truss analysis using the stiffness methodsite.iugaza.edu.ps/marafa/files/chapter-14-2019.pdf ·...

Chapter 14

Truss Analysis

Using The Stiffness Method

Stiffness 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

different procedure for each of these cases.

– The stiffness method yields the displacements & forces directly, whereas

with the flexibility method the displacements are not obtained directly.

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

of 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.

Stiffness Method

Preliminary Definitions & Concepts

• Member & node identifications– We 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

be done using an arrow written along the member with the head of the

arrow directed toward the far end.

• Global & member coordinates. – We will use two different type of coordinate systems, global or

structure 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 nodes.

– Local system x’,y’ used to specify the sense of direction of members

displacements & internal loadings.


• Degrees of freedom – The unconstrained truss has two degree of freedom or two possible

displacements 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 global

coordinate direction using an associated arrow.

• For 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 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 end

• Case II

– Positive displacement dF on the far end

• Case I + Case II

– Resultant forces caused by both displacements are

'N N

AEq d

L 'F N

AEq d

L

''N F

AEq d

L ''F F

AEq d

L

N N F

AE AEq d d

L L

F N F

AE AEq d d

L L

– These load-displacement equations written in matrix form

1 1

1 1

N N

F F

q dAE

q dL

'q k d

1 1'

1 1

AEk

L

or

where

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

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

Member Stiffness Matrix

Transformation Matrices• Displacement & force transformation

– We will now develop a method for

transforming the member forces q and

displacements d defined in local coordinate

to global coordinates

cos F Nx x

x x

L

cos F Ny y

y y

L

2 2

cos F Nx x

F N F N

x x

x x y y

2 2

cos F Ny y

F N F N

y y

x x y y

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

– Also joint F has displacements DFx & DFy

– Displacements at N & F

cos cosF Fx x Fy yd D D

F Fx x Fy yd D D

or

N Nx x Ny yd D D

F Fx x Fy yd D D

0. 0.

0. 0.

Nx

x y NyN

x y FxF

Fy

D

Dd

Dd

D

d TDor

0. 0.

0. 0.

x y

x y

T

where

Transformation Matrices

• Force transformation matrix

– Force qN applied to the near end of the member

– Force qF applied to the far end of the member

– Rewrite in a matrix form:

cosNx N xQ q or

cosNy N yQ q

Nx N xQ q Ny N yQ q

cosFx F xQ q or

cosFy F yQ q

Fx F xQ q Fy F yQ q

0.

0.

0.

0.

Nx x

Ny y N

Fx x F

Fy y

Q

Q q

Q q

Q

TQ T qor where

0.

0.

0.

0.

x

yT

x

y

T

Transformation Matrices

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. 'q 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

0. 0. 0.1 1

0.

x

y x y

x x y

y

AEk

L

Member Global Stiffness Matrix

2 2

2 2

2 2

2 2

xx x y x x y

yy x y y x y

xx x y x x y

yy x y y x y

N

NAEk

FL

F

x y x yN N F F

'Tk T k T

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 01

3x

0 00

3y

2 2

2 2

2 2

2 2

xx x y x x y

yy x y y x y

xx x y x x y

yy x y y x y

N

NAEk

FL

F

x y x yN N F F

1

0.333 0. 0.333 0. 1

0. 0. 0. 0. 2

0.333 0. 0.333 0. 3

0. 0. 0. 0. 4

k AE

1 2 3 4

Dividing each element by L = 3ft

22 1

0.3333

x

L

Member 2:Determine x & y, where L = 5ft

3 00.6

5x

4 00.8

5y

2 2

2 2

2 2

2 2

xx x y x x y

yy x y y x y

xx x y x x y

yy x y y x y

N

NAEk

FL

F

x y x yN N F F

2

0.072 0.096 0.072 0.096 1

0.096 0.128 0.096 0.128 2

0.072 0.096 0.072 0.096 5

0.096 0.128 0.096 0.128 6

k AE

1 2 5 6

Dividing each element by L = 5ft 0.8 0.6

0.0965

Structure stiffness matrix

K = k1 + k2

0.333 0 0.333 0 0 0 1 0.072 0.096 0 0 0.072 0.096

0. 0 0. 0 0 0 2 0.096 0.128 0 0 0.096 0.128

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

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

K AE AE

1

2

3

4

5

.096 0.128 6

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

0.405 0.096 0.333 0. 0.072 0.096

0.096 0.128 0. 0. 0.096 0.128

0.333 0. 0.333 0. 0. 0.

