Δικτυώματα (Δικτυωτοί Φορείς)
1
Ορισμοί
• Το δικτύωμα είναι ένα σύστημα ευθύγραμμων
ράβδων, οι οποίες συνδέονται στα άκρα τους με
αρθρώσεις, έτσι ώστε να αποτελούν ένα στερεό
σώμα.
• Το σημείο στο οποίο συνδέονται δύο ή
περισσότερες ράβδοι ονομάζεται κόμβος.
• Επίπεδο δικτύωμα, είναι ένα δικτύωμα στο
οποίο οι άξονες των ράβδων και οι εξωτερικές
δυνάμεις που το φορτίζουν, βρίσκονται στο ίδιο
επίπεδο.
2
• Κάθε ράβδος, σαν δεσμικό στοιχείο του
δικτυώματος που βρίσκεται μεταξύ δύο κόμβων
όπου ασκούνται συγκεντρωμένες δυνάμεις,
αναλαμβάνει φορτία μόνο κατά την έννοια του
μήκους της, και επομένως καταπονείται αξονικά (σε
θλίψη ή εφελκυσμό).
• Το πρόβλημα συνεπώς που εμφανίζεται στα επίπεδα
δικτυώματα είναι ο προσδιορισμός της αξονικής
αυτής δύναμης που αναπτύσσεται σε κάθε ράβδο,
τόσο σαν ένταση όσο και σαν είδος (εφελκυσμός ή
θλίψη).
3
• Για να είναι εκμεταλλεύσιμη μια δικτυωτή κατασκευή θα πρέπει να
πληροί δύο προϋποθέσεις:
• α) Να είναι σταθερή ή γεωμετρικά ορισμένη. Τούτο σημαίνει ότι ο
αριθμός των ράβδων της θα πρέπει να είναι ορισμένος σε σχέση
με τον αριθμό των κόμβων της. Γιατί αφαίρεση ακόμη και μιας
ράβδου έχει σαν αποτέλεσμα να αποκτά κινητότητα ο φορέας, ή
όπως θα λέμε, μετατρέπεται σε μονοτρόχιο μηχανισμό.
• β) Να είναι στατικά ορισμένος, δηλαδή να είναι δυνατός ο
υπολογισμός των δυνάμεων των ράβδων της. Για να συμβαίνει
αυτό θα πρέπει ο αριθμός των ράβδων του δικτυώματος να έχει
συγκεκριμένη τιμή, που εξαρτάται από τον αριθμό των κόμβων
του. Αν η τιμή αυτή είναι μεγαλύτερη από εκείνη που για το φορέα
είναι στατικά απαραίτητη, τότε το δικτύωμα χαρακτηρίζεται
στατικά αόριστο με βαθμό στατικής αοριστίας τον αριθμό των
πλεοναζουσών ράβδων.
4
• Για να πληρούνται και οι δύο αυτές προϋποθέσεις, δηλαδή
για να είναι ένα δικτύωμα στατικά ορισμένο και σταθερό, θα
πρέπει να ισχύει η σχέση:
ρεσ + ρεξ = 2κ ,
όπου:
ρεσ είναι ο αριθμός των εσωτερικών ράβδων του
δικτυώματος, δηλαδή αυτές από τις οποίες αποτελείται το
δικτύωμα,
ρεξ είναι ο αριθμός των εξωτερικών ράβδων, που
θεωρούνται απαραίτητες για τη στήριξη του δικτυώματος και
είναι πάντα τρεις (2 για άρθρωση και 1 για κύλιση).
• Αν το δικτύωμα είναι στο χώρο, θα πρέπει αντίστοιχα να
είναι: ρεσ + ρεξ = 3κ .
5
Παράδειγμα Δικτυώματος
6
Διάγραμμα Ελευθέρου Σώματος
7
Κόμβος Β
Εφαρμόζουμε τις συνθήκες
ΣFx=0, ΣFy=0 για να
υπολογίσουμε τις S1, S2
8
Για την επίλυση των δικτυωμάτων
επισημαίνεται ότι:
• α) το ίδιο (νεκρό) βάρος των ράβδων του θεωρείται
αμελητέο.
• β) ∆ικτυώματα που είναι γεωμετρικά όμοια και
δέχονται ίσες δυνάμεις σ’ αντίστοιχους κόμβους, θα
εμφανίζουν ίσες δυνάμεις και στις αντίστοιχες
ράβδους. Τούτο σημαίνει ότι η ένταση των ράβδων
ενός δικτυώματος δεν εξαρτάται από το πραγματικό
του μέγεθος, αλλά από την ένταση των εξωτερικών
φορτίων και τη γεωμετρία του.
