Structural Design II - ETH Z frames_ER...Notation - Force Diagram Cremona- Construction without...
Transcript of Structural Design II - ETH Z frames_ER...Notation - Force Diagram Cremona- Construction without...
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Structural Design II
http://www.block.arch.ethz.ch/eq/
Philippe Block ∙ Joseph Schwartz
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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
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>>
Bending frames
Rotational equilibrium
Bending moments in systems
Introduction
Shaping Frames
3
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Introduction 4
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Introduction 5
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Introduction 6
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Introduction
Design intent
Where the forces want to go
7
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RR
R = w • l
B
A
w
h
h
l
BA
Introduction 8
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B
A
R = w • l
RR
A Bl
w
Introduction 9
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l
R
= w • l
i
Ri i
C
A
A
w
C
i
i
Introduction
Free body diagram
10
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>>
Bending frames
Rotational equilibrium
Bending moments in systems
Introduction
Shaping Frames
11
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Rotational equilibrium
In equilibrium Not in equilibrium
12
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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
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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
=
=
=
=
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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
=
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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
=
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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
=
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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
=
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eQ Moment from force pair
http://www.block.arch.ethz.ch/eq/drawing/view/49
Rotational equilibrium 19
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= F • d1 1
1
M1
F
F 1
d1
Rotational equilibrium
Not in equilibrium
20
∑ M = - M1 ≠ 0
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= F • dM21 1MF
F12
2= F • d1
1F
F
22
d21d
Rotational equilibrium
In equilibrium
21
∑ M = - M1 + M2 = 0
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>>
Bending frames
Rotational equilibrium
Bending moments in systems
Introduction
Shaping Frames
22
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l
R
= w • l
i
Ri i
C
A
A
w
C
i
i
Bending moments in systems
Free body diagram
23
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V
R
CH
C
Ri
l
i
C
A
CH
CV
A
w
i
i
Bending moments in systems 24
∑ Fx = 0
∑ Fy = 0
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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
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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
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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
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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
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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
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Bending moments in systems 30
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>>
Bending frames
Rotational equilibrium
Bending moments in systems
Introduction
Shaping Frames
31
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A = F
F
d
∑ Fx = 0
∑ M ≠ 0
∑ Fy = 0
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F
M = F • d
A = F
d
Shaping frames 33
∑ Fx = 0
∑ M ≠ 0
∑ Fy = 0
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A
FF
B
F
A
F
AA B
Shaping frames 34
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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
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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
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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
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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
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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
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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
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Shaping frames
Michael Jackson: Smooth criminal, 1987
41
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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
?
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Shaping frames
Michael Jackson: Smooth criminal, 1987
43
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F
B M = B • x
A
AM
B
dx
Shaping frames 44
∑ M = - F • d + M
= - F • d + B • x
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Shaping frames 45
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Shaping frames 46
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Shaping frames 47
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Shaping frames
Taq-i Kisra Palast (Ayvan-e Kasra), Asbanbar, near Baghdad, 550 n.Chr.
48
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Shaping frames 49
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Shaping frames 50
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Shaping frames
Gustav Eiffel, Theophile Seyring: Maria Pia bridge, Porto, 1877
51
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Shaping frames
Sir Horace Jones, Sir John Wolfe Barry: Tower Bridge, London, 1894
52
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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
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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
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Shaping frames
Marcel Breuer, Bernhard Zehrfuss, Pier Luigi Nervi: Head quarters, UNESCO, Paris, 1957
55
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Shaping frames
Marcel Breuer, Bernhard Zehrfuss, Pier Luigi Nervi: Headquarters, UNESCO, Paris, 1957
56
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M
R
V
R
N
C
i
A
i
C
i
i
w
V
i
l
A
i
l
i
Ni
i
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Shaping frames
Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993
58
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Shaping frames
Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993
59
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Shaping frames
Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993
60
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Shaping frames
Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993
61
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Shaping frames
Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993
62
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Shaping frames
Sir Nicholas Grimshaw (Arch.) , Anthony Hunt (Eng.): Waterloo International Terminal, London UK, 1993
63
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GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006
64
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GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006
65
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GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006
66
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GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006
67
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GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006
68
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GMP Architects, Schlaich Bergermann und Partner (Eng.): Central Station, Berlin, 2006
69
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Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002
70
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Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002
71
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Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002
72
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Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002
73
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Zaha Hadid Architects: Bergisel ski jump, Innsbruck, Austria, 2002
74
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eQ Free-form thrust lines
http://www.block.arch.ethz.ch/eq/drawing/view/45
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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