![10
𝐾11 = 𝐾𝑏𝑐 =
2𝐸𝐼
𝐿
[2𝜃 𝑏 + 𝜃𝑐] =
2𝐸(2𝐼)
8
2 × 1 + 0 =EI
𝐾21 = 𝐾𝑐𝑏 =
2𝐸𝐼
𝐿
[2𝜃𝑐 + 𝜃 𝑏] =
2𝐸(2𝐼)
8
2 × 0 + 1 = 0.5 EI
𝐾31 = 0](https://image.slidesharecdn.com/stiffnessmethodproblem-200427072841/85/ANALYSIS-OF-CONTINUOUS-BEAM-USING-STIFFNESS-METHOD-10-320.jpg)
![12
𝐾12 = 𝐾𝑏𝑐 =
2𝐸𝐼
𝐿
[2𝜃 𝑏 + 𝜃𝑐] =
2𝐸(2𝐼)
8
2 × 0 + 1 = 0.5EI
3𝐾22 = 𝐾𝑐𝑏 + 𝐾𝑐𝑑
=
2𝐸𝐼
𝐿 𝑐𝑏
[2𝜃𝑐 + 𝜃 𝑏] +
2𝐸𝐼
𝐿 𝑐𝑑
2𝜃𝑐 + 𝜃 𝑑
0 =
2𝐸(2𝐼)
8
2 × 1 + 0 +
2𝐸(1.5𝐼)
6
2 × 1 + 0 = 2𝐸𝐼
𝐾32 = 𝐾 𝑑𝑐 =
2𝐸𝐼
𝐿 𝑏𝑐
[2𝜃 𝑑 + 𝜃𝑐] =
2𝐸(1.5𝐼)
6
2 × 0 + 1 = 0.5EI](https://image.slidesharecdn.com/stiffnessmethodproblem-200427072841/85/ANALYSIS-OF-CONTINUOUS-BEAM-USING-STIFFNESS-METHOD-12-320.jpg)
![14
0𝐾13 = 𝐾𝑏𝑐 = 0
3𝐾23 = 𝐾𝑐𝑑
=
2𝐸𝐼
𝐿 𝑏𝑐
2𝜃𝑐 + 𝜃 𝑑
0 =
2𝐸(1.5𝐼)
6
2 × 0 + 1 = 0.5𝐸𝐼
𝐾33 = 𝐾 𝑑𝑐 =
2𝐸𝐼
𝐿 𝑏𝑐
[2𝜃 𝑑 + 𝜃𝑐] =
2𝐸(1.5𝐼)
6
2 × 1 + 0 = EI](https://image.slidesharecdn.com/stiffnessmethodproblem-200427072841/85/ANALYSIS-OF-CONTINUOUS-BEAM-USING-STIFFNESS-METHOD-14-320.jpg)
ANALYSIS OF CONTINUOUS BEAM USING STIFFNESS METHOD
- 1. Stiffness Method Problem Dr. Kasi Rekha B.Tech., M.Tech., Ph.D., MIE 1 Presented by
- 2. • Analyze the continuous beam shown in figure using Stiffness method. 2 40 kN 60 kN 80 kN A B C D 2m 4m 2m4m 4m 2 I 1.5 I 60 kN 80 kN B C D 4m 2m4m 4m 2 I 1.5 I 80 kN-m
- 3. • Step1: – Degree of Freedom or Kinematic Indeterminacy : 3 3 B C D 4m 2m4m 4m 2 I 1.5 I θb θc θd
- 4. 4 • Step 2: – Assign Coordinate Numbers in the direction of degrees of Freedom B C D 4m 2m4m 4m 2 I 1.5 I 1 2 3
