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- 1. MASTERS OF SCIENCE IN STRUCTURAL ENGINEERING Presented By Amrit Regmi (2019-1-71-001) Supervised By Dr. Govind Prasad Lamichhane Associate Professor POKHARA UNIVERSITY SCHOOL OF ENGINEERING 2021/12/28
- 2. 2/64 Presentation Outline 1. Introduction 2. Complex Structures in the Research 3. Objectives of Research 4. Methodology 5. Literature Review 6. Solution of Simply Supported Cylindrical Shell (Case Study I) 7. Solution of Simply Supported Continuous Shell (Case Study II) 8. Solution of Multiple Barrel Cylindrical Shell (Case Study III) 9. Parametric Analysis 10. Observations and Conclusions 11. Limitation of the Study 12. Further Research Topic to the Researchers 13. References
- 3. 3/64 1. Introduction • Complex Structures: • Has Large Scale Complexity • Use Complex Structural Analysis • Has Irregular Geometry • Cylindrical Shell Structures I. Analytical Solution 1. Use of Differential Equations of Equilibrium 2. Membrane and Moment Theory 3. ASCE Manual No 31 II. Program Block Solution 1. Analytical Solution Based Programs III. Software Solution 1. SAP 2000 Solution 2. Finite Element Analysis
- 4. 4/64 2. Complex Structures in the Research i) Simply Supported Cylindrical Shell ii) Simply Supported Continuous Shell iii) Simply Supported Multiple Barrel Shell
- 5. 5/64 3. Objectives of Research The main purpose of this thesis is to observe the structural parameters in comparative forms in SAP2000 and Program Block Code. The specific objectives of the research work are: a) To analyze and observe the various structural parameters of Simply Supported Single Barrel Cylindrical Shell, Simply Supported Continuous Shell, Simply Supported Multiple Barrel Shell using the analytical based program approach and FEM based software approach. b) To evaluate structural response of the structures for live and dead loads. c) To observe parametric variation of the multiple barrel shell using the program developed. d) To check the precision and limitations of FEM based software for the structures.
- 6. 6/64 4. Methodology Literature Review Analyze Simply Supported Cylindrical Shell Analytically Compare the Analytical Result With FEM Result using SAP2000 Increase Complexity in the Simply Supported Shell by making it Continuous and Multiple Barrel and Analyze both of them using analytical based program code Perform Numerical Solution (FEM) of the complicated structures using SAP2000 Does the Solution from the two types of Analysis matched? Conclusion and Recommendation Perform Parametric Analysis of Interior Barrel of Multiple Barrel Shell using the Program Developed Yes No
- 7. 7/64 5. Literature Review i) Membrane Theory Forces on a small element based on Membrane Theory Source: Fig 37 of ASCE Manual
- 8. 8/64 i) Membrane Theory (Cont…) 𝑇𝜑 = −𝑝𝑢𝑟𝐶𝑜𝑠2 𝜑𝑘 − 𝜑 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 ……………………………………………………….(70c) 𝑆 = −𝑝𝑢𝑟 3 𝑛𝜋 𝑙 𝑟 cos 𝜑𝑘 − 𝜑 sin 𝜑𝑘 − 𝜑 cos 𝑛𝜋𝑥 𝑙 ……………………………………..(70b) 𝑇𝑥 = −𝑝𝑢𝑟 3 𝑛2𝜋2 𝑙 𝑟 2 [cos2( 𝜑𝑘 − 𝜑)- sin2( 𝜑𝑘 − 𝜑)] 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 ………………………….(70c) 𝑢 = 𝑝𝑢𝑟 𝑟 𝐸𝑡 𝑙 𝑟 2 3 𝑛3𝜋3 [cos2( 𝜑𝑘 − 𝜑)- sin2( 𝜑𝑘 − 𝜑)] cos 𝑛𝜋𝑥 𝑙 …………………………..(70d) 𝑣 = −𝑝𝑢𝑟 𝑟 𝐸𝑡 𝑙 𝑟 4 6 𝑛4𝜋4 cos(𝜑𝑘 − 𝜑) × sin 𝜑𝑘 − 𝜑 2 + 𝑛𝜋𝑟 𝑙 2 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 ……………...(70e) 𝑤 = −𝑝𝑢𝑟 𝑟 𝐸𝑡 𝑙 𝑟 4 6 𝑛4𝜋4 {[cos2( 𝜑𝑘 − 𝜑)− sin2( 𝜑𝑘 − 𝜑)] × [2 + 𝑛𝜋𝑟 𝑙 4 + 1 6 nπ𝑟 𝑙 4 cos2 (𝜑𝑘−𝜑)} 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 ………………………………………………..(70f) For Uniform Transverse Load Eq. Source: ASCE Manual
- 9. 9/64 i) Membrane Theory (Cont…) 𝑇𝜑 = −𝑝𝑑𝑟𝐶𝑜𝑠 𝜑𝑘 − 𝜑 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 ……………………………………………………….(71a) 𝑆 = −𝑝𝑑𝑟 2 𝑛𝜋 𝑙 𝑟 sin 𝜑𝑘 − 𝜑 cos 𝑛𝜋𝑥 𝑙 ………………………………………………….(71b) 𝑇𝑥 = −𝑝𝑑𝑟 2 𝑛2𝜋2 𝑙 𝑟 2 𝑐𝑜𝑠( 𝜑𝑘 − 𝜑)𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 …………………………………………….(71c) 𝑢 = 𝑝𝑑𝑟 𝑟 𝐸𝑡 𝑙 𝑟 2 2 𝑛3𝜋3 𝑐𝑜𝑠( 𝜑𝑘 − 𝜑) cos 𝑛𝜋𝑥 𝑙 …………………………………………….(71d) 𝑣 = −𝑝𝑑𝑟 𝑟 𝐸𝑡 𝑙 𝑟 4 2 𝑛4𝜋4 sin 𝜑𝑘 − 𝜑 1 + 2𝑛2𝜋2 𝑟 𝑙 2 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 ………………………..(71e) 𝑤 = −𝑝𝑑𝑟 𝑟 𝐸𝑡 𝑙 𝑟 4 2 𝑛4𝜋4 [1 + 2𝑛2𝜋2 𝑟 𝑙 2 + 𝑛4𝜋4 2 𝑟 𝑙 4 ]𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 ………………………..(71f) For Dead Weight Load Eq. Source: ASCE Manual
- 10. 10/64 ii) Symmetrical and Antisymmetrical Line Loads 1. A radial Shearing Force 2. A longitudinal Shearing Force 3. A tangential transverse force 4. A moment • Eight line loads, four on each edge can satisfy any edge requirement. Source: Fig 15 of ASCE Manual • Unlike arches, shells are curved at the support. So, boundary conditions are only satisfied by application of line loads along the shell.
