truss element with geometric...

Professor Terje Haukaas The University of British Columbia, Vancouver terje.civil.ubc.ca

Truss Element with Geometric Nonlinearity Updated May 15, 2020

Truss Element with Geometric Nonlinearity

Total Lagrangian Approach The Total Lagrangian approach, employing finite strain expressions, is a common approach for including geometric nonlinearity. The coordinate system remains fixed, and the strain is expressed relative to that system.

Kinematics There exists no unique way of defining finite strains, but Green’s strain, sometimes referred to as Green Lagrange strain, is typical. Several avenues are available for deriving it. Figure 1 is here used as the first starting point. The shown truss element experiences a rotation, q, but no horizontal displacement of the right-most node.

Figure 1: Elongation of truss element.

One way to derive an expression that includes geometric nonlinearity, using Figure 1, starts with the strain definition “change in length divided by original length.” Employing the theorem of Pythagoras to express the new deformed length yields:

(1)

Now employing the Taylor series approximation

(2)

yields

(3)

By including the regular axial strain, du/dx, due to longitudinal displacement, the sought expression for Green’s strain is obtained:

dx

dx.cos(q)

q

dxdw

Elongation, du

ε = dx2 + dw2 − dxdx

= dx2 + dw2

dx−1= 1+ dw

dx⎛⎝⎜

⎞⎠⎟2

−1

1+ x2 ≈1+ 12⋅ x2

ε = 1+ dwdx

⎛⎝⎜

⎞⎠⎟2

−1≈ 12⋅ dwdx

⎛⎝⎜

⎞⎠⎟2



(4)

Another way to approach the problem is the following trigonometric consideration of the gray-shaded triangle in Figure 1:

(5)

Next, a series expansion of cos(q) is considered:

(6)

Neglecting higher order terms yields

(7)

which means that Eq. (5) turns into

(8)

Substituting the approximation q=dw/dx and dividing through by dx, as well as including the regular strain, du/dx due to axial displacement, yields the same Green’s strain as in Eq. (4). Another way to look at the first approach is to reconsider the “engineering strain” expression

(9)

where Lo is the original length and Ln is the new length of a rotated truss member. The Green strain takes another step, defining strain as

(10)

That expression reveals that the Green strain essentially adds a second-order term to the engineering strain:

(11)

Eq. (10) can now be used to arrive at the same strain as in Eq. (4), after adding the regular axial strain due to elongation, while noticing that no Taylor approximation is made here with the more accurate strain expression:

(12)

ε = dudx

+ 12⋅ dwdx

⎛⎝⎜

⎞⎠⎟2

du ≈ dx − dx ⋅cos(θ )

cos(θ ) = 1− θ

2

2+ θ

4

4!− θ

6

6!+

cos(θ ) ≈1− θ2

2

du ≈ 12⋅θ 2 ⋅dx

ε = Ln − LoLo

= LnLo

−1

ε = 12⋅ Ln

2 − Lo2

Lo2 = 1

2⋅ Ln

Lo

⎛⎝⎜

⎞⎠⎟

2

−1⎛

⎝⎜

⎞

⎠⎟

ε = 12⋅ Ln

Lo

⎛⎝⎜

⎞⎠⎟

2

−1⎛

⎝⎜

⎞

⎠⎟ =

12⋅ ε +1( )2 −1( ) = ε + 1

2ε 2

Second-order!

ε = 12⋅ Ln

Lo

⎛⎝⎜

⎞⎠⎟

2

−1⎛

⎝⎜

⎞

⎠⎟ =

12⋅ dx2 + dw2

dx

⎛

⎝⎜

⎞

⎠⎟

2

−1⎛

⎝⎜⎜

⎞

⎠⎟⎟= 12⋅ dx2 + dw2

dx2−1

⎛⎝⎜

⎞⎠⎟= 12⋅ dwdx

⎛⎝⎜

⎞⎠⎟2



Yet another approach for deriving Green’s strain for a truss member is to start with the generic expression for Green’s strain from continuum mechanics, written in index notation as

(13)

Using that expression, the axial strain in the x-direction is

(14)

because uz=w. The second term is much smaller than the other terms and is neglected, leading to the same strain as in Eq. (4).

Material Law (15)

Shape Functions The finite element method is, with few exceptions, displacement-based. That means that displacement interpolation, i.e., shape functions for the displaced shape of the element, is central. To include geometric nonlinearity for a 2D truss element it is necessary to consider all four degrees of freedom shown in Figure 2. This is an extended version of the “local” degrees of freedom of ordinary linear structural analysis. In the notes on this website, the displacement interpolation is generally written as

(16) where =is the vector of displacement fields, N=matrix with shape functions, and u=degrees of freedom.

Figure 2: Degrees of freedom for 2D truss element.

