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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