9
• Για την επίλυση των επιπέδων δικτυωμάτων,
δηλαδή τον προσδιορισμό των δυνάμεων που
αναπτύσσονται στις ράβδους του, χρησιμοποιούνται
κυρίως οι παρακάτω μέθοδοι:
1. Η αναλυτική μέθοδος ισορροπίας των κόμβων.
2. Η γραφική μέθοδος του διαγράμματος Cremona.
3. Η αναλυτική μέθοδος των τομών ή μέθοδος Ritter.
10
Ανάλυση Δικτυωμάτων
Truss Analysis
External equilibrium
Internal equilibrium
To find the reaction
forces
To find the force in
each member
Method of
joints
Method of
sections
11
External equilibrium
• Draw the FBD of the entire truss,
• Consider all the forces and moments (known and
unknown),
• Write all the dimensions,
• Consider the two equilibrium equations (forces and
moments).
12
Internal equilibrium (Method of
joints)
• Draw FBD of entire truss and solve for support reactions.
• Draw FBD of a joint with at least one known force and at most two
unknown forces.
• Either assume all unknown member forces are tensile. Positive results
indicate tension and negative results indicate compression.
• Otherwise determine the correct sense for unknowns by inspection.
Positive results indicate correct assumption and negative results indicate
incorrect assumption.
• Continue selecting joints where there are at least one known force and at
most two unknown forces.
• Tension pulls on a member, compression pushes on (compresses) a
member.
• Present member forces as positive numbers with (T) or (C) indicating
tension or compression.
•
13
Παράδειγμα - 1
1.
2.
Να προσδιορίσετε τις αντιδράσεις
στήριξης των κόμβων στο
δικτύωμα.
Να προσδιορίσετε τη δύναμη σε
κάθε μέλος του δικτυώματος και
αν είναι σε εφελκυσμό (tension) ή
Θλίψη (compression)
B
500 N
 Fx  0
2m
 Fy  0
500  A x  0 C y  A y  0
A
45o
C
A x  500 N
A y  Cy
 MA  0
Ax
Ay
2m
Cy
500  2   C y  2   0
A y  C y  500 N
14
Solution
• 2 unknown member forces at joint B
• 1 unknown reaction force at joint C
• 2 unknown member forces and 2 unknown
reaction forces at point A
For Joint B,
   Fx  0;
500 N  FBC sin 45 N  0  FBC  707.1N (C )
   Fy  0;
FBC cos 45 N  FBA  0  FBA  500 N (T )
15
Solution
For Joint C,
   Fx  0;
 FCA  707.1cos 45 N  0  FCA  500 N (T )
   Fy  0;
C y  707.1sin 45 N  0  C y  500 N
For Joint A,
   Fx  0;
500 N  Ax  0  Ax  500 N
   Fy  0;
500 N  Ay  0  Ay  500 N
16
Solution
For Joint C,
   Fx  0;
 FCA  707.1cos 45 N  0  FCA  500 N (T )
   Fy  0;
C y  707.1sin 45 N  0  C y  500 N
For Joint A,
   Fx  0;
500 N  Ax  0  Ax  500 N
   Fy  0;
500 N  Ay  0  Ay  500 N
17
Αποτελέσματα
18
Παράδειγμα - 2
SOLUTION:
• Based on a free-body diagram of the
entire truss, solve the 3 equilibrium
equations for the reactions at E and C.
• Joint A is subjected to only two unknown
member forces. Determine these from the
joint equilibrium requirements.
Using the method of joints, determine
the force in each member of the truss.
• In succession, determine unknown
member forces at joints D, B, and E from
joint equilibrium requirements.
• All member forces and support reactions
are known at joint C. However, the joint
equilibrium requirements may be applied
to check the results.
19
SOLUTION:
• Based on a free-body diagram of the entire truss,
solve the 3 equilibrium equations for the reactions
at E and C.
 MC  0
 2000 lb24 ft   1000 lb12 ft   E 6 ft 
E  10,000 lb 
 Fx  0  C x
Cx  0
 Fy  0  2000 lb - 1000 lb  10,000 lb  C y
C y  7000 lb 
20
• Joint A is subjected to only two unknown
member forces. Determine these from the
joint equilibrium requirements.
