Week 11 Challenge

1.

Given - Vertical force = 125 KN.

• TEFLON pad is used between shoe and pipe support.

 Procedure:

• Friction force = pipe gravity load x friction coefficiant.

• Friction force = 125 x 0.15.

• Stainless friction coefficiant = 0.15.

• Friction force = 18.75 KN.

• Friction force is temporarily acting friction load.

• So friction load = 18.75 KN.

• Friction coefficiant for teflon = 0.1.

• Friction force due to teflon = 125 x 0.1 12.5 KN.

• Friction force due to teflon = 12.5 KN.

2.

Figure 1 -

• Fixed support beacuse both ends are jointed or fixed to members.

Figure 2

• Pinned support.

Figure 3

• Fixed support.

Figure 4 -

• Fixed and pinned both.

Figure 5 -

• Pinned support.

Figure 6 -

• Pinned support because both jointed by pin support.

3.

Given -

Spacing between two column i.e. width = 7.5 m.

• Total height = 9.0 m.

• 7 spans at 6.0 m.

• Cable tray = 150 mm. Pipe diameter = 24, 12, 18, 8 inches.

• 2 tier spacing = 1.5 m.

• Center to center distance between column= 0.75 m.

• Procedure:

• Open staad -> create new model - > Define nodes and beams as per given data.

Properties -> add section -> assign property.

Support -> create support -> assign support.

Loading add load defination -> - add load case details -> assign loads to members.

• Uniform operating dead load = 1.9

KN/m^2. Pipe load = 1.9 x 7.5 14.28 KN/m. 

• 60% of it = 14.28 x 0.6 = 8.5 KN/m.

• Point load = 15 KN.

• 60% of it = 9 KN.

Load combination -> add>envelope -> add combinations and loads.

Design parameters -> add parameters.

Run analysis -> check results.