0. 0. 0. 0. 0. 0.

0.072 0.096 0. 0. 0.072 0.096

0.096 0.128 0. 0. 0.096 0.128

K AE

Example 2

Determine the structure stiffness matrix for the truss shown. AE is constant

Member 1:

Determine x & y, where L = 10ft

10 01

10x

0 00

10y

2 2

2 2

2 2

2 2

xx x y x x y

yy x y y x y

xx x y x x y

yy x y y x y

N

NAEk

FL

F

x y x yN N F F

1

0.1 0 0.1 0 1

0 0 0 0 2

0.1 0 0.1 0 6

0 0 0 0 5

k AE

1 2 6 5

Dividing each element by L = 10ft

2

10.1

10

Member 2:

Determine x & y, where L = 14.14ft10 0

0.70714.14

x

10 0

0.70714.14

y

2 2

2 2

2 2

2 2

x x y x x y

y x y y x y

x x y x x y

y x y y x y

AEk

L

2

0.035 0.035 0.035 0.035 1

0.035 0.035 0.035 0.035 2

0.035 0.035 0.035 0.035 7

0.035 0.035 0.035 0.035 8

k AE

1 2 7 8

Member 3:

Determine x & y, where L = 10ft0 0

0.10

x

10 0

110

y

3

0 0 0 0 1

0 0.1 0 0.1 2

0 0 0 0 3

0 0.1 0 0.1 4

k AE

1 2 3 4


10 01

10x

10 100.

10y

2 2

2 2

2 2

2 2

x x y x x y

y x y y x y

x x y x x y

y x y y x y

AEk

L

4

0.1 0 0.1 0 3

0 0 0 0 4

0.1 0 0.1 0 7

0 0 0 0 8

k AE

3 4 7 8

Member 5:

Determine x & y, where L = 14.14ft10 0

0.70714.14

x

0 10

0.70714.14

y

5

0.035 0.035 0.035 0.035 3

0.035 0.035 0.035 0.035 4

0.035 0.035 0.035 0.035 6

0.035 0.035 0.035 0.035 5

k AE

3 4 6 5


10 100.

