Statics Trusses 2

1

A framework of members interconnected at their ends to form a rigid structure is known as a truss.

An assembly of short straight elements arranged in a stable, triangulated pattern.

Resulting structure is rigid i.e. only relatively minor overall deformations.

Individual members need to distort before significant overall truss movement.

Timber trusses have been around since Roman times. Today trusses used for bridges, roof supports, cranes,

transmission pylons, etc.

Trusses are popular, common and efficient.

Pin-jointed structures - trusses

2

Individual members may be I-sections, channel sections, angle sections, bars, etc. Connected by welding, bolting or pins.

Joints between elements assumed to be frictionless pins (often not in reality e.g. two bolts a pin).

Connections are often made through gusset plates.

Elements are kept short long elements could buckle (next year).

WeldedConnection

BoltedConnection

Flat Plate ChannelSection

Angle SectionGussetPlate


3

Roof Truss

Fink Roof Truss

Howe Roof Truss

Pratt Roof Truss

Pratt Bridge Truss

Warren Bridge Truss

Examples

Examples)


Basic element of a plane truss is a triangle. Three bars joined by pins at the ends form a rigid shape. External loads are applied at the joints of the structure.

This ensures internal forces are axial. Otherwise bending of

elements occurs. In such cases internal forces become complicated.

Compression members: Strut Tension members: Tie

4

5

Consider a pinned quadrilateral panel with load: P

P P

Mechanism**

Unstable,**Large*displacements*

P

P

Triangulate**

Stable*Rigid*Very*small*displacements*

Stability & triangulation

6

Consider different situation:

P

P

NO PIN

Further*triangula

Determinacy

7

Unstable Stable

Stablebut statically indeterminatei.e. more information required for solution

(a) (b)

(c)

Equations of staticsdo not apply

Can apply equationsof statics - statically determinate

A*useful*(but*not*fool*proof)*determinacy*test:* 2m j r= m**number*of*members*for*a*sta

9

Example:

H

V1 V2

22 4 3 5

m j r= = =

Actual =*6

Actual > required

Sta

11

**External*push*therefore*internal*compression*

**

*For*equilibrium*at*joints:**

**

*Internal*forces:*c

**External*pull*therefore*internal*tension*

**

*For*equilibrium*at*joints:**

**

*Internal*forces:* t

STRUT:)

TIE:)

Note*conven

13

2 2 7 4 10m j r= = =(a) Two pin support.

Determinacy test:

Actual members = 11

Therefore Statically Indeterminate

PinSupport

PinSupport

5 kN 10 kN 15 kN

AV BVBHAH

(a)*

Support Conditions

14

(b) One pin, one roller support (simply supported)

Therefore Statically Determinate 2 2 7 3 11m j r= = =

@

0 : 0

0 : 5 10 15 0 30

0 : 6 5 1 10 3 15 5 0

110 6 18.33 & 11.67

x A

y A B A B

A B

B A

F H

F V V V V

M V

V kN V kN

= =

= + = + =

= =

= = =

Pin Support Roller Support

5 kN 10 kN 15 kN

AV BVAH

1m

6m

(b)*

Support Conditions

15

1. Identify reactions (pin, roller, etc.) and construct free-body diagram.

2. Apply equations of equilibrium and solve for unknown reactions.

3. Use either: a) Method of Joints

to solve for all unknown internal member forces. or b) Method of sections

to solve for a selected number of internal forces

4. Redraw truss and show forces acting in their correct direction.

Numerical Analysis

16

( ) ( )

1 4 4 1

1 1

@4 1

1

4

0 : 0

0 : 10 0 10

0 : 1 10 1 0

10

10

x

y

F X X X X

F Y Y kN

M X

X kN

X kN

= + = =

= = =

= =

=

=

Example 1:

10 kN

1m

1m

1 2

34

12.211:. -1 21

10 kN

1m

1m

1X1Y

4Xx

y

10 kN

1m

1m

Method of joints

17

3. Method of joints Examine each joint in turn, taken as a free-body and apply equations of equilibrium.

Note: forces acting on a joint form a concurrent force system. Therefore

Moment equilibrium is satisfied There are only two equations of equilibrium which can be used:

Therefore there can be only two unknown forces; This effects where we start the analysis:

Joint 1 and 3 have 3 unknown forces.

Joint 2 and 4 have 2 unknown forces. Assume all internal forces are in tension.

0 0x yF F= =

18

13

13

0 : 10 cos45 014.14

xF ff kN

= =

=

34

34

14

0 : 10 010

0 : 0

x

y

F ff

F f

= + =

=

= =

Joint 2. y

23f12f

x12

23

0 : 0

0 : 0

x

y

F f

F f

= =

= =

Joint 4. y

34f14f

x10y

13f

x1010

045oJoint 3.

