Structural Design II - ETH Z frames_ER...Notation - Force Diagram Cremona- Construction without...

Page 1

Structural Design II

http://www.block.arch.ethz.ch/eq/

Philippe Block ∙ Joseph Schwartz

Page 2

Structural design I+II

1. Introduction

2. Equilibrium and graphic static

3.+4. Cables

13. Cable-net and membrane structures

5.+7. Arches

14.+15. Vaults, domes and shells

16. Spatial arch-cable-structures

6.+8. Arch-cable-structures

12. Materials and dimensioning

9. Trusses

17. Spacial trusses

10.+11. Beams and frames

16. Shear walls and plates

20. Columns

19. Bending

Structural design I

Structural design II

Course overview

Page 3

>>

Bending frames

Rotational equilibrium

Bending moments in systems

Introduction

Shaping Frames

3

Page 4

Introduction 4

Page 5

Introduction 5

Page 6

Introduction 6

Page 7

Introduction

Design intent

Where the forces want to go

7

Page 8

RR

R = w • l

B

A

w

h

h

l

BA

Introduction 8

Page 9

B

A

R = w • l

RR

A Bl

w

Introduction 9

Page 10

l

R

= w • l

i

Ri i

C

A

A

w

C

i

i

Introduction

Free body diagram

10

Page 11

>>

Bending frames

Rotational equilibrium

Bending moments in systems

Introduction

Shaping Frames

11

Page 12

Rotational equilibrium

In equilibrium Not in equilibrium

12

Page 13

M

F

F F

y

M

x

y

x

F

Rotational equilibrium 13

∑ Fx = 0 ∑ Fx = 0

∑ M ≠ 0∑ M = 0

∑ Fy = 0 ∑ Fy = 0

Page 14

M = F • d

F

FF

F

F

F

F

F

F

F e

d d d d

d

Rotational equilibrium

Moment

Force (couple)

moment arm

14

M = F • d

M

F

d

=

=

=

=

Page 15

M = F • d

F

FF

F

F

F

F

F

F

F e

d d d d

d M = F • d

F

FF

F

F

F

F

F

F

F e

d d d d

d

Rotational equilibrium 15

M = F • 0 + F • d

M = F • d

=

Page 16

M = F • d

F

FF

F

F

F

F

F

F

F e

d d d d

d M = F • d

F

FF

F

F

F

F

F

F

F e

d d d d

d

Rotational equilibrium 16

M = F • 0 + F * d M = F • d + F • 0

M = F • d M = F • d

=

Page 17

M = F • d

F

FF

F

F

F

F

F

F

F e

d d d d

d M = F • d

F

FF

F

F

F

F

F

F

F e

d d d d

d

Rotational equilibrium 17

M = F • 0 + F • d M = F • d + F • 0 M = F • d/2 + F • d/2

M = F • d M = F • d M = F • d

=

Page 18

M = F • d

F

FF

F

F

F

F

F

F

F e

d d d d

d

Rotational equilibrium 18

M = F • 0 + F * d M = F • d + F • 0 M = F • d/2 + F • d/2 M = F • (d+e) - F • e

M = F • d M = F • d M = F • d M = F • d

=

Page 19

eQ Moment from force pair

http://www.block.arch.ethz.ch/eq/drawing/view/49

Rotational equilibrium 19

Page 20

= F • d1 1

1

M1

F

F 1

d1

Rotational equilibrium

Not in equilibrium

20

∑ M = - M1 ≠ 0

Page 21

= F • dM21 1MF

F12

2= F • d1

1F

F

22

d21d

Rotational equilibrium

In equilibrium

21

∑ M = - M1 + M2 = 0

Page 22

>>

Bending frames

Rotational equilibrium

Bending moments in systems

Introduction

Shaping Frames

22

Page 23

l

R

= w • l

i

Ri i

C

A

A

w

C

i

i

Bending moments in systems

Free body diagram

23

Page 24

V

R

CH

C

Ri

l

i

C

A

CH

CV

A

w

i

i

Bending moments in systems 24

∑ Fx = 0

∑ Fy = 0

Page 25

C'

C

A

iR

CV

R

l

i

HC

C'

C

C

V

H

H

w

A

V

i

i

Bending moments in systems 25

∑ Fx = 0

∑ Fy = 0

∑ M ≠ 0

Page 26

C'

C

A

iR

CV

R

l

i

HC

C'

C

C

V

H

H

w

A

V

i

i

Bending moments in systems 26

∑ Fx = 0

∑ Fy = 0

∑ M ≠ 0

Page 27

M

M

R11

R1

1

R

C

R = w • l

R

R

w

1

A

R

R

H

C

R

x

C

F

1

8

1

y

F

1

1 M

9

C

C

H

H

CV

i

1

w

i

H

i

v

w

= F • d

M

= F • d

H H

1

F

F

H

F

C

i

H

R = w • l

A = F

R

H

H

A

H

A

R

R

1

F

A B

l

R

FB

F

R

C'

R

i

i

C

l

i

N

C'

V

V

H

i

C

R

V

N

R

M

i

2

i

MM = B • x

C'

i

AA

Q

Q

R

H

A

V

w

C

C

R

R

C

H

A

l

2

v

= F • d1

M

w

A

F

V

F

A B

w

A

R

HC

2

A

9

M

i

l

3

F

A = F

i

l

R

Step 3: separate the line of the dimension from drawing line a distance of 3.6 mm.

