TAliTruss Analysis Using The Stiffness MethodUsing The ...site.iugaza.edu.ps/ajubeh/files/2015/02/Chapter-14_5-5-2010.pdf · Chapter 14 TAliTruss Analysis Using The Stiffness MethodUsing

Chapter 14Chapter 14

T A l iTruss Analysis Using The Stiffness MethodUsing The Stiffness Method

Stiffness MethodStiffness Method• Fundamentals of the stiffness method

– There are essentially two ways in which structures can be analyzed using matrix methods.

• Flexibility method

• Stiffness method

– The stiffness method can be used to analyzed both statically determinate & indeterminate structures, whereas the flexibility method required a diff t d f h f thdifferent procedure for each of these cases.

– The stiffness method yields the displacements & forces directly, whereas ith th fl ibilit th d th di l t t bt i d di tlwith the flexibility method the displacements are not obtained directly.

Stiffness Method

– Application of this method requires subdividing the structure into a series

Stiffness Method

pp q gof discrete finite elements & identifying their end points as nodes.

– The force-displacement properties of each element are determined & then related to one another using the force equilibrium equations written at the nodes

– These relationships, for the entire structure, are then grouped together into what is called the structure stiffness matrix K.

– Once the K is established, the unknown displacements of the nodes can then be determined for any given loading on the structure.

– When the displacements are known, the external & internal forces in the structure can be calculated using the force-displacement relations for each member.

Preliminary Definitions & Concepts• Member & node identifications

– We will specify each member by a number enclosed within a squareWe will specify each member by a number enclosed within a square, & use a number enclosed within a circle to identify the nodes.

– Also the ‘near’ & ‘far’ ends of the member must be identified, this will b d i i l h b i h h h d f hbe done using an arrow written along the member with the head of the arrow directed toward the far end.

Preliminary Definitions & Concepts• Global & member coordinates.

– We will use two different type of coordinate systems, global or yp y , gstructure coordinate system and local or member coordinate system.

– Global system x, y, used to specify the sense of each of the external force & displacement components at the nodesforce & displacement components at the nodes.

– Local system x’,y’ used to specify the sense of direction of members displacements & internal loadings.

Preliminary Definitions & Concepts• Degrees of freedom

– The unconstrained truss has two degree of freedom or two possible g pdisplacements for each joint (node).

– Each degree of freedom will be specified on the truss using a code number shown at the joint or node & referenced to its positive globalnumber, shown at the joint or node, & referenced to its positive global coordinate direction using an associated arrow.

• For exampleFor example– The truss has eight degree of freedom or eight possible displacements.– 1 through 5 represent unknown or

unconstrained degree of freedom.– 6 through 8 represent constrained

degree of freedomdegree of freedom

Lowest code numbers will always be used to identify the unknown displacements.

Highest code numbers will be used to identify the known displacements

Member Stiffness Matrix• Case I

– Positive displacement dN on the near endAE AE

• Case II'N N

AEq dL

'F NAEq dL

– Positive displacement dF on the far end

'' AEq d '' AEq d

• Case I + Case II

N Fq dL

F Fq dL

• Case I + Case II– Resultant forces caused by both displacements are

AE AEN N F

AE AEq d dL L

AE AEq d dF N Fq d dL L

Member Stiffness Matrix

– These load-displacement equations written in matrix form

1 11 1

N N

F F

q dAEq dL

F Fq

or

'q k dwhere

1 1'

1 1AEkL

The matrix, k` is called the member stiffness matrix.

It is of the same form for each member of the truss.

Transformation MatricesDi l t & f t f ti• Displacement & force transformation– We will now develop a method for

transforming the member forces q and g qdisplacements d defined in local coordinate to global coordinates

cos F Nx x

x xL

y ycos F Ny y

y yL

2 2cos F N

x x

F N F N

x x

x x y y

2 2cos F N

y y

F N F N

y y

x x y y

F N F N

Transformation Matrices• Displacement transformation matrix

– In global coordinate each end of the member can have two independent displacements.

