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Week6_qqqTruss forces calculation (1).pdf
- 1. sibt.nsw.edu.au navitas.com ENGN150 Fundamentals ofEngineering Studies Truss forces calculation methods Lecturer Dr saad odeh, Program Convener – Science and Engineering
- 2. sibt.nsw.edu.au navitas.com Lecture outcome: Learnhow to calculate truss members forces using joint method. Learn how to calculate truss members forces using section method.
- 3. sibt.nsw.edu.au navitas.com Calculation oftruss forces using joint method Example: Analyse the internal forces in the members of the truss. E D C A 2 kN ƩMc = 0 + Ay (6) - 2 (3) = 0 Ay = 1 kN The assumed direction of Ay is right!! Ay Cx Cy ƩFy = 0 Ay + Cy = 0 1 + Cy = 0 Cy = -1 kN The assumed direction of Cy is wrong! ƩFx = 0 -2 - Cx = 0 Cx = - 2 kN The assumed direction of Cx is wrong! First draw FBD for the complete system. Second Apply ƩM = 0 at the point of highest number unknown forces to find the forces at system boundary.
- 4. Ay = 1kN Cy = 1 kN Cx = 2 kN ƩFy = 0 Fcd - 1 = 0 Fcd = 1 kN The assumed direction of Fcd is right! ƩFx = 0 - Fcb + 2 = 0 Fcb = 2 kN The assumed direction of Fcb is right! Fcd 2 kN Consider joint C Draw FBD for this joint Fcb 1 kN Start at a boundary joint with minimum number of unknown forces. Apply ƩFy = 0, ƩFx = 0 at that point
- 5. Move to internaljoints Ay = 1 kN Cy = 1 kN Cx = 2 kN ƩFy = 0 - Fdb cos 45 - 1 = 0 Fdb = - 1.41 kN The assumed direction of Fdb is wrong. Fdb = 1.41 kN (C ) ƩFx = 0 - Fed - 2 – Fdb cos 45 = 0 - Fed - 2 - (-1.41 cos 45) = 0 Fed = -1 kN The assumed direction of Fed is wrong! Fed = 1 kN (C ) Fed 2 kN 45o Consider joint D draw FBD for this joint Fdb 1 kN Consider joint E Draw FBD for this joint 1 kN 45o Fea Feb ƩFx = 0 - Fea cos 45 - 1 = 0 Fea = -1.414 kN The assumed direction of Fea is wrong. Fea = 1.414 kN (C) ƩFy = 0 - Fea cos 45 – Feb = 0 - (-1.414) cos 45 - Feb = 0 Feb = 1 kN The assumed direction of Feb is right! Feb = 1 kN (T )
- 6. Ay = 1kN Cy = 1 kN Cx = 2 kN ƩFx = 0 Fab - 1.414cos 45 = 0 Fab = 1 kN The assumed direction of Fab is right! Fab = 1 kN (T ) Summary: Fcd = 1 kN (T), Fcb = 2kN (T), Fdb = 1.41 kN (C), Fed = 1 kN (C ), Fea = 1.41 kN (C), Feb = 1 kN (T ), Fab = 1 kN (T ). Fab 45o Consider joint A Draw FBD for this joint 1 kN 1.414 kN
- 7. Example: Method ofjoints – Bridge model Analyse the internal forces in the members of the truss. All members have the same length. If all members has the same length then we have equal angles (60° ) triangles . 8 kN 4 kN 2 kN Ax Ay Cx 60°
- 8. 8 kN 4 kN2 kN Ax Ay Cx Let the length of the member be x ƩMa = 0 8x + 4(x/2) + 2(3x/2) - Cx cos 60 (2x) = 0 8x + 2x +3x – Cx (x) = 0 13x = Cx (x) Cx = 13 kN ƩFy = 0 -Ay – 4 – 2 – 8 + Cx cos 60 = 0 -Ay – 14 + 13 cos 60 = 0 Ay = - 7.5 kN ƩFx = 0 Ax - Cx cos 30 = 0 Ax - 13 cos 30 = 0 Ax = 11.26 kN Consider joint A 7.5 kN Fab 11.26 kN Fae 60o ƩFy = 0 7.5 – Fae cos30 = 0 Fae = 8.66kN The assumed direction of Fae is right! Fae = 8.66 kN (T ) ƩFx = 0 Fab + 11.26 + Fae cos 60 = 0 Fab + 11.26 +8.66 cos 60 = 0 Fab = -15.59 kN The assumed direction of Fab is wrong! Fab = 15.59 (C)
