ϕc ϕc a = + Q Ω x x3 Q 0, Q 0 : { = } 3 AG ϕc3 ϕc3 a3 = + Q3 Ω x x3 Q 0, Q 0 : { } 3 a AG ϕc ϕc + Q Ω x x3 dω dω AG ϕc3 ϕc3 AG ϕc3 ϕ c3 3 = dω + x Q Ω x3 Q Ω x x3 dω + dω (49)

50 THEORY OF UNIFORM SHEAR SHEAR DEFORMATION COEFFICIENTS STRESS FIELD { = } Q 0, Q 0 : α 3 Α = Θ e Θ e B { = } Q 0, Q 0 : α 3 { } ( ) ( ) Α = Φ d Φ d B 3 Ω Ω Q 0, Q 0 : α 3 ( ) ( ) Α = ( Φ d) ( Θ e) dω B 3 Ω dω dω (50)

51 Example Rectangular section: Q z =1kN bxh=60x30 (cm) 1.14ka 8.33ka h y b y ka Shear stresses τ Ω (ka) Λόγος oisson v=0.0 Λόγος oisson v= oisson ratio: ν= oisson ratio: ν=0.3 ΤBT: max Qz τ xz = 15. = 833. ka A Centroidal Axis yy Discrepancy 31% (51)

52 Example h b=30cm Shear Locking (Artificial, Spurious (Shear) Rigidity induced) GA sz EI yy b=30cm h Ενεργειακή Μέθοδος (Προτεινόμενη) Timoshenko Cowper Τεχνική θεωρία bh (5)

53 b=0.10m b b A b Example z Δ C Γ B m y Qz = 1kN v = Qy = 1kN v = b b T-shaped cross section b b τ Ω (ka) 0.4 Segment ΓΔ Qz = 1kN Qz = 1kN oisson ratio ν Does not affect shear center Λόγος oisson v=0.0 Λόγος oisson v=0.3 Τεχνική θεωρία δοκού Segment ΑΒ Λόγος oisson v=0.0 Λόγος oisson v=0.3 Τεχνική θεωρία δοκού τ Ω (ka) (53)

54 Example Thin walled L-shaped cross section h=15.5cm 1.495cm t=1cm z C rincipal Bending rincipal Shear φ B = (rad) y 8.6cm kn v=0.0:max τ Γ = cm kn v=0.3:max τ Γ = cm kn v=0.0:maxτ Γ = cm kn v=0.3:maxτ Γ = cm Qy = 1kN 3.995cm Qz = 1kN S φ S t=1cm b=10.5cm 4.1cm Small influence of oisson ratio ν in stresses and Shear Center at thin walled cross section bars (54)

55 ASSUMTIONS OF TIMOSHENKO BEAM THEORY The bar is straight. The bar is prismatic. Distortional deformations of the cross section are not allowed (γyz( yz=0, distortion neglected). The material of the bar is homogeneous, isotropic, continuous (no o cracking) and linearly elastic: Constitutive relations of linear elasticity are e valid. External transverse forces pass through the cross section s s shear center. Torsional and axial forces are not considered (torsionless( bending loading conditions). Deflections and bending rotations are considered to be small (geometrically linear theory). Cross sections remain plane after deformation. The distribution of stresses at the bar ends is such so that all the aforementioned assumptions are valid. Y y C S z Z L m Y m Z C p Y p Z S m Y p Y dx p Z m Z X C: centroid S: shear center x (55)

56 Y y C S z Z L m Y m Z m Z C p Z S p Y TIMOSHENKO BEAM THEORY m Y p Y dx p Z X C: centroid S: shear center x Γ = K j = 0 Γ j s yv, YV, t n Γ 1 α q r = q ω C Z, W z C y C S zw, Γ Κ Γ 0 Use of the principal shear system CXYZ passing through the centroid C Displacement Field: (Arising from the plane sections hypothesis) ( ) = θ ( )( ) θ ( )( y y ) = θ ( ) Z θ ( x) u x,y,z x z z x x v x,z ( ) ( ) w x,y = = v x ( ) ( ) Y C Z C Y Z w x ( ) ( ) θ Y x w x ( ) ( ) θ Z x v x Y (56)