- 11. 11/64 iii) Moment Theory Forces on a small element based on Moment Theory Source: Fig 38 of ASCE Manual
- 12. 12/64 ii) Moment Theory (Cont…) Eq. Source: ASCE Manual
- 13. 13/64 iv) Formulation of a Shell Element Fig: Shell Element Formulation in FEM Based Program
- 14. 14/64 v) Stiffness Matrix Formulation of a Membrane Element Shape Functions of Four Noded Membrane Element 𝑁1 𝑁2 𝑁3 𝑁4 = (1−ξ)(1−η) 4 (1+ξ)(1−η) 4 (1+ξ)(1+η) 4 (1−ξ)(1−η) 4 Fig: Four Noded Rectangular Element
- 15. v) Stiffness Matrix Formulation of a Membrane Element (cont…) 15/64 εx εy γxy = 𝐽11 ∗ 𝐽12 ∗ 0 0 0 0 𝐽21 ∗ 𝐽22 ∗ 𝐽21 ∗ 𝐽22 ∗ 𝐽11 ∗ 𝐽12 ∗ 𝜕𝑁1 𝜕ξ 0 𝜕𝑁2 𝜕ξ 0 𝜕𝑁3 𝜕ξ 0 𝜕𝑁4 𝜕ξ 0 𝜕𝑁1 𝜕η 0 𝜕𝑁2 𝜕η 0 𝜕𝑁3 𝜕η 0 𝜕𝑁4 𝜕η 0 0 𝜕𝑁1 𝜕ξ 0 𝜕𝑁2 𝜕ξ 0 𝜕𝑁3 𝜕ξ 0 𝜕𝑁4 𝜕ξ 0 𝜕𝑁2 𝜕η 0 𝜕𝑁2 𝜕η 0 𝜕𝑁3 𝜕η 0 𝜕𝑁4 𝜕η 𝑢1 𝑣1 𝑢2 𝑣2 𝑢3 𝑣3 𝑢4 𝑣4 = 𝐵 [𝑑] Strain-displacement relationship: Stiffness Matrix: 𝑘 = 𝑡 𝐵 𝑇 𝐷 𝐵 𝑑𝑥 𝑑𝑦 = 𝑡 𝐵 𝑇 𝐷 𝐵 𝐽 𝑑ξ 𝑑η 𝐷 = 𝐸 1 − μ2 1 μ 0 μ 1 0 0 0 1 − μ 2 Stress-Strain Matrix for Plane Stress Condition: (2.7.12) (2.7.14) (2.7.15)
- 16. 16/64 v) Stiffness Matrix Formulation of a Plate Bending Element Shape Functions of Four Noded Plate Element 𝑁1 𝑁2 𝑁3 𝑁4 = (1−ξ)(1−η) 4 (1+ξ)(1−η) 4 (1+ξ)(1+η) 4 (1−ξ)(1−η) 4 Fig: Four Noded Plate Element
- 17. v) Stiffness Matrix Formulation of a Plate Bending Element (cont…) 17/64 Strain-displacement relationship For 𝜀𝑥, 𝜀𝑦 𝑎𝑛𝑑 𝛾𝑥𝑦: εx εy γxy = −𝑧 𝐽11 ∗ 𝐽12 ∗ 0 0 0 0 𝐽21 ∗ 𝐽22 ∗ 𝐽21 ∗ 𝐽22 ∗ 𝐽11 ∗ 𝐽12 ∗ 0 − 𝜕𝑁1 𝜕ξ 0 0 − 𝜕𝑁2 𝜕ξ 0 0 − 𝜕𝑁3 𝜕ξ 0 0 𝜕𝑁4 𝜕ξ 0 0 − 𝜕𝑁1 𝜕η 0 0 − 𝜕𝑁2 𝜕η 0 0 − 𝜕𝑁3 𝜕η 0 0 − 𝜕𝑁4 𝜕η 0 0 0 𝜕𝑁1 𝜕ξ 0 0 𝜕𝑁2 𝜕ξ 0 0 𝜕𝑁3 𝜕ξ 0 0 𝜕𝑁4 𝜕ξ 0 0 𝜕𝑁2 𝜕η 0 0 𝜕𝑁2 𝜕η 0 0 𝜕𝑁3 𝜕η 0 0 𝜕𝑁4 𝜕η 𝑤1 θ𝑦1 θ𝑥1 𝑤2 θ𝑦2 θ𝑥2 𝑤3 θ𝑦3 θ𝑥3 𝑤4 θ𝑦4 θ𝑥4 = 𝐵𝑘 [𝑑] (2.8.12)
- 18. v) Stiffness Matrix Formulation of a Plate Bending Element (cont…) 18/64 Strain-displacement relationship for 𝛾𝑥𝑧 & 𝛾𝑦𝑧: γxz γyz = 𝐽11 ∗ 𝐽12 ∗ 1 0 𝐽21 ∗ 𝐽22 ∗ 0 1 𝜕𝑁1 𝜕ξ 0 0 𝜕𝑁2 𝜕ξ 0 0 𝜕𝑁3 𝜕ξ 0 0 𝜕𝑁4 𝜕ξ 0 0 𝜕𝑁1 𝜕η 0 0 𝜕𝑁2 𝜕η 0 0 𝜕𝑁3 𝜕η 0 0 𝜕𝑁4 𝜕η 0 0 0 𝑁1 0 0 𝑁2 0 0 𝑁3 0 0 𝑁4 0 0 0 −𝑁1 0 0 −𝑁2 0 0 −𝑁3 0 0 −𝑁4 𝑤1 θ𝑦1 θ𝑥1 𝑤2 θ𝑦2 θ𝑥2 𝑤3 θ𝑦3 θ𝑥3 𝑤4 θ𝑦4 θ𝑥4 = 𝐵𝛾 [𝑑] (2.8.14)
- 19. v) Stiffness Matrix Formulation of a Plate Bending Element (cont…) 19/64 The element stiffness matrix will become: 𝐾 = 𝐵𝑘 𝑇 𝐶𝑏 𝐵𝑘 + 𝐵γ 𝑇 𝐶𝑠 𝐵γ 𝐽 𝑑ξ 𝑑η (2.8.16) Where, 𝐶𝑏 = 𝐸ℎ3 12(1−μ2) 1 μ 0 μ 1 0 0 0 1−μ 2 (2.8.17) 𝐶𝑠 = 𝐸ℎ𝑘 2(1+μ) 1 0 0 1 (2.8.18)
- 20. vi) Stiffness Matrix Formulation of a Shell Element 20/64 (2.9.1) 𝑁1 = (1 − ξ)(1 − η)(−ξ − η − 1) 4 𝑁5 = (1 + ξ)(1 − ξ)(1 − η) 2 𝑁2 = (1 + ξ)(1 − η)(ξ − η − 1) 4 𝑁6 = (1 + ξ)(1 + η)(1 − η) 2 𝑁3 = (1 + ξ)(1 + η)(ξ + η − 1) 4 𝑁7 = (1 + ξ)(1 − ξ)(1 + η) 2 𝑁4 = (1 − ξ)(1 + η)(−ξ + η − 1) 4 𝑁8 = (1 − ξ)(1 − η)(1 + η) 2
- 21. vi) Stiffness Matrix Formulation of a Shell Element (cont…) 21/64 𝜕𝑢 𝜕ξ 𝜕𝑣 𝜕ξ 𝜕𝑤 𝜕ξ 𝜕𝑢 𝜕η 𝜕𝑣 𝜕η 𝜕𝑤 𝜕η 𝜕𝑢 𝜕ζ 𝜕𝑣 𝜕ζ 𝜕𝑤 𝜕ζ = 𝑖=1 8 𝜕𝑁𝑖 𝜕ξ 𝜕𝑁𝑖 𝜕η 0 𝑢𝑖 𝑣𝑖 𝑤𝑖 − 𝑖=1 8 𝑡𝑖𝑣2𝑖 2 ζ 𝜕𝑁𝑖 𝜕ξ 𝜕𝑁𝑖 𝜕η 𝑁𝑖 × β1 β2 β3 𝑇 + 𝑖=1 8 𝑡𝑖𝑣2𝑖 2 ζ 𝜕𝑁𝑖 𝜕ξ 𝜕𝑁𝑖 𝜕η 𝑁𝑖 × α1 α2 α3 𝑇 (Eq. source: http://nptel.iitm.ac.in 𝑘 = 𝐵 𝑇 𝐷 𝐵 𝑑𝜔 𝑘 = 𝑘 𝑏 + 𝑘 𝑠 Stiffness matrix will become: 𝑘 𝑏 = 𝐵 𝑏 𝑇 𝐷 𝑏 𝐵 𝑏𝑑𝜔 𝑘 𝑠 = 𝐵 𝑠 𝑇 𝐷 𝑠 𝐵 𝑠𝑑𝜔
- 22. 6. Solution of a Simply Supported Cylindrical shell (Case Study I) 22/64 Thickness of shell (t)=100mm Radius of shell (r)=10m Length of shell (l)=20m Angle subtended by the edge of shell measured from the centerline axis (φk) = 40𝑜 𝐷𝑒𝑎𝑑 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑠ℎ𝑒𝑙𝑙 𝑃𝑤𝑑 = 25𝑘𝑁/𝑚3. Live Load on the shell 𝑃𝑢 = 3𝑘𝑁/𝑚2 Fig: Simply Supported Cylindrical Shell
- 23. i) Analytical Solution According to ASCE Code no 31 (Step1) Find out 𝑇𝑥, 𝑇𝜑, 𝑆, from Table 1B (Membrane Theory) (Step2) Compute Unbalanced Force at 𝜑 = 0 (Step3) 𝐶𝑜𝑚𝑝𝑢𝑡𝑒 𝐻𝐿, 𝑉𝐿 𝑎𝑛𝑑 𝑆𝐿 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑢𝑛𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 (𝐸𝑑𝑔𝑒 𝐿𝑜𝑎𝑑𝑖𝑛𝑔) (Step4) Find out 𝑇𝑥, 𝑇𝜑, 𝑆, 𝑀𝜑, from Table 2A (Moment Theory) (Step5) Add Solution From Step 1 and Step 4 to Get The Final Solution 23/64