– Joint N has displacements DNx & DNy in global coordinate

cos cosN Nx x Ny yd D D

N Nx x Ny yd D D or

N Nx x Ny y

Transformation Matrices– Also joint F has displacements DFx & DFy

cos cosF Fx x Fy yd D D

Di l N & F

y y

F Fx x Fy yd D D or

– Displacements at N & F

N Nx x Ny yd D D

F Fx x Fy yd D D

0. 0.Nx

x y NyN

DDd 0. 0. x y FxF

Fy

DdD

d TDor

d TD

0. 0.x yT

where

0. 0.x y

x yT

f i iTransformation Matrices

• Force transformation matrix– Force qN applied to the near end of the member

cosN NQ q cosN NQ q

Force q applied to the far end of the member

cosNx N xQ q or

cosNy N yQ q

Nx N xQ q Ny N yQ q

– Force qF applied to the far end of the membercosFx F xQ q

orcosFy F yQ q

Fx F xQ q Fy F yQ q

– Rewrite in a matrix form:

Fx F xQ q Fy F yQ q

0.Nx xQ 0.

0.0

Nx x

Ny y N

Fx x F

QQ qQ qQ

0.Fy yQ

TQ T qor where

0.0.

0

x

yTT

Q T q 0.

0.x

y

Member Global Stiffness Matrix• Stiffness matrix

– We will determine the stiffness matrix for a member which relates the member’s global force components Q to its global displacements D.

'k d d TDd 'k T Dq k d d TDand q k T D

TQ T q 'TQ T k TD Q kD

'Tk T k T0.0. 0. 0.1 1

x

y x yAEk

0. 0. 0.1 1

0.

y y

x x y

y

kL

Member Global Stiffness Matrix

'Tk T k TTk T k T

2 2 N x y x yN N F F

2 2

2 2

2 2

xx x y x x y

yy x y y x y

xx x y x x y

NNAEkFL

2 2

xx x y x x y

yy x y y x y F

Example 1 Determine the structure stiffness matrix for the two-member truss

shown. AE is constant

Solution:

Establish the x, y global system

Identify each joint & member numerically.

Member 1:Determine x & y, where L = 3ft

3 0 0 03 0 13x

0 0 0

3y

x y x yN N F F2 2

2 2

2 2

xx x y x x y

yy x y y x y

NNAEkFL

2 2

2 2xx x y x x y

yy x y y x y

FLF

0 333 0 0 333 0 1

1 2 3 4Dividing each element by L = 3ft

22 10 333x

1

0.333 0. 0.333 0. 10. 0. 0. 0. 2

0.333 0. 0.333 0. 3k AE

0.3333L

0. 0. 0. 0. 4


3 0 4 03 0 0.65x

4 0 0.8

5y

x y x yN N F F2 2

2 2

2 2

xx x y x x y

yy x y y x y

NNAEkFL

2 2

2 2xx x y x x y

yy x y y x y

FLF

0 072 0 096 0 072 0 096 1

1 2 5 6Dividing each element by L = 5ft

0.8 0.60.096

5

2

0.072 0.096 0.072 0.096 10.096 0.128 0.096 0.128 20.072 0.096 0.072 0.096 5

k AE

5

0.096 0.128 0.096 0.128 6

Structure stiffness matrixK = k1 + k2

0.333 0 0.333 0 0 0 1 0.072 0.096 0 0 0.072 0.0960. 0 0. 0 0 0 2 0.096 0.128 0 0 0.096 0.128

12

1 2 3 4 5 6 1 2 3 4 5 6

0.333 0 0.333 0 0 0 3 0. 0. 0 0 0. 0.0. 0 0. 0 0 0 4 0. 0. 0 0 0. 0.0 0 0 0 0 0 5 0 072 0 096 0 0 0 072 0 096

K AE AE

345

0. 0 0. 0 0 0 5 0.072 0.096 0 0 0.072 0.096

0. 0 0. 0 0 0 6 0.096 0.128 0 0 0

5.096 0.128 6

0.405 0.096 0.333 0. 0.072 0.0960.096 0.128 0. 0. 0.096 0.1280 333 0 0 333 0 0 0

0.333 0. 0.333 0. 0. 0.