57 ε ε xx yy TIMOSHENKO BEAM THEORY Components of the Infinitesimal Strain Tensor (Geometrically linear theory) u dθy dθ = = Z Z Y x dx dx v w = = 0 ε zz = = 0 y z γ yz v w = + = 0 z y γ γ xy xz u v dv = + = θ y x dx Z u w dw = + = + θy z x dx (57)

58 TIMOSHENKO BEAM THEORY Components of the Cauchy Stress Tensor (ν=0) E d Y d σ Z xx ν εxx ν εyy ε θ θ = + + zz = E Z Y 1 1 dx dx ( )( ) ( 1 ) ( ) + ν ν E σyy = 1 ν εyy ν εxx εzz 0 ν ν + + = ( )( ) ( ) ( ) 1+ 1 E σzz = 1 v εzz + v εxx + ε yy = 0 ( )( ) ( ) ( ) 1+ v 1 v τ yz = G γ = 0 yz xy xy Z ( θ v ) τ = G γ = G + τ xz xz Y ( θ + w ) = G γ = G (58)

59 TIMOSHENKO BEAM THEORY Differential Equilibrium Equations of 3D Elasticity (Body forces neglected) Not Satisfied! Inconsistency of Timoshenko Beam Theory: Overall equilibrium of the bar is satisfied (energy principle). The violation of the longitudinal equilibrium equation (along x) and of the associated boundary condition is due to the unsatisfactory distribution of the shear stresses arising from the plane sections hypothesis. Thus, in order to correct at the global level this unsatisfactory distribution of shear stresses, we introduce shear correction factors in the cross sectional shear rigidities at the global equilibrium equations = G = G ( θ + v ) τ = G γ = G ( θ + w ) τ γ xy xy Z xz xz Y constant distribution: unsatisfactory (59)

60 Shear stress resultants: TIMOSHENKO BEAM THEORY Stress Resultants dv Q = τ dω = GA θ y xy y Z Ω dx dw Qz τxzdω GA z θ A,A = = Y + y z: Shear areas with respect Ω dx to the y,z axes 1 1 Ay = κ y A = A Az = κ z A = A ay az κy, κz: shear correction factors ( < 1) a,a : shear deformation coefficients( >1) y z From the assumed displacement field we would have obtained shear rigidities GA which are larger than the actual ones Since we are working with the principal shear system of axes a = 0 Thus the relations of shear stress resultants with respect to the kinematical components are decoupled yz (60)

61 In general we would have a 0: dv Q = τ dω = GA θ y xy y Z Ω dx Q = τ dω = GA θ + z xz z Y Ω A y A 1 TIMOSHENKO BEAM THEORY Stress Resultants yz dw dx = κ y = A Az κ z ay U exact Ω 1 = A = A 1 Ayz = κ yz A = A az a yz From uniform shear beam Uappr theory From this theory α y y αzq α z yz y z τxy + τxz Q Q Q dω = + + G AG AG AG (61)

62 Bending moments: MY = σ xx Ω TIMOSHENKO BEAM THEORY Stress Resultants ZdΩ Bending moments are defined with respect to the principal shear system of axes passing through the centroid of the cross section dθy dθz dθy dθz MY = σxxzdω = EZ dω EYZ dω = EIY EIYZ dx dx dx dx Ω Ω Ω Z xx Ω Ω dx Ω M Z = σ Ω xx YdΩ dθ dθy dθz dθy MZ = σ YdΩ = EY dω EYZ dω = EIZ EIYZ dx dx dx Y Z YZ Ω Ω Ω I = Z d Ω, I = Y d Ω, I = YZdΩ: Moments of inertia with respect to the centroid of the cross section (6)

63 TIMOSHENKO BEAM THEORY Global Equilibrium Equations & Boundary conditions Method of Equilibrium or Energy Method y,v x z,w M Y Q z S w C dx p z m Y w dw M Y +dm Y Q z +dq z x,u dw TOTAL OTENTIAL ENERGY F d F d F 0 θ θ + θ = 1 dx 1 1 dx 1 1 (Euler Lagrange eqns) x dx z,w v p y v dv M Z +dm Z x,u M Z Q y S C m Z Q y +dq y dv y,v (63)