- 5. 5 • Step 3: – Restrain the structure at all coordinates B C D 4m 2m4m 4m 2 I 1.5 I 1 2 3
- 6. 6 • Step 4: – Determination of Fixed End Moments 60 kN 80 kN B C D 4m 2m4m 4m 2 I 1.5 I
- 7. • 𝑀 𝐹 𝑏𝑐 = − 𝑤𝑎𝑏2 𝑙2 = − 60×4×42 82 = −60 𝑘𝑁 − 𝑚 • 𝑀 𝐹 𝑐𝑏 = 𝑤𝑎2 𝑏 𝑙2 = 60×4×42 82 = 60 𝑘𝑁 − 𝑚 • 𝑀 𝐹 𝑐𝑑 = − 𝑤𝑎𝑏2 𝑙2 = − 80×4×22 62 = −35.55 𝑘𝑁 − 𝑚 • 𝑀 𝐹 𝑑𝑐 = 𝑤𝑎2 𝑏 𝑙2 = 80×22×4 62 = 71.11𝑘𝑁 − 𝑚 7
- 8. • Step 5: – Calculation of forces in co ordinate numbers 𝑃1𝑙 = 𝑀 𝐹 𝑏𝑐 = - 60 kN-m 𝑃2𝑙 = 𝑀 𝐹 𝑐𝑏 + 𝑀 𝐹 𝑐𝑑 = 60-35.55= 24.45 kN-m 𝑃3𝑙 = 𝑀 𝐹 𝑑𝑐 = 71.11 kN-m 8 B C D 4m 2m4m 4m 2 I 1.5 I 1 2 3
- 9. • Step 6: – Assembly of Stiffness Matrix i. Applying Unit Rotation in co ordinate direction 1 9 B C D 4m 2m4m 4m 2 I 1.5 I 1 2 3 θb = 1 θc = 0 θd = 0
- 10. 10 𝐾11 = 𝐾𝑏𝑐 = 2𝐸𝐼 𝐿 [2𝜃 𝑏 + 𝜃𝑐] = 2𝐸(2𝐼) 8 2 × 1 + 0 =EI 𝐾21 = 𝐾𝑐𝑏 = 2𝐸𝐼 𝐿 [2𝜃𝑐 + 𝜃 𝑏] = 2𝐸(2𝐼) 8 2 × 0 + 1 = 0.5 EI 𝐾31 = 0
- 11. 11 • Step 6: – Assembly of Stiffness Matrix ii. Applying Unit Rotation in co ordinate direction 2 8m 6m θb = 0 1 2 3 2 I 1.5 I θc = 1 θd = 0
- 12. 12 𝐾12 = 𝐾𝑏𝑐 = 2𝐸𝐼 𝐿 [2𝜃 𝑏 + 𝜃𝑐] = 2𝐸(2𝐼) 8 2 × 0 + 1 = 0.5EI 3𝐾22 = 𝐾𝑐𝑏 + 𝐾𝑐𝑑 = 2𝐸𝐼 𝐿 𝑐𝑏 [2𝜃𝑐 + 𝜃 𝑏] + 2𝐸𝐼 𝐿 𝑐𝑑 2𝜃𝑐 + 𝜃 𝑑 0 = 2𝐸(2𝐼) 8 2 × 1 + 0 + 2𝐸(1.5𝐼) 6 2 × 1 + 0 = 2𝐸𝐼 𝐾32 = 𝐾 𝑑𝑐 = 2𝐸𝐼 𝐿 𝑏𝑐 [2𝜃 𝑑 + 𝜃𝑐] = 2𝐸(1.5𝐼) 6 2 × 0 + 1 = 0.5EI
- 13. 13 • Step 6: – Assembly of Stiffness Matrix iii. Applying Unit Rotation in co ordinate direction 3 2 I 1.5 I 8m 6m 1 2 3 θb = 0 θc = 0 θd = 1