D

w

Free-Body diagram

In case you draw a free body diagram, make the lines of the bars manually thicker--> 0.35 mm

C

A

R

A

w

F

F

l

C

w

A

A B

w

C

1

1

R

lBA

Q

8

E

5

6

C'

w

l

A

R

S

C

R

F

V

i

A

R

R

Q

F

w

l

F

SC IC

= w • l

IISC

F

2

N 4

A

R

A

N

C

H

8

x

R

N

9

F

i

N7

C

A

P

V

N

P

3

A

w

H

9

w

A

C

B

A

P

F

H

H

Ri

C

C

l

A

F

C

B

i

A

C

A

f

B

F

A

F

A

F

C

A

H

F

C

V

V

F

H

l

l

i

CV

mmm

22

F

i

A

1

F

1

1 1

F

R

w

D2

1D

C 8

9C

C 7

C 6

5C

C 4

3C

C 2

1C

B9

8

cable

F

A

2

C

w

5

6

B4

B2

B7

B

B

B

B

A

B

A 6

A 5

4

F

A 3

A 2

1A

3D

4

Step 4: Pick the text box from the library which fits to the text you wish to write and modify the text

E

2E

3E

E 4

1

1

2F

F3

4

F

G1

G 2

3G

G 4

1H

H2

3H

4H

A C D

l

G

1

1R

1

F

1

8D

9D

5E

6E

7E

1

E

R

9

R

F

N

F

II

w

IV

C

C

AH

V

A

VIII

C'

H

IX

d

M = F • d

B

H

A

H6

7

D

VIIVI

A h

Bv

hB

IIII

hCv

Dh

Q

Q

C

Q

Q

Q

A

3

4

5

6

7

8

9

5N N5

M = F • d

1

2R

R3

4R

III

1

i

N

N

4

3 3

N

C'

N

N2

1NR N

C

5

6

7R

H

R

9

H

6

1

N

N

N

5

9N

N8

N

parallel lines

6

8

7

P

P

P

P

P

P

P

P

G

1

F

9

4

5

6

7

8

9

G 8

7

6

5

4

3

2

1

PP

P

P

P

P

P

F

P

P

Step 6: Copy the perpendicular line and place it through the corner of the text margin which is more near to the dimension line. Then, move the text box to the dimension line keeping the center of the box on the first perperdicular. Then group the text and the dimension line.

8

Ri

B

F

y

C'

F

R

G

w

2

F

Q

w

Text for dimensions 2 mm Arial Narrow

2.3 mm text Arial Narrow

generic security factormeaning unitssymbol 3.4 mm text Arial Narrow / subscript 2.3 mm

γγ = 1,35

= 1,5γsecurity factor materialsecurity factor external loadsecurity factor self-weight

M

EthblA

CBR

I

A'A

hinge

chosen point by the designerresulting point (intersection etc.)

44m /mmmoment of inertia

direction of rotation in cremona

section mark

symmetry

sliding support

hinge support

density kg/m3ρ

2N/mmgeneric strength of material

Elastic/Young’s modulus 2N/mmmthickness

depth mmwidth

length

7

2

G

2marea

subsystems (force elements where necessary)III III IV nodes in diagrams (where necessary)

support reaction force kNkNresultant force

area dead load kN/m2

kN/mlinear dead loadgarea life load kN/m2

q linear life loaddead load kN

kN/mG

kNlife loadQprestress force kNP

kNinternal compression force

N kNinternal tension forceF generic force kN

εσ 2kN/cmMPa2N/mm

mm/mmgeneric straingeneric stress

MγExternal force, 1 mm distance to structure / forces

Support forces, use margin symbol support

support force is 1.5 cm long, but in case it its represented in horizontal and vertical components it is relatively related to the magnitude

Uniformly distributed load: 0.5 cm and arrowhead 2mm

Resultant Force: 2 cm long and arrowhead 4.5 mm

External point force and support force: 1.5 cm long and arrowhead 3.5 mm

node numbers (Roman) (I,II, etc) from the library.if possible, place the node number to the right of the nodealso if possible below the nodeotherwise use your eye.

Position of arrowheads

element / subsystem numbers, from library. always in the center of the lineNotation - Form Diagram

Arrows & Arrowheads - Lengths

Notation - Force DiagramCremona- Construction without notationonly external forces..full arrow head

Cremona- Construction without notationinner forces always precise without arrowhead, external with an offset of 1.0 mm and with half arrow heads when internal eq is also shown

Cremona- Construction with notationexternal eq..corner point of name goes to the center of the line

Dimensioning

Cremona- Construction with notationinternal eq..offset from inner to external 1.5 mmNode number goes to barycenter of the triangle

Case 1: vertical forces --> name always on the right side of the force, in the centerpick text from the library!

Case 2: diagonal forces --> name always in the interior of the structure. Corner point in the center of the arrow. pick text from the library!

R

B7

H

6

H

8

F

w

A

I

G

A B

F

BA

F

arch

BA

F

I

I

F

A B

R

A

F

1

BA

F

I

F

A B

A

BF

5

A

A

B

I

1/32/3

I

line of the drawing

dimension from the library

9

Step 2: manually adapt it to the line of the drawing by rotating and streching one of its sides.

Step 1: Pick the dimension from the library

i

F

F

i

8

1

M

C

Step 5: Ungroup and erase the old text.Build a perpendicular line through the mid point of the dimension line and place the center of the text on it.