0. 0. 0. 0. 0. 0.0.072 0.096 0. 0. 0.072 0.096

K AE

0.096 0.128 0. 0. 0.096 0.128

Example 2 Determine the structure stiffness matrix for the truss shown. AE is

constant


10 0 0 010 0 110x

0 0 010y

x y x yN N F F2 2

2 2

2 2

xx x y x x y

yy x y y x y

NNAEkFL

2 2

2 2xx x y x x y

yy x y y x y

FLF

0 1 0 0 1 0 1 1 2 6 5

Dividing each element by L = 10ft

210 1

1

0.1 0 0.1 0 10 0 0 0 20.1 0 0.1 0 6

k AE

0.110

0 0 0 0 5

Member 2:

Determine x & y, where L = 14.14ft10 0 0.70714 14x

10 0 0.70714 14y

14.14x 14.14y

0.035 0.035 0.035 0.035 1

1 2 7 8

2

0.035 0.035 0.035 0.035 20.035 0.035 0.035 0.035 70 035 0 035 0 035 0 035 8

k AE

2 2x x y x x y

0.035 0.035 0.035 0.035 8


2 2

2 2

2 2

y x y y x y

x x y x x y

AEkL

0 0 0.10x

10 0 1

10y

1 2 3 4y x y y x y

3

0 0 0 0 10 0.1 0 0.1 20 0 0 0 3

k AE

0 0 0 0 30 0.1 0 0.1 4

M b 4Member 4:Determine x & y, where L = 10ft

10 0 110x

10 10 0.

10y

10x 10y

0.1 0 0.1 0 3

3 4 7 8

4

0 0 0 0 40.1 0 0.1 0 70 0 0 0 8

k AE

2 2x x y x x y

0 0 0 0 8

Member 5:Determine x & y, where L = 14.14ft

2 2

2 2

2 2

y x y y x y

x x y x x y

AEkL

10 0 0.70714.14x

0 10 0.70714.14y

3 4 6 5y x y y x y

5

0.035 0.035 0.035 0.035 30.035 0.035 0.035 0.035 40 035 0 035 0 035 0 035 6

k AE

0.035 0.035 0.035 0.035 6

0.035 0.035 0.035 0.035 5

M b 6Member 6:Determine x & y, where L = 10ft

10 10 0.10x

10 0 1

10y

10x 10y

0 0 0 0 6

6 5 7 8

6

0 0.1 0 0.1 50 0 0 0 70 0 1 0 0 1 8

k AE

2 2

2 2x x y x x y

0 0.1 0 0.1 8

Structure stiffness matrix2 2

2 2

2 2

y x y y x y

x x y x x y

AEkL

K = k1 + k2 + k3 + k4 + k5 + k6

y x y y x y

0.1 0 0 0 0 0.1 0 0 10 0 0 0 0 0 0 0 2

1 2 3 4 5 6 7 8

0 0 0 0 0 0 0 0 30 0 0 0 0 0 0 0 40 0 0 0 0 0 0 0 5

K AE

0.1 0 0 0 0 0.1 0 0 6

0 0 0 0 0 0 0 0 70 0 0 0 0 0 0 0 8

0.035 0.035 0 0 0 0 0.035 0.035 1

1 2 3 4 5 6 7 8

0.035 0.035 0 0 0 0 0.035 0.035 20 0 0 0 0 0 0 0 30 0 0 0 0 0 0 0 4

AE

0 0 0 0 0 0 0 0 50 0 0 0 0 0 0 0 6

0.035 0.035 0 0 0 0 0.035 0.035 7

AE

0.035 0.035 0 0 0 0 0.035 0.035 8

1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 80 0 0 0 0 0 0 0 10 0.1 0 0.1 0 0 0 0 20 0 0 0 0 0 0 0 3

