Week 11 Challenge

1. Friction load determination

Given vertical force =125 kN

Friction load = 40% of a single largest pipe weight = 0.4*125KN = 50KN

Friction Coefficient =0.1 (Between Teflon and Stainless Steel )

Friction Force = 0.1*50 = 5 KN

(2)

Figure-1:Spring Support

Figure-2:Stop Support

Figure-3:Stop Support

Figure-4:Semi-Anchor

Figure-5:Guyed Support

Figure-6:Guyed Support

(3)Geometric Model

Properties:

Materials:

Specifications:

Support:

Loading:

Design Parameters:

Utilization ratio

Note:STAAD model has been appended for Question-3