Point types - hinges / selected points / resulting pointsA hinge is a hinge --cicrle with no hatchA choosen geometric point is a thin black circle with grey hatchA resulting point is black point

---> take them from the library

7

A

C

F

H

Dimensioning

Please add the units (m) (mm) like in the following way.

25 m or 25.12 m250 mm or 25 cm25.1 cm or 251 mm

C

V

C

F

F

V

example:

III

II

I

R

R

B

A

3F

i

2

F1

F3

2F

1FIII

II

I

B

A

intersection point --> resultpoint selected on the closing string

Force 0.28 mm Bar 0.35 mm

Force- change line properties 0.28 mm

Bars - keep line properties

Drawing Conventions - Latest Update 17.09.2015

CS ISC closing string

nm

i intersection point of closing string and line of action of resultant

geometric planes

o' o'' o'''

r' r'' r''' rise point (form diagram)

trial pole (force diagram)o

E F

B

X

h

H

R

D

h

5

6

w

D

R

7

w

D

1R

R9

8

7

6

5

4

3

n

n

n

n

n

n

n

1

2n

n

i

i

80.2 kN

54.9 kN

54.9 kN

13 m

13 m

m88 m 13

kN8888 kN

kN88

i

0

1.50

i

yi

i

1.00

y

i

i

15.00

1.00

15.00

d

15.00

15.00

15.00

20.00

15.00

15.00

5.00

d

3

2

d

1

6f

m

y

cmmm

y

kNN

3

88.88 m

1

88 m

y

8.8 m

2

m8

d

m8.88

i

m88.8

d

m888

8 cm8.8 cm

8.88 cm88 cm

88.8 cm

888 cm88.88 mm

88 mm

8.8 mm

d

mm8

mm8.88

d

f

mm

f 1

88.8

f 2

kNf

f88.88

d

f

f

kN

f

f

88

3

4

kN

5

6

8.8

7

8

kN

f 9

8

8.88 kN

kN88.888.88 N

e

88 N

8.8 N

x

N8

N8.88

d

N88.8cm88.88

mm888

d

N888 888 kN888

kN88

kN88

i

f 6

i

88 kN kN88

88.8888.8

88

i

8.888.8

8

i

l

i i

i

i

i

d

i

i

2dd1

1d

l

d

d

2

5

2

2 3

322

4 9

1

5

2

6

1

7

2

3

4

2

81

1

1

1

1

1

Bending moments in systems 27

Page 28

MC

H

V

H

i

V

R

i

A

l

C

w

R

H

V

H

A

C

C

C

C'

C'

iy

i

i

Bending moments in systems 28

M = C’H ∙ yi + C’V ∙ 0

= H ∙ yi

Page 29

i

A

l

Ri

H

R

C

H

l

C

CV

w

B

A

i

2y

1

3

y1

2

3

y

0

Bending moments in systems 29

M3 = H ∙ y3

M2 = H ∙ y2

M1 = H ∙ y1

M0 = H ∙ 0

Page 30

Bending moments in systems 30

Page 31

>>

Bending frames

Rotational equilibrium

Bending moments in systems

Introduction

Shaping Frames

31

Page 32

A = F

F

d

∑ Fx = 0

∑ M ≠ 0

∑ Fy = 0

Shaping frames 32

Page 33

F

M = F • d

A = F

d

Shaping frames 33

∑ Fx = 0

∑ M ≠ 0

∑ Fy = 0

Page 34

A

FF

B

F

A

F

AA B

Shaping frames 34

Page 35

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

Shaping frames 35

F • d = F • 0 = 0

Page 36

A

FF

B

F

A

F

AA B

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

Shaping frames 36

F • d = F • 0 = 0

Page 37

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

Shaping frames 37

F • d = F • 0 = 0 - F • d + B • l = 0

Page 38

A

FF

B

F

A

F

AA B

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

Shaping frames 38

F • d = F • 0 = 0 - F • d + B • l = 0

Page 39

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

Shaping frames 39

F • d = F • 0 = 0 - F • d + B • l = 0 - F • d + C • l = 0

Page 40

A

FF

B

F

A

F

AA B

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

Shaping frames 40

F • d = F • 0 = 0 - F • d + B • l = 0 - F • d + C • l = 0

Page 41

Shaping frames

Michael Jackson: Smooth criminal, 1987

41

Page 42

A

FF

B

F

A

F

AA B

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

Shaping frames 42

F • d = F • 0 = 0 - F • d + B • l = 0 - F • d + C • l = 0

?

Page 43

Shaping frames

Michael Jackson: Smooth criminal, 1987

43

Page 44

F

B M = B • x

A

AM

B

dx

Shaping frames 44

∑ M = - F • d + M

= - F • d + B • x

Page 45

Shaping frames 45

Page 46

Shaping frames 46

Page 47

Shaping frames 47

Page 48

Shaping frames

Taq-i Kisra Palast (Ayvan-e Kasra), Asbanbar, near Baghdad, 550 n.Chr.

