S. 1 / 4Arch-cable structuresIntroduction to Structural Design

Name:EX 2Draw the force diagrams for the given cases. Draw the tension forces in red, the compression forces in blue and the external forces in green.Arch-cable Structures 1Task 1

S. 2 / 4Arch-cable structuresIntroduction to Structural Design

Name:EX 2

9f8

7

6

5

4

3

f

f

f

f

f

f2f1ff

6f

f 6

m13

2

kN54.9

ff

kN54.9

ff

2

kN80.2

ff

l

5.00

15.00

ff

15.00

20.00

15.00

kN88

mcmmmkNN

15.00

ff

88.88 m

88 m

8.8 mm8

m8.88

m88.8

m888

3

l /

8 cm8.8 cm

8.88 cm88 cm

88.8 cm

888 cm

15.00

l /

l /

3

88.88 mm

88

15.00

mm

8.8 mmmm8

mm8.88

mm88.8kN88.88

kN

2

88

kN8.8kN8

8.88 kN

kN88.8

1.00

l /

88.88 N

88 N

8.8

2

NN8

N

15.00

3

N88.8cm88.88

mm888N888 888 kN888

l /

kN88

kN88 88 kN

88.8888.8

888.888.8

8

l

l

1.00 1.50

88 kN

8.88

kN88

l /

88 kN

13m88 m

m13

l /

VIII

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

Dimensioning

Text for dimensions 2 mm Arial Narrow

2.3 mm text Arial Narrow

generic security factormeaning units

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

symbol

V

= 1,5γsecurity factor material

VII

security factor external loadsecurity factor self-weight

fEthbl

VI

A

I

hinge

chosen point by the designerresulting point (intersection etc.)

44m /mmmoment of inertia

section mark

symmetry

sliding support

hinge support

density kg/m3ρ

2N/mmgeneric strength of material

Elastic/Young’s modulus 2N/mmmthickness

depth mmwidth

length m

IV

2mm2marea

subsystems (force elements where necessary)

nodes in diagrams (where necessary)

support reaction force kN

I

kNresultant force

area dead load kN/m2

kN/mlinear dead load

III

area life load kN/m2

linear life load

II

dead load kNkN/m

I

kNlife load

prestress force kNkNinternal compression force

kNinternal tension force

generic force kN

εσ

I

2kN/cmMPa2N/mmmm/mmgeneric strain

generic stress

Mγ

example:

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

2

1q

intersection point --> result

q

point selected on the closing string

2.8 mm text Arial Narrow

I

SC ICS II

S

I

C III

I

VIVIIIIII

AA'

I

I

o

I

I

II

III

γ = 1,35

γ

I

II

III

IV

I

I

parallel lines

closing string

I

nm

III

i

V

V

intersection point of closing string and line of action of resultant

I

geometric planes

o' o'' o'''

I

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

trial pole (force diagram)

IV

CS

C

III

ISC

C

direction of rotation in cremona

III

Trako Gang Drawing Conventions - Latest Update 14.1.2016

A'A

I

I

I

I

intersection point of closing string and line of action of resultant

I

trial pole (funicular construction)pole (funicular construction)

hinge support

V

sliding support

symmetry

IV

section mark

I

III

direction of rotation in cremona

III

geometric planes

M

I

γ

ε

σ

γ=1,5

γ=1,35

γ

A

bh

I

t

V

E

f

ρ

IV

support reaction forceresultant force

I

linear life loaddead load

II

Free-Body diagram

life load

L

linear dead load

max

area life load

area dead load

nodes in diagrams (where necessary)

III

L

subsystems (force elements where necessary)

min

arealengthwidth

A

depth

erf

thickness

N

II

d

Elastic/Young’s modulusgeneric strength of materialdensitymoment of inertia

kN

I

i

kN

I

A

N

kN/m

i

kN/m

IV

kN/m

9

8

kN/m

7

6

kN

5

4

kN

3

m

2

III

n

2

2

n

mm2

n

mm

n

mm

I

n

2N/mm

n

N/mm2

n

1

kg/m

2n

3

n

m /mm4 4

generic security factor

f a1

III

+

security factor self-weightsecurity factor external loadsecurity factor material

g1f +

generic stressgeneric strain

f

prestress force

+

II

internal compression forceinternal tension force

s1

c1

generic force

meaning

+

units

f

symbol

mm/mm

I

+r1

kN

f

f

kN

mx+

kNkN

N/mm2

IV

MPa kN/cm2

f -mx

f r1-

III

-

II

f c1

letters with subscription and bounding boxalphabet

s1f -

l

resulting point (intersection etc.)

I

f

chosen point by the designer

-

hinge

II

g1

III IV

III

V

-

II

a1,df

I

g1,d-f

-f s1,d

c1,d -ff

III

+c1,d

II

s1,df +

f +

I

g1,d

a1f -

I

f a1,d

III

+

f

II

mx,d+

+f t ,d -f t ,d

I

-mx,df

1/32/3

I

f t + -f t

line of the drawing

t ,d

dimension from the library

f -t ,df +ll llf t ll + t llf -

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

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.

External force, 1 mm distance to structure / forcesSupport forces, use margin symbol support

III

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

IV

Uniformly distributed load: 0.5 cm and arrowhead 2mm

Resultant Force: 2 cm long and arrowhead 4.5 mm

V

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

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

C

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

DA

I IIB

III

C

BIII

AIV

III

II

I

IV

III

I II

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

II

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

III

I

Position of arrowheads

III

II

III II

I

I

element / subsystem numbers, from library. always in the center of the line

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.