- 24. i) Analytical Solution According to ASCE Code no 31(Cont…) 24/64 SN Loading Type Force Functio n Force in kN/m at different values of 𝛗 (Maximum) 40 30 20 10 0 1 Live 𝑇∅𝑀𝑈 -38.197 -37.044 -33.728 -28.648 -22.414 2 Dead 𝑇∅𝑀𝐷 -31.831 -31.347 -29.912 -27.566 -24.383 3 𝑉𝐿 𝑇φ𝑉𝐿 -10.769 -32.156 -62.026 -36.517 19.342 4 𝐻𝐿 𝑇φ𝐻𝐿 -7.887 13.013 50.725 54.776 27.46 5 𝑆𝐿 𝑇φS𝐿 -4.803 -3.625 0.167 4.834 0 Summation 𝑻𝛗 -93.487 -91.159 -74.774 -33.121 0 6 Live 𝑆𝑀𝑈 0 -12.475 -23.445 -31.627 -35.921 7 Dead 𝑆𝑀𝐷 0 -7.035 -13.866 -20.264 -26.05 8 𝑉𝐿 𝑆𝑉𝐿 0 134.278 45.542 -210.743 0 9 𝐻𝐿 𝑆𝐻𝐿 0 -135.435 -106.685 72.915 0 10 𝑆𝐿 𝑆S𝐿 0 -8.465 -17.451 -9.94 61.971 Summation 𝑺 0 -29.132 -115.91 -199.66 0 SN Loading Type Force Function Force in kN/m at different values of 𝛗 (Maximum) 40 30 20 10 0 11 Live 𝑇𝑥𝑀𝑈 -46.448 -43.636 -35.584 -23.224 -8.067 12 Dead 𝑇𝑥𝑀𝐷 -25.796 -25.414 -29.091 -26.799 -23.713 13 𝑉𝐿 𝑇𝑥𝑉𝐿 658.398 168.931 -801.46 -700.991 3143 14 𝐻𝐿 𝑇𝑥𝐻𝐿 -614.01 -263.7 478.36 601.678 -1745 15 𝑆𝐿 𝑇xS𝐿 -28.333 -34.902 -21.145 104.558 472.964 Summation 𝑻𝐱 -56.189 -198.72 -408.92 -44.778 1839.18 13 𝑉𝐿 𝑀φ𝑉𝐿 -39.826 -40.729 -37.721 -22.59 0 14 𝐻𝐿 𝑀φ𝐻𝐿 24.449 27.03 29.36 21.258 0 15 𝑆𝐿 𝑀φS𝐿 -2.107 -2.107 -0.868 0.186 0 Summation 𝑴𝛗 -17.484 -15.806 -9.229 -1.146 0 Table: Final Forces Computation using Analytical Solution
- 25. ii) FEM Solution Using SAP2000 V20 (Step1) Initialize of Model in SAP2000 ie 𝐿 = Length of Shell=20m 𝑁𝐴 = 𝑁𝑜. 𝑜𝑓 𝐷𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠, 𝐴𝑥𝑖𝑎𝑙=40 𝑇 = 𝑅𝑜𝑙𝑙 𝐷𝑜𝑤𝑛 𝐴𝑛𝑔𝑙𝑒 = 400 𝑁𝐺 = 𝑁𝑜. 𝑜𝑓 𝐷𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠, 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 = 16 𝑅 = 𝑅𝑎𝑑𝑖𝑢𝑠 = 10𝑚 Type of Shell= Single Barrel (Step2) Initialization of Shell Material (M20 Concrete) 𝑃𝑤𝑑 = 𝑈𝑛𝑖𝑡 𝑊𝑒𝑖𝑔ℎ𝑡 = 25𝑘𝑁/𝑚3 𝐸 = 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝐸𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 = 2.236 × 107 𝑘𝑁/𝑚2 𝑈 = 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 = 0.2 𝐺 = 𝑆ℎ𝑒𝑎𝑟 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 = 9.316 × 106 𝑘𝑁/𝑚2 (Step3) Initialization of Shell Section Data 𝑆ℎ𝑒𝑙𝑙 𝑇𝑦𝑝𝑒 = 𝑆ℎ𝑒𝑙𝑙 − 𝑇ℎ𝑖𝑛 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑎𝑡 𝑀𝑒𝑚𝑏𝑟𝑎𝑛𝑒 = 0.1𝑚 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑎𝑡 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 = 0.1𝑚 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝐴𝑠𝑠𝑖𝑔𝑛𝑒𝑑 = 𝑀20 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑 𝐴𝑠𝑠𝑖𝑔𝑛𝑒𝑑 = 3𝑘𝑁/𝑚2 (Step4) Make Load Combination: Live+Dead and Run the Analysis to find the final Forces 𝐹11 ≡ 𝑇𝑥 𝐹22 ≡ 𝑇φ 𝐹12 ≡ 𝑆 𝑀22 ≡ 𝑀φ 25/64
- 26. ii) FEM Solution Using SAP2000 V20 (Cont…) 26/64 SN Resultant Type Maximum Force Components at Various Angles in degree for Load Combination: Dead Load + Live Load (kN/m) 40 30 20 10 0 1 𝐹11 ≡ 𝑇𝑥 at x=l/2 100.68 -151.73 -541.75 -168.63 2074.21 2 𝐹22 ≡ 𝑇𝜑 at x=l/2 -76.6 -76.047 -67.529 -47.353 142.4 3 𝐹12 ≡ 𝑆 at x=0 0 -12.035 2.103 65.872 698.812 4 𝑀22 ≡ 𝑀𝜑at x=l/2 -19.673 -17.238 -10.087 -1.595 0 Simply Supported Cylindrical Shell (Initialization in SAP2000) Table: Final Forces Computation using SAP2000
- 27. iii) Comparison of Forces From Analytical and SAP2000 Solution Observations: 1. Tx value on the simply supported shell matched in both analytical and software solution with some discrepancy. 2. T𝛗 deviates at the edges in software solution because of the edge zone effect. 27/64 -1000 -500 0 500 1000 1500 2000 2500 0 5 10 15 20 25 30 35 40 Longitudinal Force in kN/m Angle (𝛗) Tx Comparision Software Solution Analytical Solution -150 -100 -50 0 50 100 150 200 0 5 10 15 20 25 30 35 40 Transverse Force in kN/m Angle (𝛗) T𝛗 Comparision Software Solution Analytical Solution