48

Page 49

Shaping frames 49

Page 50

Shaping frames 50

Page 51

Shaping frames

Gustav Eiffel, Theophile Seyring: Maria Pia bridge, Porto, 1877

51

Page 52

Shaping frames

Sir Horace Jones, Sir John Wolfe Barry: Tower Bridge, London, 1894

52

Page 53

M

M

R11

R1

R

i

R

R = w • l

R

R

R

R

w

F

M

y

C

2

1

F

F

M = F • d

H

F

R

1

M 2= F • d

B

H

C

1

C

i

A

A

Ri

R

C

2

1

1

H

8

F

R

w

F

2

F

5

C

A

F

R = w • l

4

2

N

A

w

A

1

1

1

F

A

Q

C

H

A

C'

M = B • x

i

B

i

X

i

Ai

N

C'

C

i

C

H

H

i

l

R

A

i

i

l

C

i

N

x

w

R

A

F

i

C'

A

w

Q

V

H

y

R

A

A

V

V

i

l

R

H

V

l

= F • d

A

M

1

F

H

w

F

C

V

A B

w

A

R

1F

H

C

F

i

l

w

A

R

A = F

l

3

H

H

Point types - hinges / selected points / resulting pointsA hinge is a hinge --cicrle with no hatchA choosen geometric point is a thin black circle with grey hatchA resulting point is black point

---> take them from the library

3

C

R

V

A

w

F

F

l

C

w

A

A

A

1

1

1

R

w

lBA

Q

E

9

F

R

C'

w

C

F

A

A

V

F

R

5

l

i

V

F

C

v

II

A

A

SC ICS IISC III

F

C

N

C

A

N

C

V

C'

C'

9

6

N

V

F

6

P

R

V

M

1

3

5

H

9

i

w

M

A

P

B

M = F • d

R

C

C'

ii

A

w

l

F

H

F

l

B

R

A

C

A

C

B

w

R

F

A

C

C

F

H

C

C

F

V

C

H

F

l

F

length

A

2area

i

2M

VII

F

= F • d11

kN/m

i

w

H

D2

1D

C 8

9C

generic force

7

C 6

5C

C 4

3C

C 2

1C

B9

8

C

7A

8

A

A 9

A

6

B4

B2

B7

B

B

B3

1B

A 6

A

5

4

R

A 3

A 2

dimension from the library

A = F

Step 1: Pick the dimension from the library

D

Step 4: Pick the text box from the library which fits to the text you wish to write and modify the text

1

i

Free-Body diagram

In case you draw a free body diagram, make the lines of the bars manually thicker--> 0.35 mm

H

E

25 m or 25.12 m250 mm or 25 cm25.1 cm or 251 mm

1

1

2F

F3

4F

G1

G 2

3G

G 4

1H

H2

3H

4H

A C D

H

R

B

R1

R

1

D

8D

9D

5E

6E

7E

8

1

E

1

5

A

6

I

C

A

B

V

C

w

C

VI

R

VIII IX

B

9

H

H

1

d

H

H

6

7

Dv

A h

w

v

h

IV

v

III

hC

C

Dh

Q

Q

Q

Q

Q

Q

w

2

3

4

5

6

7

8

9

5N N5

R1

2R

R3

4

R

N1

2N

N

4N3 3

H4

N

N2

1NR

H

N

A

R

6

7

parallel lines

8R

5

R

1

N

N

N

F

9 9N

N8

N

N

G

7

8

7

P

P

F

P

P

P

P

P

PP

9

2

x

4

G

6

7

88

9

8

7

6

5

4

3

2

1

M

P

P

P

P

P

P

P

P

P

Step 6: Copy the perpendicular line and place it through the corner of the text margin which is more near to the dimension line. Then, move the text box to the dimension line keeping the center of the box on the first perperdicular. Then group the text and the dimension line.

= w • l

F

F

B

R

G

R

F

C

w

C

1

Q

Text for dimensions 2 mm Arial Narrow

2.3 mm text Arial Narrow

generic security factormeaning unitssymbol 3.4 mm text Arial Narrow / subscript 2.3 mm

γγ = 1,35

= 1,5γsecurity factor materialsecurity factor external loadsecurity factor self-weight

fEthblA

CB

I

A'A

hinge

chosen point by the designerresulting point (intersection etc.)

44m /mmmoment of inertia

direction of rotation in cremona

section mark

symmetry

sliding support

hinge support

density kg/m3ρ

2N/mmgeneric strength of material

Elastic/Young’s modulus 2N/mmmthickness

depth mmwidth

7

m

2mm

i

m

G

subsystems (force elements where necessary)III III IV nodes in diagrams (where necessary)

support reaction force kNkNresultant force

area dead load kN/m2

6

linear dead loadgarea life load kN/m2

q linear life loaddead load kN

kN/mG

kNlife loadQprestress force kNP

kNinternal compression force

N kNinternal tension forceF

i

kN

εσ 2kN/cmMPa2N/mm

mm/mmgeneric straingeneric stress

MγExternal force, 1 mm distance to structure / forces

Support forces, use margin symbol support

support force is 1.5 cm long, but in case it its represented in horizontal and vertical components it is relatively related to the magnitude

Uniformly distributed load: 0.5 cm and arrowhead 2mm

Resultant Force: 2 cm long and arrowhead 4.5 mm

External point force and support force: 1.5 cm long and arrowhead 3.5 mm

node numbers (Roman) (I,II, etc) from the library.if possible, place the node number to the right of the nodealso if possible below the nodeotherwise use your eye.

Position of arrowheads

element / subsystem numbers, from library. always in the center of the lineNotation - Form Diagram

Arrows & Arrowheads - Lengths

Notation - Force DiagramCremona- Construction without notationonly external forces..full arrow head

Cremona- Construction without notationinner forces always precise without arrowhead, external with an offset of 1.0 mm and with half arrow heads when internal eq is also shown

Cremona- Construction with notationexternal eq..corner point of name goes to the center of the line

Dimensioning

Cremona- Construction with notationinternal eq..offset from inner to external 1.5 mmNode number goes to barycenter of the triangle

Case 1: vertical forces --> name always on the right side of the force, in the centerpick text from the library!

