Δικτυώματα (Δικτυωτοί Φορείς) Δικτυώματα (Δικτυωτοί Φορείς)

Truss structures constitute a special class of structures in which individual straight members are connected at joints. The members are assumed to be connected to the joints in a manner that permit rotation, and thereby it follows from equilibrium considerations, to be detailed in the following, that the individual structural members act as bars, i.e. structural members that can only carry an axial force in either tension or compression. Often the joints do not really permit free rotation, and the assumption of a truss structure then is an approximation. Even if this is the case the layout of a truss structure implies that it can carry its loads under the assumption that the individual members act as bars supporting only an axial force. This greatly simplifies the analysis of the forces in the structure by hand calculation and undoubtedly contributed to their popularity e.g. for bridges, towers, pavilions etc. up to the middle of the twentieth century. The layout of the structural members in the form of a truss structure also finds use with rigid or semi-rigid joints, e.g. space truss roofs, girders for suspension bridges, or steel offshore structures. The rigid joints introduce bending effects in the structural members, but these effects are easily included by use of numerically based computational methods.

In a statically determinate truss all the bar forces can be determined by the equilibrium equations, applied to the bars and joints of the truss. There are several strategies for carrying out the corresponding calculations.

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.

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).

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.

Για να πληρούνται και οι δύο αυτές προϋποθέσεις, δηλαδή για να είναι ένα επίπεδο 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 (χωρικό δικτύωμα), θα πρέπει αντίστοιχα να είναι: ρεσ + ρεξ = 3κ .

Ράβδοι Μηδενικής Φόρτισης The following rules are useful in identifying zero-force members. 1. If two members are not collinear at an unloaded joint (Fig. a), then both members are zero-force members. 2. Let two members be connected at a loaded joint (Fig. b). If the action line of the external force F coincides with the direction of one of the members, then the other member is a zero-force member. 3. Let three members be connected at an unloaded joint (Fig. c). If two members have the same direction, the third member is a zero-force member.

Δικτύωμα , Διάγραμμα Ελευθέρου Σώματος - Ράβδοι μηδενικής φόρτισης, Ανάλυση με τη μέθοδο των κόμβων

Παράδειγμα Δικτυώματος

Διάγραμμα Ελευθέρου Σώματος

Κόμβος Β Εφαρμόζουμε τις συνθήκες ΣFx=0, ΣFy=0 για να υπολογίσουμε τις S1, S2

Για την επίλυση των δικτυωμάτων επισημαίνεται ότι: το ίδιο (νεκρό) βάρος των ράβδων του θεωρείται αμελητέο. ∆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.

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.

To find the reaction forces To find the force in each member 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

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).

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. 

Παράδειγμα - 1 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)

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,

Solution For Joint C, For Joint A,

Αποτελέσματα

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. In succession, determine unknown member forces at joints D, B, and E from joint equilibrium requirements. Using the method of joints, determine the force in each member of the truss. All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

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 Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements. There are now only two unknown member forces at joint D.

There are now only two unknown member forces at joint B There are now only two unknown member forces at joint B. Assume both are in tension. There is one unknown member force at joint E. Assume the member is in tension.

All member forces and support reactions are known at joint C All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

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.

Παράδειγμα - 3 SOLUTION: Take the entire truss as a free body. Apply the conditions for static equilib-rium 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.

SOLUTION: Take the entire truss as a free body. Apply the conditions for static equilib-rium 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.

Παράδειγμα - 4 12.5 kN 12.5 kN 12.5 kN 12.5 kN Using the method of joints, determine the force in each member of the truss shown. 2 m 2 m 2 m A B C D 2.5 m G F E

Solving Problems on Your Own 12.5 kN 12.5 kN 12.5 kN 12.5 kN Using the method of joints, determine the force in each member of the truss shown. 2 m 2 m 2 m A B C D 1. Draw the free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports. 2.5 m G F E 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 compression.

Solving Problems on Your Own 12.5 kN 12.5 kN 12.5 kN 12.5 kN Using the method of joints, determine the force in each member of the truss shown. 2 m 2 m 2 m A B C 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.

S MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m) Draw the free-body diagram of the entire truss, and use it to determine reactions at the supports. 2 m 2 m 2 m A Ax B C D Ay 2.5 m G 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 E = 60 kN + S Fy = 0: Ay - (4)(12.5 kN) = 0 Ay = 50 kN S Fx = 0: Ax - E = 0 Ax= 60 kN +

S Fy = 0: FGD - 12.5 kN = 0 FGD = 32.5 kN C S Fx = 0: FGD - FCD = 0 A B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN Locate a joint connecting only two members, and draw the free-body diagram for its pin. Use the free-body diagram to determine the unknown forces in each of the two members. Joint D 12.5 kN 2.5 6.5 + S Fy = 0: FGD - 12.5 kN = 0 FCD FGD = 32.5 kN C 6.5 2.5 6 6.5 6 S Fx = 0: FGD - FCD = 0 + FGD FCD = 30 kN T

S F = 0: FCG = 0 S F = 0: FFG - 32.5 kN = 0 FFG = 32.5 kN C A B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN 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 S F = 0: FCG = 0 32.5 kN S F = 0: FFG - 32.5 kN = 0 FFG = 32.5 kN C FFG