- 28. iii) Comparison of Forces From Analytical and SAP2000 Solution Observations: 3. S is 0 in the analytical solution at the edges as per moment theory because of vanishing moment at the support. But it is highly deviating in the software solution because of concentration of stress at the support. 4. M𝛗 deviates at the crown for software solution because software doesn’t account for curvature. 28/64 -300 -200 -100 0 100 200 300 400 500 600 700 800 0 10 20 30 40 Shear Force in kN/m Angle (𝛗) S Comparision Software Solution Analytical Solution -25 -20 -15 -10 -5 0 0 10 20 30 40 50 Transverse Moment in kNm/m Angle (𝛗) M𝛗 Comparision Software Solution Analytical Solution
- 29. 7. Solution of a Simply Supported Continuous Shell (Case Study II) 29/64 • Parameters of the Shell Used: Same as Simply Supported Cylindrical Shell of Case Study I • Increased Complexity: Continuous over Support Fig: Simply Supported Continuous Shell
- 30. i) Analytical Program Solution According to ASCE Code no 31 30/64 • Analysis of the shell relies upon the values obtained for a simply supported shell. • Use of Equations of Table 5 of ASCE manual which are derived in Appendix II of ASCE: 𝑇𝑥, S and 𝑀𝜑 require correction factors to the forces of simply supported shell. 𝑇𝜑 doesn’t change when simply supported shell changes to continuous. • Development of Analytical Based program for the computation of forces.
- 31. i) Analytical Program Solution According to ASCE Code no 31(Cont…) 31/64 Fig: Code Developed in Matlab for Solution of Continuous Shell
- 32. i) Analytical Program Solution According to ASCE Code no 31(Cont…) 32/64 Tx 𝑇𝜑 S 𝑀𝜑 Output From Matlab
- 33. ii) FEM Solution Using SAP2000 V20 (Step1) Initialization of Model in SAP2000 Replicate model of simply supported shell used in case study I to 20m along x-axis. (Step2) Run the Analysis to find the final Forces 𝐹11 ≡ 𝑇𝑥 𝐹22 ≡ 𝑇φ 𝐹12 ≡ 𝑆 𝑀22 ≡ 𝑀φ 33/64 Fig: Initialization of Model in SAP2000
- 34. ii) FEM Solution Using SAP2000 V20 (Cont…) 34/64 Angle Maximum Force (kN/m) T𝛗 Tx S M𝛗 40.00 -57.703 138.32 0 -10.45 30.00 -66.163 -23.73 -11.94 -10.64 20.00 -71.675 -315.9 -3.23 -8.484 10.00 -50.97 -184.57 49.341 -2.31 0.00 88.85 1207.11 520.12 0 Angle Force at intersecting Line (kN/m) T𝛗 Tx S M𝛗 40.00 -57.76 -220.45 0 2.47 30.00 -21.757 -6.093 0.06 3.92 20.00 25.78 519.1 -13.402 1.72 10.00 -217.05 564.94 -66.896 -14.274 0.00 -2166.76 -3757.38 171.191 0 Table: Computation of Final Forces from SAP2000
- 35. iii) Comparison of Forces From Program and SAP2000 Solution 35/64 Comparison of Maximum Forces -600 -400 -200 0 200 400 600 800 1000 1200 1400 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 Longitudinal Force in kN/m Angle (𝛗) Tx Comparision at x=l/2 Software Solution Program Solution -150 -100 -50 0 50 100 150 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 Transverse Force in kN/m Angle (𝛗) T𝛗 Comparision at x=l/2 Software Solution Program Solution
- 36. iii) Comparison of Forces From Program and SAP2000 Solution 36/64 Comparison of Maximum Forces -100 0 100 200 300 400 500 600 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 Shear Force in kN/m Angle (𝛗) S Comparision at x=0 Software Solution Program Solution -16 -14 -12 -10 -8 -6 -4 -2 0 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 Moment in kNm/m Angle (𝛗) M𝛗Comparision at x=0 Software Solution Program Solution On providing the intermediate support in the FEM solution in the continuous shell, the values of structural parameters are near to the value of the analytical solution for maximum value of forces. Observation:
- 37. iii) Comparison of Forces From Program and SAP2000 Solution 37/64 Comparison of Forces at Intersecting Line -3000 -2500 -2000 -1500 -1000 -500 0 500 1000 0 5 10 15 20 25 30 35 40 Longitudinal Force in kN/m Angle (𝛗) Tx Comparision Software Solution Program Solution -600 -500 -400 -300 -200 -100 0 100 200 0 5 10 15 20 25 30 35 40 Transverse Force in kN/m Angle (𝛗) T𝛗 Comparision Software Solution Program Solution