In the present case, Eq. (16) is spelled out as

ε ij =12⋅ ui, j + uj ,i + uk ,i ⋅uk , j( )

ε xx =12⋅ ux,x + ux,x + ux,x ⋅ux,x + uz,x ⋅uz,x( )

≡ 12⋅ dudx

+ dudx

+ dudx

⎛⎝⎜

⎞⎠⎟2

+ dwdx

⎛⎝⎜

⎞⎠⎟2⎛

⎝⎜⎞

⎠⎟

= dudx

+ 12⋅ dudx

⎛⎝⎜

⎞⎠⎟2

+ 12⋅ dwdx

⎛⎝⎜

⎞⎠⎟2

σ = E ⋅ε

!u = Nu!u

3

421

z, w

x, u

L



(17)

where

(18)

Principle of Virtual Displacements This principle expresses equilibrium in an average sense:

(19)

where p is a two-dimensional vector containing distributed loads in the x- and z-directions and is defined in Eq. (17). Substitution of material law and integrating over the cross-section area yields

(20)

The virtual strain, de, requires special attention in face of geometric nonlinearity. In the linear case, matters are simple: de=du'. That changes in light of the strain in Eq. (4). To address this issue, it is helpful to think of d as a “variation” instead of “virtual.” Calculus of variation, i.e., executing the variation, yields

(21)

With de addressed, at least for now, attention turns to e, still in light of Green’s strain in Eq. (4). The first objective is to express the derivatives du/dx and dw/dx in that strain expression in terms of the shape functions. To that end, Eq. (17) is first split up:

(22)

(23)

That means the sought derivatives are

(24)

(25)

u(x)w(x)

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

N1 0 N2 00 N1 0 N2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

u1u2u3u4

⎧

⎨

⎪⎪

⎩

⎪⎪

⎫

⎬

⎪⎪

⎭

⎪⎪

N1(x) = 1− xL

and N2 (x) = xL

σ δε dVV∫ = pT δ !udx

0

L

∫

!u

EA ⋅ εδε dx0

L

∫ = pT δ !udx0

L

∫

δε = δ ′u + 12⋅ ′w 2⎛

⎝⎜⎞⎠⎟ = δ ′u + ′w ⋅δ ′w

u(x) = N1 0 N2 0⎡⎣

⎤⎦u

w(x) = 0 N1 0 N2⎡⎣

⎤⎦u

dudx

= − 1L

0 1L

0⎡

⎣⎢⎢

⎤

⎦⎥⎥u ≡ Bu

dwdx

= 0 − 1L

0 1L

⎡

⎣⎢⎢

⎤

⎦⎥⎥u ≡ Cu



where the row vectors B and C are defined. That means the strain in Eq. (4) is written

(26)

That notation also means that the virtual strain in Eq. (21) reads

(27)

Substitution of Eqs. (16), (26), and (27) into the internal virtual work expression in Eq. (20) yields

(28)

Doing the rearranging that we always do in these type of finite element derivations, i.e., changing the order of multiplication of the scalars within the parenthesis, taking the transpose of a scalar, and extracting du, assuming arbitrary virtual displacements yields

(29)

Now notice that EA multiplied by the second parenthesis is the axial force in the element because

(30)

That means Eq. Error! Reference source not found. can be written

(31)

That is a useful expression for nonlinear structural analysis, where the axial force, N, is determined from the trial displacements so that the internal resisting forces, Pint, can be calculated. Eq. (31) represents equilibrium in the displaced shape, but only for moderate displacements. It is not valid for large rotations of the truss element. To complement Eq. (31) it is useful to derive the tangent stiffness matrix:

(32)

where the product rule of differentiation is employed, and where, in light of Eq. (30):

(33)

That means the tangent stiffness is

ε = Bu+ 12⋅ Cu( )T Cu = Bu+ 1

2⋅uTCTCu

δε = δ ′u + ′w ⋅δ ′w = Bδu+ uTCTCδu

EA ⋅ Bu+ 12⋅uTCTCu⎛

⎝⎜⎞⎠⎟ Bδu+ u

TCTCδu( )dx0

L

∫ = pT Nδu( )dx0

L

∫

EA ⋅ BT +CTCu( ) Bu+ 12 ⋅uTCTCu⎛

⎝⎜⎞⎠⎟ dx

0

L

∫ = NTpdx0

L

∫

N = A ⋅σ = EA ⋅ε = EA ⋅ Bu+ 12⋅uTCTCu⎛

⎝⎜⎞⎠⎟

BT ⋅N dx0

L

∫

Linear term! "# $#

+ CTCu ⋅N dx0

L

∫

Geometric nonlinearity! "## $##

Pint! "#### $####

= NTpdx0

L

∫Pext

!"# $#

KT =∂Pint∂u

= BT ⋅ ∂N∂u

dx0

L

∫ + CTCu ⋅ ∂N∂u

dx0

L

∫ + CTC ⋅I ⋅N dx0

L

∫

∂N∂u

= EA ⋅ B+ uTCTC( )



(34)

where the elastic and geometric stiffness matrices from more elementary structural analysis are identified.

KT = EA ⋅BTBdx0

L

∫

Ko! "## $##

+ EA ⋅BTuTCTCdx0

L

∫

+ EA ⋅CTCuBdx0

L

∫ + EA ⋅CTCuuTCTCdx0

L

∫ + CTC ⋅N dx0

L

∫KG

! "# $#

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