2000 lb FAB FAD


4
3
5
FAB  1500 lb T
FAD  2500 lb C
• There are now only two unknown member
forces at joint D.
FDB  FDA
FDE  2 53 FDA

FDB  2500 lb T
FDE  3000 lb C
21
• There are now only two unknown member
forces at joint B. Assume both are in tension.
 Fy  0  1000  54 2500  54 FBE
FBE  3750 lb
FBE  3750 lb C
 Fx  0  FBC  1500  53 2500  53 3750
FBC  5250 lb
FBC  5250 lb T
• There is one unknown member force at joint
E. Assume the member is in tension.
 Fx  0  53 FEC  3000  53 3750
FEC  8750 lb
FEC  8750 lb C
22
• All member forces and support reactions are
known at joint C. However, the joint equilibrium
requirements may be applied to check the results.
 Fx   5250  53 8750  0 checks
 Fy  7000  54 8750  0 checks
23
Analysis of Trusses by the Method of Sections
• When the force in only one member or the
forces in a very few members are desired, the
method of sections works well.
• To determine the force in member BD, pass a
section through the truss as shown and create
a free body diagram for the left side.
• With only three members cut by the section,
the equations for static equilibrium may be
applied to determine the unknown member
forces, including FBD.
24
Παράδειγμα - 3
SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static equilibrium to solve for the reactions at A and L.
• Pass a section through members FH,
GH, and GI and take the right-hand
section as a free body.
• Apply the conditions for static
equilibrium to determine the desired
member forces.
Determine the force in members FH,
GH, and GI.
25
SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static equilibrium to solve for the reactions at A and L.
 M A  0  5 m 6 kN   10 m 6 kN   15 m 6 kN 
 20 m 1 kN   25 m 1 kN   25 m L
L  7.5 kN 
 Fy  0  20 kN  L  A
A  12.5 kN 
26
• Pass a section through members FH, GH, and GI
and take the right-hand section as a free body.
• Apply the conditions for static equilibrium to
determine the desired member forces.
MH  0
7.50 kN 10 m   1 kN 5 m   FGI 5.33 m   0
FGI  13.13 kN
FGI  13.13 kN T
27
FG 8 m

 0.5333
  28.07
GL 15 m
 MG  0
7.5 kN 15 m   1 kN 10 m   1 kN 5 m 
  FFH cos 8 m   0
tan  
FFH  13.82 kN
tan  
GI
5m
2
 0.9375
HI
8 m 
3
FFH  13.82 kN C
  43.15
ML  0
1 kN 10 m   1 kN 5 m   FGH cos  10 m   0
FGH  1.371 kN
FGH  1.371 kN C
28
Παράδειγμα - 4
12.5 kN 12.5 kN
12.5 kN 12.5 kN
2m
2m
2m
A
C
B
D
2.5 m
Using the method of
joints, determine the
force in each member
of the truss shown.
G
F
E
29
Solving Problems on Your Own
12.5 kN 12.5 kN
12.5 kN 12.5 kN
Using the method of
2m
2m
2m
joints, determine the
force in each member
A
C
B
of the truss shown.
D
2.5 m
G
F
E
1. Draw the free-body diagram of
the entire truss, and use this
diagram to determine the
reactions at the supports.
2. Locate a joint connecting only two members, and draw the
free-body diagram for its pin. Use this free-body diagram to
determine the unknown forces in each of the two members.
Assuming all members are in tension, if the answer obtained
from SFx = 0 and SFy = 0 is positive, the member is in
tension. A negative answer means the member is in
30
compression.
Solving Problems on Your Own
12.5 kN 12.5 kN
12.5 kN 12.5 kN
2m
2m
2m
A
C
B
Using the method of
joints, determine the
force in each member
of the truss shown.
D
2.5 m
G
F
E
3. Next, locate a joint where the forces in only two of the
connected members are still unknown. Draw the free-body
diagram of the pin and use it as indicated in step 2 to
determine the two unknown forces.
4. Repeat this procedure until the forces in all the members of
the truss have been determined.
31
12.5 kN 12.5 kN
12.5 kN 12.5 kN
2m
2m
2m
Ax A
C
B
D
Ay
2.5 m
G
Draw the free-body
diagram of the entire
truss, and use it to
determine reactions
at the supports.
