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 lb24 ft 1000 lb12 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