1. Senior Project Presentation
Zakariya Saleh Al-Helal
ID#

2. Outline Problem Description Objective Function Design Stage
Design Constrains
Excel Model
Model Importance
Results Concluded

3. Problem Description Project Title: Pipe-Line Anchor Block Optimization
Anchor block! Why?
restraints for pressurized pipes

4. Problem Description

5. Problem Description

6. Problem Description

7. Objective Function To optimize an anchor block for a given force How ?
Minimize the volume A

8. Design Stage
Normal forces analysis
Frictional forces
Moments

9. Design Stage
Normal forces analysis
In the horizontal direction

10. Design Stage Normal forces analysis Where k is ka for active soil
kp for passive soil
ka =

11. Design Stage
Normal forces analysis

12. Design Stage
Normal forces analysis

13. Design Stage
Normal forces analysis

14. Design Stage
Arms of actions: 2

15. Design Stage
Arms of actions:
For resultant force:

16. Design Stage
Normal forces analysis

17. Design Stage
Frictional forces

18. Design Constrains Factors of Safety Soil bearing capacity
Against sliding ≥ 1.25
Against overturning ≥1.5
Soil bearing capacity
≤ 3 ksf
≥ 0

19. Design Constrains
Factors of Safety
Against sliding ≥ 1.25

20. Design Constrains
Factors of Safety
Against overturning ≥1.5

21. Design Constrains
Soil bearing capacity
≤ 3 ksf
≥ 0

22. Design Constrains
Soil bearing capacity
≤ 3 ksf
≥ 0

23. Excel Model
Features:
Easy to Handle
Flexible
Minimum Cost
Time Saving

24. Verification & Comparison
Excel Model
Construction
Optimization
Verification & Comparison

25. Model Importance Material Saving: For a pipe force: Fp=1502 lb;
Current Technique:
V=30*16*16=7680 ft^3
Excel Model:minimzed to V=4288 ft^3

26. Model Importance Material Saving: For a pipe force: Fp=1502 lb;
Typical Solution Obtained by Saudi Aramco Engineer:
V=30*16*16=7680 ft^3
Excel Model:minimzed to V=4288 ft^3
% Saving: =45%

27. Results Concluded
The dimensions for a given force can be obtained using:
Model (Solver)
Tables
Charts
Equations

28. Results Concluded Equations: with h1 = 3 ft For W=4H:
H= x B= x0.4072
W = x d= x0.3794
For W=8H:
W = x H= x0.3905,
B= x d= x0.4162

29. Results Concluded Equations: For x=1000 For W=4H: H= 10.6 B= 6.6
W = d= 6.4

30. Results Concluded Important Notice
The dimensions are exponentially related to the pipe’s force magnitude
Approximately Min 40% saving in material

31. THANKS