analisa struktur indo

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analisa struktur indo

  1. 1. II MadeMade GatotGatot KarohikaKarohika, ST. MT., ST. MT. Mechanical EngineeringMechanical Engineering UdayanaUdayana UniversityUniversity
  2. 2. Contents Introduction Definition of a Truss Simple Trusses Analysis of Trusses by the Method of Joints Joints Under Special Loading Trusses Made of Several Simple Trusses Sample Problem 6.3 Analysis of Frames Frames Which Cease to be Rigid When Detached From Their 6 - 2 Joints Under Special Loading Conditions Space Trusses Sample Problem 6.1 Analysis of Trusses by the Method of Sections When Detached From Their Supports Sample Problem 6.4 Machines
  3. 3. Introduction • Untuk keseimbangan struktur yang terdiri dari beberapa bagian batang yang bersambungan, maka gaya dalam (internal forces) seperti halnya gaya luar harus diperhitungkan juga. • Dalam interaksi antara bagian yang terhubung, Newton’s 3rd Law menyatakan bahwa gaya aksi dan reaksi antara benda dalam keadaan kontak mempunyai besar yang sama, garis aksi yang sama, dan berlawanan arah. • Tiga kategori dari struktur teknik: a) Portal (Frames): terdiri dari paling kurang satu batang 6 - 3 a) Portal (Frames): terdiri dari paling kurang satu batang dengan pelbagai gaya (multi-force member), i.e., batang yang mengalami tiga atau lebih gaya yang umumnya tidak tidak searah sumbu batang. b) Rangka batang(Trusses): dibentuk dari batang dengan dua gaya [two-force members], i.e., batang lurus dengan ujung-ujung berhubungan, batang mengalami dua gaya sama besar dan berlawanan yang searah dengan sumbu batang c) Machines: struktur yang terdiri dari bagian-bagian yang bergerak dirancang untuk menyalurkan dan mnengubah gaya-gaya.
  4. 4. 2 - 4
  5. 5. Definisi Truss [ rangka batang ] • Truss terdiri dari bagian berbentuk lurus dan terhubung pada sambungan. Tidak ada bagian yang menembus sambungan. • Sebagaian besar struktur dibentuk dari beberapa truss yang dihubungkan bersama membentuk kerangka ruang (space framework). Masing-masing truss menumpu beban yang beraksi pada bidangnya sehingga dapat diperlakukan sebagai struktur. 6 - 5 • Sambungan Baut atau las diasumsikan disambung dengan memakai pin (pasak). Gaya yang beraksi pada ujung- ujung bagian tereduksi menjadi gaya tunggal dan tidak ada kopel. Hanya dipandang sebagai bagian dua gaya (two-force members). diperlakukan sebagai struktur. • Ketika gaya cenderung untuk menarik bagian batang, dikatakan mengalami tegang (tension) [T]. Ketika gaya cenderung untuk menekan batang, batang dalam keadaan tekan (compression) [C].
  6. 6. Definition of a Truss 6 - 6 Bagian-bagian truss berbentuk batang (slender) dan hanya bisa mendukung beban ringan dalam arah lateral. Beban harus diterapkan pada sambungannya bukan langsung pada bagian-bagiannya.
  7. 7. Definition of a Truss 6 - 7
  8. 8. 2 - 8
  9. 9. Simple Trusses/ Rangka batang sederhana • Truss tegar (rigid truss) tidak akan ambruk collapse karena aplikasi beban. • A simple truss disusun dengan menambahkan berturutan dua bagian dan satu sambungan pada segitiga dasar (basic) truss. 6 - 9 dasar (basic) truss. • In a simple truss, m = 2n - 3 dimana m adalah jumlah total members and n adalahjumlah sambungan.