64 y,v x z,w TIMOSHENKO BEAM THEORY Global Equilibrium Equations & Boundary conditions Method of Equilibrium M Y Q z S w C dx p z m Y w dw M Y +dm Y Q z +dq z x,u dw Equilibrium of bending moments dmy Q z m Y 0 dx + = dmz Q y m Z 0 dx + + = Equilibrium of transverse shear forces x z,w M Z Q y v S C dx p y m Z v dv M Z +dm Z Q y +dq y x,u dv dqy p Y 0 dx + = dqz p Z 0 dx + = y,v (64)

65 TIMOSHENKO BEAM THEORY Global Equilibrium Equations & Boundary conditions Equilibrium of bending moments z Y 0 Y θy d θz GA dw YZ θy dx dx az dx Y y Z 0 Z θz d θy GA dv YZ θz dx dx ay dx Z dmy d Q + m = EI EI + + m = 0 1 dx dmz d + Q + m = EI EI + + m = 0 dx Equilibrium of transverse shear forces GA d v dθ Z + = p + y = 0 3 dx a Y dx dx dqy p y 0 dqz p z 0 dx ( ) GA dθ Y d w + = + p + z = 0 4 a Z dx dx β v+ β Q = β γ1w+ γqz = γ3 1 y 3 ( ) ( ) ( ) Inside the bar interval At the bar ends β1θz+ βμz= β3 γ1θy+ γμy= γ3 Coupled system of equations due to principal shear system of axes and due to shear deformation effects (65)

66 TIMOSHENKO BEAM THEORY Global Equilibrium Equations & Boundary conditions Combination of equations may be performed in order to uncouple the problem unknowns - Solution with respect to bending rotations (or deflections) Resolution of rotations: 3 3 d θy d θz dmy Y 3 YZ 3 y ( ), ( ) into ( ) EI EI + + p = 0 1' 1 3 dx dx dx 3 3 d θz d θy dmz Z YZ z ( ), ( ) into ( ) EI EI + p = 0 ' 4 dx dx dx Resolution of deflections: β1θz+ βμz= β3 γ1θy+ γμy= γ3 d θy d θz GA dw Y YZ θy Y dx dx az dx EI EI + + m = 0 1 d θz d θy GA dv Z YZ θz Z dx dx ay dx ( ) ( ) EI EI + + m = 0 γ w+ γ Q = γ 1 z 3 β v+ β Q = β 1 y 3 (66)

67 TIMOSHENKO BEAM THEORY - EXAMLE O a L 3 A b L x 1 Torsionless bending on the Ox1x3 plane Find the elastic curve and the reactions of the beam using Timoshenko beam theory x 3 ( 1' ) STE 1 : Equation of equilibrium (forces): d dx d θ d θ3 dm d θ dx 3 dx1 dx1 1 dx1 EI EI p = 0 EI = 0 d 1 dx 1 ( 1) dθ EI = 0, 0 < x1 < a dx1 (a) ( ) dθ EI = 0, a < x1 < L dx1 (b) (67)

68 TIMOSHENKO BEAM THEORY - EXAMLE O a L 3 A b L x 1 Torsionless bending to the Ox1x3 plane Find the elastic curve and the reactions of the beam using Timoshenko beam theory x 3 Integrate three times eqns (a,b): 0 < x < a a < x < L 1 1 ( 1) ( ) ( 1) d dθ ( ) d dθ Q 3 = EI = C1 Q 3 = EI = C dx1 dx1 dx1 dx1 ( 1) ( ) ( 1) dθ ( ) dθ M =EI =C x +C M =EI =C x +C dx dx ( ) x EI θ = C +C4x 1+C6 ( 1) x EI θ 1 = C +C x +C (c) (d) (e) (68)

