Module 5 Frame and Grid Elements

Frame and Grid Elements

Finite Element Analysis (ENGR 455)

Dr. Andreas SchifferAssistant Professor, Mechanical Engineering

Tel: +971‐(0)2‐4018204

[email protected]

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Introduction

• Many structures such as buildings andbridges are composed of frame and/orgrid systems.

• In this chapter we develop the finiteelement equations of plane frames andgrids.

• First, we will develop the stiffness matrixfor a beam element arbitrarily orientedin a plane.

• Then, we will include the axial nodaldisplacement degree of freedom in thelocal beam element stiffness matrix.

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Plane Frame Definition

• A rigid plane frame is defined as a series of beam elements arbitrarily oriented in a plane and rigidly connected to each other at their joints.

• This means that the original angles made between elements at their joints remain unchanged after the deformation.

• Furthermore, forces AND moments are transmitted from one element to another at the joints. Hence, moment continuity exists at the rigid joints.

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2

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2FBD

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2D Arbitrarily Oriented Beam Element

• We can derive the stiffness matrix for an arbitrarily oriented beam element,in a manner similar to that used for the bar element.

• The local axes are located along the beam element and transverse to thebeam element, respectively; the global axes, x and y, are located to beconvenient for the total structure.

• Recall the transformation between global and local coordinates

• We need to apply this transformation to both nodes.

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2D Arbitrarily Oriented Beam Element• For the beam element, we can relate local nodal degrees of freedom to

global degree of freedom as follows. In the global system, each node has3 DOFs instead of 2 DOFs:

• Notice that the rotations are not affected by the orientation of the beam.

• Substituting the above transformation into the general form of the stiffnessmatrix gives

… transformation matrix

[k] depends on E, I, L, and θ.

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2D Arbitrarily Oriented Beam Element• Let’s now consider the effect of an axial force in the beam transformation

• Recall the simple relationship for the truss element

• Combining the axial effects with the shear force and bending momenteffects, in local coordinates gives

ˆ ˆ ˆf k d

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2D Arbitrarily Oriented Beam Element• Now, since the element has 6 DOFs instead of 4, the transformation matrix

becomes a square matrix (6 x 6)

• Then, the global stiffnessmatrix becomes

… transformation matrix

[k] depends on E, A, L, I and θ.

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Plane Frame Examples (5.1)Consider the following plane frame.

Let E = 30 x 106 psi and A = 10 in2 for all elements, and let I = 200 in4 for elements 1and 3, and I = 100 in4 for element 2.

Element 1: The angle between x and x’ is 90°

= 120 in

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Plane Frame Examples (5.1)



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Now assemble the global stiffness matrix and apply the BCs:

+

+

[K] =

[K]

Reduced system of equations:

Global stiffness matrix:

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Plane Frame Examples (5.1)Solve the reduced system of equations for the unknown DOFs:

The results indicate that the top of the frame moves to the right with negligible vertical displacement and small rotations of elements at nodes 2 and 3.

The element forces can now be obtained using for each element.

We illustrate this procedure for Element 1:

nodal DOFs in local coordinates

Stiffness matrix in local coordinates

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Plane Frame Examples (5.1)Repeat this procedure for Elements 2 and 3.

Free-body diagrams of the three elements:

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Stresses in Plane FramesThe stresses in a frame element can be computed from the nodal displacements and rotations in the local coordinate system. In general, two types of stresses are induced:

(a) Stresses in the x-direction due to bending, σxb

(b) Stresses in the x-direction due to axial effects, σxa

xy

fx fx

Superposition of σxb and σxa

:

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Stresses in Plane Frames• First, compute the bending stress, σxb. The relationship between the stresses

and nodal displacements for a beam are

where [B] = and

In order to find the maximum bending stress, evaluate the above stress formula atthe extreme fiber y = ymax.

• Then, calculate the axial (bar) stress, σxa. Here, we can use the stress/displacement relationship for a bar element

where and =

… modulus of elasticity

… modulus of elasticity

in local coordinates!

in local coordinates!

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Stresses in Plane Frames• Finally combine both stress fields and compute the maximum tensile and

compressive stress.