  10. 10. Analysis Truss dengan Methode Sambungan • Uraikan truss dan buat freebody diagram untuk masing-masing bagian dan pin. • Dua gaya yang ada pada masing-masing bagian (satu gaya pada masing-masing ujung) adalah besarnya sama,garis aksi yang sama, dan berlawanan arah. • Gaya yang ditimbulkan oleh suatu bagian pada pin atau sambungan pada ujungnya harus berarah 6 - 10 atau sambungan pada ujungnya harus berarah sepanjang bagian itu dan harus sama dan berlawanan. • Kondisi keseimbangan pada pins memberikan 2n persamaan untuk 2n besaran yang tidak diketahui. untuk simple truss, 2n = m + 3. dapat dipecahkan untuk m member forces and 3 reaction forces pada tumpuan. • Kondisi keseimbangan untuk keseluruhan truss memberikan 3 persamaan tambahan yang tidak bebas dengan persamaan pin.
  11. 11. Joints Under Special Loading Conditions • Gaya pada bagian yang berlawanan berpotongan pada dua garis lurus pada sambungan harus sama • Gaya pada dua bagian yang berlawanan harus sama. Jika beban P dihubungkan dengan bagian ketiga, gaya pada bagian ketiga harus sama dengan dengan beban P ( termasuk jika beban nol). • Gaya pada dua bagian terhubung pada sambungan adalah sama jika bagian itu segaris 6 - 11 sambungan adalah sama jika bagian itu segaris (aligned) dan nol begitu sebaliknya. • Dengan mengetahui sambungan dalam kondisi pembebanan khusus akan menyederhanakan analisa truss .
  12. 12. Space Trusses • An elementary space truss consists of 6 members connected at 4 joints to form a tetrahedron. • A simple space truss is formed and can be extended when 3 new members and 1 joint are added at the same time. • In a simple space truss, m = 3n - 6 where m is the 6 - 12 • Equilibrium for the entire truss provides 6 additional equations which are not independent of the joint equations. • In a simple space truss, m = 3n - 6 where m is the number of members and n is the number of joints. • Conditions of equilibrium for the joints provide 3n equations. For a simple truss, 3n = m + 6 and the equations can be solved for m member forces and 6 support reactions.
  13. 13. 2 - 13
  14. 14. Sample Problem 6.1 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. 6 - 14 Using the method of joints, determine the force in each member of the truss. joint equilibrium requirements. • 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.
  15. 15. Sample Problem 6.1 SOLUTION: • Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. ( )( ) ( )( ) ( )ft6ft12lb1000ft24lb2000 0 E MC −+= =∑ 6 - 15 ↑= lb000,10E ∑ == xx CF 0 0=xC ∑ ++−== yy CF lb10,000lb1000-lb20000 ↓= lb7000yC
  16. 16. Sample Problem 6.1 • Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements. 6 - 16 joint equilibrium requirements. 534 lb2000 ADAB FF == CF TF AD AB lb2500 lb1500 = = • There are now only two unknown member forces at joint D. ( ) DADE DADB FF FF 5 32= = CF TF DE DB lb3000 lb2500 = =
  17. 17. Sample Problem 6.1 • There are now only two unknown member forces at joint B. Assume both are in tension. ( ) lb3750 250010000 5 4 5 4 −= −−−==∑ BE BEy F FF CFBE lb3750= ( ) ( )3750250015000 5 3 5 3 −−−==∑ BCx FF 6 - 17 lb5250 55 +=BCF TFBC lb5250= • There is one unknown member force at joint E. Assume the member is in tension. ( ) lb8750 375030000 5 3 5 3 −= ++==∑ EC ECx F FF CFEC lb8750=
  18. 18. Sample Problem 6.1 • All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results. ( ) ( ) ( ) ( )checks087507000 checks087505250 5 4 5 3 =+−= =+−= ∑ ∑ y x F F 6 - 18
  19. 19. 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. 6 - 19 • With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including FBD.
  20. 20. Trusses Made of Several Simple Trusses • Compound trusses are statically determinant, rigid, and completely constrained. 32 −= nm • Truss contains a redundant member and is statically indeterminate. 32 −> nm 6 - 20 32 −> nm • Necessary but insufficient condition for a compound truss to be statically determinant, rigid, and completely constrained, nrm 2=+ non-rigid rigid 32 −< nm • Additional reaction forces may be necessary for a rigid truss. 42 −< nm
  21. 21. Sample Problem 6.3 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. 6 - 21 Determine the force in members FH, GH, and GI. section as a free body. • Apply the conditions for static equilibrium to determine the desired member forces.