69 TIMOSHENKO BEAM THEORY - EXAMLE STE : Resolve constants C -C by exploiting the boundary conditions (rotations, moments): 1 6 ( 1 ) ( 1 ( ) ) relations (e) 1= θ ( ) = 5 = ( ) ( ( ) ) relations (d) at 1 = : ( ) 4 5 ( 1 ) ( ( ) ) 1 = θ θ ( ) relations (e) a Ca C 1 +C3a= +C4a+C6 1. θ x1 at x 0 : 0 0 C 0 (f). M x1 x L M L =0 CL+CC = 0 3. Rotational continuity condition at x a : a = a (h) 4. Equilibrium conditions (forces, moments) at x = a : ( 1 ) ( -) ( ) ( + ) 3 3 Q a =Q a + ( ) ( ) 1 ( ) ( ) M a =M a relations (c,d) 3 C1= C+ 3 (l) Ca 1 + C3 = Ca + C4 (m) From relations (g), (i), (l), (m) 1= = 3 4 = 6 = 3 C C C C L a C C L C a / 1 (g) (i) (j) (k) (n) (69)

70 TIMOSHENKO BEAM THEORY - EXAMLE STE 3: Resolve deflections from equation of equilibrium (1) (moments): d θ d θ3 GA du3 d θ du3 EI EI 3 θ+ + m = 0 EI k 3GA θ+ = 0 dx1 dx1 a3 dx1 dx1 dx1 ( 1) ( ) ( ) ( 1) 1 du 3 1 d dθ kga 3 θ + = Q3 ( x1) = EI,0 < x1 < a (o) dx1 dx1 dx 1 ( ) ( ) ( ) ( ) du 3 d dθ kga 3 θ + = Q3 ( x1) = EI,a< x1 < L (o) dx1 dx1 dx1 Substitute the values of constants (f), (n) into relations (e) and (c) and the resulting expressions in (o), we get: ( 1) ( ) + + ( + ) du3 C C x 3 CL a x = + dx k GA EI EI (p) ( ) du3 C Cx1 CLx1 a = (p) dx k GA EI EI EI 1 3 (70)

71 Integrate once relations (p): TIMOSHENKO BEAM THEORY - EXAMLE ( + ) ( + ) ( + ) ( ) 3 1 C 3 x1 C 3 x1 CL a3 x1 3 ( 1) = kga 3 6EI EI ( ) 3 Cx 1 Cx 1 CLx 1 ax ( 1) = k3ga 6EI EI EI u x C u x C STE 4: Resolve constants C,C,C by exploiting the boundary conditions (translations): 7 8 ( 1 ) ( 1 ( ) ) relations (q) 3 at 1 : 3 ( ) 7 1. u x1 x = 0 u 0 = 0 C = 0 ( ) ( ). u x1 at x 3 1 = ( ) relations (q) CL CL al3 L: u3 ( L ) = C8 = 0 (s) kga 3EI EI 3. Translational continuity condition at x = a : u a = u a relations (q) a 3 a 3 C8 = kga 6EI (q) (q) 3 ( 1 ) ( ( ) ) ( ) a 3 3aL a + kl 3EI From relations (t), (s) C =, k u3( x1)... (u) 3 = 1 k L + λ3gal (r) (t) (71)

72 TIMOSHENKO BEAM THEORY - EXAMLE bh 3h 1 For a rectangular cross section b h and ν=1/3 we have I =, k= 1 4L ( 1) a 3 3a a Q3 = C = C + = + k + L( 1+ k) L L ( ) a 3 3a a Q3 = C = + k L( 1+ k) L L ( 1) M = C1x1+ C3= C( L x1) 3( a x1) ( ) M = C x + C = C L x 3 ( +ν ) Frοm relations (n),(u),(c),(d) we get the expressions of shear forces and bending moments: ( ) (7)

73 TIMOSHENKO BEAM THEORY - EXAMLE The reactions of the beam are: ( ) 0 1 a 3 3a a R3 = Q3 ( 0) = + k + 3 L( 1+ k) L L O ( ) L a 3 3a a 0 R3 = Q3 ( L) = + k M a L( 1+ k) L L x ( ) a 3 3a a M = M ( 0) = CL a 3 = + k a 3 1 ( + k) L L L ( ) ( ) M = M L = 0 0 R3 L 3 A b L L R 3 x 1 For small h/l ratios (h/l<10) which usually occur in practice, shear deformations influence negligibly internal actions and reactions of beams (73)

74 Thank You (74)