![9
θc= 0
A
B C
D
θb= 1
𝐾12 = -
𝟔𝑬𝑰Δ
𝑳 𝟐 = -
𝟔𝑬𝑰
𝟑 𝟐 = - 0.667 EI
𝐾22 = 𝐾𝑏𝑎 + 𝐾𝑏𝑐
=
2𝐸𝐼
𝐿 𝑏𝑎
[2𝜃 𝑏 + 𝜃 𝑎] +
2𝐸𝐼
𝐿 𝑏𝑐
2𝜃 𝑏 + 𝜃𝑐
=
2𝐸𝐼
3
2 × 1 + 0 +
2𝐸𝐼
4
2 × 1 + 0 = 2.33EI
𝐾32 = 𝐾𝑐𝑏 =
2𝐸𝐼
𝐿 𝑏𝑐
[2𝜃𝑐 + 𝜃 𝑏]
– =
2𝐸𝐼
4
2 × 0 + 1 = 0.5EI
𝟔𝑬𝑰Δ
𝑳 𝟐](https://image.slidesharecdn.com/analysisofswaytypeportalframe-200511163719/85/Analysis-of-sway-type-portal-frame-using-direct-stiffness-method-9-320.jpg)
![𝐾13 = -
𝟔𝑬𝑰Δ
𝑳 𝟐 = -
𝟔𝑬𝑰
𝟑 𝟐 = - 0.667 EI
𝐾23 = 𝐾𝑏𝑐 =
2𝐸𝐼
𝐿 𝑏𝑐
2𝜃 𝑏 + 𝜃𝑐
– =
2𝐸𝐼
4
2 × 0 + 1 = 0.5EI
𝐾33 = 𝐾𝑐𝑏 +𝐾𝑐𝑑
=
2𝐸𝐼
𝐿 𝑏𝑐
[2𝜃𝑐 + 𝜃 𝑏] +
2𝐸𝐼
𝐿 𝑐𝑑
2𝜃𝑐 + 𝜃 𝑑
=
2𝐸𝐼
4
2 × 1 + 0 +
2𝐸𝐼
3
2 × 1 + 0 = 2.33EI
10
θc= 1
A
B C
D
θb= 0
𝟔𝑬𝑰Δ
𝑳 𝟐](https://image.slidesharecdn.com/analysisofswaytypeportalframe-200511163719/85/Analysis-of-sway-type-portal-frame-using-direct-stiffness-method-10-320.jpg)
Analysis of sway type portal frame using direct stiffness method
- 1. By Dr. K. Rekha B.Tech., M.Tech., Ph.D., MIE ANALYSIS OF SWAY TYPE PORTAL FRAME USING DIRECT STIFFNESS MATRIX METHOD
- 2. • Analyze the portal frame shown in figure using displacement (stiffness) method. A B C D 20 kN/m 4 m 3m 50 kN 3m
- 3. • Step 1: Determination of Degrees of Freedom • The given portal frame is sway type portal frame. • The joints (B& C) are to be considered as internal hinges • The degrees of freedom at each Joint is one, i.e., rotations θb and θc. • The frame also have sway Δ in the direction of Horizontal force. • At supports A & D , the degrees of freedom is equal to zero, since these are fixed supports. • Hence, the total degree of freedom of the given frame is THREE (θb , θc & Δ).