- 9. ƩFy = 0 8.66cos 30 + Feb cos30 - 4 = 0 7.5 + Feb (0.866) - 4 = 0 Feb = - 4.04 kN The assumed direction of Feb is wrong! Feb = 4.04 kN (C ) Feb Fed Consider joint E 8.66 kN 4kN 60o 60o ƩFx = 0 -8.66 cos 60 - Feb cos60 + Fed = 0 - 4.33 - 4.04 (0.5) + Fed = 0 Fed = 6.35 kN The assumed direction of Fed is right! Fed = 6.35 kN (T ) 15.59 kN Consider joint B 4.04 kN Fbd 60o 60o 60o 8kN ƩFy = 0 4.04 cos30 - Fbd cos30 - 8 = 0 Fbd cos30 = - 4.5 kN Fbd = - 5.2 kN The assumed direction of Fbd is wrong! Fbd = 5.2 kN (C ) ƩFx = 0 15.59 + 4.04 cos60 - Fbd cos 60 + Fbc = 0 15.59 + 2.02 – 5.2 (0.5) + Fbc = 0 Fbc = -15.01 kN The assumed direction of Fbc is wrong! Fbc = 15.01 kN (C )
- 10. ƩFx = 0 15.01cos 60 – Fcd = 0 Fcd = 7.5 kN The assumed direction of Fcd is right! Fcd = 7.5 (T ) 15.01 kN 13 Consider joint C Fcd 60o Summary: Fae = 8.66 kN (T), Fab = 15.59 kN (C), Feb = 4.04 kN (C ), Fed = 6.35 kN (T ), Fbd = 5.2 kN (C), Fbc = 15.01 kN (C ), Fcd = 7.5 kN (T ).
- 11. Example: Method ofjoints Analyse the internal forces in the members of the truss. 3 kN Ax Ay Ex Consider joint C Fbc 3kN 30o ƩFx = 0 -Fde cos 30 – 6 cos 30 = 0 Fde = - 6 kN The assumed direction of Fde is wrong! Fde = 6 kN (C ) ƩFy = 0 - Fde cos 60 – Fcd cos 60 + Fbd = 0 - 6 cos 60 – 6 cos 60 + Fbd = 0 Fbd = 6 kN The assumed direction of Fbd is right! Fbd = 6 kN (T ) Fbd Fde Fcd = 6 kN (C ) 60o ƩFx = 0 -Fde cos 30 – 6 cos 30 = 0 Fde = - 6 kN The assumed direction of Fde is wrong! Fde = 6 kN (C ) Consider joint D ƩFy = 0 - Fde cos 60 – Fcd cos 60 + Fbd = 0 - 6 cos 60 – 6 cos 60 + Fbd = 0 Fbd = 6 kN The assumed direction of Fbd is right! Fbd = 6 kN (T )
- 12. ƩFy = 0 Fbacos 60 – Fbd = 0 Fba (0.5) – 6 = 0 Fba = 12kN The assumed direction of Fba is right! Fba = 12 kN (T ) Fbe Fbd = 6 kN (T) Consider joint B ƩFx = 0 - Fba cos 30 – Fbe + Fbc = 0 - 12 cos 30 – Fbe + 5.2 = 0 Fbe = - 5.19 kN The assumed direction of Fbe is wrong! Fbe = 5.19 kN (C ) Fbc = 5.2 kN (T ) 30o Fba Consider joint E ƩFy = 0 Fae + Fed cos 60 = 0 Fae + 6 (0.5) = 0 Fae = - 3kN The assumed direction of Fae is wrong! Fae = 3 kN (C ) Summary: Fcd = 6 kN (C ), Fbc = 5.2 kN (T), Fde = 6 kN (C), Fbd = 6 kN (T ), Fba = 12 kN (T), Fbe = 5.19 kN (C ), Fae = 3 kN (C ). Fae Feb = 5.2 kN (C ) 30o Fed = 6 kN (C ) Ex
- 13. Trusses Zero force members Isit possible to have a zero-force member in a truss? How to identify zero-force members? 2 conditions 1. If two members form a truss joint and NO external load or support reaction is applied to the joint, the two members must be zero force members. 2. If three members form a truss joint for which two of the members are collinear, the third member is a zero-force member provided NO external force or support reaction is applied to the joint. Why do we need to design zero-force members? 1. To increase stability of truss during construction and to provide added support if the loading is changed. 2. Truss analysis using the method of joints is greatly simplified if we can first identify those members which support no loading.
