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# 8. analysis of truss part ii, method of section, by-ghumare s m

Analysis of Truss-Method of Section

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### 8. analysis of truss part ii, method of section, by-ghumare s m

1. 1. Analysis of Truss - Part-II (Method of Section ) Sub- Engg. Mechanics By: Mr. Ghumare S. M. Truss
2. 2. Stability of Truss Depending on no. of members, stability of the truss is checked by using following equation. m = 2j-r r = No. of Reactions = 3 m = 2j-3 Where, m = Number of members, j = Number of Joints 1. m = 2j-3 -----Satisfied Eq. ----Stable Truss 1. m 2j-3 -----Not Satisfied Eq.----Unstable
3. 3. Methods of Section Use: When forces in few members are required, method of section is used. Assumptions in the Analysis of Truss 1. Truss should be perfect truss 2. Self wt. of the truss members are neglected 3. External load acts on the joints only 4. All joints are assumed as hinge or pin joints 5. Members carrying axial loads only, Tension or Compression in nature. Sign Convention: Assume Tensile force +ve (T) and Compression force – ve (C)
4. 4. Analysis of Truss using Method of Section Steps: 1. Draw F.B.D. of the given Truss, check stability if reqd. 2. Apply conditions of equilibrium to entire truss and find external reactions at supports 3. Take section or pass section through the members where forces to be determined. 4. At a time, section should not pass through more than three members. 5. Consider any one side of section i.e. L.H.S. or R.H.S. for analysis. 6. Apply conditions of Equilibrium to selected section. 7. Find the forces in the corresponding members.
5. 5. Example 1. Determine the forces in the members BC, HC and HG for the truss shown in Fig. State weather the forces are in tension or compression. To check stability of the truss: m = 2j-r , r = 3, m =13 , j =8 13 = 2x 8-3 , 13=16-3 , 13=13 Given truss is stable
6. 6. Draw F.B.D. of the given Truss Solution: Draw F.B.D. Find angles E  -1 H H θ = tan (3/2) θ = 56.30 Apply conditions of equim. and find reactions at supports 0 A = 0,xxF  0 Y E Y E A + R -6 - 10 - 6 = 0, A + R = 22 KN = 0, -(1) yF   E A E E E -6 x 3 - 10 x 5 - 6 x 7 + R x10 =0, -18 - 50 - 42 + 10 x R =0 10 R = 110, R = 11 M =0 Anticlockwise KN( ), A = 11 N -V ( e K )y  
7. 7. L. H. S. R. H. S.
8. 8. Consider L.H.S. Assume all are tensile Apply conditions of equim. 01. Y HC HC HC A - 6 - F sin56.3= 0, 11- 6 - F sin56.3 = 0, F = 6 KN (T) yF  2. , 3 @HG HC BC BC C H B F x 3 - 11 x = 0, F x 3 = Mome 33, F nt of F ,F 6 H = 11 = KN M = 0 0 (T) 03. HG BC HC HG HG HG F + F + F cos56.3 = 0, F + 11 + 6 x 0.554 = 0, F = -14.32 KN F = 14.32 KN (C) xF  HC BC HGF = 6 KN (T), F = 11 KN (T) F = 14.32 KNAns: (C)
9. 9. Example 2. Determine the forces in the members GF and GB for the truss shown in Fig. State nature of the forces. To check stability of the truss: m = 2j-r , r = 3, m =11 , j =7 11 = 2x 7-3 , 11=14-3 , 11=11 Given truss is stable
10. 10. Draw F.B.D. of the given Truss and find support reactions Apply conditions of equillibrium. 0 A = 0,xxF  0 Y D Y D A + R -6 - 8 = 0, A + R = 14 KN = 0, (1) yF    D A D D D -6 x 3.05 - 8 x 5.49 + R x 8.54 =0, -62.22 + 8.54 R = 0 8.54 R = 62.22, R = 7.285 KN( M =0 Anticlockwise ), A = 6.715 -V ( e KNy  
11. 11. Take section 1-1 Consider L.H.S.
12. 12. Consider L.H.S. Assume all are tensile Apply conditions of equim. 01. Y GB GB GB A - F = 0, 6.715 - F = 0, F = 6.715 KN (T) yF  2. AB G AB AB GB GF F x 3.05 - Ay x 3.05 = 0, F x 3.05 - 6.715 x 3.05 = 0 Moment of F , , F = 6 F @ .71 G = 5 KN M =0 0 (T)  6.715 . 03 AB GF x GF GF HG F + F + A = 0, + F + 0 = 0, F = -6.715 KN F = 6.715 KN(C) xF  GB AB HGF = 6.715 KN (T), F = 6.An 715 KN (T) F = 6.715 KNs: (C)
13. 13. Example 3. Determine the forces in the members DF, DG and EG for the truss shown in Fig. State nature of the forces. To check stability of the truss: m = 2j-r , r = 3, m =21 , j =12 21 = 2x 12-3 , 21= 24-3 , 21=21 Given truss is stable
14. 14. Draw F.B.D and Find support reactions y L y L y L Given loa A + R -2- d is Sy 2-2-2-2 =0 A + mmetric R = 10--------(1) A = 5KN( ), R = a 5 KN ) 0, ( l =yF    0 0 ,x x A F  
15. 15. To find forces in members pass section through DF, DG and EG Considering the L.H.S to find the forces in the members DF, DG and EG -1 D 0 D θ = tan (3/4) θ = 36.86
16. 16. To find forces in members DF, DG and EG DG DG DG DG DG 5 - 2 - 2 - F sin36.86 =0, 1 - F sin36.86 =0, 1 - F sin36.86 =0 F = 1 F =0, / = 1.67KN(T sin36.86, ) yF EG EG EG EG EG D -A x 7 + 2 x 4 + F x 3 =0, -5 x 7 + 2 x 4 + F x 3 =0, -35 + 8 + F x 3=0, -27 + F x 3=0 F = 9K M =0 ) , N (T y 
17. 17. To find force in member DF X EG DF DG DF DF D F F F D D A + F + F + F cos36.86 = 0, 0 + 9 + F +1.67cos36.86 = 0, 9 + F + 1.67cos36.86 = 0, F + 1.67cos36.86 = -9, F = -10.33 F = 10.3 =0, 3 K KN N(C) XF
18. 18. Thank You