- 4. A B C D 4 m 3m θb θc Δ
- 5. • Step 2: Assign coordinate numbers in the direction of DoF A B C D 2 3 1
- 6. Step 3: Determination of fixed end Moments • 𝑀 𝐹 𝑎𝑏 = 𝑀 𝐹 𝑏𝑎 = 𝑀 𝐹 𝑐𝑑 = 𝑀 𝐹 𝑑𝑐 = 0 • 𝑀 𝐹 𝑏𝑐 = − 𝑤𝑙2 12 = − 20×42 12 = −26.67𝑘𝑁 − 𝑚 • 𝑀 𝐹 𝑐𝑏 = 𝑤𝑙2 12 = 20×42 12 = 26.67𝑘𝑁 − 𝑚 A B C D 20 kN/m 4 m 3m 50 kN
- 7. 7 A B C D 4m 𝑴 𝑭 𝒃𝒄 𝑴 𝑭 𝒄𝒃 𝑴 𝑭 𝒃a 𝑴 𝑭 ab 𝑴 𝑭 𝒅𝒄 𝑴 𝑭 𝒄𝒅 P1L P2L P3L
- 8. Step 5: Assembly of Stiffness Matrix 1. Applying Unit rotation in Coordinate Direction 1 𝐾11 = 𝟏𝟐𝑬𝑰Δ 𝑳 𝟑 + 𝟏𝟐𝑬𝑰Δ 𝑳 𝟑 = 𝟏𝟐𝑬𝑰 𝟑 𝟑 + 𝟏𝟐𝑬𝑰 𝟑 𝟑 = 0.888 EI (Since Δ = 1) 𝐾21 = - 𝟔𝑬𝑰Δ 𝑳 𝟐 = - 𝟔𝑬𝑰 𝟑 𝟐 = - 0.667 EI 𝐾31 = - 𝟔𝑬𝑰Δ 𝑳 𝟐 = - 𝟔𝑬𝑰 𝟑 𝟐 = - 0.667 EI Δ= 1 Δ= 1 𝟏𝟐𝑬𝑰Δ 𝑳 𝟑 𝟏𝟐𝑬𝑰Δ 𝑳 𝟑 𝟏𝟐𝑬𝑰Δ 𝑳 𝟑 𝟏𝟐𝑬𝑰Δ 𝑳 𝟑 𝟔𝑬𝑰Δ 𝑳 𝟐 𝟔𝑬𝑰Δ 𝑳 𝟐 𝟔𝑬𝑰Δ 𝑳 𝟐 𝟔𝑬𝑰Δ 𝑳 𝟐
- 9. 9 θc= 0 A B C D θb= 1 𝐾12 = - 𝟔𝑬𝑰Δ 𝑳 𝟐 = - 𝟔𝑬𝑰 𝟑 𝟐 = - 0.667 EI 𝐾22 = 𝐾𝑏𝑎 + 𝐾𝑏𝑐 = 2𝐸𝐼 𝐿 𝑏𝑎 [2𝜃 𝑏 + 𝜃 𝑎] + 2𝐸𝐼 𝐿 𝑏𝑐 2𝜃 𝑏 + 𝜃𝑐 = 2𝐸𝐼 3 2 × 1 + 0 + 2𝐸𝐼 4 2 × 1 + 0 = 2.33EI 𝐾32 = 𝐾𝑐𝑏 = 2𝐸𝐼 𝐿 𝑏𝑐 [2𝜃𝑐 + 𝜃 𝑏] – = 2𝐸𝐼 4 2 × 0 + 1 = 0.5EI 𝟔𝑬𝑰Δ 𝑳 𝟐
- 10. 𝐾13 = - 𝟔𝑬𝑰Δ 𝑳 𝟐 = - 𝟔𝑬𝑰 𝟑 𝟐 = - 0.667 EI 𝐾23 = 𝐾𝑏𝑐 = 2𝐸𝐼 𝐿 𝑏𝑐 2𝜃 𝑏 + 𝜃𝑐 – = 2𝐸𝐼 4 2 × 0 + 1 = 0.5EI 𝐾33 = 𝐾𝑐𝑏 +𝐾𝑐𝑑 = 2𝐸𝐼 𝐿 𝑏𝑐 [2𝜃𝑐 + 𝜃 𝑏] + 2𝐸𝐼 𝐿 𝑐𝑑 2𝜃𝑐 + 𝜃 𝑑 = 2𝐸𝐼 4 2 × 1 + 0 + 2𝐸𝐼 3 2 × 1 + 0 = 2.33EI 10 θc= 1 A B C D θb= 0 𝟔𝑬𝑰Δ 𝑳 𝟐