- 14. 14 0𝐾13 = 𝐾𝑏𝑐 = 0 3𝐾23 = 𝐾𝑐𝑑 = 2𝐸𝐼 𝐿 𝑏𝑐 2𝜃𝑐 + 𝜃 𝑑 0 = 2𝐸(1.5𝐼) 6 2 × 0 + 1 = 0.5𝐸𝐼 𝐾33 = 𝐾 𝑑𝑐 = 2𝐸𝐼 𝐿 𝑏𝑐 [2𝜃 𝑑 + 𝜃𝑐] = 2𝐸(1.5𝐼) 6 2 × 1 + 0 = EI
- 15. • GLOBAL STIFFNESS MATRIX 𝐾 = 𝐾11 𝐾12 𝐾13 𝐾21 𝐾22 𝐾23 𝐾31 𝐾32 𝐾33 = EI 1 0.5 0 0.5 2 0.5 0 0.5 1 15
- 16. • STEP7: – Calculation of Actual Forces 𝑃 − 𝑃𝐿 = 𝑃1 − 𝑃1𝐿 𝑃2 − 𝑃2𝐿 𝑃3 − 𝑃3𝐿 = −80 − (−60) 0 − 24.45 0 − 71.11 = −20 −24.45 −71.11 16
- 17. 17 • GLOBAL STIFFNESS MATRIX K Δ = P-𝑃𝐿 EI 1 0.5 0 0.5 2 0.5 0 0.5 1 Δ1 Δ2 Δ3 = −20 −24.45 −71.11 Δ1 + 0.5 Δ2 = -20 0.5 Δ1 + 2 Δ2 + 0.5 Δ3 = -24.45 0.5Δ2 + Δ3 = -71.11 − 27.035 𝐸𝐼 Δ1 = 𝜃 𝑏 = 14.07 𝐸𝐼 Δ2 = 𝜃𝑐 = − 78.145 𝐸𝐼 Δ3 = 𝜃 𝑑 =
- 18. • Step 8: – Final Moments 𝑀 𝑎𝑏 = ̶ 80 kN-m 𝑀 𝑏𝑐 = 𝑀 𝐹 𝑏𝑐 + 2𝐸𝐼 𝐿 𝑏𝑐 2𝜃 𝑏 + 𝜃𝑐 − 3𝐸𝐼∆ 𝐿 𝑏𝑐 = -60+ 2𝐸(2𝐼) 8 ( 2×(−27.035) 𝐸𝐼 + (14.07) 𝐸𝐼 ) = -80 kN-m 18
- 19. 𝑀𝑐𝑏 = 𝑀 𝐹 𝑐𝑏 + 2𝐸𝐼 𝐿 𝑏𝑐 2𝜃𝑐 + 𝜃 𝑏 − 3𝐸𝐼∆ 𝐿 𝑏𝑐 = 60+ 2𝐸(2𝐼) 8 ( 2×(14.07) 𝐸𝐼 + (−27.035 𝐸𝐼 ) = 60.55kN-m 𝑀𝑐𝑑 = 𝑀 𝐹 𝑐𝑑 + 2𝐸𝐼 𝐿 𝑐𝑑 2𝜃𝑐 + 𝜃 𝑑 − 3𝐸𝐼∆ 𝐿 𝑐𝑑 = -35.55+ 2𝐸(1.5𝐼) 6 ( 2×(14.07) 𝐸𝐼 + (−78.145) 𝐸𝐼 ) = -60.55 kN-m 𝑀 𝑑𝑐 = 𝑀 𝐹 𝑑𝑐 + 2𝐸𝐼 𝐿 𝑐𝑑 2𝜃 𝑑 + 𝜃𝑐 − 3𝐸𝐼∆ 𝐿 𝑐𝑑 = 71.11+ 2𝐸(1.5𝐼) 6 ( 2×(−78.145) 𝐸𝐼 + (14.07) 𝐸𝐼 ) = 0 19
- 20. 20 40 kN 60 kN 80 kN A B C D 2m 4m 2m4m 4m 2 I 1.5 I 80 kN-m 60.55 kN-m 120 kN-m 106.66kN-m
- 21. • Assessment question: • Analyze the continuous beam shown in figure using stiffness matrix method. 21 70kN A B C 6m3m 5m 80kN/ m 100kN 2m D 2 I 1.5 I I
- 22. THANK YOU 22