2.5 Beam Elements
2.5.1 Beam Element Shape Functions
Although solving the global stiffness matrix for the applied loads, the resulting values only give information about displacements at supplied nodes. For information about how the displacements lookinsideeach element, the (sub-element) displacements must be interpo-lated from the nodal displacements. A way of achieving this is by applying an assumed displacement field (Saouma, 1999). The displacement field is an assumed polynomial which aims to approximate the deformation shape of the element. Mathematically, this may be expressed as:
∆=
n
X
i=1
Ni(x)∆el,i=N(x)∆el (Eq. 2.5.1) where
1. ∆l= local generalized displacement 2. ∆el = element’s local nodal displacement 3. N(x)i= shape functions
4. N(x)= displacement field
∆el is defined as:
∆el =h
ux,1 uy,1 uz,1 θx,1 θy,1 θz,1 ux,2 uy,2 uz,2 θx,2 θy,2 θz,2iT The number of shape functions are dependent on the number of dofs, as well as desired continuity. Continuity pertains to the reproduction of deflection and curvature. The degree of continuity decides whether the displacements are constant or requires continuity of slopes.
Note that the nodal displacement vector for element e,∆e, which is calculated by Cholesky Banachiewicz as shown in Ch. 2.9 must be transformed from global to local coordinates before multiplied with the displacement field. The transformation matrix is a 12x12 matrix like the ones from Eq. 2.7.27-2.7.29.
After calculating the generalized deformations by Eq. 2.5.1, the resulting displacements for each new sub element, such asux,uy,uz,θx,θyandθz, must then be transformed back to global coordinates using the following equation
∆=TT∆l (Eq. 2.5.3)
Axial and Torsional Shape Function
Both axial force and torsion is constant along the length of the element (since St. Venant’s Torsion is assumed). This means that both axial and torsional displacements are linear and can be approximated using the same shape function. Deriving these shape functions are done by starting from the linear polynomial
u=ax+b (Eq. 2.5.4)
Coefficients can be found by applying boundary conditions
u(x= 0) =u1= 0 +b=b (Eq. 2.5.5) u(x=L) =u2=aL+b=aL+u1 (Eq. 2.5.6) L is the element’s local length along the X-axis. Solving for a and b yields
a=u2−u1
L = u2
L −u1
L b=u1 (Eq. 2.5.7)
Substituting Eq. 2.5.7 into Eq. 2.5.4 gives u=ax+b
= (u2
L −u1
L)x+u1 (Eq. 2.5.8)
=u2 Lx−u1
Lx+u1 (Eq. 2.5.9)
= (1− x
L)u1+ x
Lu2 (Eq. 2.5.10)
=N1u1+N2u2 (Eq. 2.5.11)
The shape functions for axial and torsional displacement are then defined as N1= 1−x
L N2= x
L (Eq. 2.5.12)
With this there are shape functions representing two out of six dofs.
The procedure can be sped up by use of matrix notation. Going back to Eq. 2.5.4,u(x) can be described by the polynomial vectorpand the coefficient vectorΨ
u=ax+b=h
MultiplyingΨwith the boundary condition matrixΥconstructed from Eq. 2.5.5-2.5.6 gives the displacements
Here∆aare the nodal axial and torsional parts of∆el. The exact same procedure can be done for the rotational parts∆r.
By invertingΥ,Ψis now defined fromΥandu
The coefficient values are thus given as
Ψ=Υ−1∆a= 1
By multiplying the polynomialspwith the coefficientsΨcalculated from Eq. 2.5.17, the interpolated displacementsucan be found. Substituting Eq. 2.5.17 into Eq. 2.5.14 gives
u=pΨ=pΥ−1∆a (Eq. 2.5.18)
As can be observed,Ncan quickly be found by solving
N=pΥ−1 (Eq. 2.5.20)
Eq. 2.5.20 can be used to easily derive shape functions for flexural dofs as well.
Flexural Shape Functions
Since axial and torsional dofs now are in place, the remaining displacements areuy,uz, θyandθz. four more dofs need their associated shape functions, hence four more shape functions need to be found. Four boundary conditions are used to find these shape functions, meaning that the polynomial must be of order three (Saouma, 1999), the polynomial is assumed as follows for displacements
u=ax3+bx2+cx+d=h
where the rotational displacements are defined as θ= du
dx = 3ax2+ 2bx+c (Eq. 2.5.22) Applying boundary conditions gives
u(x= 0) =u1 du
Converting Eq. 2.5.21 to matrix notation using the notation from Eq. 2.5.23-2.5.24 yields the following boundary condition matrixΥ
∆f =
Here∆f is the nodal flexural part of∆el. The invertedΥ-matrix is The Complete Shape Functions
Now all shape functions are found, representing all six dofs (in each node):
N1= 1− x
Notice that several shape functions are almost identical, meaning shortcuts can be made to avoid recalculating them in the program. Addition and subtraction operations have shorter time execution costs than division and exponentiation. Some simplification can be made as
N1= 1−N2 N5=−N3+ 1
The first order derived shape functions can be useful for finding strains, stresses and internal forces related to the axial and torsional deformations. Deriving the shape functions yields the following equations
dN1=−1
L (Eq. 2.5.35)
dN2= 1
L (Eq. 2.5.36)
dN3=−6 x L2 + 6x2
L3 (Eq. 2.5.37)
dN4= 1−4x L+ 3x2
L2 (Eq. 2.5.38)
dN5= 6 x L2 −6x2
L3 (Eq. 2.5.39)
dN6= 3x2 L2 −2x
L (Eq. 2.5.40)
Similarly toN2andN4, dN2anddN5can freed from recalculation.
dN2=−dN1 dN5=−dN3
Remember thatθyandθzare defined as θy =duz
dx θz=duy
dx (Eq. 2.5.41)
This means thatθyandθzare calculated from the derived displacement field that will be presented later.
The second order derivative shape functions are useful for finding strains and stresses, but follows the same logic as before so are not shown.
2.5BeamElements The shape functions are used to construct adisplacement fieldNwhich is used to approximate a displacement pattern, as per Eq. 2.5.1.
When multiplied by the nodal displacements per element,∆el, the displacement fieldNrepresents the general deformations ofux,uy,uz
andθx. The first order derived displacement fielddNcan be used to findθyandθz, see Eq. 2.5.41.
If there is a nodal displacement of 1 in the Y-direction (uy,2= 1), as in Fig 2.3a, the (transposed) displacement vector∆el will look like
∆el =
By using Eq. 2.5.1 on bothNanddN, then retrieving appropriate values, the displacement
AssumingL = 1 and incrementing values for x at intervals of 0.05, the resultinguy
displacement looks much like expected, see Fig 2.3b
(a) Assumed deformation (b) Interpolated deformation
Figure 2.3: Nodal displacement of 1 in Y-direction
An example following the same procedure for a displacement situation like on Fig. 2.4a, whereuz,2= 1andθy =−1results in a displacement pattern like on Fig. 2.4b. Notice that this case, whereθy=−1, illustrates whyN3,5andN3,11are negative in the displacement matrix since rotation about the Y-axis contributes negatively to theuzvalue.
(a) Assumed deformation (b) Interpolated deformation Figure 2.4: Nodal displacement of 1 in Z-direction and -1 about the Y-axis