chapter6-stiffnessmethod

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53:134 Structural Design II

Determinant: det( .)m m A ×

Inverse: 1

m m A−× .

Solution of linear algebraic equations:

1 1m n n m A x b× × ×= .

3. Basic Procedure of the Stiffness Method

Use a two-member spring example to illustrate these steps.

♦ Assign a coordinate system for the structure. Assign nodenumbers for the structure.

♦ Define degrees of freedom for the structure and assign numbersfor them.

♦ Assign numbers to the members of the structure.♦ Break-up the structure into smaller pieces called elements or

members. Define element nodal displacements and forces.♦ Identify boundary conditions for the structure in terms of

displacements.♦ Write compatibility conditions between the structural nodal

displacements and element nodal displacements for eachmember.

♦ Identify external loads for each degree of freedom♦ Write equilibrium equations for each element in terms of

displacements. kd = f ♦ Combine elements equilibrium equations to form equilibrium

equations for the entire structure. KD = F ♦ Apply boundary conditions to the system equilibrium equations

and solve for the system nodal displacements.♦ Finally, consider equilibrium equation for each element and

solve for the element nodal forces.

4. Direct Stiffness Method for Truss Analysis

J.S. Arora/Q. Wang Chapter6-StiffnessMethod.doc2




• A truss is a structural system that satisfies the followingrequirements:a. The members are straight, slender, and prismatic. The cross-

sectional dimensions are small in comparison to the member

lengths. The weights of the members are small compared to theapplied loads and can be neglected. Also when constructing thetruss model for analysis, we treat the members as a one-dimensional entity (having length and negligible cross-sectionaldimensions).

b. The joints are assumed to be frictionless pins (or internalhinges).

c. The loads are applied only at the joints in the form of concentrated forces.

As a consequence of these assumptions, the members are two-forcemembers, meaning that they carry only axial forces. In very manyways, a truss member is quite similar to the typical linear spring. Twonodes define a typical truss element.

• For a two-node truss element shown below, governing equationswith respect to the local coordinate system x′ can be expressed as:

1

2 2

1 1

1 1

d f AE d f L

′− ⎧ ⎫ ⎧ ⎫⎡ ⎤ =⎨ ⎬ ⎨ ⎬⎢ ⎥ ′−⎣ ⎦⎩ ⎭ ⎩1′′⎭

2 2 2 1 2 1or ′× × ×′ ′=k d f .

• From the transformation matrix between the local and globalcoordinate systems shown below, the relationship between thelocal nodal displacements and global nodal displacements isderived as





1

1 2

2 3

4

0 0

0 0

d

d d l m

d d l m

d

⎧ ⎫⎪ ⎪′⎧ ⎫ ⎡ ⎤⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥′⎩ ⎭ ⎣ ⎦⎪ ⎪⎪ ⎪⎩ ⎭

or 2 1 2 4 4 1×′ = × × d T d .

The corresponding nodal force relationship reads

1

2

3 2

4

0

0

0

0

f l

1 f f m

f f l

f m

⎧ ⎫ ⎡ ⎤

⎪ ⎪ ⎢ ⎥ ′⎧ ⎫⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬′⎢ ⎥⎩ ⎭⎪ ⎪⎢ ⎥

⎪ ⎪⎩ ⎭ ⎣ ⎦

orT

4 1 4 2 2 1× × ×′T f .=

• The global governing equations for a truss element are therefore

written as

2 21 1

2 22

2 23 3

2 24 4

d f l lm l lmd f lm m lm m AE d f L l lm l lmd f lm m lm m

⎡ ⎤− −

2

⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪− − ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥− − ⎪ ⎪ ⎪ ⎪⎢ ⎥

⎪ ⎪ ⎪ ⎪− −⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦

or 4 4 4 1 4 1× × ×=k d f .

k ij = force required in the direction of dof i to produce unitdisplacement along the dof j.

• The major steps in solving planar truss problems using the directstiffness method:

Step 1 : Select the problem units. Set up the coordinate system.Identify and label the nodes and the elements. For each





element select a start node (node 1) and an end node(node2). We use an arrow along the member to indicatethe direction from the start node to the end node. Thisestablishes the local coordinate system for each element.Label the two global dof at each node starting at node 1and proceeding sequentially.

Step 2 : Construct the equilibrium-compatibility equations for atypical element.

Step 3 : Using the problem data, construct the element equationsfrom Step 2 for all the elements in the problem .

Step 4 : Assemble the element equations into the systemequations, 2 2 2 1 2 1 j j j j× × ×=k d f where j is the number of

joints in the truss. Step 5 : Impose the boundary conditions. Step 6 : Solve the system equations KD = F for the nodal

displacements D. Step 7 : For each element using the nodal displacements, compute

the element nodal forces.

• Example 1 : The figure shows a planar truss. The material is steelwith elastic modulus 200 E GPa= and the cross-sectional area of both members is 20.01 A m= . Use the direct stiffness method tosolve for nodal displacements and member forces. (Rajan’s book page 351-353, Example 6.2.1)

• Example 2 : The figure shows a planar truss. The material is steelwith elastic modulus and the cross-sectional area of each members is . Use the direct stiffness method to solvefor nodal displacements and member forces. (Rajan’s book page354-358, Example 5.2.5)

630 10 Psi E = ×21.2in A =

5. Direct Stiffness Method for Frame Analysis

• A planar frame is a structural system that satisfies the followingrequirements:a. The members are slender and prismatic. They can be straight or

curved, vertical, horizontal, or inclined. The cross-sectional





dimensions are small in comparison to the member lengths.Also when constructing the frame model, we treat the membersas one-dimensional entities (having length and negligible cross-sectional dimensions).

b. The joints can be assumed to be rigid connection, frictionlesspins (or internal hinges), or typical connections.

c. The loads can be concentrated forces or moments that act at joints or on the frame members, or distributed forces acting onthe members.