- 14. Trusses Method of section Methodof section is based on the principle that if truss is in equilibrium, then any segments of the truss is also in equilibrium. The method of section is used when we need to find the force in only a few members of a complex truss seystem. Consider the following truss system, what will happen if it is sectioned into two segments? a. We can find forces in members FE, BE or BC using the left or right side of the sectioned truss. b. Note that the force in FE, BE and BC are the same in both sides of the sectioned truss.
- 15. Method of section Atruss can be sectioned in many different ways. SECTION 1 SECTION 2
- 16. Example :Determine theforce in member BE of the loaded truss by the two methods, method of joints and method of section Ax Ay Gy 4 kN 5 kN 6 kN ƩMa = 0 4 (2) + 5( 4) + 6( 6 ) - Gy (3 ) = 0 64 – Gy (3) = 0 Gy = 21.33 kN ƩFy = 0 Ay + 21.33 = 0 Ay = -21.33 kN ƩFx = 0 Ax – 4 – 5 - 6 = 0 Ax - 15 = 0 Ax = 15 kN
- 17. 15 kN 21.33 kN Gy= 21.33 kN 4 kN Fbc Fbe Fef Fbe Fef Fbc 5 kN 6 kN Bottom section of truss Top section of truss
- 18. Determine the forcein member BE of the loaded truss. Fbe Fef Fbc 5 kN 6 kN ƩMd = 0 - 5(2) + Fbe cos 63.44 (2) + Fbe sin 63.44 (1 ) = 0 - 10 + Fbe (0.894) + Fbe (0.894) = 0 Fbe = 5.59 kN The assumed direction of Fbe is right! Fbe = 5.59 kN (T ) 1m 63.44o
- 19. Calculate the forcein members AB, BG and GF. O 6 kN Y = 12-2.4 = 9.6m 5/x = 2/(x-7.2) 5 (x – 7.2) = 2x 5x – 2x = 36 x = 12m x O Fab Fbg Fgf Can you think of the easiest way to solve the problem?
- 20. Calculate the forcein members AB, BG and GF. 6 kN Y = 12-2.4 = 9.6m O ƩMo = 0 - Fbg (9.6) – 6 (12- 2.4 – 2.4 – 2.4) = 0 -9.6 Fbg – 28.8 = 0 Fbg = - 3 kN The assumed direction of Fbe is wrong! Fbg = 3 kN (C ) Fab Fbg Fgf ƩMf = 0 What is the length of Cf? -Fab (3) + Fbg (2.4) + 6 (2.4) = 0 - 3 Fab + 3(2.4) + 14.4 = 0 Fab = 7.2 kN (T)
- 21. Calculate the forcein members AB, BG and GF. 6 kN Y = 12-2.4 = 9.6m O Fab Fbg Fgf 67.38o ƩFx = 0 - Fab – Fgf cos 22.62 = 0 - 7.2 – Fgf cos 22.62 = 0 Fgf = - 7.8 kN Fgf = 7.8 kN (C)
- 22. Calculate the forcein members JK, CJ and CD. Fkj 10.33 kN Determine the reaction force Ay, Ay = 10.33 kN Fcd Fcj 4 kN 5 kN ƩMc = 0 10.33 (4) + Fkj (3) – 4 (2) = 0 3 Fkj = -33.33 kN Fkj = - 11.11 kN Fkj = 11.11 kN ( C)
- 23. Calculate the forcein members JK, CJ and CD. Fkj 10.33 kN Determine the force Fcj, Fcd Fcj 4 kN 5 kN ƩFx = 0 Fkj + Fcd + Fci cos (56.31) = 0 -11.11 + Fcd -1.6 cos 56.31 = 0 Fcd = 12 kN Fcd = 12 kN ( T) ƩFy = 0 10.33 – 4 – 5 + Fcj cos 33.69 = 0 Fcj = -1.6 kN Fcj = 1.6 kN (C ) Determine the force Fcd,