0 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 20 0 0 1 0 0 0 0 1 0 3

1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8

0 0 0 0 0 0 0 0 30 0.1 0 0.1 0 0 0 0 40 0 0 0 0 0 0 0 5

AE

0 0 0.1 0 0 0 0.1 0 30 0 0 0 0 0 0 0 40 0 0 0 0 0 0 0 5

AE

0 0 0 0 0 0 0 0 6

0 0 0 0 0 0 0 0 70 0 0 0 0 0 0 0 8

0 0 0 0 0 0 0 0 60 0 0.1 0 0 0 0.1 0 70 0 0 0 0 0 0 0 8

0 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 0 1

1 2 3 4 5 6 7 81 2 3 4 5 6 7 8

0 0 0 0 0 0 0 0 20 0 0.035 0.035 0.035 0.035 0 0 30 0 0.035 0.035 0.035 0.035 0 0 4

AE

0 0 0 0 0 0 0 0 20 0 0 0 0 0 0 0 30 0 0 0 0 0 0 0 4

AE

0 0 0.035 0.035 0.035 0.035 0 0 50 0 0.035 0.035 0.035 0.035 0 0 60 0 0 0 0 0 0 0 7

AE

0 0 0 0 0.1 0 0 0.1 50 0 0 0 0 0 0 0 60 0 0 0 0 0 0 0 7

AE 0 0 0 0 0 0 0 0 7

0 0 0 0 0 0 0 0 8

0 0 0 0 0 0 0 0 70 0 0 0 0.1 0 0 0.1 8

0.135 0.035 0 0 0 0.1 0.035 0.0350 035 0 135 0 0 1 0 0 0 035 0 035

12

1 2 3 4 5 6 7 8

0.035 0.135 0 0.1 0 0 0.035 0.0350 0 0.135 0.035 0.035 0.035 0.1 00 0.1 0.035 0.135 0.035 0.035 0 0

K AE

234

0 0 0.035 0.035 0.135 0.035 0 0.10.1 0 0.035 0.035 0.035 0.135 0 0

0 035 0 035 0 1 0 0 0 0 1

K AE

56

35 0 035 7

0.035 0.035 0.1 0 0 0 0.1 35 0.035 7

0.035 0.035 0 0 0.1 0 0.035 0.135 8

0. 0. 0.1 0. 0.035 0. 0.135

0. 0. 0. 0. 0.035 0. 0.035

Application of Stiffness Method• Truss analysis

– The global force components Q acting on the truss can be l t d t it l b l di l t D irelated to its global displacements D using

k 11 12Q K K D

Q KD

k 11 12 u

u 21 22 k=

Q K K DQ K K D

Q D = know external loads & displacements TheQk, Dk = know external loads & displacements. The loads here exist on the truss as part of the problem

Qu, Du = unknown loads & displacements. The loads here represent the unknown support reactions

K = structure stiffness matrix

11 12k u kQ K D K D

Q K D K D 21 22u u kQ K D K D

Application of Stiffness Method– The member forces can be determined using

'q k TD

0. 0.1 10 01 1

Nx

x y NyN

DDq AEDq L

0. 0.1 1 x y FxF

Fy

Dq LD

D Nx

NyF x y x y

Fx

DDAEqDL

Fx

FyD

Example 3• Determine the force in each member of the two member truss

shown. AE is constant.

Structure stiffness matrix: from previous example

0.405 0.096 0.333 0. 0.072 0.0960 096 0 128 0 0 0 096 0 128

0.096 0.128 0. 0. 0.096 0.128

0.333 0. 0.333 0. 0. 0.0. 0. 0. 0. 0. 0.

K AE

0.072 0.096 0. 0. 0.072 0.0960.096 0.128 0. 0. 0.096 0.128

Displacements and loads

Q KD

1 1

2 2

0.405 0.096 0.333 0. 0.072 0.0960.096 0.128 0. 0. 0.096 0.128

Q DQ D

3 3

4 4

5 5

0.333 0. 0.333 0. 0. 0.0. 0. 0. 0. 0. 0.