Superposition of σxb and σxa:

σx,max

σx,miny = -R

R

y = R

y = 0

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Example• In example 5.1, the nodal displacements for element 1 in local coordinates were

calculated as

• Assume a circular cross-section of radius = 1.78 in. Then, the bending stress aty = 1.78 in is

• Then compute the axial stress using

• Total compressive and tensile stresses:

=

axial effect

bending effect

σxb = - 1.78 E = 3333 psi

σxa = E = 370 psi

σx,max = 3333 + 370 = 3703 psiσx,min = -3333 + 370 = -2963 psi

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Plane Frame Examples (5.2)Consider the frame shown in the figure below.

• The frame is fixed at nodes 1 and 3 and subjected to a distributed load of 1,000 lb/ft appliedalong element 2.

• Let E = 30 x 106 psi and A = 100 in2 for all elements, and let I = 1,000 in4 for all elements.

• First we need to replace the distributed load with a set of equivalent nodal forces andmoments acting at nodes 2 and 3.

• For a beam subjected to a uniform distributed load, w, the equivalent nodal forces and moments are (Appendix D):

Equivalent nodal forces

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Plane Frame Examples (5.2)• For the sake of simplicity, we consider only the parts of the stiffness matrix associated with the

three degrees of freedom at node 2 (nodes 1 and 2 fixed, zero displacement and rotation). This reduces the size of the elements stiffness matrices to 3 x 3.

• Element 1: θ = 45º

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Plane Frame Examples (5.2)• Element 2: θ = 0º

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Plane Frame Examples (5.2)• Superimposing the stiffness matrices of the elements using the nodal forces and moments at

node 2 (node 3 fixed) and solving the equations yields

• Now determine the local forces in each element using

• Local forces for Element 1: θ = 45º

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Plane Frame Examples (5.2)• Local forces for Element 2: no transformation required since θ = 0º

• For Element 2 we need to subtract from the above result the equivalent nodal forces used toreplace the distributed load.

Equivalent nodal forces

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• Free-body diagrams of the two elements

x’y’

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Plane Frame Examples (5.4)The frame shown below is fixed at nodes 2 and 3 and subjected to a concentrated load of 500 kN applied at node 1. For the bar, A = 1 x 10-3 m2, for the beam, A = 2 x 10-3 m2, I = 5 x 10-5 m4, and L = 3 m. Let E = 210 GPa for both elements.

Again, for brevity’s sake, since nodes 2 and 3 are fixed, we keep only the parts of the parts of the elements stiffness matrices that are needed to obtain the global [K]-matrix necessary for solution of the nodal degrees of freedom.

Bar element: θ = 45º

Beam element (including axial effects): θ = 0º

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Plane Frame Examples (5.4)We now assemble the stiffness matrix for the whole structure, apply the loading and write the system of governing equations for node 1 only:

On solving the latter set of equations we obtain

Calculate the axial forces in the Bar Element:

[K]

Force at node 1:

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Plane Frame Examples (5.4)Calculate the forces and moments in the Beam Element for node 1:

Similarly, at node 2 we have

Free-body diagram

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Grid Analysis• A grid is a structure on which loads are

applied perpendicular to the plane of the structure, as opposed to a plane frame, where loads are applied in the plane of the structure.

• Due to the out-of-plane loading, both torsional and bending moment continuity are maintained at each node in a grid element.

• Typical grid structures include floors of buildings and bridge decks.

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Grid Element Stiffness Matrix• We will now develop the stiffness element of a grid element.

• The grid element can also support torsional moments; therefore, the torsional rotation needs

to be considered in the stiffness equation.

• The DOFs are: vertical displacement diy (normal to the grid), torsional rotation about the

x-axis, and a bending rotation about the z-axis.

• For the sake of simplicity, any effects of axial displacement are ignored (no dix); hence, our

grid elements do not resist axial loading.

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Grid Element Stiffness MatrixStep 1: Consider the sign convention for torsional rotation

Step 2: Assume a linear angle-of-twist distribution along the element

• Applying the boundary conditions and solving for the unknown coefficients gives

Step 3: Establish the strain/displacement and stress/strain relations

Note that the choice of a linear polynomial was not arbitrary. Since there are two rotational DOFs on the grid element, a linear polynomial provides the right number of unknown coefficients a1 and a2.