Notation - Form Diagram

Arrows & Arrowheads - Lengths

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

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

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

XIX

N

N

Dh

AF1

2F B

F3

F

A

F

1F

F B2

F

1F

F

A

2

A

F

2F

F1

A

A

F

B

F1

R

F1

2

B

R

FF

1q = 5kN/m

A B

F

R

B

A

g

B

q

F

A

R

GQP

F

q = 5kN/m1

A B

R

R

A

B

2

1q = 5kN/m

A B

R

A

R

A

F = 60kN

B

B

F

BA

B

F

A

q = 5kN/m

1q = 5kN/m

A

A

B

R

B

B

R

A

BA

q = 5kN/m

1

1

BA

B

F

F

A

R

B

B

A

A

BF

F

F = 60kN1

B

A

B

A

1F

2F

A

F

HA

VA

BA

F

BA

R

BA

q = 5kN/m1

3

2

F

B

F

A

q = 5kN/m1

1

1A

F

C

BA

BA

1q = 5kN/m

R'

R''

B

A

A

B

1q = 5kN/m

A

1

B

F F3

5

R

4

1

2

3

A = 37.5kN B = 37.5kN

F

1

F

3

BA

4

2

5

B

B

B

F

A

q = 5kN/m1

BA

q = 5kN/m1

BA

1q = 5kN/m

1F = 30kN

1q = 5kN/m

A B

F = 30kN

3F = 30kN

2F = 30kN

F = 30kN3

21 F = 30kN

A = 45kN B = 45kN

21F = 30kN F = 30kN

BA

F = 60kN

hA

vA

B

1

A

1 2F = 30kNF = 30kN

A

A

F = 30kN1

2

B

1F = 30kN F = 30kN

1

A

F

F3

4F

F = 60kNBA

1

F = 60kN

F

5F

B

6

A

F

1

7

F

8F

F9

B

B

B

B

F = 30kN3

Q

Gq

g

R

F

P

2F = 30kN1F = 30kN

BA

AB

2

B

F = 30kN1F = 30kN

hAAv

B

A

1F = 60kN

A

1

F = 30kN2F = 30kN1

A

A B C

1F = 30kN2F = 30kN

F = 60kN1

A

B

F = 60kN1

F

QQ 1

2Q 3

Q 4Q 5

F 5

F 4

F 2

1F

D

3

2

F

1D

C 8

A

9

B

C

C 7

C 6

5C

C 4

3C

C1C

B9

8B7A

8

B = 45kNA = 45kN

A

A

F = 30kN

9

1

5

2

6

B4

B2

B7

B

B

B3

1B

A 6

A 5

4A

A 3

A 2

3D

4D

1

F = 30kN

E

3

F = 30kN

2E

3E

E 4

F

R

BA

F

21F = 30kN F = 30kN

hA

vA

B

F = 60kN1

A

1 2F = 30kNF = 30kN

B

B

B

C

A

D

A

E

A

D5

6D

7D

8D

9D

5E

6E

7E

8E

E 9

D

2

F = 60kN

F = 60kN

1

1

1

F = 30kN F = 30kN

A v

A

B

A h

Bv

hBv

vA

C

h

h

A

C

F1

v

F

N

N

i

o'

o

m n

3

2 1

1

3

1

3

1

3

3

1

5

4 7

6

2

6 1

2

3 5

1

7

3

4

4

3

4

1

2

3

1

3

32

21

2 4

5

51 3

2 1

2

3

1

3

25

3

14

3

21

2

3

1

3

2 1

13 5 6

2

4 7

2

4

3 1

1

2

3

5

4

3

2

1

54 21

2 1

3

3

3

2 1

3

2

1

6

5

4

1

2

3

7

1

3

42

5

4

2

1

3

5

4

2

1

3

3

2

1

3

1

2

12

3

3

1

2

1

2

12

3

1

3

2

12

3

2

3

1

12

3

3

1

3

2 1

2

3

1

3

2

1

3

1

2

6

654321 7

7

3

4 7 2

6

2

1

1 2 3 4 5 6

71 2 3 4 5

i

6

7

3

5

4

1

5

22

3

15

5

3

6

4

6 7

2

341 2

72

3

1

9

2

876542

1

2

2

1

1

1

1

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17

17

10

10

14

14

13

13

11

11

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16

16

15

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65

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6

force diagram 1cm ≙ 7kNform diagram 1:100

force diagram 1cm ≙ 7kNform diagram 1:100

force diagram 1cm ≙ 7kNform diagram 1:100

force diagram 1cm ≙ 7kNform diagram 1:100

Draw the force diagrams for the given cases and determine the magnitude of A and B. Draw the tension forces in red, the compression forces in blue and the external forces in green.

Arch-Cable Structures 2Task 2

S. 3 / 4Arch-cable structuresIntroduction to Structural Design

Name:EX 2

Q3Q2Q1 = 100 kN = 100 kN = 75 kN

0.75 m 0.75 m 0.75 m 0.75 m

force diagram 1cm ≙ 25kN

Find the form of a possible arch-cable structure in equilibrium which is able to support the given loads using graphic statics. Draw the tension forces in red, the compression forces in blue and the external forces in green.

Task 3 Form-finding

S. 4 / 4Arch-cable structuresIntroduction to Structural Design

Name:EX 2

Cathédrale Saint-Pierre de Beauvais

Location: Beauvais, France Year: 1272

Atlantida Church, Eladio Dieste

Location: Atlantida, UruguayYear: 1952

Task 4Compare the two structures given below describing all the most relevant features in terms of architecture, structural concept, use of material and construction aspects.Write a short text also including sketches, where relevant.

Comparing Structures