- 38. iii) Comparison of Forces From Program and SAP2000 Solution 38/64 Comparison of Forces at Intersecting Line On the intersecting line of the continuous shell, FEM solution has highly deviated in comparison to the analytical program solution. Observation: -300 -250 -200 -150 -100 -50 0 50 100 150 200 0 5 10 15 20 25 30 35 40 Shear Force in kN/m Angle (𝛗) S Comparision Software Solution Program Solution -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 0 5 10 15 20 25 30 35 40 Moment in kNm/m Angle (𝛗) M𝛗 Comparision Software Solution Program Solution
- 39. 8. Solution of a Simply Supported Multiple Barrel Cylindrical Shell (Case Study III) 39/64 • Parameters of the shell used: Same as Simply Supported Cylindrical Shell of Case Study I • Increased Complexity: Barrels added to the original shell on either side. • Barrel Analyzed: Interior Barrel Fig: Simply Supported Multiple Barrel Shell
- 40. i) Analytical Program Solution According to ASCE Code no 31 (Step1) Compute coefficients required by the equations given in table 4B (Step2) Find out 𝑇𝑥, 𝑇𝜑, 𝑆, from eqs 70 and 71 of Appendix II (Membrane Theory) (Step3) Put respective coefficients of Step 1 in the equations of table 4B (Moment Theory) (Step4) Apply 4 Boundary Conditions to find the values of Arbitrary Constants B1, B2,B3 & B4 (Step5) Put the Constants in respective equations to find 𝑇𝑥, 𝑇𝜑, 𝑀𝜑 & 𝑆 (Step6) Sum up Moment and Membrane Solution to find the Final Solution 40/64
- 41. i) Analytical Program Solution according to ASCE Code no 31(cont…) 1. Vertical line load at edge must be equal to the vertical component of the membrane transverse force. 2. Horizontal displacement at edge must be zero. 3. Shearing line load at edge must be equal to the membrane shearing force 4. Rotation of edge must be zero. Four Boundary Condition: 41/64
- 42. i) Analytical Program Solution according to ASCE Code no 31(cont…) 42/64 Fig: Code Developed in Matlab for Solution of Interior Barrel of Multiple Barrel Shell
- 43. i) Analytical Program Solution According to ASCE Code no 31(Cont…) 43/64 Tx 𝑇𝜑 S 𝑀𝜑 Output From Matlab
- 44. ii) FEM Solution Using SAP2000 V20 (Step1) Initialize of Model in SAP2000 ie 𝐿 = Length of Shell=20m 𝑁𝐴 = 𝑁𝑜. 𝑜𝑓 𝐷𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠, 𝐴𝑥𝑖𝑎𝑙=40 𝑇 = 𝑅𝑜𝑙𝑙 𝐷𝑜𝑤𝑛 𝐴𝑛𝑔𝑙𝑒 = 400 𝑁𝐺 = 𝑁𝑜. 𝑜𝑓 𝐷𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠, 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 = 16 𝑅 = 𝑅𝑎𝑑𝑖𝑢𝑠 = 10𝑚 Type of Shell= Multi Bay Cylindrical Shell Number of Bays, Y=3 (Step2) Initialization of Shell Material (M20 Concrete) 𝑃𝑤𝑑 = 𝑈𝑛𝑖𝑡 𝑊𝑒𝑖𝑔ℎ𝑡 = 25𝑘𝑁/𝑚3 𝐸 = 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝐸𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 = 2.236 × 107 𝑘𝑁/𝑚2 𝑈 = 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 = 0.2 𝐺 = 𝑆ℎ𝑒𝑎𝑟 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 = 9.316 × 106 𝑘𝑁/𝑚3 (Step3) Initialization of Shell Section Data 𝑆ℎ𝑒𝑙𝑙 𝑇𝑦𝑝𝑒 = 𝑆ℎ𝑒𝑙𝑙 − 𝑇ℎ𝑖𝑛 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑎𝑡 𝑀𝑒𝑚𝑏𝑟𝑎𝑛𝑒 = 0.1𝑚 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑎𝑡 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 = 0.1𝑚 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝐴𝑠𝑠𝑖𝑔𝑛𝑒𝑑 = 𝑀20 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑 𝐴𝑠𝑠𝑖𝑔𝑛𝑒𝑑 = 3𝑘𝑁/𝑚2 (Step4) Make Load Combination: Live+Dead and Run the Analysis to find the final Forces 𝐹11 ≡ 𝑇𝑥 𝐹22 ≡ 𝑇φ 𝐹12 ≡ 𝑆 𝑀22 ≡ 𝑀φ 44/64
