Chapter 5 Structural Elements: The truss & beam elements

Institute of Structural Engineering Page 1 Chapter 5 Structural Elements: The truss & beam elements

Institute of Structural Engineering Page 2 Chapter Goals Learn how to formulate the Finite Element Equations for 1D elements, and specifically The bar element (review) The Euler/Bernoulli beam element The Timoshenko beam (brief mention) What is the Weak form? What order of elements do we use? What is the isoparametric formulation? How is the stiffness matrix written? How are the external loads approximated?

Institute of Structural Engineering Page 3 Today s Lecture Contents The truss element The Euler/Bernoulli beam element

Institute of Structural Engineering Page 4 FE Classification

Institute of Structural Engineering Page 5 Assumptions Uniaxial Element i. The longitudinal direction is sufficiently larger than the other two ii. The bar resists an applied force by stresses developed only along its longitudinal direction Prismatic Element i. The cross-section of the element does not change along the element s length

Institute of Structural Engineering Page 6 Variational Formulation Prismatic Element: Uniaxial Element: Only the longitudinal stress and strain components are taken into account. The traction loads degenerate to distributed line loads

Institute of Structural Engineering Page 7 Finite Element Idealization Truss Element Shape Functions (in global coordinates)

Institute of Structural Engineering Page 8 Strain-Displacement relation define The truss element stress displacement matrix Constant Variation of strains along the element s length

Institute of Structural Engineering Page 9 Or we may use the isoparametric formulation: 1 N1 ξ 1 ξ 2 ( ) = ( ) 1 u u( ξ) = N ( ξ) N ( ξ) Isoparametric Shape Functions Once again: 1 2 2 u ( ) ( ξ) ξ ( ξ) 1 N2 ξ 1 ξ 2 N x N N B = = = J = x ξ x ξ L ( ) = ( + ) 1 1 [ 1 1 ] -1-1 1 The truss element stress displacement matrix 1 Constant Variation of strains along the element s length

Institute of Structural Engineering Page 10 From the Continuous to the Discrete form from Strong to Weak form It is convenient to re-write the Principle of Virtual Work in matrix form Elastic material: where: and: and for concentrated loads on the element end nodes 1, 2, or: L T { } ( ) ( ) F = N x q x dx 0 for distributed loads q(x)

Institute of Structural Engineering Page 11 From the Continuous to the Discrete form Therefore the relation turns to Rearranging terms:

Institute of Structural Engineering Page 12 From the Continuous to the Discrete form Therefore the relation turns to Rearranging terms:

Institute of Structural Engineering Page 13 From the Continuous to the Discrete form Finally the discrete form of the truss element equilibrium equation reduces to or more conveniently where: Performing the integration, the following expression is derived

Institute of Structural Engineering Page 14 Example 1: Truss Element under concentrated tensile force Strong Form Solution Finite Element Solution Boundary Conditions Boundary Conditions Solution Solution

Institute of Structural Engineering Page 15 Example 2: Truss Element under uniform distributed tensile force Strong Form Solution Boundary Conditions Solution

Institute of Structural Engineering Page 16 Example 2: Truss Element under uniform distributed tensile force Finite Element Solution Boundary Conditions But now we need to account for the distributed loading Solution Displacement at node 2@ x=l

Institute of Structural Engineering Page 17 Example 2: Truss Element under uniform distributed tensile force Strong Form Solution Finite Element Solution Displacement at node 2@ x=l

Institute of Structural Engineering Page 18 Example 2: Truss Element under uniform distributed tensile force Strong Form Solution Finite Element Solution

Institute of Structural Engineering Page 19 Example 2: Truss Element under uniform distributed tensile force 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 Length (m)

Institute of Structural Engineering Page 20 Example 2: Truss Element under uniform distributed tensile force 2 1.5 1 0.5 0 0 0.5 1 1.5 2 Length (m)

Institute of Structural Engineering Page 21 The 2-node Euler/Bernoulli beam element Assumptions Uniaxial Element i. The longitudinal direction is sufficiently larger than the other two Prismatic Element i. The cross-section of the element does not change along the element s length Euler/ Bernoulli assumption i. Upon deformation, plane sections remain plane AND perpendicular to the beam axis

Institute of Structural Engineering Page 22 The 2-node Euler/Bernoulli beam element Uniaxial Element i. The longitudinal direction is sufficiently larger than the other two Prismatic Element i. The cross-section of the element does not change along the element s length Euler/ Bernoulli assumption i. Upon deformation, plane sections remain plane AND perpendicular to the beam axis

Institute of Structural Engineering Page 23 Finite Element Idealization We are looking for a 2-node finite element formulation

Institute of Structural Engineering Page 24 Finite Element Idealization Carlos Felippa

Institute of Structural Engineering Page 25 Kinematic Field Two deformation components are considered in the 2-dimensional case 1. The axial displacement 2. The vertical displacement dw θ = = w dx

Institute of Structural Engineering Page 26 Kinematic Field Plane sections remain plane AND perpendicular to the beam axis

Institute of Structural Engineering Page 27 Kinematic Field Plane sections remain plane AND perpendicular to the beam axis Point A displacement (that s because the section remains plane) (that s because the plane remains perpendicular to the neutral axis)

Institute of Structural Engineering Page 28 Kinematic Field Plane sections remain plane AND perpendicular to the beam axis Therefore, the Euler/ Bernoulli assumptions lead to the following kinematic relation

