Senior Project Presentation Zakariya Saleh Al-Helal ID# 212543

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

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

Problem Description

Problem Description

Problem Description

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

Design Stage Normal forces analysis Frictional forces Moments

Design Stage Normal forces analysis In the horizontal direction

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

Design Stage Normal forces analysis

Design Stage Normal forces analysis

Design Stage Normal forces analysis

Design Stage Arms of actions: 2

Design Stage Arms of actions: For resultant force:

Design Stage Normal forces analysis

Design Stage Frictional forces

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

Design Constrains Factors of Safety Against sliding ≥ 1.25

Design Constrains Factors of Safety Against overturning ≥1.5

Design Constrains Soil bearing capacity ≤ 3 ksf ≥ 0

Design Constrains Soil bearing capacity ≤ 3 ksf ≥ 0

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

Verification & Comparison Excel Model Construction Optimization Verification & Comparison

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

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%

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

Results Concluded Equations: with h1 = 3 ft For W=4H: H= 0.8039 x0.3736 B= 0.3968 x0.4072 W = 3.2154 x0.3736 d= 0.4661x0.3794 For W=8H: W = 4.3888 x0.3905 H= 0.5486 x0.3905, B= 0.4239 x0.3593 d= 0.2743x0.4162

Results Concluded Equations: For x=1000 For W=4H: H= 10.6 B= 6.6 W = 42.5 d= 6.4

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

THANKS