S. 1 / 3TrussesIntroduction to Structural Design

Name:EX 3

9f8

7

6

5

4

3

f

f

f

f

f

f2f1ff

6f

f 6

kN88

mcmmmkNN

88.88 m

88 m

8.8 mm8

m8.88

m88.8

m888

8 cm8.8 cm

8.88 cm88 cm

88.8 cm

888 cm88.88 mm

88 mm

8.8 mmmm8

mm8.88

mm88.8kN88.88

kN88

kN8.8kN8

8.88 kN

kN88.888.88 N

88 N

8.8 NN8

N8.88

N88.8cm88.88

mm888N888 888 kN888

kN88

kN88 88 kN

88.8888.8

888.888.8

8

5.00

15.00

15.00

20.00

15.00

15.00

15.00

15.00

1.00

15.00

1.00 1.50

88 kN

kN8888 kN

13m88 m

m13

m13

kN54.9

kN54.9

kN80.2

ff

ff

ff

ff

ff

C

V

VI II

C

III

C

I

C

IV

I

I

I

I

I II

FDE

I

I

II

II

III

I

III

II

III

I

F

III

BC

II

HI

III

IV

F

III

SL

II

IV

o'

III

IV

IV

IIIII

I

IIIIII

III

VII

III

IIIIII

I

I II

IIIIV

V

II

I II

IVIII

N1

N2

N3

IV

VIII III IV

21N N

V

IVIII

VI

I

III

C

III

III

III

VIVIIIIII

II

I II

Free-Body diagram

L max

L min

A erf

Nd

iA

Ni

9

8

7

6

5

4

3

n

n

n

n

n

n

n

1

2n

n

I

f a1+

g1f +

f +s1

c1+f

+r1f

f mx+ f -mx

f r1-

-f c1

s1f -

f -g1

-a1,df

g1,d-f

-f s1,d

c1,d -ff +c1,d

s1,df +

f +g1,d

Schlusslinie SL

a1f - f a1,d+

f mx,d+

+f t ,d -f t ,d

-mx,df

f t + -f

M

t

I

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

IV

III

Text for dimensions 2 mm Arial Narrow

2.3 mm text Arial Narrow

generic security factormeaning unitssymbol

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

fEthblA

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

2mm2marea

subsystems (force elements where necessary)

nodes in diagrams (where necessary)

support reaction force kNkNresultant force

area dead load kN/m2

kN/mlinear dead load

area life load kN/m2

linear life loaddead load kN

kN/m

kNlife load

prestress force kNkNinternal compression force

kNinternal tension force

generic force kN

εσ 2kN/cmMPa2N/mm

mm/mmgeneric straingeneric stress

Mγ

2

1q

q

2.8 mm text Arial Narrow

SC ICS II

SC III

II

VIVIII

I

III

IV

III

AA'

IIIII

IVI

parallel lines

closing string

n

SL

m

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

geometric planes

o' o'' o'''

r' rise point (form diagram)

trial pole (force diagram)

CS

I

ISC

direction of rotation in cremona

IV

subsystems (force elements where necessary)

Trako Gang Drawing Conventions - Latest Update 14.1.2016

A'A

SL

I

I

intersection point of closing string and line of action of resultant

trial pole (funicular construction)pole (funicular construction)

hinge support

sliding support

symmetry

section mark

direction of rotation in cremona

II

geometric planes

Mγ

ε

σ

γ=1,5

γ=1,35

γ

A

bh

t

E

III

f

ρ

support reaction forceresultant force

linear life loaddead loadlife load

linear dead loadarea life load

area dead load

nodes in diagrams (where necessary)

arealengthwidth

IV

depththicknessElastic/Young’s modulusgeneric strength of materialdensitymoment of inertia

kN

III

kNkN/mkN/m

kN/mkN/m

kNkN

m

2

II

2

2 mm2

mmmm

2N/mmN/mm2

kg/m3

m /mm4 4

generic security factorsecurity factor self-weightsecurity factor external loadsecurity factor materialgeneric stressgeneric strain

prestress forceinternal compression forceinternal tension forcegeneric force

meaning unitssymbol

I

mm/mmkNkNkNkN

N/mm2 MPa kN/cm2

letters with subscription and bounding boxalphabet

l

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

hinge

II III IV V

o

γ = 1,35

γ

VII

VI

V

IV

IV

III

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.