F
E E
+ S MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m)
- (12.5 kN)(6 m) = 0
+ S Fy = 0: Ay - (4)(12.5 kN) = 0
S
F
=
0:
A
E
=
0
x
x
+
E = 60 kN
Ay = 50 kN
Ax= 60 kN
32
12.5 kN
12.5 kN
2m
60 kN
A
12.5 kN 12.5 kN
2m
B
2m
C
D
50 kN
2.5 m
G
F
60 kN E
Locate a joint connecting only two
members, and draw the free-body
diagram for its pin. Use the freebody diagram to determine the
unknown forces in each of the two
members.
Joint D
12.5 kN
FCD
6.5
6
FGD
2.5
+ S Fy = 0:
2.5
F - 12.5 kN = 0
6.5 GD
FGD = 32.5 kN C
6
S
F
=
0:
FGD - FCD = 0
+
x
6.5
FCD = 30 kN T
33
12.5 kN
12.5 kN
2m
60 kN
A
12.5 kN 12.5 kN
2m
B
2m
C
D
50 kN
2.5 m
G
F
60 kN E
Next, locate a joint where the forces
in only two of the connected
members are still unknown. Draw
the free-body diagram of the pin and
use it to determine the two unknown
forces.
Joint G
FCG
32.5 kN
FFG
S F = 0:
FCG = 0
S F = 0:
FFG - 32.5 kN = 0
FFG = 32.5 kN C
34
12.5 kN
2m
60 kN
12.5 kN 12.5 kN
12.5 kN
A
2m
B
2m
C
D
50 kN
2.5 m
G
F
60 kN E
12.5 kN
FBC

FCD = 30 kN
Repeat this procedure until the
forces in all the members of the
truss have been determined.
Joint C
2
BF = (2.5 m) = 1.6667 m
3
BF
-1
 = BCF = tan
= 39.81o
2
+ S Fy = 0: - 12.5 kN - FCF sin  = 0
- 12.5 kN - FCF sin 39.81o = 0
FCF = 19.53 kN C
S Fx = 0: 30 kN - FCF cos  - FBC = 0
+
30 kN - (-19.53) cos 39.81o - FBC = 0
FBC = 45.0 kN T
35
FCF
12.5 kN
2m
60 kN
12.5 kN 12.5 kN
12.5 kN
A
2m
B
2m
Joint F
C
FBF
=39.81o
D
50 kN
2.5 m
G
FEF
F
60 kN E
+
FCF = 19.53kN
6.5
2.5
FFG = 32.5 kN
6
S Fx = 0: -
6
6
F
(32.5
6.5 EF 6.5
kN) - FCF cos  = 0
o F
FEF = -32.5 kN - ( 6.5
)
(19.53)
cos
39.81
EF = 48.8 kN C
6
2.5
+ S Fy = 0: FBF - 6.5 FEF - 6.5 (32.5 kN) - (19.53) sin  = 0
2.5
FBF -
2.5
6.5
(-48.8 kN) - 12.5 kN - 12.5 kN = 0
FBF = 6.25 kN T
36
12.5 kN
2m
60 kN
12.5 kN 12.5 kN
12.5 kN
A
2m
B
2m
C
D
50 kN
2.5 m
12.5 kN
Joint B
G
F
FAB
FBC = 45.0 kN
g
FBE
60 kN E
2.5 m
tan g = 2 m
FBF = 6.25kN
; g = 51.34o
+ S Fy = 0: -12.5 kN -6.25 kN - FBE sin 51.34o = 0
FBE = -24.0 kN
FBE = 24.0 kN C
+ S F = 0: 45.0 kN - F + (24.0 kN) cos 51.34o = 0
x
AB
FAB = 60.0 kN
FAB = 60.0 kN T
37
12.5 kN
12.5 kN
2m
60 kN
A
12.5 kN 12.5 kN
2m
B
2m
Joint E
C
D
50 kN
2.5 m
FBE = 24 kN
G
FAE
F
60 kN E
g
FEF = 48.75 kN
6.5
60 kN
+ S Fy = 0: FAE - (24 kN) sin
FAE = 37.5 kN
51.34o
2.5
6
g = 51.34o
- (48.75 kN)
2.5
6.5
=0
FAE = 37.5 kN T
38
Παράδειγμα - 5
3 kN
3 kN
3 kN
F
3 kN
D
H
3 kN
1.5 kN
6.75 m
1.5 kN
B
J
L
A
C
3m
E
G
3m
3m
I
3m
K
3m
A Pratt roof truss
is loaded as
shown. Using
the method of
sections,
determine the
force in
members FH ,
FI, and GI.