Case 2: diagonal forces --> name always in the interior of the structure. Corner point in the center of the arrow. pick text from the library!

cable

B

H

G

C F

C

H

A

H

I

5

R

B

F

BA

F

arch

BA

F

I

I

F

A B

BA

F

9

BA

F

I

F

F

B

A

BF

F

A

A

B

I

1/32/3

I

line of the drawing

R

1A

Step 2: manually adapt it to the line of the drawing by rotating and streching one of its sides.

F

Step 3: separate the line of the dimension from drawing line a distance of 3.6 mm.

A

R

8

F

B

D

M

H

E

Step 5: Ungroup and erase the old text.Build a perpendicular line through the mid point of the dimension line and place the center of the text on it.

7F

E

C

C

H

Dimensioning

Please add the units (m) (mm) like in the following way.

w

E 4

F

V

F

example:

III

II

I

R

R

B

A

3F

F2

F1

F3

2F

1FIII

II

I

B

A

intersection point --> resultpoint selected on the closing string

Force 0.28 mm Bar 0.35 mm

Force- change line properties 0.28 mm

Bars - keep line properties

Drawing Conventions - Latest Update 17.09.2015

CS ISC closing string

nm

i intersection point of closing string and line of action of resultant

geometric planes

o' o'' o'''

r' r'' r''' rise point (form diagram)

trial pole (force diagram)o

E

B

F

l

G

R

H

h

D5

h

6

R

D

w

7

R

1R

R9

8

7

6

5

4

3

n

n

n

n

n

n

n

1

2n

n

i

i

80.2 kN

54.9 kN

54.9 kN

13 m

13 m

m88 m 13

kN88

0

88 kN

i

kN88

i

1.50

yi

i

1.00

y

i

i

d

15.00

1.00

15.00

15.00

d

15.00

15.00

20.00

15.00

15.00

5.00

3

2

d

1

6f

y

mcm

y

mmkN

3

N

88.88

1

m

88

y

m

8.8

2

d

mm8

m

i

d

8.88

m88.8

m888

8 cm8.8 cm

8.88 cm88 cm

88.8 cm

888 cm88.88 mm

88

d

mm

8.8 mmmm

d

8

mm8.88

mm

f

f

88.8

1

f

kN

d

2

f88.88

f

f

kN

f

f

88

f

3

kN

4

5

8.8

6

7

kN

8

f

8

9

8.88 kN

kN

e

88.888.88 N

88 N

8.8

x

NN8

N

d

8.88

N88.8cm88.88

mm

d

888N888 888 kN888

kN88

kN88

f

i

6

i

88

i

kN kN88

88.8888.8

888.888.8

8

i

l

i i

i

i

d

ii

i

2d

l

d1

1d

d

d

2

5

2

2 3

322

4 9

1

5

2

6

1

7

2

3

4

2

81

1

1

1

1

1

Shaping frames 53

Page 54

M

M

R11

R1

R

i

R

R = w • l

R

R

R

R

w

F

M

y

C

2

1

F

F

M = F • d

H

F

R

1

M 2= F • d

B

H

C

1

C

i

A

A

Ri

R

C

2

1

1

H

8

F

R

w

F

2

F

5

C

A

F

R = w • l

4

2

N

A

w

A

1

1

1

F

A

Q

C

H

A

C'

M = B • x

i

B

i

X

i

Ai

N

C'

C

i

C

H

H

i

l

R

A

i

i

l

C

i

N

x

w

R

A

F

i

C'

A

w

Q

V

H

y

R

A

A

V

V

i

l

R

H

V

l

= F • d

A

M

1

F

H

w

F

C

V

A B

w

A

R

1F

H

C

F

i

l

w

A

R

A = F

l

3

H

H

Point types - hinges / selected points / resulting pointsA hinge is a hinge --cicrle with no hatchA choosen geometric point is a thin black circle with grey hatchA resulting point is black point

---> take them from the library

3

C

R

V

A

w

F

F

l

C

w

A

A

A

1

1

1

R

w

lBA

Q

E

9

F

R

C'

w

C

F

A

A

V

F

R

5

l

i

V

F

C

v

II

A

A

SC ICS IISC III

F

C

N

C

A

N

C

V

C'

C'

9

6

N

V

F

6

P

R

V

M

1

3

5

H

9

i

w

M

A

P

B

M = F • d

R

C

C'

ii

A

w

l

F

H

F

l

B

R

A

C

A

C

B

w

R

F

A

C

C

F

H

C

C

F

V

C

H

F

l

F

length

A

2area

i

2M

VII

F

= F • d11

kN/m

i

w

H

D2

1D

C 8

9C

generic force

7

C 6

5C

C 4

3C

C 2

1C

B9

8

C

7A

8

A

A 9

A

6

B4

B2

B7

B

B

B3

1B

A 6

A

5

4

R

A 3

A 2

dimension from the library

A = F

Step 1: Pick the dimension from the library

D

Step 4: Pick the text box from the library which fits to the text you wish to write and modify the text

1

i

Free-Body diagram

In case you draw a free body diagram, make the lines of the bars manually thicker--> 0.35 mm