II

External force, 1 mm distance to structure / forcesSupport 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

I

III V

V

SL

Schlusslinie SL

IVIII

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.

Position of arrowheads

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

II

Notation - Form Diagram

Arrows & Arrowheads - Lengths

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

I

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

C

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

I

I

I

I

I

I

I

I

o'

I

I

IV

VII

III

SLI

SL

I

XIX

C

VIIIV VIIVI

arch

IVIIIIII

V

I

I

IV

I

I

1/32/3

I

line of the drawing

dimension from the library

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

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

III

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

o'

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

II

example:

intersection point --> resultpoint selected on the closing string

I

C

SL

I

I

I

I

I

V

r'''r''

IV

I

II

III

I

II

III

IIIII

SL

I

C

N

N

A

A

B

B

F

R

F

F1

A

B

R

A

A

B

vA

hA

F1

F

F = 10kN

BA

F = 25kN

B

R

Av

Ah

BA

q = 5kN/m1

2

F1 2

F

F

A

3F F4

B

F

A

1

R

q = 5kN/m

A

R

B

LR R

B

R

B

RB

F

A

1q = 5kN/m

A B

RR

F

1q = 5kN/m

A B

q = 8kN/m

g = 4kN/m

q = 5kN/m1

F

g = 4kN/m

F

A B

B

R

R

R

R

R

B

1

RL

F

A

B

LB

B

LA

B

B

R

A

R

R

F

A

B

RA

B

L

A

B

B

LA

2

R

F

L

R

R

R

B

A

B

1F

R

F

A

B

R

A

B

A

F1F

H

I

E

F

D

C

B

BA

G

1F = 30kN 2F = 30kN

F

DEF

V

A

B

A

F

F = 30kN1

DE

HIF

F = 30kN2

GA

I

H

G

BC

E

D

C

B

A

F

F 1F

1F = 60kN

A

A

A

A

H

F

F

B

B

F

1F = 60kN

F

A = 45kN B = 45kN

3F = 30kN

F = 30kN2

B

F

F

F = 30kN

M

1

A

A B

A

F

B

F = 30kN2F = 30kN1

B

R

B

A

H

F

h

1

A

2

F

F

F

3F

Av

4F

1F = 60kN

A

s = 6kN/m

R = 60kNA

A

F

B

F

F

A

B

R

F

A

B

L

AR

RB

LB

H

RR

RL

F

F

AA +A = RL

B =

F

LB + RB

F

A

V

B

F

H

R =

B

RR +

LR

A

F

VAA

D2

1D

C 8

9C

C 7

C 6

5

F

C

C 4

3C

C1C

B9

8B7A

8A

A 9

5

6

B4

B2

B7

B

B

B3

1B

A 6

A 5

4A

A 3

A 2

3D

4D

1E

F

R

2E

3E

E 4

1F

F

H

B

A

F

B C D E

D5

6

1

D

7D

8

F = 70kN

D

9D

5

C

E

6

B

E

7E

8E

E 9

F

DA v

A h

Bv

hBvC

hCv

D

Dh

E

I

H

H

DEF

FBC

A

FHI

VA

H

R

R

A

A

1F

F

B

g

q

B

GQ

A

P

F

F

F

F

A

F

B

2

V

B

A

A

B

F

B

A

F = 70kN

VA

A

2

F

F

4

B

VA

F1

B

A

F 3

1F

F 2

VA

B

1F = 30kN

2

B

F = 50kN

A V

A H

432 F = 15kNF = 15kNF = 15kNF = 15kN

F

1

2

V

2

34

4

3

1

4321

F

A

F = 20kN

F = 30kN

F = 16kN

A

F = 16kN

F = 30kN

F = 20kN

3

2

B

1

R

A B

Q

G

F = 50kN

q

g

R

F

P

F = 50kN

B

R

2F

A

HA

VA

F = 40kN

H

A B

1

C

A

CBA

R'

R''