0.072 0.096 0. 0. 0.072 0.096

Q DAE

Q DQ D

5 5

6

0.072 0.096 0. 0. 0.072 0.0960.096 0.128 0. 0. 0.096 0.128

Q DQ

6D

Th k l di l D D D D 0The known external displacements are D3 = D4 = D5 = D6 = 0.The known external loads are Q1 = 0., Q2 = -2k

Rewrite the matrixRewrite the matrix

1

2

0 0.405 0.096 0.333 0. 0.072 0.0962 0.096 0.128 0. 0. 0.096 0.128

DD

3

4

0.333 0. 0.333 0. 0. 0. 00. 0. 0. 0. 0. 0. 0

0 072 0 096 0 0 0 072 0 096 0

QAE

QQ

Determine the unknown displacements by

5

6

0.072 0.096 0. 0. 0.072 0.096 00.096 0.128 0. 0. 0.096 0.128 0

QQ

p y

1

2

0 0.405 0.096 02 0.096 0.128 0

DAE

D

1 20 0.405 0.096AE D D

1 22 0.096 0.128AE D D 1

4.505DAE

219.003DAE

The support reactions are now obtained by3 0.333 0 04.05

0 0 0QQ

4

5

6

0 0 00.072 0.096 19.003 00.096 0.128 0

Q AEAEQ

AEQ

6

3 0.333(4.505) 1.5Q K

0Q4 0Q

5 0.072(4.505) 0.096( 19.003) 1.5Q K

6 0.096(4.505) 0.128( 19.003) 2.0Q K

The force in each member obtained by

6 0.096( .505) 0. 8( 9.003) .0Q

Nx

NyF x y x y

F

DDAEqDL

Fx

Fy

DLD

Member 1:

1x 0y 3L ft

4.505

19.0031 0 1 0 1 5

AEAE k

1 1 0 1 0 1.5

300

q kAE

0

Member 2:0 6 0 8 5L ft

4.505

19 003AE

0.6x 0.8y 5L ft

2

19.0030.6 0.8 0.6 0.8 2.5

50

AEq kAE

0

Example 4

• Determine the support reactions and the force in member 2 of the truss shown in figure. AE is constantg

Example 4

• The Stiffness matrix has been determine in Example 2 using the same notation as shown.

Example 5 Determine the force in member 2 of the assembly if the support at joint 1

settles downward 25mm. AE = 8103 kN

Members ForcesMember 1: 0, 1, 3

0x y L m

1

0.00258000 0 1 0 1 8.3330.0055630 021875

q kN 0.021875

Member 2: 0.8, 0.6, 5

0 00556x y L m

2

0.005560.0218758000 0.8 0.6 0.8 0.6 13.9

05q kN

0

Member 3: 1, 0, 4L m Member 3:

1, 0, 4

008000 1 0 1 0 11 11

x y L m

q kN

3 1 0 1 0 11.11

0.0055640.021875

q kN

Nodal Coordinate

If the support is an inclined roller The zero deflection cannot be defined using single horizontal andThe zero deflection cannot be defined using single horizontal and vertical global coordinate system

To solve this problemA set of nodal coordinate system x’’, y’’ located at the inclined y ysupport will be used

Nodal Coordinate

0 0NxD

Dd

The Nodal Displacements

'' '' ''

''

0. 0.0. 0.

x y NyN

x y FxF

Fy

DdDdD

The Nodal Forces

0.Nx xQ

'' ''

0.0.0

Ny y N

Fx x F

Q qQ qQ

'' ''0.Fy yQ

Nodal CoordinateThe Stiffness Matrix

'Tk T k T

0.0. 0. 0.1 1

x

y x yAEk

'' '' ''

''

0. 0. 0.1 10.

x x y

y

kL

2'' '' xN

'' ''x y x yN N F F

''