- 45. ii) FEM Solution Using SAP2000 V20 (Cont…) 45/64 Fig: Multi Barrel Cylindrical Shell (Initialization in SAP2000) Table: Final Forces Computation (Interior Panel) using SAP2000 Maximum Force Angle in Degree 0 10 20 30 40 Tx 443.00 -81.98 -279.61 -201.47 -120.20 T 𝛗 46.56 -1.58 -36.17 -65.36 -75.63 S -1.30 -66.99 -4.62 -1.54 0.00 M 𝛗 -8.41 5.13 3.70 -2.23 -4.86 Force in Intersecting Line Distance x in meter 0 5 10 15 20 Tx -2410.00 243.53 445.05 243.53 - 2410.00 T 𝛗 -2211.12 31.52 46.74 31.52 - 2211.12 S 1.30 0.57 0.00 -0.57 -1.30 M 𝛗 34.41 -0.77 -8.40 -0.77 34.41
- 46. iii) Comparison of Forces From Program and SAP2000 Solution 46/64 Comparision of Maximum Forces -400.00 -300.00 -200.00 -100.00 0.00 100.00 200.00 300.00 400.00 500.00 600.00 700.00 0 10 20 30 40 Longitudinal Force in kN/m Angle (𝛗) Tx Comparision at x=l/2 Software Solution Program Solution -100.00 -80.00 -60.00 -40.00 -20.00 0.00 20.00 40.00 60.00 0 5 10 15 20 25 30 35 40 Longitudinal Force in kN/m Angle (𝛗) T𝛗 Comparision at x=l/2 Software Solution Program Solution
- 47. iii) Comparison of Forces From Program and SAP2000 Solution 47/64 Comparison of Maximum Forces On further increasing the complexity of the simply supported shell by making it multiple barrels, the FEM solution is nearer to the value of analytical solution. This is because intermediate support has been provided on the shell. The maximum shear force value at the edges in the FEM solution has also matched with the analytical solution in this case. It was observed deviating in the case of simply supported and continuous shells. So, the intermediate support has decreased the discrepancy in the FEM solution. Observations: -120.00 -100.00 -80.00 -60.00 -40.00 -20.00 0.00 0 10 20 30 40 Longitudinal Force in kN/m Angle (𝛗) S Comparision at x=0 Software Solution Program Solution -10.00 -8.00 -6.00 -4.00 -2.00 0.00 2.00 4.00 6.00 8.00 0 10 20 30 40 Longitudinal Force in kN/m Angle (𝛗) M𝛗 Comparision at x=l/2 Software Solution Program Solution
- 48. iii) Comparison of Forces From Program and SAP2000 Solution 48/64 Comparison of Forces at Intersecting Line -3000.00 -2500.00 -2000.00 -1500.00 -1000.00 -500.00 0.00 500.00 1000.00 0 5 10 15 20 Longitudinal Force in kN/m x in meter Tx Comparision at intersecting line Analytical Program Solution Software Solution -2500.00 -2000.00 -1500.00 -1000.00 -500.00 0.00 500.00 0 5 10 15 20 Longitudinal Force in kN/m x in meter T𝛗 Comparision at intersecting line Analytical Program Solution Software Solution
- 49. iii) Comparison of Forces From Program and SAP2000 Solution 49/64 Comparison of Forces at Intersecting Line • In the case of intersecting lines as in the continuous shell, the solution in the FEM solution has highly deviated from the analytical solution Observation: -1.50 -1.00 -0.50 0.00 0.50 1.00 1.50 0 5 10 15 20 Longitudinal Force in kN/m x in meter S Comparision at intersecting line Analytical Program Solution Software Solution -15.00 -10.00 -5.00 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 0 5 10 15 20 Longitudinal Force in kN/m x in meter M𝛗 Comparision at intersecting line Analytical Program Solution Software Solution
- 50. 9) Parametric Analysis 50/64 • Parametric Analysis is performed on the interior barrel of the multiple barrel long cylindrical shell. • Varying parameters are width and height. • Analysis is done by the program developed for Case study III where variables can be changed easily. Constant input Parameters for all models under analysis: Length of shell L=30m Thickness of shell t=100mm Boundary condition: simply supported at both interior and exterior barrels. Uniform Loading on the shell= 2kN/m2 Dead Weight Loading on the shell = 25 kN/m3
