Plane Truss Stiffness Matrix - Cleveland State Universityacademic.csuohio.edu/duffy_s/511_13.pdf · Lecture 13: Trusses & Grids –Stiffness Method Washkewicz College of Engineering

Lecture 13: Trusses & Grids – Stiffness Method

Washkewicz College of Engineering

Plane Truss – Stiffness Matrix

The distinguishing feature of a plane truss is that loads are applied in the plane of the

structure whereas in a space truss they are not.

We now wish to outline the procedure of formulating the joint stiffness matrix [SJ] for a

plane truss structure. The extension to a three dimensional space truss will be intuitively

obvious. 1



2

Consider an arbitrary member, i. isolated from a

generalized plane truss depicted below:

The joints at the end of truss member i are denoted j and k. The plane truss lies in the x-y

plane. The joint translations are the unknown displacements and these displacements are

expressed in terms of their x and y components.

We will relax the requirement that truss members are two force members. This allows for

loads that are applied between joints to a truss member, and it allows consideration of the

weight of the truss member.

i

k

j

x

y

Primary load carrying capabilities

are in the plane of the truss –

hence the need for shear walls.



The positive directions of the four displacement components (two translations at either

end) of member i are depicted in the figure below

It will be convenient to utilize the direction cosines associated with this arbitrary

member. In terms of the joint coordinates the direction cosines are

L

xx

C

jk

X

1cos

L

yy

C

jk

Y

2cos

22

jkjk yyxxL

with

3



The beam member stiffness matrix developed in the previous section of notes can be

easily adapted for use in the case of a plane truss. The member stiffness matrix [SM] for

an arbitrary truss member with member axes Xm and Ym oriented along the member and

perpendicular to the member can be obtained by considering Case #1 and Case #7 from

the previous section of notes.

Using the numbering joint numbering system and the member axes depicted in the

following figure

then the member stiffness matrix for

a truss member is as follows

4



Note that [SJ] is based on axes oriented to the

structure. Truss member stiffnesses may be

obtained in one of two ways. Either the

stiffnesses are directly computed using the figure

to the left, or the second method consists of first

obtaining the stiffness matrix relative to the

member oriented axes and then imposing a

suitable matrix transformation that transforms

these elements to axes relative to the structure.

We will focus on the direct method first to help

develop an intuition of how the structure

behaves. Unit displacement in both the x and y

directions are applied at each end of the member.

If a unit displacement in the x direction is applied

to the j end of the member, the member shortens

and an axial compression force is induced. The

magnitude of the force is

xx C

L

EA

5



Restraint actions at the ends of the truss member in the x and y directions are required.

They are equal to the components of the axial force induced in the member, and are

identified here as elements of the [SMD] matrix in order to distinguish them from elements

of the [SM] matrix. The numbering of these elements are shown in the previous figure.

Thus

yx

xMDMD

xx

MDMD

yxx

MD

xx

MD

CCL

EASS

CL

EASS

CCL

EAS

CL

EAS

2141

2

1131

21

2

11

6



In a similar fashion, a unit displacement in the y direction at the j end of the member yields

22242

1232

2

22

12

yx

MDMD

yxx

MDMD

yx

MD

yxx

MD

CL

EASS

CCL

EASS

CL

EAS

CCL

EAS

7



In a similar fashion, a unit displacement in the x direction at the k end of the member yields

yx

xMDMD

xx

MDMD

yxx

MD

xx

MD

CCL

EASS

CL

EASS

CCL

EAS

CL

EAS

2343

2

1333

23

2

13

8



In a similar fashion, a unit displacement in the y direction at the k end of the member yields

22444

1434

2

24

14

yx

MDMD

yxx

MDMD

yx

MD

yxx

MD

CL

EASS

CCL

EASS

CL

EAS

CCL

EAS

9



We have just developed the four rows of the [SMD] matrix, i.e.,

10



Example 13.1

Consider the plane truss with four bars meeting at a common joint E. This truss

only has two degrees of freedom from a kinematic standpoint. It is a convenience

to identify the bars of the truss numerically. The bars have lengths L1, L2, L3 and

L4 and axial rigidities EA1, EA2, EA3 and EA4

The loads consist

of two

concentrated

forces P1 and P2

action at joint E.

We will consider

the bar weights

identified here as

w1, w2, w3 and w4

(force/length).

The unknown displacements at joint E are identified as D1 and D2. We seek to

calculate member end actions AM1, AM2, AM3 and AM4. 11



Because the weight of each truss member is included, the axial forces at either end

of a truss member will be different at joints A, B, C and D then the axial force at

joint E. The axial forces at joint E could be computed as well as the shear stresses

at the end of each truss member, however they are omitted in this example for

simplicity.

The loads P1 and P2 correspond to unknown displacements D1 and D2, thus

We next consider the restrained

structure shown at the right. Here joint

E is fixed with a pin support that

produce loads ADL1 and ADL2 associated

with D1 and D2.

2

1

P

PAD

12



Each truss member can be considered loaded as shown below. The points of

support are indicated as A and E for the purpose of discussion and do not

correspond to actual joints in labeled in the original truss. One could use the Greek

alphabet, but the nomenclature should be transparent given the context where it

used.

13



Since the weights of the truss members produce no horizontal reactions, the actions

ADL1 must be zero and ADL2 must be equal to half the weight of all the truss elements,

i.e.,

2

0

2222

0

44332211 WLwLwLwLwADL

The quantity W is the total weight of the truss. For the purpose of calculating end

actions for the vector AML, consider that from the previous figure

iii

MLi

LwA sin

2

444

333

222

111

sin

sin

sin

sin

2

1

Lw

Lw

Lw

Lw

AMLor

14



The next step is formulating the stiffness matrix by imposing unit displacement

associated with D1 and D2 on the restrained structure as indicated below

To obtain the stiffness values it is necessary to compute the forces in the truss

elements when the unit displacements are applied to joint E.15



When the upper joint of the element

moves to the right, the lower joint stays

fixed.