H

E

25 m or 25.12 m250 mm or 25 cm25.1 cm or 251 mm

1

1

2F

F3

4F

G1

G 2

3G

G 4

1H

H2

3H

4H

A C D

H

R

B

R1

R

1

D

8D

9D

5E

6E

7E

8

1

E

1

5

A

6

I

C

A

B

V

C

w

C

VI

R

VIII IX

B

9

H

H

1

d

H

H

6

7

Dv

A h

w

v

h

IV

v

III

hC

C

Dh

Q

Q

Q

Q

Q

Q

w

2

3

4

5

6

7

8

9

5N N5

R1

2R

R3

4

R

N1

2N

N

4N3 3

H4

N

N2

1NR

H

N

A

R

6

7

parallel lines

8R

5

R

1

N

N

N

F

9 9N

N8

N

N

G

7

8

7

P

P

F

P

P

P

P

P

PP

9

2

x

4

G

6

7

88

9

8

7

6

5

4

3

2

1

M

P

P

P

P

P

P

P

P

P

Step 6: Copy the perpendicular line and place it through the corner of the text margin which is more near to the dimension line. Then, move the text box to the dimension line keeping the center of the box on the first perperdicular. Then group the text and the dimension line.

= w • l

F

F

B

R

G

R

F

C

w

C

1

Q

Text for dimensions 2 mm Arial Narrow

2.3 mm text Arial Narrow

generic security factormeaning unitssymbol 3.4 mm text Arial Narrow / subscript 2.3 mm

γγ = 1,35

= 1,5γsecurity factor materialsecurity factor external loadsecurity factor self-weight

fEthblA

CB

I

A'A

hinge

chosen point by the designerresulting point (intersection etc.)

44m /mmmoment of inertia

direction of rotation in cremona

section mark

symmetry

sliding support

hinge support

density kg/m3ρ

2N/mmgeneric strength of material

Elastic/Young’s modulus 2N/mmmthickness

depth mmwidth

7

m

2mm

i

m

G

subsystems (force elements where necessary)III III IV nodes in diagrams (where necessary)

support reaction force kNkNresultant force

area dead load kN/m2

6

linear dead loadgarea life load kN/m2

q linear life loaddead load kN

kN/mG

kNlife loadQprestress force kNP

kNinternal compression force

N kNinternal tension forceF

i

kN

εσ 2kN/cmMPa2N/mm

mm/mmgeneric straingeneric stress

MγExternal force, 1 mm distance to structure / forces

Support forces, use margin symbol support

support force is 1.5 cm long, but in case it its represented in horizontal and vertical components it is relatively related to the magnitude

Uniformly distributed load: 0.5 cm and arrowhead 2mm

Resultant Force: 2 cm long and arrowhead 4.5 mm

External point force and support force: 1.5 cm long and arrowhead 3.5 mm

node numbers (Roman) (I,II, etc) from the library.if possible, place the node number to the right of the nodealso if possible below the nodeotherwise use your eye.

Position of arrowheads

element / subsystem numbers, from library. always in the center of the lineNotation - Form Diagram

Arrows & Arrowheads - Lengths

Notation - Force DiagramCremona- Construction without notationonly external forces..full arrow head

Cremona- Construction without notationinner forces always precise without arrowhead, external with an offset of 1.0 mm and with half arrow heads when internal eq is also shown

Cremona- Construction with notationexternal eq..corner point of name goes to the center of the line

Dimensioning

Cremona- Construction with notationinternal eq..offset from inner to external 1.5 mmNode number goes to barycenter of the triangle

Case 1: vertical forces --> name always on the right side of the force, in the centerpick text from the library!

Case 2: diagonal forces --> name always in the interior of the structure. Corner point in the center of the arrow. pick text from the library!

cable

B

H

G

C F

C

H

A

H

I

5

R

B

F

BA

F

arch

BA

F

I

I

F

A B

BA

F

9

BA

F

I

F

F

B

A

BF

F

A

A

B

I

1/32/3

I

line of the drawing

R

1A

Step 2: manually adapt it to the line of the drawing by rotating and streching one of its sides.

F

Step 3: separate the line of the dimension from drawing line a distance of 3.6 mm.

A

R

8

F

B

D

M

H

E

Step 5: Ungroup and erase the old text.Build a perpendicular line through the mid point of the dimension line and place the center of the text on it.

7F

E

C

C

H

Dimensioning

Please add the units (m) (mm) like in the following way.

w

E 4

F

V

F

example:

III

II

I

R

R

B

A

3F

F2

F1

F3

2F

1FIII

II

I

B

A

intersection point --> resultpoint selected on the closing string

Force 0.28 mm Bar 0.35 mm

Force- change line properties 0.28 mm

Bars - keep line properties

Drawing Conventions - Latest Update 17.09.2015

CS ISC closing string

nm

i intersection point of closing string and line of action of resultant

geometric planes

o' o'' o'''

r' r'' r''' rise point (form diagram)