- 51. 9) Parametric Analysis (Cont…) 51/64 Varying Input Parameters: Width of barrels, B=6m, 8m, 10m, 12m, 14m, 16m, 18m,20m,22m Height of barrels, h=2m, 2.5m, 3m, 3.5m Solution of the barrel shell with all possible combination of B and h has been done. Dependent Parameters: Dependent parameters are radius of shell R, subtended angle 𝛗k. R/L ratios and R/t ratios. Output Parameters Considered: Transverse moment M𝛗 (kNm/m) at transverse mid-section. Longitudinal normal force Tx (kN/m) at transverse mid-section.
- 52. 9) Parametric Analysis (Cont…) 52/64 Values of R B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m h=2m 3.250 5.000 7.250 10.000 13.250 17.000 21.250 26.000 31.250 h=2.5m 3.050 4.450 6.250 8.450 11.050 14.050 17.450 21.250 25.450 h=3m 3.000 4.167 5.667 7.500 9.667 12.167 15.000 18.167 21.667 h=3.5m - 4.036 5.321 6.893 8.750 10.893 13.321 16.036 19.036 Values of 𝛗k B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m h=2m 67.380 53.130 43.603 36.870 31.891 28.072 25.058 22.621 20.610 h=2.5m 79.611 64.011 53.130 45.240 39.308 34.708 31.049 28.073 20.710 h=3m 90.000 73.740 61.928 53.130 46.397 41.112 36.871 33.399 24.544 h=3.5m - 82.372 69.984 60.513 53.130 47.259 42.502 38.581 28.217 Table: Dependent Variable for Parametric Analysis
- 53. i) Variation of M𝛗 for Constant Height 53/64 -15.000 -10.000 -5.000 0.000 5.000 10.000 15.000 20.000 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for h=2m B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m -15.000 -10.000 -5.000 0.000 5.000 10.000 15.000 20.000 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for h=2.5m B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m -15.000 -10.000 -5.000 0.000 5.000 10.000 15.000 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for h=3m B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m -12 -10 -8 -6 -4 -2 0 2 4 6 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for h=3.5m B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m
- 54. ii) Variation of M𝛗 for Constant Breadth 54/64 -10.0 -5.0 0.0 5.0 10.0 15.0 20.0 0.0 20.0 40.0 60.0 80.0 100.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=6m h=2m h=2.5m h=3m h=3.5m -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 0.0 20.0 40.0 60.0 80.0 100.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=8m h=2m h=2.5m h=3m h=3.5m -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 0.0 20.0 40.0 60.0 80.0 100.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=10m h=2m h=2.5m h=3m h=3.5m -5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 0.0 20.0 40.0 60.0 80.0 100.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=12m h=2m h=2.5m h=3m h=3.5m -8.000 -6.000 -4.000 -2.000 0.000 2.000 4.000 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=14m h=2m h=2.5m h=3m h=3.5m -10.000 -8.000 -6.000 -4.000 -2.000 0.000 2.000 4.000 6.000 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=16m h=2m h=2.5m h=3m h=3.5m
- 55. ii) Variation of M𝛗 for Constant Breadth (Cont…) 55/64 -12.000 -10.000 -8.000 -6.000 -4.000 -2.000 0.000 2.000 4.000 6.000 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=18m h=2m h=2.5m h=3m h=3.5m -15.000 -10.000 -5.000 0.000 5.000 10.000 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=20m h=2m h=2.5m h=3m h=3.5m -15.000 -10.000 -5.000 0.000 5.000 10.000 0.0 20.0 40.0 60.0 80.0 100.0 120.0 M𝛗 in kNm/m % of 𝛗k Variation of M𝛗 for B=22m h=2m h=2.5m h=3m h=3.5m
- 56. iii) Influence of M𝛗 (Observations) 56/64 For constant values of h, Values of M𝛗 decreases at the edges but it becomes constant at the crown when B increases. For constant values of h, values of M𝛗 changes its sign moving from edge towards the crown. Lesser the value of B for constant values of h, M𝛗 becomes positive at the edge. For constant value of B, M𝛗 decreases at the edge when h decreases. Lowering the values of both B and h, M𝛗 becomes dominant at the intersecting lines. For higher value of breadth, M𝛗 is constant at every section ie from edge to the crown irrespective of height.