S Fx = 0: 30 kN - FCF cos b - FBC = 0 A B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN Repeat this procedure until the forces in all the members of the truss have been determined. Joint C 2 3 BF = (2.5 m) = 1.6667 m 12.5 kN BF 2 b = BCF = tan-1 = 39.81o FBC FCD = 30 kN b + S Fy = 0: - 12.5 kN - FCF sin b = 0 - 12.5 kN - FCF sin 39.81o = 0 FCF = 19.53 kN C FCF S Fx = 0: 30 kN - FCF cos b - FBC = 0 30 kN - (-19.53) cos 39.81o - FBC = 0 FBC = 45.0 kN T +

S Fx = 0: - FEF - (32.5 kN) - FCF cos b = 0 A B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN FCF = 19.53kN Joint F FBF b=39.81o 6.5 FEF FFG = 32.5 kN 2.5 6 + 6 6.5 6 6.5 S Fx = 0: - FEF - (32.5 kN) - FCF cos b = 0 6.5 6 FEF = -32.5 kN - ( ) (19.53) cos 39.81o FEF = 48.8 kN C 2.5 6.5 2.5 6.5 + S Fy = 0: FBF - FEF - (32.5 kN) - (19.53) sin b = 0 2.5 6.5 FBF - (-48.8 kN) - 12.5 kN - 12.5 kN = 0 FBF = 6.25 kN T

S Fy = 0: -12.5 kN -6.25 kN - FBE sin 51.34o = 0 FBE = -24.0 kN A B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN 12.5 kN Joint B FAB FBC = 45.0 kN g FBF = 6.25kN FBE 2.5 m 2 m tan g = ; 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 Fx = 0: 45.0 kN - FAB + (24.0 kN) cos 51.34o = 0 FAB = 60.0 kN FAB = 60.0 kN T

S Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0 B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN Joint E FBE = 24 kN FAE g FEF = 48.75 kN 6.5 2.5 60 kN 6 g = 51.34o 2.5 6.5 + S Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0 FAE = 37.5 kN FAE = 37.5 kN T

Παράδειγμα - 5 3 kN A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI. 3 kN 3 kN F 3 kN 3 kN D H 6.75 m 1.5 kN 1.5 kN B J A L C E G I K 3 m 3 m 3 m 3 m 3 m 3 m

Solving Problems on Your Own 3 kN Solving Problems on Your Own 3 kN 3 kN F 3 kN A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI. 3 kN D H 6.75 m 1.5 kN 1.5 kN B J A L C E G I K 3 m 3 m 3 m 3 m 3 m 3 m 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.

Solving Problems on Your Own 3 kN Solving Problems on Your Own 3 kN 3 kN F A Pratt roof truss is loaded as shown. Using the method of sections, determine the force in members FH , FI, and GI. 3 kN 3 kN D H 6.75 m 1.5 kN 1.5 kN B J A L C E G I K 3 m 3 m 3 m 3 m 3 m 3 m 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.

S Fx = 0: Ax = 0 Ay = L = (18kN) = 9 kN Draw the free-body diagram of the entire truss, and use it to determine reactions at the supports. F 3 kN 3 kN D H 1.5 kN 1.5 kN B J A L S Fx = 0: Ax = 0 Ax C E G I K + Ay L 3 m 3 m 3 m 3 m 3 m 3 m Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry 1 2 Ay = L = (18kN) = 9 kN

Pass a section through three members of the truss, one of 3 kN 3 kN 3 kN F Pass a section through three members of the truss, one of which is the desired member. 3 kN 3 kN D H 1.5 kN 1.5 kN B J A L C E G I K 9 kN 9 kN 3 m 3 m 3 m 3 m 3 m 3 m

tan a = = a = 66.04o Free Body: Portion HIL 5 3 Free Body: Portion HIL 4 3 kN F FFH Select one of the two portions of the truss you have obtained, and draw its free-body diagram. 3 kN H FFI 6.75 m 1.5 kN J a L Slope of FHJL G I FGI K 6.75 9.00 3 4 5 9 kN 3 = 4 3 m 3 m 3 m FG GI 6.75 3.00 tan a = = a = 66.04o Now write three equilibrium equations which can be solved for the forces in the three intersected members.

S MI = 0: FFH ( x 6.75 m) + (9 kN)(6 m) 3 Free Body: Portion HIL 4 3 kN F FFH Write the three equilibrium equations. 3 kN H FFI 6.75 m 1.5 kN J 66.04o L G I FGI K 9 kN Force in FH: 3 m 3 m 3 m 4 5 2 3 S MI = 0: + FFH ( x 6.75 m) + (9 kN)(6 m) - (1.5 kN)(6m) - (3 kN)(3m) = 0 4 5 FFH (4.5 m) + 36 kN-m = 0 FFH = -10.0 kN FFH = 10.0 kN C

FFH = 10.0 kN C S ML = 0: -FFI sin 66.04o(6 m) + (3 kN)(6 m) 5 3 Free Body: Portion HIL 4 3 kN F FFH Write the three equilibrium equations. 3 kN H FFH = 10.0 kN C FFI 6.75 m 1.5 kN J 66.04o L G I FGI K 9 kN Force in FI: 3 m 3 m 3 m 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

-FGI (6.75 m) - (3 kN)(3 m) - (3 kN)(6 m) Free Body: Portion HIL 4 3 kN F FFH Write the three equilibrium equations. 3 kN H FFH = 10.0 kN C FFI = 4.92 kN T FFI 6.75 m 1.5 kN J 66.04o L G I FGI K 9 kN Force in GI: 3 m 3 m 3 m 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