- 11. K = 𝐾11 𝐾12 𝐾13 𝐾21 𝐾22 𝐾23 𝐾31 𝐾32 𝐾33 K = EI 0.888 −0.667 −0.667 −0.667 2.33 0.5 −0.667 0.5 2.33
- 12. • STEP 6: Calculation of Actual Forces 𝑃 − 𝑃𝐿 = 𝑃1 − 𝑃1𝐿 𝑃2 − 𝑃2𝐿 𝑃3 − 𝑃3𝐿 = 50 − 0 0 − (−26.66) 0 − 26.66 = 50 26.66 −26.66 12
- 13. 13 • STEP 6: Application Stiffness Equation Δ1 = Δ = 87.17/EI Δ2 = 𝜃 𝑏 = 35.11/𝐸𝐼 Δ3 = 𝜃𝑐 = 5.976/𝐸𝐼 EI 0.888 −0.667 −0.667 −0.667 2.33 0.5 −0.667 0.5 2.33 Δ1 Δ2 Δ3 = 50 26.66 − 26.66 K Δ = P-𝑃𝐿
- 14. • Step 8: Final Moments 𝑀 𝑎𝑏 = 𝑀 𝐹 𝑎𝑏 + 2𝐸𝐼 𝐿 𝑎𝑏 2𝜃 𝑎 + 𝜃 𝑏 − 3∆ 𝐿 𝑎𝑏 = 0+ 2𝐸𝐼 3 0 + 35.11 𝐸𝐼 − 3( 87.17 𝐸𝐼 ) 3 = -34.70 kN-m 𝑀 𝑏𝑎 = 𝑀 𝐹 𝑏𝑎 + 2𝐸𝐼 𝐿 𝑏𝑎 2𝜃 𝑏 + 𝜃 𝑎 − 3𝐸𝐼∆ 𝐿 𝑏𝑐 = 0+ 2𝐸𝐼 3 2 35.11 𝐸𝐼 + 0 − 3( 87.17 𝐸𝐼 ) 3 =-11.3 kN-m 𝑀 𝑏𝑐 = 𝑀 𝐹 𝑏𝑐 + 2𝐸𝐼 𝐿 𝑏𝑐 2𝜃 𝑏 + 𝜃𝑐 − 3𝐸𝐼∆ 𝐿 𝑏𝑐 = -26.66+ 2𝐸𝐼 4 2 35.11 𝐸𝐼 + 5.97 𝐸𝐼 + 0 = 11.4 kN-m
- 15. • Step 8: Final Moments 𝑀𝑐𝑏 = 𝑀 𝐹 𝑐𝑏 + 2𝐸𝐼 𝐿 𝑐𝑏 2𝜃𝑐 + 𝜃 𝑏 − 3∆ 𝐿 𝑐𝑏 = 26.66+ 2𝐸𝐼 4 2 5.97 𝐸𝐼 + 35.11 𝐸𝐼 − 0 = 50.18 kN-m 𝑀𝑐𝑑 = 𝑀 𝐹 𝑐𝑑 + 2𝐸𝐼 𝐿 𝑐𝑑 2𝜃𝑐 + 𝜃 𝑑 − 3𝐸𝐼∆ 𝐿 𝑐𝑑 = 0+ 2𝐸𝐼 3 2 5.976 𝐸𝐼 + 0 − 3( 87.17 𝐸𝐼 ) 3 = -50.15 kN-m 𝑀 𝑑𝑐 = 𝑀 𝐹 𝑑𝑐 + 2𝐸𝐼 𝐿 𝑑𝑐 2𝜃 𝑑 + 𝜃𝑐 − 3𝐸𝐼∆ 𝐿 𝑑𝑐 = 0+ 2𝐸𝐼 3 0 + 5.976 𝐸𝐼 − 3( 87.17 𝐸𝐼 ) 3 = -54.12 kN-m
- 16. Thank You