3m
39
3 kN
3 kN
Solving Problems on Your Own
3 kN
F
3 kN
D
H
3 kN
1.5 kN
B
J
6.75 m
1.5 kN
L
A
C
3m
E
G
3m
3m
I
3m
K
3m
3m
A Pratt roof truss
is loaded as
shown. Using the
method of
sections,
determine the
force in members
FH , FI, and GI.
1. Draw the free-body diagram of the entire truss, and use
this diagram to determine the reactions at the supports.
2. Pass a section through three members of the truss, one of
which is the desired member.
40
3 kN
3 kN
Solving Problems on Your Own
F
3 kN
A Pratt roof truss
D
H
6.75 m is loaded as
1.5 kN
1.5 kN
shown. Using the
B
J
method of
L
A
sections,
I
C
E
G
K
determine the
force in members
3m
3m
3m
3m
3m
3m
FH , FI, and GI.
3. Select one of the two portions of the truss you have obtained,
and draw its free-body diagram. This diagram should include
the external forces applied to the selected portion as well as
the forces exerted on it by the intersected members before
these were removed.
4. Now write three equilibrium equations which can be solved for
the forces in the three intersected members.
41
3 kN
3 kN
3 kN
3 kN
3 kN
F
3 kN
D
H
3 kN
1.5 kN
B
J
L
A
Ax
1.5 kN
C
E
I
G
K
Ay
Draw the free-body
diagram of the entire
truss, and use it to
determine reactions
at the supports.
+ S Fx = 0: Ax = 0
L
3m
3m
3m
3m
3m
3m
Total load = 5(3 kN) + 2(1.5 kN) = 18 kN
By symmetry
Ay = L =
1
2
(18kN) = 9 kN
42
3 kN
3 kN
3 kN
F
3 kN
D
H
3 kN
1.5 kN
B
J
1.5 kN
L
A
C
E
I
G
Pass a section
through three
members of the
truss, one of
which is the
desired member.
K
9 kN
9 kN
3m
3m
3m
3m
3m
3m
43
3
5
Free Body: Portion HIL
4
F
3 kN
FFH
H
3 kN
FFI
6.75 m
J
a
G
FGI
1.5 kN
L
I
3m
Slope of FHJL
K
9 kN
3m
Select one of the two portions of
the truss you have obtained,
and draw its free-body diagram.
3m
3
6.75
=
4
9.00
6.75
FG
tan  = GI = 3.00
3
5
4
 = 66.04o
Now write three equilibrium equations which can be solved for
the forces in the three intersected members.
44
3
5
Free Body: Portion HIL
4
F
3 kN
Write the three equilibrium equations.
FFH
3 kN
H
FFI
6.75 m
J
G
66.04o
FGI
1.5 kN
L
I
K
9 kN
3m
3m
Force in FH:
3m
4
2
+ S MI = 0: FFH (
x 6.75 m) + (9 kN)(6 m)
5
3
- (1.5 kN)(6m) - (3 kN)(3m) = 0
4
F
(4.5
m)
+
36
kN-m
=
0
FH
5
FFH = -10.0 kN
FFH = 10.0 kN C
45
3
5
Free Body: Portion HIL
4
F
3 kN
Write the three equilibrium equations.
FFH
H
3 kN
FFI
6.75 m
J
G
66.04o
FGI
1.5 kN
L
I
K
9 kN
3m
FFH = 10.0 kN C
3m
Force in FI:
3m
+ S ML = 0: -FFI sin 66.04o(6 m) + (3 kN)(6 m)
+ (3 kN)(3 m) = 0
FFI sin 66.04o(6 m) = 27 kN-m
FFI = + 4.92 kN
FFI = 4.92 kN T
46
3
5
Free Body: Portion HIL
4
F
3 kN
Write the three equilibrium equations.
FFH
H
3 kN
FFI
6.75 m
J
G
1.5 kN
66.04o
FGI
FFH = 10.0 kN C
FFI = 4.92 kN T
L
I
K
9 kN
3m
3m
Force in GI:
3m
+ S MF = 0: -FGI (6.75 m) - (3 kN)(3 m) - (3 kN)(6 m)
- (1.5 kN)(9 m) + (9 kN)(9 m) = 0
FGI (6.75 m) = +40.5 kN-m
FGI = + 6.00 kN
FGI = 6.00 kN T
47