S. 1 / 4Equilibrium, cables and archesIntroduction to Structural Design

Name:EX 1 LösungsvorschlagProposal for solution

R

A BF1

2F

1F

F2

F3

o'

o'

F1

2F

3F

F4

R

2F

4F

F31F

C

F'

F'1

2

RF1

2F

F'

F'1

2

R

R

R

321F FF

123

4

1

2

3

4

1

23

4

5

1

23

4

5

R

A BF1

2F

1F

F2

F3

o'

o'

F1

2F

3F

F4

R

2F

4F

F31F

C

F'

F'1

2

RF1

2F

F'

F'1

2

R

R

R

321F FF

123

4

1

2

3

4

1

23

4

5

1

23

4

5

R

A BF1

2F

1F

F2

F3

o'

o'

F1

2F

3F

F4

R

2F

4F

F31F

C

F'

F'1

2

RF1

2F

F'

F'1

2

R

R

R

321F FF

123

4

1

2

3

4

1

23

4

5

1

23

4

5

force diagram 1cm ≙ 15kN

force diagram 1cm ≙ 15kN

force diagram 1cm ≙ 15kN

Find the resultant force of the forces F1 and F2.1.1 Resultant of two forces

1.2 Resultant of a set of forces acting in any direction

Task 1

Find the resultant force of the forces F1, F2, F3 and F4.

F1 = 45 kNF2 = 30 kNF3 = 15 kNF4 = 30 kN

R = 120 kN

F1 = 45 kNF2 = 30 kN

R = 70.5 kN

1.3 Resultant of parallel forcesFind the resultant force of the forces F1, F2 and F3.

F1 = 45 kNF2 = 15 kNF3 = 30 kN

R = 90 kN

S. 2 / 4Equilibrium, cables and archesIntroduction to Structural Design

Name:EX 1 LösungsvorschlagProposal for solution

F1 = 100 kN A = 58.4 kN B = 58.4 kNNmax= N2 = N1 = 58.4 kN

F1 = 100 kN A = 130 kN B = 130 kNNmax= N2 = N1 = 130 kN

F1 = 100 kN A = 78 kN B = 78 kNNmax= N2 = N1 = 78 kN

2hh/2

h

1F

1F

F1

F1

1F

F

F1

F1

1F

1

B B

A

A

B

A

A

B

A

A

B

B

B

A

A

B

A

B

I

I

I

I

I

III

II

III

II

III

II

I

I

I

II III

II III

IIIII

I

1

1

1

1

1

1

1

1

1

1

1

1

2

2

2

2

2

2

2

2

2

2

2

2

2hh/2

h

1F

1F

F1

F1

1F

F

F1

F1

1F

1

B B

A

A

B

A

A

B

A

A

B

B

B

A

A

B

A

B

I

I

I

I

I

III

II

III

II

III

II

I

I

I

II III

II III

IIIII

I

1

1

1

1

1

1

1

1

1

1

1

1

2

2

2

2

2

2

2

2

2

2

2

2

force diagram 1cm ≙ 20kN

form diagram 1:50

force diagram 1cm ≙ 20kN

form diagram 1:50

force diagram 1cm ≙ 20kN

form diagram 1:50

Calculate the internal forces and reaction forces in the cables using graphic statics.Task 2 Drawing force diagrams

a)

b)

c)