Institute of Structural Engineering Page 29 Basic Beam Theory Δx p q a b y x M σ E σ = y, ε =, σ = I E R where σ : normal stress M : bending moment R : curvature y Reminder: from the geometry of the beam, it holds that: θ w Therefore: 2 dw 2 dx M = EI slope : θ = 1 curvature : κ = = R dw dx 2 dw dx (that s because the plane remains perpendicular to the neutral axis) 2

Institute of Structural Engineering Page 30 Basic Beam Theory Assume an infinitesimal element: Force Equilibrium: ( ) ( ) ( ) 0 ( + ) ( ) x 0 p( x) V x + p x x V x+ x = V x x V x x = dv dx = ( ) p x 2 4 dm dw Finally, = p( x) EI = p( x) 2 4 dx dx Moment Equilibrium: x M ( x) V ( x) x+ M ( x+ x) p( x) x = 0 2 M ( x+ x) M ( x) x x 0 dm = V ( x) + p( x) V x x 2 dx = ( )

Institute of Structural Engineering Page 31 Interpolation Scheme Beam homogeneous differential equation EI 4 dw p( x) 0 4 dx + = Boundary Conditions Dirichlet : w = w on Γ dw dx u = θ = θ on Γ θ 2 dw Neumann : EI = M on Γ 2 dx 3 dw EI = S on Γ dx S M

Institute of Structural Engineering Page 32 Interpolation Scheme Beam homogeneous differential equation That s a fourth order differential equation, therefore a reasonable assumption for the interpolation field would be at least a third order polynomial expression: Therefore the rotation would be a second order polynomial expression: or in matrix form: (I)

Institute of Structural Engineering Page 33 Interpolation Scheme The relation must hold for the arbitrary displacements at the nodal points: Therefore, by solving w.r.t the polynomial coefficients : So now we can derive the 2-dimensional Euler/Bernoulli finite element interpolation scheme (i.e. a relation between the continuous displacement field and the beam nodal values):

Institute of Structural Engineering Page 34 Interpolation Scheme Therefore by substituting in (I): Or more conveniently: where:

Institute of Structural Engineering Page 35 Interpolation Scheme in global coordinates After some algebraic manipulation the following expressions are derived for the shape functions 1.0 N2 0.14 N3 0.8 0.6 0.4 0.2 0.12 0.10 0.08 0.06 0.04 0.02 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0 1.0 0.8 0.6 0.4 0.2 N5 0.2 0.4 0.6 0.8 1.0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 N6 0.2 0.4 0.6 0.8 1.0

Institute of Structural Engineering Page 37 Interpolation Scheme in isoparametric coordinates Hermite Polynomials 1 w1 N2 = 1 2 2 + 1 θ1 N3 ξ = 1 ξ 4 1 + ξ 1 w2 N5 = 1 + 2 2 1 θ2 N6 ξ = 1 + ξ 4 ξ 1 and 2 ( ξ) ( ξ) ( ξ) 2 ( ) ( ) ( ) 2 ( ξ) ( ξ) ( ξ) 2 ( ) ( ) ( ) l l dx l w( ξ) = N2( ξ) w1+ N3( ξ) θ1+ N5( ξ) w2 + N6( ξ) θ2 with = J = 2 2 dξ 2 e l l N = N N N N d = w w 2 2 e ( ξ) ( ξ) ( ξ), [ θ θ ] 2 3 5 6 1 1 2 2

Institute of Structural Engineering Page 38 Interpolation Scheme What about the axial displacement component?????? Since the axial and bending displacement fields are uncoupled Shape Function Matrix [N]

Institute of Structural Engineering Page 39 Strain-Displacement compatibility relations 2-dimensional beam element The normal strain : Remember that the beam s curvature is The shear strain : The normal strain : Euler/ Bernoulli Theory The Euler/ Bernoulli assumptions predict zero variation of both the shear strain and the vertical component of the normal strain

Institute of Structural Engineering Page 40 Strain-Displacement relations in matrix form We can re-write the strain displacement relation in the following form and we can easily substitute the matrix interpolation form

Institute of Structural Engineering Page 41 Strain-Displacement relations in matrix form So the strain-displacement matrix assumes the following form where: and: and by substituting the expressions of the shape functions, the strain-displacement matrix B assumes the following form or in isoparametric coordinates: 1 6ξ 6ξ dw dw d w B J Bd L L L dξ dx dξ 2 e e e = 1 3ξ 1 1 3ξ + 1 with = & = 2

Institute of Structural Engineering Page 42 The evaluation of the Beam Element Stiffness Matrix The beam element stiffness matrix is readily derived as: ( ) = Ω e T K B EI Bd Performing the integration with respect to x: where: is the cross-sectional moment of inertia and is

Institute of Structural Engineering Page 43 The evaluation of the Beam Element force vector The beam element force vector is readily derived as: e Ω e T ( ) T T d N = ( ) ( ) + + dx Ω e e e f N p x dx M N S f S ΓM e f Γ Γ Assuming constant distributed load p: T ( ) ( ξ) = = ξ = e e e fω N x pdx p N Jd Ω ξ pl 2 1 L 6 1 L 6 p L

Institute of Structural Engineering Page 44 The Timoshenko Beam Element Carlos Felippa

Institute of Structural Engineering Page 45 The Timoshenko Beam Element The transverse shear strain now also comes into play: dw γ = θ + = θ + ψ dx Carlos Felippa

Institute of Structural Engineering Page 46 The Timoshenko Beam Element

Institute of Structural Engineering Page 47 The Timoshenko Beam Element Carlos Felippa