trial pole (force diagram)o

E

B

F

l

G

R

H

h

D5

h

6

R

D

w

7

R

1R

R9

8

7

6

5

4

3

n

n

n

n

n

n

n

1

2n

n

i

i

80.2 kN

54.9 kN

54.9 kN

13 m

13 m

m88 m 13

kN88

0

88 kN

i

kN88

i

1.50

yi

i

1.00

y

i

i

d

15.00

1.00

15.00

15.00

d

15.00

15.00

20.00

15.00

15.00

5.00

3

2

d

1

6f

y

mcm

y

mmkN

3

N

88.88

1

m

88

y

m

8.8

2

d

mm8

m

i

d

8.88

m88.8

m888

8 cm8.8 cm

8.88 cm88 cm

88.8 cm

888 cm88.88 mm

88

d

mm

8.8 mmmm

d

8

mm8.88

mm

f

f

88.8

1

f

kN

d

2

f88.88

f

f

kN

f

f

88

f

3

kN

4

5

8.8

6

7

kN

8

f

8

9

8.88 kN

kN

e

88.888.88 N

88 N

8.8

x

NN8

N

d

8.88

N88.8cm88.88

mm

d

888N888 888 kN888

kN88

kN88

f

i

6

i

88

i

kN kN88

88.8888.8

888.888.8

8

i

l

i i

i

i

d

ii

i

2d

l

d1

1d

d

d

2

5

2

2 3

322

4 9

1

5

2

6

1

7

2

3

4

2

81

1

1

1

1

1

Shaping frames 54

Page 55

Shaping frames

Marcel Breuer, Bernhard Zehrfuss, Pier Luigi Nervi: Head quarters, UNESCO, Paris, 1957

55

Page 56

Shaping frames

Marcel Breuer, Bernhard Zehrfuss, Pier Luigi Nervi: Headquarters, UNESCO, Paris, 1957

56

Page 57

M

R

V

R

N

C

i

A

i

C

i

i

w

V

i

l

A

i

l

i

Ni

i

Shaping frames 57

Page 58

Shaping frames

Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993

58

Page 59

Shaping frames

Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993

59

Page 60

Shaping frames

Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993

60

Page 61

Shaping frames

Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993

61

Page 62

Shaping frames

Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993

62

Page 63

Shaping frames

Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993

63

Page 64

Shaping frames

GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006

64

Page 65

Shaping frames

GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006

65

Page 66

Shaping frames

GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006

66

Page 67

Shaping frames

GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006

67

Page 68

Shaping frames

GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006

68

Page 69

Shaping frames

GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006

69

Page 70

Shaping frames

Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002

70

Page 71

Shaping frames

Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002

71

Page 72

Shaping frames

Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002

72

Page 73

Shaping frames

Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002

73

Page 74

Shaping frames

Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002

74

Page 75

eQ Free-form thrust lines

http://www.block.arch.ethz.ch/eq/drawing/view/45

Shaping frames 75

Page 76

ADD FORCE DIAGRAMS

Shaping frames 76

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

C

CB A M

A

A

FF

A

B

A

F FF

F

A

F

A

ld

l

d d

M

M

R11

R1

R

i

R

R = w • l

R

R

R

R

w

F

M

y

C

2

1

F

F

M = F • d

H

F

R

1

M 2= F • d

B

H

C

1

C

i

A

A

Ri

R

C

2

1

1

H

8

F

R

w

F

2

F

5

C

A

F

R = w • l

4

2

N

A

w

A

1

1

1

F

A

Q

C

H

A

C'

M = B • x

i

B

i

X

i

Ai

N

C'

C

i

C

H

H

i

l

R

A

i

i

l

C

i

N

x

w

R

A

F

i

C'

A

w

Q

V

H

y

R

A

A

V

V

i

l

R

H

V

l

= F • d

A

M

1

F

H

w

F

C

V

A B

w

A

R

1F

H

C

F

i

l

w

A

R

A = F

l

3

H

H

Point types - hinges / selected points / resulting pointsA hinge is a hinge --cicrle with no hatchA choosen geometric point is a thin black circle with grey hatchA resulting point is black point

---> take them from the library

3

C

R

V

A

w

F

F

l

C

w

A

A

A

1

1

1

R

w

lBA

Q

E

9

F

R

C'

w

C

F

A

A

V

F

R

5

l

i

V

F

C

v

II

A

A

SC ICS IISC III

F

C

N

C

A

N

C

V

C'

C'

9

6

N

V

F

6

P

R

V

M

1

3

5

H

9

i

w

M

A

P

B

M = F • d

R

C

C'

ii

A

w

l

F

H

F

l

B

R

A

C

A

C

B

w

R

F

A

C

C

F

H

C

C

F

V

C

H

F

l

F

length

A

2area

i

2M

VII

F

= F • d11

kN/m

i

w

H

D2

1D

C 8

9C

generic force

7

C 6

5C

C 4

3C

C 2

1C

B9

8

C

7A

8

A

A 9

A

6

B4

B2

B7

B

B

B3

1B

A 6

A

5

4

R

A 3

A 2

dimension from the library

A = F

Step 1: Pick the dimension from the library

D

Step 4: Pick the text box from the library which fits to the text you wish to write and modify the text

1

i

Free-Body diagram

In case you draw a free body diagram, make the lines of the bars manually thicker--> 0.35 mm

H

E

25 m or 25.12 m250 mm or 25 cm25.1 cm or 251 mm

1

1

2F

F3

4F

G1

G 2

3G

G 4

1H

H2

3H

4H

A C D

H

R

B

R1

R

1

D

8D

9D

5E

6E

7E

8

1

E

1

5

A

6

I

C

A

B

V

C

w

C

VI

R

VIII IX

B

9

H

H

1

d

H

H

6

7

Dv

A h

w

v

h

IV

v

III

hC

C

Dh

Q

Q

Q

Q

Q

Q

w

2

3

4

5

6

7

8

9

5N N5

R1

2R

R3

4

R

N1

2N

N

4N3 3

H4

N

N2

1NR

H

N

A

R

6

7

parallel lines

8R

5

R

1

N

N

N

F

9 9N

N8

N

N

G

7

8

7

P

P

F

P

P

P

P

P

PP

9

2

x

4

G

6

7

88

9

8

7

6

5

4

3

2

1

M

P

P

P

P

P

P

P

P

P

Step 6: Copy the perpendicular line and place it through the corner of the text margin which is more near to the dimension line. Then, move the text box to the dimension line keeping the center of the box on the first perperdicular. Then group the text and the dimension line.