S. 3 / 4Equilibrium, cables and archesIntroduction to Structural Design

Name:EX 1 LösungsvorschlagProposal for solution

6f

f 6

ff 1

f 2

f

f

f

f

f

f

3

4

5

6

7

8

f 9

l /

0.75 m

3

0.75 m0.75 m

l / 33l /

2 2l /l /

l

l / 3 l / 3 l / 3

l

2 2l /l /

ff

l

80.2 kN

54.9 kN

ff

54.9 kN

13 m

13 m

l

m88 m 13

kN8888 kN

l /

l /

3

kN88

l / 3

3

1.50

l /

3l /

3

1.00

l / 3

15.00

1.00

l/2l/2

15.00

15.00

15.00

15.00

20.00

15.00

15.00

5.00

l

1.6m 1.6m 1.6m 1.6m 1.6m 1.6m 1.6m 1.6m 1.6m 1.6m

ll / 3l / 3

l/2

3

l/2

l / l / 3 3l / 3l /l

l/2 l/2

88.8

8.8888

88.888.88

kN8888 kN

88 kN

888 kN888888 N 888 mm88.88 cm

88.8 N

8.88 N

8 NN8.8

N88

N88.8888.8 kN

kN8.88

8 kN8.8 kN

88 kN

88.88 kN88.8 mm

8.88 mm

8 mmmm8.8

mm88

mm88.88cm888

cm88.8cm88cm8.88cm8.8cm8

888 m

88.8 m

8.88 m

8 mm8.8

m88

m88.88

N kN mm cm m

88 kN

2l /l

3l / 3

ll / 3 l / 3 l /

ff

3

l / l /

2l /

22

l3l / 3

ll /

l

3 l / 3 l / 3

ll / 3l / 32l / 3l /l /

l3 3l /3 3l / l /

l3

ll / 3 l / 3 l / 3 l / 3 3l / 3l /

l

l / l / 22

ll / 3l /

l

33l / l / 3 3l / 3

l /

l /

3

l

3l /l 3l /l / 3l / 3

l

3l /

ff

l/2l/2

2l /l

3

ll /

l /

3l / 33

3

l /

ll / 3 l /

ff

3 l /

l/2 l/2

3

1.6m1.6m1.6m1.6m1.6m1.6m1.6m1.6m1.6m1.6m

ff

l/2l/2

0.75 m0.75 m0.75 m

0.75 m0.75 m0.75 m

l

0.75 m

0.75 m

II

I

II

III

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

I II III IV VI VIIV VIII IX X

I II

III

III

II

I

III

II

I

point selected on the closing stringintersection point --> result

example:

III

II

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

Dimensioning

I

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

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.

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

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

Step 1: Pick the dimension from the library Step 2: manually adapt it to the line of the drawing by rotating and streching one of its sides.

dimension from the library

line of the drawing

I

2/3 1/3

I

I

I

I

arch

III

II

I

I

cable

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!

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

Dimensioning

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

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 without notationonly external forces..full arrow head

Notation - Force Diagram

Arrows & Arrowheads - Lengths

Notation - Form Diagramelement / subsystem numbers, from library. always in the center of the line

Position of arrowheads

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.

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

Resultant Force: 2 cm long and arrowhead 4.5 mm

Uniformly distributed load: 0.5 cm and arrowhead 2mm

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

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

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.

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

VIVIIIII

hinge

chosen point by the designerresulting point (intersection etc.)

l

alphabetletters with subscription and bounding box

2kN/cmMPa2N/mm

kNkNkNkNmm/mm

symbol unitsmeaning

generic forceinternal tension forceinternal compression forceprestress force

generic straingeneric stresssecurity factor materialsecurity factor external loadsecurity factor self-weightgeneric security factor

44m /mm

3kg/m

2N/mmN/mm2

mmmm

2mm2

2

2

m

kNkN

kN/mkN/m

kN/mkN/mkNkN

moment of inertiadensitygeneric strength of materialElastic/Young’s modulusthicknessdepthwidth lengthareasubsystems (force elements where necessary)nodes in diagrams (where necessary)

area dead load

area life loadlinear dead load

life loaddead loadlinear life load

resultant forcesupport reaction force

ρ

f

E

t

hb

A

γ

γ=1,35

γ=1,5

σ

ε

γM

geometric planes

direction of rotation in cremona

section mark

symmetry

sliding support

hinge support

pole (funicular construction)trial pole (funicular construction)intersection point of closing string and line of action of resultant

I

I

AA'

Trako Gang Drawing Conventions - Latest Update 14.1.2016

direction of rotation in cremona

CS ISC

trial pole (force diagram)

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

o'''o''o'

geometric planes

intersection point of closing string and line of action of resultant

i

m nclosing string

parallel lines

A'A

I II III IV V

IIICSIISCICS

2.8 mm text Arial Narrow

q

q1

2

γM

generic stressgeneric strain mm/mm

N/mm2 MPa kN/cm2σε

kNgeneric force

internal tension force kNinternal compression force kN

kNprestress force

life load kN

kN/mkNdead load

linear life load2kN/marea life load

linear dead load kN/m2kN/marea dead load

resultant force kNkNsupport reaction force

nodes in diagrams (where necessary)

subsystems (force elements where necessary)

area m2 mm2

mlength

width mmdepth

thickness mN/mm2Elastic/Young’s modulus

generic strength of material N/mm2

ρ 3kg/mdensity

hinge support

sliding support

symmetry

section mark

moment of inertia m /mm4 4

resulting point (intersection etc.)chosen point by the designer

hinge

I

AlbhtEf

security factor self-weightsecurity factor external loadsecurity factor material

γ = 1,5

symbol unitsmeaninggeneric security factor

2.3 mm text Arial Narrow

Text for dimensions 2 mm Arial Narrow

III

II

I

I

II

III

-f t ll+t llf llll +f t ,d -f t ,d

t f -+t f

f mx,d -

t ,df -t ,df +

+mx,df

+a1,df-f a1

g1,d+f

+f s1,d

c1,d +f f -c1,d

s1,df -

II

f -g1,d

I

III

f a1,d-

g1-f

-f s1

c1f -

-r1f

mx -f+mxf

f r1+

f +c1

s1+f

+f g1

+a1fn

n2

1

n

n

n

n

n

n

n

3

4

5

6

7

8

9

iN

A i

dN

ferA

minL

III

maxL

Free-Body diagram

III

II

I II III IV V

I

III

I II

III

II

IV

I

V

V

III

III IV

II

I

γ

= 1,35γ

o

A'B'

