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Vukazich CE 160 Beam Direct Stiffness Lab 11 [L11] 1 CE 160 Lab – Beam Analysis by the Direct Stiffness Method Beam Element Stiffness Matrix in Local Coordinates Consider an inclined bending member of moment of inertia I and modulus of elasticity E subjected shear force and bending moment at its ends. We will consider only bending and not include axial force for this lab. Define a local coordinate system x’y’ where the x’ axis is attached to the long dimension of the member and runs from the i (initial) end of the member to the j (terminal) end of the member. Note that the SAP 2000 local 1 axis is analogous to the x’ axis. We seek a relationship between the shear and bending moment at the member ends and the transverse displacement and rotation at the ends of the form: where the sixteen terms in the 4x4 matrix are the stiffness influence coefficients which make up the element stiffness matrix in local coordinates. φ x’ y’ y x i j L Δ j Δ i θ i θ j V i V j M j EI M i V i M i V j M j ! " # # $ # # % & # # ' # # = 11 k 12 k 13 k 14 k 21 k 22 k 23 k 24 k 31 k 32 k 33 k 34 k 41 k 42 k 43 k 44 k ( ) * * * * * * + , - - - - - - Δ i θ i Δ j θ j ! " # # $ # # % & # # ' # #

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VukazichCE160BeamDirectStiffnessLab11[L11] 1

CE 160 Lab – Beam Analysis by the Direct Stiffness Method

Beam Element Stiffness Matrix in Local Coordinates Consider an inclined bending member of moment of inertia I and modulus of elasticity E subjected shear force and bending moment at its ends. We will consider only bending and not include axial force for this lab. Define a local coordinate system x’y’ where the x’ axis is attached to the long dimension of the member and runs from the i (initial) end of the member to the j (terminal) end of the member. Note that the SAP 2000 local 1 axis is analogous to the x’ axis. We seek a relationship between the shear and bending moment at the member ends and the transverse displacement and rotation at the ends of the form:

where the sixteen terms in the 4x4 matrix are the stiffness influence coefficients which make up the element stiffness matrix in local coordinates.

φ

x’

y’ y

x

i

j

L

Δj

Δi

θi

θj

Vi

Vj

Mj

EI

Mi

ViMi

Vj

M j

!

"

##

$

##

%

&

##

'

##

=

11k 12k 13k 14k21k 22k 23k 24k31k 32k 33k 34k41k 42k 43k 44k

(

)

******

+

,

------

Δi

θiΔ j

θ j

!

"

##

$

##

%

&

##

'

##

VukazichCE160BeamDirectStiffnessLab11[L11] 2

We can find this relationship from the analysis of a fixed-fixed beam:

ViMi

Vj

M j

!

"

##

$

##

%

&

##

'

##

=

12EIL3

6EIL2

−12EIL3

6EIL2

6EIL2

4EIL

−6EIL2

2EIL

−12EIL3

−6EIL2

12EIL3

−6EIL2

6EIL2

2EIL

−6EIL2

4EIL

)

*

++++++++++

,

-

.

.

.

.

.

.

.

.

.

.

Δi

θiΔ j

θ j

!

"

##

$

##

%

&

##

'

##

(1)

we can write Eqn. (1) in shorthand form:

𝑄 = 𝑘 ∆

where the 4x4 beam element stiffness matrix in local coordinates is

!k[ ] =

12EIL3

6EIL2

−12EIL3

6EIL2

6EIL2

4EIL

−6EIL2

2EIL

−12EIL3

−6EIL2

12EIL3

−6EIL2

6EIL2

2EIL

−6EIL2

4EIL

#

$

%%%%%%%%%%

&

'

((((((((((

(2)

Note that, as was the case in the truss element, the [k ] matrix is symmetric (kpq = kqp for p ≠ q). Note that all stiffness matrices are symmetric. Example of Direct Stiffness Assembly of the Beam Structure System of Equations In order to illustrate the concept of the assembly of a system of equations for the entire beam structure, consider the beam that is made of two elements labeled 1 and 2 as shown:

VukazichCE160BeamDirectStiffnessLab11[L11] 3

Note that the global degrees of freedom are labeled related to the joint numbering. For a general joint with number n;

For this example, the beam has 3 joints and so there are a total of 6 global DOF for the structure and so, similar to the truss problem, the structure system of equations will be of the form:

𝐹 = 𝐾 ∆ (3) where for this example;

[K] = 6x6 global stiffness matrix; {F} = 6x1 vector of applied joint forces (and support reactions); {Δ} = 6x1 vector of joint displacements.