When the upper joint of the element

moves up, again the lower joint stays

fixed. Both actions elongate the truss

elements. The geometry of the elongation

is determined by the translation of joint E.

cos

AE

PL

L

EAP cos

16



cosL

EA

sinL

EA

When joint E is subjected to a unit translation to the right the truss element elongates

an amount

When joint E is subjected to a unit translation vertically the truss element elongates

an amount

The formulas given above are suitable for use in analyzing this plane truss. In a later

lecture a more systematic approach to the development of member stiffnesses is

developed that works for trusses and all types of structures.

The stiffness S11 is composed of contributions from various elements of the truss.

Consider the contribution to S11 from member 3, i.e.,

3

2

3

311

3 cos L

EAS

17



Thus

3

2

3

32

2

2

2

1

1

4

43

2

3

32

2

2

2

1

1

4

2

4

43

2

3

32

2

2

21

2

1

1

11

4

11

3

11

2

11

1

11

coscos

0coscos1

coscoscoscos

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

SSSSS

The final expression results from the fact that truss element 1 is horizontal and truss

element 4 is vertical.

18



21

4

21

3

21

2

21

1

21 SSSSS

Similarly the stiffness S21 is composed of contributions from various elements of the

truss. Consider the contribution to S21 from member 3, i.e.,

33

3

321

3 sincos L

EAS

Thus

33

3

322

2

2

4

433

3

322

2

2

1

1

44

4

433

3

322

2

211

1

1

sincossincos

10sincossincos01

sincossincossincossincos

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

19



By an analogous procedure S12 and S22 are

33

3

322

2

212 sincossincos

L

EA

L

EAS

4

43

2

3

32

2

2

222 sinsin

L

EA

L

EA

L

EAS

The two expressions on this page as well as the two from the previous page

constitute the stiffness matrix [S]. The next step would be inverting this matrix and

performing the following matrix computation to find the displacement D1 and D2.

DLD AASD 1

The vector {AD} and the matrix {ADL} were established earlier.

20



Since the vector {AML} was determined earlier as well, we need only identify the elements

of the matrix {AMD}. This matrix contains the member end-actions due to unit

displacements associated with the displacements D1 and D2, but the end actions are

computed using the restrained structure. Thus for ith member using a previous figure

i

i

iMDi

L

EAA cos1 i

i

iMDi

L

EAA sin2

4

4

44

4

4

3

3

33

3

3

2

2

22

2

2

1

1

11

1

1

sincos

sincos

sincos

sincos

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

L

EA

AMD

thus

And we can now solve

DAAA MDMLM 21



22

Example 13.2



23

Structural Grids (Need a theory section on - Stiffness Method as it applies to

grids

Examples of two dimensional grids are depected below:

Grids behave more like frames. However, loads are typically applied perpendicular

to the plane of the structure. Moments are allowed in the plane of the structure and

perpendicular to the structure.



24

A space grid – the structure is not in a plane, but the loads are in essence perpendicular to

the structure.



Example 13.3

The grid shown below consists of two members (AB and BC) that are rigidly joined at

B. Each member is assumed to have flexural rigidity EI and torsional rigidity GJ.

Kinematically, the only unknowns are the displacements at B. Since axial rigidities of

the members is assumed to be quite large relative to EI and GJ, the displacements at B

consist of one translation (D1) and two rotations (D2 and D3). Determine these

unknown displacements.

25



When analyzing a grid by the stiffness method, an artificial restraint is

provided at joint B, i.e.,

It is easier to see what the reactions are if we break the structure above into two

substructures such that

26



From the last figure it is easy to see that

or

L

PADL 0

4

8

80

2321

PLAA

PA DLDLDL

and in a matrix format

80

2321

PLAA

PA DLDLDL

000 321 DLDLDL AAA

27



0

0

0

DA

The vector {AD} represents actions in the unrestrained structure associated

with the unknown displacement D1, D2 and D3. Since there are no loads

associated with these displacements {AD} is a null vector and in a matrix

format

DLD AASD 1

We have {ADL} and {AD} the next step is the solution of the superposition

expression

for the unknown displacements. To do that we need to formulate the stiffness

matrix and find its inverse.

28



From the figures above

23121311

60

12

L

EISS

L

EIS

0612

31221311 SL

EIS

L

EIS

231221311

6624

L

EIS

L

EIS

L

EIS

The stiffness matrix is found by analyzing the restrained structure for the effects of unit

translations and rotations associated with the unknown displacements. In the following

figure the grid structure is once again split into two substructures.

29



To obtain the second column of the stiffness matrix utilize the following figure

00 322212 SL

GJSS

046

3222212 SL

EIS

L

EIS

046

3222212 SL

GJ

L

EIS

L

EIS


30



To obtain the third column of the stiffness matrix utilize the following figure


L

EISS

L

EIS

40

63323213

J

GJSSS

332313 00

L

GJ

L

EISS

L

EIS

40

63323213

31



Define

EI

GJ

2

2

3

406

046

6624

LL

LL

LL

L

EIS

2

2

2

1

2

21

1

406

046

66

24

1

LL

LL

LLCL

CEICS

then

and inverting this stiffness matrix leads to

25

1

4

3

2

1

C

C

C

where

32



DLD AASD 1

18

256

254

4196

2

L

EI

PLD

Solving

leads to

33



34

Example 13.4

Documents

Plane Truss Stiffness Matrix - Cleveland State Universityacademic.csuohio.edu/duffy_s/511_13.pdf · Lecture 13: Trusses & Grids –Stiffness Method Washkewicz College of Engineering