10x

10 01

10y

2 2

2 2

2 2

2 2

x x y x x y

y x y y x y

x x y x x y

y x y y x y

AEk

L

6

0 0 0 0 6

0 0.1 0 0.1 5

0 0 0 0 7

0 0.1 0 0.1 8

k AE

6 5 7 8

Structure stiffness matrix

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

0.1 0 0 0 0 0.1 0 0 1

0 0 0 0 0 0 0 0 2

0 0 0 0 0 0 0 0 3

0 0 0 0 0 0 0 0 4

0 0 0 0 0 0 0 0 5

0.1 0 0 0 0 0.1 0 0 6

0 0 0 0 0 0 0 0 7

0 0 0 0 0 0 0 0 8

K AE

1 2 3 4 5 6 7 8

0.035 0.035 0 0 0 0 0.035 0.035 1

0.035 0.035 0 0 0 0 0.035 0.035 2

0 0 0 0 0 0 0 0 3

0 0 0 0 0 0 0 0 4

0 0 0 0 0 0 0 0 5

0 0 0 0 0 0 0 0 6

0.035 0.035 0 0 0 0 0.035 0.035 7

0.035 0.035 0 0 0 0 0.035 0.035 8

AE

1 2 3 4 5 6 7 8

0 0 0 0 0 0 0 0 1

0 0.1 0 0.1 0 0 0 0 2

0 0 0 0 0 0 0 0 3

0 0.1 0 0.1 0 0 0 0 4

0 0 0 0 0 0 0 0 5

0 0 0 0 0 0 0 0 6

0 0 0 0 0 0 0 0 7

0 0 0 0 0 0 0 0 8

AE

0 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 0 2

0 0 0.1 0 0 0 0.1 0 3

0 0 0 0 0 0 0 0 4

0 0 0 0 0 0 0 0 5

0 0 0 0 0 0 0 0 6

0 0 0.1 0 0 0 0.1 0 7

0 0 0 0 0 0 0 0 8

AE

0 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 0 2

0 0 0.035 0.035 0.035 0.035 0 0 3

0 0 0.035 0.035 0.035 0.035 0 0 4

0 0 0.035 0.035 0.035 0.035 0 0 5

0 0 0.035 0.035 0.035 0.035 0 0 6

0 0 0 0 0 0 0 0 7

0 0 0 0 0 0 0 0 8

AE

0 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 0 2

0 0 0 0 0 0 0 0 3

0 0 0 0 0 0 0 0 4

0 0 0 0 0.1 0 0 0.1 5

0 0 0 0 0 0 0 0 6

0 0 0 0 0 0 0 0 7

0 0 0 0 0.1 0 0 0.1 8

AE

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

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

0.135 0.035 0 0 0 0.1 0.035 0.035

0.035 0.135 0 0.1 0 0 0.035 0.035

0 0 0.135 0.035 0.035 0.035 0.1 0

0 0.1 0.035 0.135 0.035 0.035 0 0

0 0 0.035 0.035 0.135 0.035 0 0.1

0.1 0 0.035 0.035 0.035 0.135 0 0

0.035 0.035 0.1 0 0 0 0.1

K AE

1

2

3

4

5

6

35 0.035 7

0.035 0.035 0 0 0.1 0 0.035 0.135 8

1 2 3 4 5 6 7 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

related to its global displacements D using

Qk, 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 12

21 22

k u k

u u k

Q K D K D

Q K D K D

k 11 12 u

u 21 22 k

=

Q KD

Q K K D

Q K K D

Application of Stiffness Method

– The member forces can be determined using

0. 0.1 1

0. 0.1 1

Nx

x y NyN

x y FxF

Fy

D

Dq AE

Dq L

D

'q k TD

Nx

Ny

F x y x y

Fx

Fy

D

DAEq

DL

D

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 0.096

0.096 0.128 0. 0. 0.096 0.128

0.333 0. 0.333 0. 0. 0.

0. 0. 0. 0. 0. 0.

0.072 0.096 0. 0. 0.072 0.096

0.096 0.128 0. 0. 0.096 0.128

K AE

Displacements and loads

Q KD

1 1

2 2

3 3

4 4

5 5

6

0.405 0.096 0.333 0. 0.072 0.096

0.096 0.128 0. 0. 0.096 0.128

0.333 0. 0.333 0. 0. 0.

0. 0. 0. 0. 0. 0.

0.072 0.096 0. 0. 0.072 0.096

0.096 0.128 0. 0. 0.096 0.128

Q D

Q D

Q DAE

Q D

Q D

Q

6D

The known external displacements are D3 = D4 = D5 = D6 = 0.

Determine the unknown displacements by

The known external loads are Q1 = 0., Q2 = -2k

Rewrite the matrix

1

2

3

4

5

6

0 0.405 0.096 0.333 0. 0.072 0.096

2 0.096 0.128 0. 0. 0.096 0.128

0.333 0. 0.333 0. 0. 0. 0

0. 0. 0. 0. 0. 0. 0

0.072 0.096 0. 0. 0.072 0.096 0

0.096 0.128 0. 0. 0.096 0.128 0

D

D

QAE

Q

Q

Q

1

2

0 0.405 0.096 0

2 0.096 0.128 0

DAE

D

1 20 0.405 0.096AE D D

1 22 0.096 0.128AE D D 1

4.505D

AE

2

19.003D

AE

The support reactions are now obtained by

The force in each member obtained by

3

4

5

6

0.333 0 04.05

0 0 0

0.072 0.096 19.003 0

0.096 0.128 0

Q

Q AEAE

Q

AEQ

3 0.333(4.505) 1.5Q K

4 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

Nx

Ny

F x y x y

Fx

Fy

D

DAEq

DL

D

Member 1:

1

4.505

19.0031 0 1 0 1.5

3

0

0

AE

AEq k

AE

1x 0y 3L ft

Member 2:

2

4.505

19.0030.6 0.8 0.6 0.8 2.5

5

0

0

AE

AEq k

AE

0.6x 0.8y 5L ft

Example 4

• Determine the support reactions and the force in member 2 of

the truss shown in figure. AE is constant

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

Member 1:

1

0, 1, 3

0

0.002580000 1 0 1 8.333

0.005563

0.021875

x y L m

q kN

Member 2:

2

0.8, 0.6, 5

0.00556

0.02187580000.8 0.6 0.8 0.6 13.9

05

0

x y L m

q kN

Member 3:

3

1, 0, 4

0

080001 0 1 0 11.11

0.005564

0.021875

x y L m

q kN

Members Forces

Nodal Coordinate

To solve this problem

A set of nodal coordinate system x’’, y’’ located at the inclined

support will be used

If the support is an inclined roller

The zero deflection cannot be defined using single horizontal and

vertical global coordinate system

Nodal Coordinate

'' ''

'' ''

0.

0.

0.

0.

Nx x

Ny y N

Fx x F

Fy y

Q

Q q

Q q

Q

The Nodal Forces

'' '' ''

''

0. 0.

0. 0.

Nx

x y NyN

x y FxF

Fy

D

Dd

Dd

D

The Nodal Displacements

Nodal Coordinate

''

''

2

'' ''

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

x y y y x y y y

N

NAEk

FL

F

'' ''x y x yN N F F

'Tk T k T

The Stiffness Matrix

'' '' ''

''

0.

0. 0. 0.1 1

0. 0. 0.1 1

0.

x

y x y

x x y

y

AEk

L

– The member forces can be determined using

'' '' ''

''

0. 0.1 1

0. 0.1 1

Nx

x y NyN

F x y Fx

Fy

D

Dq AE

Dq L

D

'q k TD

''

''

'' ''

Nx

Ny

F x y x y

Fx

Fy

D

DAEq

DL

D

Nodal Coordinate

The Member Forces

Example 6

• Determine the support reaction for the truss shown

Member 1:

'' ''

1

1, 0, 0.707, 0.707, 4

0

011 0 0.707 0.707 22.5

127.34

0

x y x yL m

EAq kN

EA

Member 2:

'' ''

2

0, 1, 0.707, 0.707, 3

352.5

157.510 1 0.707 0.707 22.5

127.33

0

x y x yL m

EAq kN

EA

Member 3:

3

0.8, 0.6, 5

0

010.8 0.6 0.8 0.6 37.5

352.55

157.5

x y L m

EAq kN

EA

Members Forces

Thermal Changes and Fabrication Errors

Thermal Effects

If a truss member of length L is subjected to a temperature

increase T, The member undergo an increase in length of

L TL

0

0

N

F

q AE T

q AE T

Then

Transforming into global coordinate

0

0

0

0

0.

0. 1

0. 1

0.

Nxx x

Ny y y

x xFx

y yFy

Q

QAE T AE T

Q

Q


Fabrication Errors

If 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

0

0

N

F

AE Lq

L

AE Lq

L

In global coordinates

0

0

0

0

Nxx

Ny y

xFx

yFy

Q

Q AE L

LQ

Q


0Q KD Q

Q0 : is the column matrix for the entire truss of the

initial fixed-end force caused by the temperaturechanges and fabrication errors of the members

defined in previous equations

kk 11 12 u 0

u 21 22 k u 0

=QQ K K D

Q K K D Q

Matrix Analysis


The Member forces

'

0q k TD q

0

Nx

Ny

F x y x y F

Fx

Fy

DAE T

DAEq q

DLLAED L

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

was fitted into place. Take AE=8(103)

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.