Matrix form

Shear strain as a function of the twist angle (see Textbook):

Hooke’s law for shear stresses: G … shear modulus

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Grid Element Stiffness MatrixStep 4: Derive the elements stiffness matrix

• From elementary mechanics, we have the shear stress related to the applied torque by

• Nodal torque sign convention:

• Then, the governing equations for the torsional moments become

• Combining the torsional effects with the bending effects, we obtain the local stiffness matrix equationsfor a grid element:

J … polar moment of inertia

where

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Grid Element Stiffness Matrix• The transformation matrix for the grid element is

• Note that transverse displacements are not affected by in-plane rotations.

• The local stiffness matrix in the global coordinates is given by

• Now that we have formulated the global stiffness matrix for the grid element, the procedure for solution then follows in the same manner as that for the plane frame.

• We shall illustrate the use of the developed questions in the following example problems.

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Example Problem (5.5)

Consider the grid shown in the figure below. The frame is fixed at nodes 2, 3, and 4,and is subjected to a load of 100 kips applied at node 1. Assume I = 400 in4, J = 110in4, G = 12 x 103 ksi, and E = 30 x 103 ksi for all elements. To facilitate a timelysolution, apply the boundary conditions at nodes 2, 3, and 4 to the local stiffnessmatrices at the beginning of the solution process.

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Example Problem (5.5)Only node 1 needs to be considered since all DOFs associated with the remaining nodes areconstrained to zer0.

Stiffness matrix of Element 1:

Calculate the stiffness matrix using

x-coordinate of node 2 minus x-coordinate of node 1 =

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Example Problem (5.5)Stiffness matrix of Element 2:

Stiffness matrix of Element 3:

Calculate the stiffness matrix using

Using we obtain

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Example Problem (5.5)Upon superposition of the element stiffness matrices in the global coordinate system, we obtainthe total stiffness matrix of the whole structure for node 1

The grid matrix equation then becomes

The results indicate that the y displacement at node 1 is downward as indicated by the minus sign, the rotation about the x axis is positive, and the rotation about the z axis is negative. Based on the downward loading location with respect to the supports, these results are expected.

Having solved for the unknown displacement and the rotations, we can obtain the local element forces on formulating the element equations in a manner similar to that for the beam and the plane frame.

Solution

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Consider the frame shown in the figure below, where a load of 22 kN is applied at node 2. All other nodes are fully fixed. Assume I = 16.6 x 10-5 m4, J = 4.6 x 10-5 m4, G = 84 GPa, and E = 210 GPa for all elements. To facilitate a timely solution, apply the boundary conditions at nodes 1 and 3 to the local stiffness matrices at the beginning of the solution process.

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• Recall the stiffness matrix of a grid element:

• Stiffness matrix of Element 1 in global coordinates:

Transformation matrix:local coordinates global coordinates

Using we obtain

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Example Problems (5.6)

• Stiffness matrix of Element 2 in global coordinates:

• Calculate the total stiffness matrix by superposition of k(1) and k(2):

• The grid matrix equation then becomes

• Solving for the unknown displacements yields

Using we obtain

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• Now we employ to compute the local forces/moments.

• Element 1:

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• We can use the same procedure for Element 2. The results for the local forces/moments are

• Draw the free body diagrams

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Practice Problem 5.4 (Frames)

For the rigid frame shown in the figure below, determine (1) the nodal displacements and rotation at node 4, (2) the forces and moments in each elements, and (3) provide a free-body diagram of nodal forces and moments for each element. (4) Then check equilibrium at node 4.

Let E = 30 x 106 psi, A = 8 in2, and I = 800 in4 for all elements.

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For the rigid frame shown in the figure below, determine the displacements and rotations of node 2 and the element forces/moments. Also provide a free-body diagram of each element and the whole structure.

The values of E, A, and I to be used are listed next to the figure.

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For the rigid frame shown in the figure below, determine equivalent nodal forces/moments and solve for the unknown displacements and rotations.

The values of E, A, and I to be used are listed next to the figure.

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Practice Problem 5.51 (Grids)

For the grid shown in figure below, determine the nodal displacements and the local element forces.

Let E = 210 GPa, G = 84 GPa, I = 2 x 10-4 m4, J = 1 x 10-4 m4, and A = 1 x 10-2 m2.

Module 5 Frame and Grid Elements

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