Assembly of the Truss Structure Stiffness Matrix We will assemble the 6x6 structure stiffness matrix from the 4x4 element stiffness matrices (Eqn. 7) for elements 1 and 2 A connectivity table is constructed that maps each element DOF with its corresponding global DOF (shaded)

Element DOF

y, y’

x, x’

P 3

4

2 2 1 3

L1 L2

1

1

2

6

5

2n - 1

2n

n

VukazichCE160BeamDirectStiffnessLab11[L11] 4

1 (Δi) 2 (θi) 3 (Δj) 4 (θj) Associated global DOF for element 1 1 2 3 4 Associated global DOF for element 2 3 4 5 6

The structure (global) stiffness matrix is assembled from the two element contributions

(4) Note that the rows and columns of the element stiffness matrices are labeled with their corresponding global DOF in order to aid the assembly of the structure system of equations which yields the 6x6 structure (global) stiffness matrix

(5) Beam Structure (Global) System of Equations Next the beam structure system of equations can be assembled. Note that global DOF 3, 4 and 6 are unrestrained (free) and DOF 1, 2 and 5 are restrained (supported). We can partition the structure system of equations by restrained and unrestrained DOF. With the unrestrained partitions shaded below

1

k[ ] =

1 2 3 41 11

1k 121k 13

1k 141k

2 211k 22

1k 231k 24

1k3 31

1k 321k 33

1k 341k

4 411k 42

1k 431k 44

1k

2

k[ ] =

3 4 5 63 11

2k 122k 13

2k 142k

4 212k 22

2k 232k 24

2k5 31

2k 322k 33

2k 342k

6 412k 42

2k 432k 44

2k

K[ ] =

1 2 3 4 5 61 k11

1 k121 k13

1 k141 0 0

2 k211 k22

1 k231 k24

1 0 0

3 k311 k32

1 k331 + k11

2 k341 + k12

2 k132 k14

2

4 k411 k42

1 k431 + k21

2 k441 + k22

2 k232 k24

2

5 0 0 k312 k32

2 k332 k34

2

6 0 0 k412 k42

2 k432 k44

2

VukazichCE160BeamDirectStiffnessLab11[L11] 5

(6) Note at the unrestrained DOF we know the forces or moments applied to the joints but the joint displacements and rotations are unknown. At the restrained DOF we know that the displacements (or rotations) are equal to zero at the supports but we do not know the support reactions. We can solve the following system of equations for the unknown displacements at the unrestrained DOF

−P00

"

#$

%$

&

'$

($=

K33 K34

K43

K63

K44

K64

K36

K46

K66

)

*

++++

,

-

.

.

.

.

Δ3θ4θ6

"

#$

%$

&

'$

($

or equivalently

K33 K34

K43

K63

K44

K64

K36

K46

K66

!

"

####

$

%

&&&&

Δ3θ4θ6

(

)*

+*

,

-*

.*=

−P00

(

)*

+*

,

-*

.*

(7)

onceΔ3,θ4,andθ6are found by solving the system of equations shown in Eqn. 7, the support reactions can be found by performing the matrix multiplication

V1M2

V5

!

"#

$##

%

&#

'##

=

K13 K14 0K23 K24 0K53 K54 K56

(

)

****

+

,

----

Δ3θ4θ6

!

"#

$#

%

&#

'# (8)

and Eqn. 1 can be used to find the shear and bending moments at the ends of the individual beam members.

V1M2

−P0V50

"

#

$$$

%

$$$

&

'

$$$

(

$$$

=

K11 K12 K13 K14 0 0K21 K22 K23 K24 0 0K31 K32 K33 K34 K35 K36

K41 K42 K43 K44 K45 K46

0 0 K53 K54 K55 K56

0 0 K63 K64 K65 K66

)

*

++++++++

,

-

.

.

.

.

.

.

.

.

00Δ3θ40θ6

"

#

$$$

%

$$$

&

'

$$$

(

$$$

VukazichCE160BeamDirectStiffnessLab11[L11] 6

CE 160 Direct Stiffness Beam Analysis Lab Problem

For the beam shown the properties of the elements are: Member Section I E 1 W8x10 30.8 in4 29000 ksi 2 W8x10 30.8 in4 29000 ksi Using the coordinate system given in the figure:

1. Find the 4x4 element stiffness matrices (be guided by Eqn. 1) and write the values in the spaces below. Use force units of kips and length units of inches for all calculations.

y, y’

x, x’

12 k 3

4

2 2 1 3

12 ft 6 ft

1

1

2

6

5

[k]1 =

1 2 3 4

1

2

3

4

VukazichCE160BeamDirectStiffnessLab11[L11] 7

2. Assemble the 6x6 structure stiffness matrix (be guided by Eqn. 5) from the element contributions found in Step 1 and write the values in the spaces below.

[k]2 =

3 4 5 6

3

4

5

6

[K]

=

1 2 3 4

1

2

3

4

5

6

5

6

VukazichCE160BeamDirectStiffnessLab11[L11] 8

3. From the structure system of equations write the 3x3 system of equations (Eqn.7) for the unrestrained DOF in the space below.

4. Verify that the solution to the 3x3 system of equations from Step 3 is:

∆!= −1.238155 𝑖𝑛 𝜃! = 0.0051590 𝑟𝑎𝑑 𝜃! = 0.0232154 𝑟𝑎𝑑

5. Find the support reactions (V1, M2, and V5) using Eqn. 8 and your results from Step 4.

6. Using statics, verify that the results from Step 5 satisfy equilibrium of the beam.

Δ3

θ4

θ6