Each member has across sectional area of A=0.75 in2

Springs Structures

1 1'

1 1

Where:

:is the spring stiffness

s

s

k k

k

'q k dks

ks q2q1

d2d1

Example 9

Given: For the spring system shown,

K1 = 100 N/mm, K2 = 200 N/mm,

K3 = 300 N/mm

P = 500 N.

Find:

(a) The global stiffness matrix

(b) Displacements of nodes 1 and 2

(c) The reaction forces at nodes 3 and 4

(d) The force in the spring 2

1 23 41 2 3

Solution

(a) The global stiffness matrix

1

100 0 100 0 1

100 100 3 0 0 0 0 2

100 100 1 100 0 100 0 3

0 0 0 0 4

k

2

200 200 0 0 1

200 200 1 200 200 0 0 2

200 200 2 0 0 0 0 3

0 0 0 0 4

k

3

0 0 0 0 1

300 300 2 0 300 0 300 2

300 300 4 0 0 0 0 3

0 300 0 300 4

k

Solution

The global stiffness matrix

100 0 100 0 1 200 200 0 0 1 0 0 0 0 1

0 0 0 0 2 200 200 0 0 2 0 300 0 300 2

100 0 100 0 3 0 0 0 0 3 0 0 0 0 3

0 0 0 0 4 0 0 0 0 4 0 300 0 300 4

K

K = k1 + k2 + k3

300 200 100 0 1

200 500 0 300 2

100 0 100 0 3

0 300 0 300 4

K

1 23 41 2 3

Application of Stiffness Method– The global force components Q acting on the Structure can

be related to its global displacements D using

k 11 12 u

u 21 22 k

=

Q KD

Q K K D

Q K K D

Qk, 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 12

21 22

k u k

u u k

Q K D K D

Q K D K D

(b) Displacements of nodes 1 and 2

D3 = D4 = 0., Q1 = 0 and Q2 = P = 500

1 1

2 2

10

300 200 0. 11

200 500 500 15

11

D D

D D

1 1

2 2

3 3

4 4

300 200 100 0

200 500 0 300

100 0 100 0

0 300 0 300

Q KD

Q D

Q D

Q D

Q D

1

2

3

4

0 300 200 100 0

500 200 500 0 300

100 0 100 0 0

0 300 0 300 0

D

D

Q

Q

3

3 4

4

10

0300 200 100 0 11 1000

500200 500 0 300 15 11

11100 0 100 0 4500

0 110 300 0 300

0

Q

Q Q

Q

(c) The reaction forces at nodes 3 and 4

(d) The force in the spring 2

22 2 1 1

22 2 2 2

2 2

1 1

2 2

2 2

10 1000

200 200 11 11

200 200 15 1000

11 11

k k d q

k k d q

q q

q q

Space Truss Analysis

2 2 2

2 2 2

2 2 2

cos

cos

cos

F N F Nx x

F N F N F N

F N F Ny y

F N F N F N

F N F Nz z

F N F N F N

x x x x

L x x y y z z

y y y y

L x x y y z z

z z z z

L x x y y z z

0 0 0

0 0 0

x y z

x y z

T

Stiffness Matrix

2 2

2 2

2 2

2 2

2 2

2 2

x y z x y z

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

x y x x z x x y x z x

x y y y z x y y y z y

x z y z z x z y z z z

N N N F F F

N

N

NAEk

FL

F

F

0

0

0 0 00 1 1

0 0 00 1 1

0

0

x

y

x y zz

x y zx

y

z

AEk

L

Stiffness Matrix

2

2

2

xyz xyz

xyz xyz

x x y x z

xyz x y y y z

x z y z z

C Ck

C C

AEC

L

truss analysis using the stiffness methodsite.iugaza.edu.ps/marafa/files/chapter-14-2019.pdf ·...

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