  22. 22. Sample Problem 6.3 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. 6 - 22 ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ↑= ++−== ↑= +−− −−−== ∑ ∑ kN5.12 kN200 kN5.7 m25kN1m25kN1m20 kN6m15kN6m10kN6m50 A ALF L L M y A
  23. 23. Sample Problem 6.3 • 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 6 - 23 ( )( ) ( )( ) ( ) kN13.13 0m33.5m5kN1m10kN7.50 0 += =−− =∑ GI GI H F F M • Apply the conditions for static equilibrium to determine the desired member forces. TFGI kN13.13=
  24. 24. Sample Problem 6.3 ( )( ) ( )( ) ( )( ) ( )( ) kN82.13 0m8cos m5kN1m10kN1m15kN7.5 0 07.285333.0 m15 m8 tan −= =+ −− = °==== ∑ FH FH G F F M GL FG α αα CFFH kN82.13= 6 - 24 ( ) ( )( ) ( )( ) ( )( ) kN371.1 0m10cosm5kN1m10kN1 0 15.439375.0 m8 m5 tan 3 2 −= =++ = °==== ∑ GH GH L F F M HI GI β ββ CFGH kN371.1=
  25. 25. Analysis of Frames • Frames and machines are structures with at least one multiforce member. Frames are designed to support loads and are usually stationary. Machines contain moving parts and are designed to transmit and modify forces. • A free body diagram of the complete frame is used to determine the external forces acting on the frame. • Internal forces are determined by dismembering the frame 6 - 25 • Internal forces are determined by dismembering the frame and creating free-body diagrams for each component. • Forces between connected components are equal, have the same line of action, and opposite sense. • Forces on two force members have known lines of action but unknown magnitude and sense. • Forces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components.
  26. 26. Frames Which Cease To Be Rigid When Detached From Their Supports • Some frames may collapse if removed from their supports. Such frames can not be treated as rigid bodies. • A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium 6 - 26 can not be determined from the three equilibrium conditions. • The frame must be considered as two distinct, but related, rigid bodies. • With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components. • Equilibrium requirements for the two rigid bodies yield 6 independent equations.
  27. 27. Sample Problem 6.4 SOLUTION: • Create a free-body diagram for the complete frame and solve for the support reactions. • Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown 6 - 27 Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force in link DE and the components of the force exerted at C on member BCD. has a known line of action but unknown magnitude. It is determined by summing moments about C. • With the force on the link DE known, the sum of forces in the x and y directions may be used to find the force components at C. • With member ACE as a free-body, check the solution by summing moments about A.
  28. 28. Sample Problem 6.4 SOLUTION: • Create a free-body diagram for the complete frame and solve for the support reactions. N4800 −==∑ yy AF ↑= N480yA ( )( ) ( )mm160mm100N4800 BM A +−==∑ 6 - 28 →= N300B xx ABF +==∑ 0 ←= −= N300Ax NAx 300 °== − 07.28tan 150 801 α Note:
  29. 29. Sample Problem 6.4 • Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C. ( )( ) ( )( ) ( )( ) NF mmN480mm0N300mmFM DE DEC 561 1008250sin0 −= ++==∑ α CFDE N561= 6 - 29 NFDE 561 CFDE N561= • Sum of forces in the x and y directions may be used to find the force components at C. ( ) N300cosN5610 N300cos0 +−−= +−==∑ α α x DExx C FCF N795−=xC ( ) N480sinN5610 N480sin0 −−−= −−==∑ α α y DEyy C FCF N216=yC
  30. 30. Sample Problem 6.4 • With member ACE as a free-body, check the solution by summing moments about A. 6 - 30 ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( )( ) 0mm220795mm100sin561mm300cos561 mm220mm100sinmm300cos =−−−+−= −+=∑ αα αα xDEDEA CFFM (checks)
  31. 31. Machines • Machines are structures designed to transmit and modify forces. Their main purpose is to transform input forces into output forces. • Given the magnitude of P, determine the magnitude of Q. • Create a free-body diagram of the complete machine, including the reaction that the wire 6 - 31 machine, including the reaction that the wire exerts. • The machine is a nonrigid structure. Use one of the components as a free-body. • Taking moments about A, P b a QbQaPM A =−==∑ 0

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