In this section, however, we assume that the frame is made of straight members and that the connections are rigid. We develop theelement capable of modeling a planar frame in two stages. In the firststage, the flexure effects (due to shear force and bending moments)

will be considered. In the second stage, the axial effects will beconsidered. Using the superposition principle, we can then constructthe behavior of a frame element. As before, the superposition is validonly if the displacements are small. In structural analysis terminology,members that are subjected primarily to flexural effects are said to bebeams whereas members with combined axial-flexural effects arecalled beam-columns. By the end of this section we will havedeveloped the element equations for the combined effects that canalso be used to model pure beam behavior. To avoid construction andusage of several different terminology, we refer to this element simplyas the beam element .

The planar beam element displacements (degrees of freedom) inthe local coordinate system (x’-y’)

• For a two-node planar frame element shown below, governingequations with respect to the local coordinate system x′ can be





expressed as:

without axial forces

or 4 4 4 1 4 1× × ×′ ′= k' d f

.

2 2

4 4 3

2 2

12 6 12 6

6 4 6 2

12 6 12 66 2 6 4

L L

L L L L EI

L L L L L L L

×

−⎡ ⎤

⎢ ⎥−⎢ ⎥=⎢ ⎥− − −⎢ ⎥−⎣ ⎦

k'

with axial forces

or 6 6 6 1 6 1× × ×′ ′= k' d f .





3 2 3 2

2 2

6 6

3 2 3

2 2

0 0 0 0

12 6 12 60 0

6 4 6 20 0

0 0 0 0

12 6 12 60 0

6 2 6 40 0

AE AE L L

EI EI EI EI L L L L

EI EI EI EI L L L L

AE AE L L

EI EI EI EI L L L L EI EI EI EI L L L L

×

⎡ ⎤−⎢ ⎥

⎢ ⎥

⎢ ⎥−⎢ ⎥

⎢ ⎥

⎢ ⎥−⎢ ⎥=⎢ ⎥−⎢ ⎥

⎢ ⎥

⎢ ⎥− − −⎢ ⎥

⎢ ⎥

⎢ ⎥−⎢ ⎥⎣ ⎦

k'

2

• From the transformation matrix between the local and globalcoordinate systems shown below, the relationship between thelocal nodal displacements and global nodal displacements isderived as

Beam element’s local and global coordinate systems and defrees of freedom. Z and z’ axes coincide and point out of the page.





Beam element’s local and global nodal forces

1 1

2 2

3 3

44

55

66

0 0 0 0

0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 1

d d l md d m l

d d

d d l m

d m ld

d d

′⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥′ −⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪′ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥′⎪ ⎪ ⎪ ⎪⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥′ −⎪ ⎪ ⎪ ⎪⎢ ⎥

′ ⎪ ⎪⎪ ⎪ ⎣ ⎦⎩ ⎭⎩ ⎭

or 6 1 6 6 6 1× × ×′ = d T d .

The corresponding nodal force relationship reads

11

22

33

4 4

5 5

6 6

0 0 0 0

0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 1

f f l m

f f m l

f f

f f l m

f m l f

f f

′− ⎧ ⎫⎧ ⎫ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎢ ⎥ ′⎪ ⎪⎪ ⎪ ⎢ ⎥⎪ ⎪′⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥ ′−

⎪ ⎪ ⎪ ⎪⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥ ′⎪ ⎪ ⎪ ⎪⎢ ⎥

′⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩ ⎭

or T6 1 6 6 6 1× × ×′= f T f .

• The global governing equations for a truss element are thereforewritten as

6 6 6 1 6 1× × ×= k d f .





• Element Loads . All the loads on the elements must betransformed to equivalent loads at the node points. The equivalentforce system (equivalent joint forces) is nothing but the opposite of the fixed-end forces .

• The major steps in solving any planar frame problem using thedirect stiffness method:

Step 1 : Select the problem units. Set up the coordinate system.Identify and label the nodes and the elements. For eachelement select a start node (node 1) and an end node(node 2). We use an arrow along the member to indicatethe direction from the start node to the end node. Thisestablishes the local coordinate system for each element.Label the three global dof at each node starting at node 1and proceeding sequentially.

Step 2 : Construct the equilibrium-compatibility equations for atypical element.

Step 3 : Using the problem data, construct the element equationsfrom Step 2 for all the elements in the problem . If thereare element loads, compute the equivalent joint loads andtransform them to the global coordinate system. Note thatif there is more than one element load acting on anelement, use linear superposition (algebraic sum) of all

the element loads acting on that element. Step 4 : Assemble the element equations into the systemequations.

Step 5 : Impose the boundary conditions. Step 6 : Solve the system equations KD = F for the nodal

displacements D. Step 7 : For each element using the nodal displacements, compute

the element nodal forces.

• Example 1 : The figure shows a continuous beam. The material issteel with elastic modulus 200 E GPa= and the cross-sectionalproperties are such that 20.01 A m= , and 40.0001 I m= . Use the directstiffness method to solve for nodal displacements and memberforces. (Rajan’s book page 369-373, Example 6.2.5)

• Example 2 : The figure shows a planar frame. The material is steel





with elastic modulus 200 E GPa= and the cross-sectional propertiesare such that 20.01 A m= , and 40.0001 I m= . Use the direct stiffnessmethod to solve for nodal forces and support reactions. (Rajan’sbook page 373-377, Example 6.2.6)


chapter6-stiffnessmethod

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