C'

I

II

III

o'

III

II

I

o'

BCi

ABi

i AC

I

N

N

B'

100 kN

75 kN

75 kN100 kN 100 kN

B''A''

B'A'

s = 5kN/m

A

B

R

BA

R

R

109

87

65

43

2

F

1

F

A

F

R

qR

F

F

F

F 2

F

F F

F

F

1

B

R R

R R

R R

R

2 1

R R

R

R

q

R R

R R

R R

R R

R

109

8

s = 2.5kN/m

7

BA

65

43

21

B

RL

s = 4kN/m

2F

R

3F

B

F1

B

A

R = 48kN

A

B

R = 48kN

R = 32kN

F

R = 64kN

1

F

L

F

2

3

R

A B

F

A

F

B

F

A

B

F

FB

A

F

F

B

A B

A

BA

B

A B

F

BA

F

s = 4kN/mR

R

A

B

F

C

RR

F

LR

F

R

F

q

F

A

F

F

R

F F

B

F F

q

F

s = 5kN/m

B

A

108 975 6432 RRRRRR

B

RRRR1

F

s = 5kN/m

F

A

q

s = 2.5kN/m

CBA

BA

F = 40kNF = 40kN

A

P

F

R

g

qG

Q

A

2

A

1

R q

B

s = 4kN/m

B

B

F

A

1

A

s = 4kN/m

F

23

45

67

2

89

10

s = 2.4 kN/m

F

s = 2.4 kN/m

d d

R

F

RRR

P

R

Q

R

G

R

q

RR

g

R10 9

B

8

tot

7

C

6

tot

R

2

R

5

RL

4

A

3 1 1

B

23456

R

78910

B

A

h

F

Dv

Ch

Cv

Bh

vB

hAv

A

A

d s = 2.4 kN/m

B

B

D

B

9E

C

E 8

E 7

E 6

E5

D9

D8

75 kN

D 7

D6

5D

EDCB

s = 4kN/m

F

4E

E 3

E 2

E 1

D4

D3

2A

3A

A 4

5A

6A

B

F

1

3B

B

B

7B

2B

4B

6

5

9A

A 8

A 7

B8

9B

C1

C

C 3

4C

C 5

A'

6C

7C

C 9

8C

D1

C

2D

A

RR

LR

F

B C

1

A

F

R q

F 2

s = 4kN/m

F

F

RLR = 64kN R = 32kN

A

F

B

A''

B

RR

A

9F

R

F8

R

F

7F6

F5

F

q

F

100 kN

F4

3F

F1

B

1F = 30kN

A

A

vB

A

3F = 30kN

F = 30kN2

F

RR

q

F = 30kN1

F

F

B

F

F

R

F

B

R

AB

F

R

100 kN

RL

q

A

F

F

R''

R'

A B C

A 1

F

F

B''

F

B

F

B

TASK

Solution

qR

A V

A H

A

F2

F

q

totR

qR TOTR

q

1R

R

2R

3R

R

4R

5R

6

100 kN

R

R7

R8

R9

R10

q

B

A

R = 30kN

q

F

R q

BA

R=200 kN

1

B

2

2F

R

q

TOT

R

F = 40kN F = 40kN

F

F

F 1

1

F 2

2

F

1F

1

totR

R q

R

F

F

2

F

R

s = 5kN/m

q

F

A

B

1R R

F

R R R

F

R R

F

R R

F

A

R2 3

F

A

4 65 7

F

98 10

F

F

F

F

2F

R

B

F 1

B

A

B

A

A

A

R = 80kN

B

A

A B

B

F

R q

B

R

N

N

CA

F

B CA

A C

A C

B

ff

ff

ff

nm

o

o'

i

4

2

1

2

1

3

1

3

44

35

2

2

1

7 8

1

2

1

3

9

4

10

3

11

2

4

3

1

23

CL

1

i

1'CL'

2'

i1

2

1

CL'

2

i2

2

A

CL'

B

O'

O

1'

2'

4

3

2

3

2

11

3

10

2

98

1

1

7

2

65

2

1

3

4

4

3

2

21

1

1

1

2

3

2

3

4

2

4

3

3 4

1

4

2

3

3

2

4

21

1 2

2

1

2 1

2

1

1

12

43

3

3

2

3

2

1

3

2

3

1

1

2

3

2

3

4

2

3

3

2

2

1

1

1

1

4

4

5

5

6

7

67

245

7

8

1

12

1

2

2

1

2

1

1

2

6

654321 7

7

3

4 7 2

6

2

1

1 2 3 4 5 6

71 2 3 4 5 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

1

C

C

C C

C

C

hB

1 4

Task 3Find the form of a possible cable structure in equilibrium 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.

Funicular form - Cables

S. 4 / 4Equilibrium, cables and archesIntroduction to Structural Design

Name:EX 1 LösungsvorschlagProposal for solution

Task 4Find the form of a possible arch in equilibrium 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.

Funicular form - Arches