Stiffness Matrix for a Bar Element - Memphis

CIVL 7/8117

Chapter 3 - Truss Equations - Part 2

Chapter 3b ? Development of Truss Equations

Learning Objectives

? To derive the stiffness matrix for a bar element. ? To illustrate how to solve a bar assemblage by the direct

stiffness method. ? To introduce guidelines for selecting displacement

functions. ? To describe the concept of transformation of vectors in

two different coordinate systems in the plane. ? To derive the stiffness matrix for a bar arbitrarily oriented

in the plane. ? To demonstrate how to compute stress for a bar in the

plane. ? To show how to solve a plane truss problem. ? To develop the transformation matrix in three-

dimensional space and show how to use it to derive the stiffness matrix for a bar arbitrarily oriented in space. ? To demonstrate the solution of space trusses.

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Stiffness Matrix for a Bar Element

Inclined, or Skewed Supports

If a support is inclined, or skewed, at some angle for the global x axis, as shown below, the boundary conditions on the displacements are not in the global x-y directions but in the x'-y' directions.

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Chapter 3 - Truss Equations - Part 2

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Stiffness Matrix for a Bar Element

Inclined, or Skewed, Supports We must transform the local boundary condition of v'3 = 0

(in local coordinates) into the global x-y system.

Stiffness Matrix for a Bar Element

Inclined, or Skewed, Supports

Therefore, the relationship between of the components of the

displacement in the local and the global coordinate systems

at node 3 is:

u v

'3 '3

cos sin

sin cos

vu33

We can rewrite the above expression as:

d '3 [t3 ]d3

t3

cos sin

sin cos

We can apply this sort of transformation to the entire displacement vector as:

d ' [T1]d or d [T1]T d '

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Chapter 3 - Truss Equations - Part 2

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Stiffness Matrix for a Bar Element

Inclined, or Skewed, Supports

Where the matrix [T1]T is:

[I] [0] [0]

[T1]T [0]

[I ]

[0]

[0] [0] [t3 ]

Both the identity matrix [I] and the matrix [t3] are 2 x 2 matrices.

The force vector can be transformed by using the same transformation.

f ' [T1]f

In global coordinates, the force-displacement equations are:

f [K ]d

Stiffness Matrix for a Bar Element

Inclined, or Skewed, Supports

Applying the skewed support transformation to both sides of

the equation gives: f [K ]d [T1]f [T1][K ]d

By using the relationship between the local and the global displacements, the force-displacement equations become:

d [T1]T d '

f ' [T1][K ][T1]T d '

Therefore the global equations become:

F1x

F1y

u1

v1

F2 x F2y

[T1

][K

][T1

]T

u2 v2

F F

'3 x '3 y

u v

'3 '3

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Chapter 3 - Truss Equations - Part 2

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Stiffness Matrix for a Bar Element

Example 9 ? Space Truss Problem

Consider the plane truss shown below. Assume E = 210 GPa, A = 6 x 10-4 m2 for element 1 and 2, and A = 2 (6 x 10-4)m2 for element 3.

Determine the stiffness matrix for each element.

C2

k

AE L

CS

C 2

CS

CS S2

CS S 2

C 2 CS C2 CS

CS

S

2

CS

S

2

Stiffness Matrix for a Bar Element

Example 9 ? Space Truss Problem The global elemental stiffness matrix for element 1 is:

C2

k

AE L

CS

C 2

CS

CS S2

CS S 2

C 2 CS C2 CS

CS

S

2

CS

S

2

cos (1) 0 sin (1) 1

u1 v1 u2 v 2

0 0 0 0

k(1)

(210 106 kN

/ m2 )(6 104 m2 ) 1m

0 0

1 0 1 0 0 0

0 1 0 1

CIVL 7/8117

Chapter 3 - Truss Equations - Part 2

Stiffness Matrix for a Bar Element

Example 9 ? Space Truss Problem The global elemental stiffness matrix for element 2 is:

C2

k

AE L

CS

C 2

CS

CS S2

CS S 2

C 2 CS C2 CS

CS

S

2

CS

S

2

cos (2) 1 sin (2) 0

u2 v2

1 0

k(2)

(210 106 kN

/ m2 )(6 104 m2 ) 1m

0

1

0 0

0

0

u3 v3

1 0 0 0 1 0 0 0

Stiffness Matrix for a Bar Element

Example 9 ? Space Truss Problem The global elemental stiffness matrix for element 3 is:

C2

k

AE L

CS

C 2

CS

CS S2

CS S 2

C 2 CS C2 CS

CS

S

2

CS

S

2

cos (3)

2 2

sin (3)

2 2

u1 v1 u3 v3

1 1 1 1

k(3) (210 106 kN / m2 )(6 2 2m

2

104

m2

)

1

1

1 1

1 1

1 1

1 1 1 1

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