Lecture 13 Trusses and Grids.pptLecture 13: TRUSSES & GRIDS – STIFFNESS METHOD Example Consider the plane truss with four bars meeting at a common joint E. This truss only has two

Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD

ExampleConsider 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 f P d Pforces P1 and P2action at joint E. We will consider the bar weights gidentified 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.


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 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, thus1 2 p p 1 2,

{ }

=2

1

PP

AD

We next consider the restrained structure shown at the right. Here joint E i fi d ith i t th tE is fixed with a pin support that produce loads ADL1 and ADL2 associated with D1 and D2.


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

d l j i i l b l d i h i i l O ld h G kcorrespond 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.


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 ei.e.,

{ }

=

=

00

A{ }

=

+++

=

2222244332211 WLwLwLwLw

ADL

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

MLiLwA γsin

2−

= { }

−=333

222

111

sinsinsin

21

γγγ

LwLwLw

AMLor2

444

333

sin γγ

Lw


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.


When the upper joint of the element moves to the right, the lower joint stays fixedfixed.

When the upper joint of the element moves up, again the lower joint stays fixed. Both actions elongate the truss gelements. The geometry of the elongation is determined by the translation of joint E.


EA

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

γcosL

EA

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

γsinL

EAan 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.,

A3

2

3

311

3 cos γL

EAS =


Thus

42

4

43

2

3

32

2

2

21

2

1

1

114

113

112

111

11

coscoscoscos γγγγL

EAL

EAL

EAL

EASSSSS

+++=

+++=

( ) ( )4

43

2

3

32

2

2

2

1

1

4321

0coscos1 γγ

EAEAEAL

EAL

EAL

EAL

EA+++=

32

3

32

2

2

2

1

1 coscos γγL

EAL

EAL

EA++=

The final expression results from the fact that truss element 1 is horizontal and truss element 4 is vertical.


Similarly the stiffness S21 is composed of contributions from various elements of the truss. Consider the contribution to S21 from member 3, i.e.,

333

321

3 sincos γγL

EAS =

Thus

214

213

212

211

21 SSSSS +++=

Thus

EAEAEAEA

( )( ) ( )( )433

322

21

444

433

3

322

2

211

1

1

10sincossincos01

sincossincossincossincos

γγγγ

γγγγγγγγ

EAEAEAEAL

EAL

EAL

EAL

EA

+++=

+++=

( )( ) ( )( )

333

322

2

2

433

322

21

sincossincos γγγγ

γγγγ

LEA

LEA

LLLL

+=


By an analogous procedure S12 and S22 are

333

322

2

212 sincossincos γγγγ

LEA

LEAS +=

EAEAEA

4

43

2

3

32

2

2

222 sinsin

LEA

LEA

LEAS ++= γγ

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

Th {A } d h i {A } bli h d liThe vector {AD} and the matrix {ADL} were established earlier.


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

ii

iMDi L

EAA γcos1 = ii

iMDi L

EAA γsin2 =i i

11

11

1

1 sincos γγL

EAL

EAthus

{ }

=3

33

3

22

22

2

2

11

sincos

sincos

γγ

γγ

EAEAL

EAL

EA

AMD

4

4

44

4

4

33

33

sincos

sincos

γγ

γγ

LEA

LEA

LL

And we can now solve

{ } { } { }{ }DAAA MDMLM +=


ExampleThe 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 theseconsist of one translation (D1) and two rotations (D2 and D3). Determine these unknown displacements.


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


From the last figure it is easy to see that

PLP8

02 321

PLAAPA DLDLDL −=′=′=′

or

000 321 =′′=′′=′′ DLDLDL AAA

80

2 321PLAAPA DLDLDL −===

d i t i f t

{ }

=

PADL 04

8

and in a matrix format

{ }

− L

DL 8


The vector {AD} represents actions in the unrestrained structure associated with the unknown displacement D1, D2 and D3. Since there are no loads

0

p 1, 2 3associated with these displacements {AD} is a null vector and in a matrix format

{ }

=

00DA

{ } [ ] { } { }{ }1

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

{ } [ ] { } { }{ }DLD AASD −= −1

for the unknown displacements. To do that we need to formulate the stiffness matrix and find its inversematrix and find its inverse.


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

From the figures above

231213116012LEISS

LEIS −=′=′=′

23 LL

061231221311 =′′=′′=′′ S

LEIS

LEIS

2312213116624LEIS

LEIS

LEIS −===


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


00 322212 =′=′=′ SL

GJSS


0463222212 =′′=′′=′′ S

LEIS

LEIS

0463222212 =+== S

LGJ

LEIS

LEIS


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

From the figures aboveFrom the figures above

LEISS

LEIS 406

3323213 =′=′−=′

JGJSSS =′′=′′=′′ 332313 00

LGJ

LEISS

LEIS +==−=

4063323213


Define

EIGJ

=η

[ ] ( )

+−

= 2 0466624

LLLL

EIS η

then

[ ] ( )( )

+−

+=2

3

406046

LLLL

LS

ηη

and inverting this stiffness matrix leads to

[ ]( )

( )

+−

=− 2

21

2

1 04666

241 LL

LLCL

CEICS η

( )

+− 221 406

24LL

CEICη

η4 +=Cwhere

ηηη

2514

3

2

1

+=+=+=

CCC


{ } [ ] { } { }{ }DLD AASD −= −1

Solving

{ } [ ] { } { }{ }DLD

( ) ( )

++− 254

2ηηL

leads to

{ } ( ) ( )

( ) ( )( )

−+

++=

18256

4196

2

ηηη

ηηEIPLD

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Lecture 13 Trusses and Grids.pptLecture 13: TRUSSES & GRIDS – STIFFNESS METHOD Example Consider the plane truss with four bars meeting at a common joint E. This truss only has two