'' ''2

'' ''2

'' '' '' '' ''2

xx x y x x x y

yy x y y x y y

x x y x x x y x

NAEk FL

''2

'' '' '' '' ''x y y y x y y yF

Nodal Coordinate

Th b f b d i d i

The Member Forces

– The member forces can be determined using

'q k TD

0. 0.1 1Nx

x y NyN

DDq AE

q

'' '' ''

''

0. 0.1 1F x y Fx

Fy

Dq LD

NxD

'''' ''

Nx

NyF x y x y

Fx

DAEq DLD

''FyD

Example 6

• Determine the support reaction for the truss shown

Members ForcesMember 1: '' ''0, 1, 0.707, 0.707, 4

0x y x y

L m

1

010 1 0.707 0.707 22.5127.34

0

EAq kNEA

0

Member 2: '' ''0, 1, 0.707, 0.707, 3x y x yL m

2

352.5157.510 1 0.707 0.707 22.5127.33

EAq kNEA

127.33

0EA

Member 3: 0.8, 0.6, 5L m Member 3:

0.8, 0.6, 5

0010 8 0 6 0 8 0 6 37 5

x y L m

EAq kN

3 0.8 0.6 0.8 0.6 37.5

352.55157.5

q kNEA

Thermal Changes and Fabrication Errors

Thermal EffectsIf a truss member of length L is subjected to a temperatureIf a truss member of length L is subjected to a temperature increase T, The member undergo an increase in length of

L TL

Then

0Nq AE T

q AE T

0Fq AE T Transforming into global coordinate

Q

0

0

0.0. 1

0. 1

Nxx x

Ny y y

Q

QAE T AE T

Q

0

0

0. 10.

x xFx

y yFy

Q

Q


Fabrication ErrorsIf a truss member is made too long by an amount L then the forceIf a truss member is made too long by an amount L, then the force q0 needed to keep the member at its design length L

0NAE Lq

LAE L

0FAE Lq

L

I l b l di tIn global coordinates

0Nxx

Q

0

0

x

Ny y

xFx

Q AE LLQ

0y

FyQ


Matrix Analysis

0Q KD Q

Where: Q0 is the column matrix for the entire truss of the initial fixed-end force caused by the temperature changes and fabrication errors of the members defined in previous equations

kk 11 12 u 0=QQ K K D

Q K K D Q

The Member forces

u 21 22 k u 0Q K K D Q

The Member forces'

0q k TD q

Example 7

• Determine the force in member 1 and 2 of the pin-connected assembly if the member 2 was made 0.01m too short before it ywas fitted into place. Take AE=8(103) kN.

Example 8

Member 2 of the truss shown is subjected to an increase in temperature of 150 F. Determine the force developed in member 2. E=29(106)lb/in2. p ( )Each member has across sectional area of A=0.75 in2 , α=6.5(10-6)/Fo

Space Truss AnalysisSpace Truss Analysis

cos F N F Nx x x x

2 2 2

2 2 2

cos

cos

x x

F N F N F N

F N F Ny y

L x x y y z z

y y y yL

2 2 2

2 2 2cos

y y

F N F N F N

F N F Nz z

L x x y y z z

z z z zL

F N F N F NL x x y y z z

0 0 00 0 0x y z

x y zT

Stiffness MatrixStiffness Matrix00

x 0

0 0 00 1 10 0 00 1 1

y

x y zz

x y zx

AEkL

00

x y zx

y

z

2 2

x y z x y zN N N F F F

N 2 2

2 2

2 2

x x y x z x x y x z x

y x y y z y x y y z y

x z y z z x z y z z z

NNNAEk

2 2

2 2

2 2

x y x x z x x y x z x

x y y y z x y y y z y

kFLFF

x z y z z x z y z z z

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TAliTruss Analysis Using The Stiffness MethodUsing The ...site.iugaza.edu.ps/ajubeh/files/2015/02/Chapter-14_5-5-2010.pdf · Chapter 14 TAliTruss Analysis Using The Stiffness MethodUsing