= w • l

F

F

B

R

G

R

F

C

w

C

1

Q

Text for dimensions 2 mm Arial Narrow

2.3 mm text Arial Narrow

generic security factormeaning unitssymbol 3.4 mm text Arial Narrow / subscript 2.3 mm

γγ = 1,35

= 1,5γsecurity factor materialsecurity factor external loadsecurity factor self-weight

fEthblA

CB

I

A'A

hinge

chosen point by the designerresulting point (intersection etc.)

44m /mmmoment of inertia

direction of rotation in cremona

section mark

symmetry

sliding support

hinge support

density kg/m3ρ

2N/mmgeneric strength of material

Elastic/Young’s modulus 2N/mmmthickness

depth mmwidth

7

m

2mm

i

m

G

subsystems (force elements where necessary)III III IV nodes in diagrams (where necessary)

support reaction force kNkNresultant force

area dead load kN/m2

6

linear dead loadgarea life load kN/m2

q linear life loaddead load kN

kN/mG

kNlife loadQprestress force kNP

kNinternal compression force

N kNinternal tension forceF

i

kN

εσ 2kN/cmMPa2N/mm

mm/mmgeneric straingeneric stress

MγExternal force, 1 mm distance to structure / forces

Support forces, use margin symbol support

support force is 1.5 cm long, but in case it its represented in horizontal and vertical components it is relatively related to the magnitude

Uniformly distributed load: 0.5 cm and arrowhead 2mm

Resultant Force: 2 cm long and arrowhead 4.5 mm

External point force and support force: 1.5 cm long and arrowhead 3.5 mm

node numbers (Roman) (I,II, etc) from the library.if possible, place the node number to the right of the nodealso if possible below the nodeotherwise use your eye.

Position of arrowheads

element / subsystem numbers, from library. always in the center of the lineNotation - Form Diagram

Arrows & Arrowheads - Lengths

Notation - Force DiagramCremona- Construction without notationonly external forces..full arrow head

Cremona- Construction without notationinner forces always precise without arrowhead, external with an offset of 1.0 mm and with half arrow heads when internal eq is also shown

Cremona- Construction with notationexternal eq..corner point of name goes to the center of the line

Dimensioning

Cremona- Construction with notationinternal eq..offset from inner to external 1.5 mmNode number goes to barycenter of the triangle

Case 1: vertical forces --> name always on the right side of the force, in the centerpick text from the library!

Case 2: diagonal forces --> name always in the interior of the structure. Corner point in the center of the arrow. pick text from the library!

cable

B

H

G

C F

C

H

A

H

I

5

R

B

F

BA

F

arch

BA

F

I

I

F

A B

BA

F

9

BA

F

I

F

F

B

A

BF

F

A

A

B

I

1/32/3

I

line of the drawing

R

1A

Step 2: manually adapt it to the line of the drawing by rotating and streching one of its sides.

F

Step 3: separate the line of the dimension from drawing line a distance of 3.6 mm.

A

R

8

F

B

D

M

H

E

Step 5: Ungroup and erase the old text.Build a perpendicular line through the mid point of the dimension line and place the center of the text on it.

7F

E

C

C

H

Dimensioning

Please add the units (m) (mm) like in the following way.

w

E 4

F

V

F

example:

III

II

I

R

R

B

A

3F

F2

F1

F3

2F

1FIII

II

I

B

A

intersection point --> resultpoint selected on the closing string

Force 0.28 mm Bar 0.35 mm

Force- change line properties 0.28 mm

Bars - keep line properties

Drawing Conventions - Latest Update 17.09.2015

CS ISC closing string

nm

i intersection point of closing string and line of action of resultant

geometric planes

o' o'' o'''

r' r'' r''' rise point (form diagram)

trial pole (force diagram)o

E

B

F

l

G

R

H

h

D5

h

6

R

D

w

7

R

1R

R9

8

7

6

5

4

3

n

n

n

n

n

n

n

1

2n

n

i

i

80.2 kN

54.9 kN

54.9 kN

13 m

13 m

m88 m 13

kN88

0

88 kN

i

kN88

i

1.50

yi

i

1.00

y

i

i

d

15.00

1.00

15.00

15.00

d

15.00

15.00

20.00

15.00

15.00

5.00

3

2

d

1

6f

y

mcm

y

mmkN

3

N

88.88

1

m

88

y

m

8.8

2

d

mm8

m

i

d

8.88

m88.8

m888

8 cm8.8 cm

8.88 cm88 cm

88.8 cm

888 cm88.88 mm

88

d

mm

8.8 mmmm

d

8

mm8.88

mm

f

f

88.8

1

f

kN

d

2

f88.88

f

f

kN

f

f

88

f

3

kN

4

5

8.8

6

7

kN

8

f

8

9

8.88 kN

kN

e

88.888.88 N

88 N

8.8

x

NN8

N

d

8.88

N88.8cm88.88

mm

d

888N888 888 kN888

kN88

kN88

f

i

6

i

88

i

kN kN88

88.8888.8

888.888.8

8

i

l

i i

i

i

d

ii

i

2d

l

d1

1d

d

d

2

5

2

2 3

322

4 9

1

5

2

6

1

7

2

3

4

2

81

1

1

1

1

1