A

B

F = 15kNF = 15kNF = 15kNF = 15kN

A

F

BA

B

F

B

F

A

F

1F = 30kN 2F = 30kN

F = 30kN3

F

A

A

F = 30kN1

B

1F

F3

4F

A

F

5F

6

F 7

F

8F

F9

F = 30kN

3F = 30kN

21

B

F = 30kN

F

2

1 2

21

1

F = 30kN F = 30kN

F = 30kNF = 30kN

F = 30kN F = 30kN

F = 60kN

F = 60kNF = 60kN

F = 60kN

B

B

B

B

B

B

B

BA

F1

F2

F3

1q = 5kN/m

F

BA

q = 5kN/m1

A

B

F

A

BA

q = 5kN/m

B

1

BA

q = 5kN/m1

F

1q = 5kN/m

A B

hA

vA

B

1

B

B

A

A

1

F

B

B

A

A

A

A

1

1

B

F

B

F

AA V

F

A = 45kN B = 45kN

A H

F = 40kN

A

B

B

F1

F2

F3

F

A

3F

BF2

1F A

F

2 B

1

F

F1 A

A

F = 30kN

3F = 30kN

A B

F

BA

A

A

B

B

A

BA

B

F

F

21

F

F = 30kN

A

BF

F

B

A

F

V

BA

F

B

A

A

F = 30kN

2

F = 30kN

F = 30kN

R

3

F = 30kN

1

F = 30kN

3

F = 30kN

2

2

1

F

1

F

1

1

1

F

1

1

F = 60kN

F = 60kN

2

B

2

A

1

A

H

2

B

F = 60kN

F = 60kN

1

1

B

2

F F3

R

F2

F

A

A

F

A

A

A

A

B

B

F = 50kN

F

A

F

F1

F

F

F1

A

B

A

A

B

1F

F1

F

1F = 100kN

A

q = 8kN/m

g = 4kN/m

B

R

N

N

I

H

HIF

i

o'

o

m n

F = 105kN F = 105kN

A

6

4

2

2

4

3

1

2

2

1

3

2

3

1

2

3

1

3

3

2

1

1

2

5

4

2

3

1

2

3

1

5

3

4

3

3

2

1

1

2

2

3

1

3

2

1

1

3

2

3

3

2

1

1

6

1

5

2

4

3

3

5

3

1

2

2

1

1

3

3

2

7

3

1

5

3

6

2

1

3

4

1

4

7

1

1

3

2

4

1

2

1

3

6

3

5

4

1

2

3

2

3

3

7

6

2

4

3

3

1

7

8

3

1

7

3

2

1

4

1

9

7

2

1

1

2

2

10

3

3

2

3

6

11

1

12

2

13

3

3

3

1 2

4

12

4 5

3

2

7

1

1

1

5

6

4

1

1

5

43

1

2

2

2

2

3

3

4

2

4

1

3

2

1

5

5

6

3

1

5

52

4

4

3

3

3

4

4

1

71

6

1

1

2

2

3

2

3

3

3

4

2

6

1

5

1

5

5

6

4

1

2

7

3

1

2

2

5

1

3

4

7

6

2

7

1

4

6

5

5

1

5

4

6

2

8

3

5

3

9

5

11

3

10

12

4 2

13

7

5

1

o

i

o'

5

3

2

2

4

5

2

2

2

3

34

4

5

52

3

3

1

2

23

4

4

1

1

1

15

5

5

5

5

5

5

5

3

3

3

3

4

6

6

6

6

4

4

4

4

6

6

6

6

7

7

7

7

5

4

5

5 5

6

2

3

3

3

3

4

4

4

41

1

1

1

2

2

2

2

7

7

7

72

2

2

2

3

4

4

7

7 7

1

8

9

9

99

11

10

11

11

11

11

66

10

2

7

8

31

7654

4

2

2

1

1

2

3

6

5

8

8

3 3

7

76

2 2

10102

2

43

5

1

6

1 6

5

32

5

4

3

10

9

7

8

9

2

1

8

1

1

1

1

11

1

1

1

1

1

Qualitative flow of forces in trussesTask 1Determine the direction of the reaction forces and - qualitatively - the inner flow of forces in the truss for the given cases. Draw the tension forces in red, the compression forces in blue and the external forces in green.

S. 2 / 3TrussesIntroduction to Structural Design

Name:EX 3

force diagram 1cm ≙ 25kN

3

2

1 2F = 60kN F = 20kN

F = 30kN1

F = 30kN

F = 30kN

Task 2 Quantitative flow of forces in trussesDraw the force diagram for the given case. Draw the tension forces in red, the compression forces in blue and the external forces in green.

S. 3 / 3TrussesIntroduction to Structural Design

Name:EX 3Task 3 Comparing Structures

Fink Truss, Albert Fink

Location: - Year: 1854

Compare 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 concise text also including sketches, where relevant.

Multihalle, C.Mutschler, J.Langner, F. Otto

Location: Mannheim, Germany Year: 1975