- 57. iv) Variation of Tx for Constant Height 57/64 -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 1200.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for h=2m B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m -1000.00 -500.00 0.00 500.00 1000.00 1500.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for h=2.5m B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 1200.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for h=3m B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m -600 -400 -200 0 200 400 600 800 1000 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for h=3.5m B=6m B=8m B=10m B=12m B=14m B=16m B=18m B=20m B=22m
- 58. v) Variation of Tx for Constant Breadth 58/64 -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=6m h=2m h=2.5m h=3m -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=8m h=2m h=2.5m h=3m h=3.5m -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=10m h=2m h=2.5m h=3m h=3.5m -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=12m h=2m h=2.5m h=3m h=3.5m -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 1200.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=14m h=2m h=2.5m h=3m h=3.5m -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 1200.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=16m h=2m h=2.5m h=3m h=3.5m
- 59. v) Variation of Tx for Constant Breadth 59/64 -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 1200.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=18m h=2m h=2.5m h=3m h=3.5m -600.00 -400.00 -200.00 0.00 200.00 400.00 600.00 800.00 1000.00 1200.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=20m h=2m h=2.5m h=3m h=3.5m -1000.00 -500.00 0.00 500.00 1000.00 1500.00 0.0 20.0 40.0 60.0 80.0 100.0 Tx in kN/m % of 𝛗k Variation of Tx for B=22m h=2m h=2.5m h=3m h=3.5m
- 60. vi) Influence of Tx (Observations) 60/64 For constant values of h, Values of Tx decreases at the edges but it becomes constant at the crown when B increases. For constant values of h, values of Tx change its sign moving from edge towards the crown. For constant value of h, there is no significant difference in the value of Tx at crown on increasing the value of B. Decreasing both B and h, value of Tx increases at edge. At 40% to 45% of 𝛗k, Tx always changes its sign moving from edge towards the crown for both constant values of h and B.
- 61. 10) Conclusions 61/64 From the case studies of the long barrel thin cylindrical shells and the parametric study, following are the key points observed and conclusions made. ● In interior panels of multiple barrel cylindrical shells of higher breadths, transverse moment has similar value at every section of the barrel irrespective of height. So similar reinforcements can be provided in the design of those barrels. ● Huge amount of longitudinal compressive force exists in the interior panels of multiple barrel cylindrical shell which becomes tensile with almost half its magnitude moving from edge towards the crown. The force changes its sign at 40%-45% of the percentage of the 𝛗k. So special consideration must be done in design of that section. ● With the increase in complexity of simply supported shell ie by making it continuous or multiple barrel, FEM solution deviates at the intersecting lines of the models and is observed to have a high percentage of error there. So, an analytical based program considering the continuity of equations of curves in complex structures leads to the nearest results. ● In FEM analysis, when the structure is provided with intermediate support conditions, the accuracy of the analysis increases. ● FEM doesn’t account for curvature in shell analysis as FEM solution is seen deviating at the crown of cylindrical shells. Hence, quality of meshing also determines the accuracy of FEM solution.
- 62. 11) Limitation of the Study 62/64 The classical shell theory develops the higher order differential equations to solve the problems of arbitrary geometries. Those arbitrary geometries can only be solved approximated by using the finite element method or the numerical evaluation of infinite series. So, analytical solutions of complex geometries exist only for limited number complexities. But, those solutions offer vital function in the evaluation of FEM or modern FEM based software. However, to analyze shells with arbitrary geometry that interact with various supports and edge beams for static and dynamic solution of shells, the practical approach is only provided by FEM.
- 63. 12) Further Research Topic for the Researchers 63/64 From this paper, it is seen that with the increase in complexity of the cylindrical shell structure, FEM based solution highly deviates at the intersecting line. The stiffness matrix used by the FEM solution has been derived in Chapter 2 of this paper. So, researchers can develop FEM based program and observe why FEM solution lacks proper solution at the intersecting lines.
- 64. 13) References: 1. ASCE MANUAL NO-31: Design of cylindrical concrete shell roofs, prepared by the Committee on Masonry and Reinforced Concrete of the Structural Division, through its Subcommittee on Thin Shell Design. American Society of Civil Engineers. New York, N.Y.: The Society, 1952 2. IS: 2210–1988, CRITERIA FOR DESIGN OF REINFORCED CONCRETE SHELL STRUCTURES AND FOLDED PLATES (First Revision) 3. Daryl L. Logan (2015) “First Course in the Finite Element Method” University of Wisconsin–Platteville 4. “Parametric study on the structural forces and the moments of cylindrical shell roof using ANSYS” by Ashique Jose , Ramadass S and Jayasree Ramanujan 5. Kaushalkumar M. Kansara (2004) “Development of Membrane, Plate and Flat Shell Elements in Java” 6. Nawfal Hasaine (www.mathworks.com) “Satic Structural Analysis of Shell Roof Structure” 7. http://nptel.iitm.ac.in/ 64/64