Research

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.

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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…)
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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…)
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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…)
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𝜕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…)
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-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)
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 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)
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 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
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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
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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
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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/
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