10 kN

12

34

10 kN

10 kN

10 kN

19

Joint 1. As a check.

0 : 10 14.14cos45 0

0 : 10 14.14sin 45 0

x

y

F

F

= + =

= =

y

x100

10

45o 014.14

Summary:

10 kN

10

10

10

x

y

10

14.14

10 kN

12

34

10 kN

10 kN

10 kN

Qualitative Analysis

20

10 kN

1m

1m

1 2

34

Remove 3-4:

compression

Remove 1-3:

tension

stretch

Remove 1-4:

No difference zero force

Also for 1-2 or 2-3

Qualitative Analysis

21

Change position of load

10

C

C

T

10

CT

T

10

CT

T

T

Change diagonal

Change position of load 10

10

10

By*inspec

Example 2: Qualitative Analysis

23

y

AFfABf

x

25REACTION

Symmetric!*

50 kN

2m

2m 2m

B

A

D

E

C

F

Reac

25

y

CDf x25BCf =

CFf

y

x

25 025

0

0 :

0 :

CD

CD

x

y CF

F

F

ff kN

f

=

=

+ =

=

=

B

A

D

E

C

F

25

5025 25

Joint C.

26

sin 45 35.36sin 45 50 0 35.36

cos45 35.36cos45 0

0 :

35.36cos45 35.36cos45 0 0

0 :

FD FD

FD FE

FE FE

y

x

F

F

f f kN

f f

f f

+ = = =

+ =

+ = =

=

B

A

D

E

C

F

25

5025 25

25

35.36

Joint F. Assume*Tension!*

y

FDf

x

50

45oFEf0

35.36

45o0

25

5025 25

25

35.36 35.36

25

25

Summary:

Example 3: Qualitative Analysis

27

*compression*

WL L

BA D

E

C

F

L

40o0.84L

*compression*

Central)Diagonal:*

Bo

Example 3: Numerical Analysis

WL L

BA D

E

C

F

L

40o0.84L

WL L

BA D

E

C

F

L

40o0.84L

AX

AY DY

y

x EquaBons)of)equilibrium:)

0: 0AxF X= =

( ) ( )@ 3 0 0.3330: D DA Y L LM W Y W = ==

0 6 70: 0 . 6A D Ay Y Y WF Y W+ = ==29

30

i.e.*compression*

( )

0.667 sin 40 0 1.04

cos40 0

cos40 1.

0 :

0 :

04 cos40 0.8

AF AF

AF AB

AB A

y

x

F

W f f W

f f

f f W W

F

F

+ = =

+ =

= =

=

=

=

y

ABfx

0.667W

40oAFf ASSUME

TENSION

REACTION

Joint A. B

D

E

C

F

40o0.667W 0.333WW

A

31

y

x

i.e.*compression*1.04 cos40 0 0.8

1.04 sin 40 0

0 :

0.60 : 7

FE FE

Fy B FB

xF

F

f W f W

W f f W

+ = =

= =

=

=

y

FEfx

40o1.04AF Wf =

ASSUMETENSION

FBf

BD

E

C

F

40o0.667W 0.333WW

A

1.04W

0.8W

Joint F.

32

BD

E

C

F

40o0.667W 0.333WW

A

1.04W

0.8W

y

BCfx40o

0.8W

ASSUMETENSION

W

BEf0.67W

Joint B.

( )

0.67 sin 40 0

0.51cos 40 0.8 0

0.51 cos 40 0.8 0

0 :

0 :

0.4

BE

BE

BC B

y

x E

BC

BC

W W ff Wf f W

f

F

f W WW

F + = =

+ =

+ =

=

=

=

33

y

x

0.4 0 0.40 :

0 : 0

CD Cx

y

D

CE

F

F

f W f W

f

=

==

= =

y

CDfx

0.4W

CEfB

D

E

C

F

40o0.667W 0.333WW

A

1.04W

0.8W

0.67W

0.8W

0.4W

0.51W

y

0

x40o

0.8W

0.51W40o

EDf

Joint E. sin 40 0.51 sin 40 00.51

0.8 cos40 0.51 cos40 00.8 0.51 cos40 0.5

0

1 cos40

:

:0

0

ED

ED

ED

y

x

f Wf WW f

F

WF W

W W

=

=

=

=

+ =

=

Joint)C.*

34

BD

E

C

F

40o0.667W 0.333WW

A

1.04W

0.8W

0.67W

0.8W

0.4W

0.51W

0.4W

0.51W

Summary:

Documents

Statics Trusses 2