Week 8 Challenge

Week- 8

Q.1 Applying Load Exercise

For the structure modelled in previous exercise. Apply following loads

Form a Load case as DL for Dead Load

• Apply Self weight
• Apply member load of 3 kN/m on the rafter
• Apply member load of 6 kN/m on the secondary beams
• Apply point load of 20 kN at node intersecting inclined member on the top chord.

Form a Load case as LL for Live Load

• Apply member load of 4 kN/m on the rafter
• Apply member load of 7.5 kN/m on the secondary beams

Form load combination 1.5DL+1.5LL

Procedure: -

• Form a Load case as DL for Dead Load

Step 1: - Loading – Load Case details - Add – Loading Type – Dead Load - Add

Apply Self Weight: -

• Modeling – Loading – DL – Add – Direction – Y Factor -1.05 – Add
• 5% Extra considered in self-weight for the connections and bolts etc.
• Next assign the DL- select assign to view and click assign

Apply Member Load for 3 Kn/m on rafter.

• Now, similarly under Dead Load case add 3Kn/m Member Load on Rafter
• Loading – Dead Load – Add – Member Load – Uniform Force - W1 -3 Kn/m, Direction GY – add – Now select the Rafter by group tool – Assign to selected member.

Apply member load of 6 kN/m on the secondary beams

• Similarly apply member load on secondary beams
• Loading – Dead Load – Add – Member Load – Uniform Force - W1 -6 Kn/m, Direction GY – add – Now select the Secondary beam by group tool – Assign to selected member.

Apply point load of 20 kN at node intersecting inclined member on the top chord.

• On the inclined member 8 intersecting nodes are present, So 20 KN is divided by 8 Nodes.
• 20/8 = 2.5 kN
• Loading – Dead Load – Add – Nodal Load – Node - Fy -2.5 Kn,– add – Now select the Inclined member Nodes by using node courser – Assign to selected member.
• With this all Dead load generated.

Form a Load case as LL for Live Load

• Go to Loading – Load Case Details – Add – Loading Type Live – Title Live Load – Add

Apply member load of 4 kN/m on the rafter

• Now similarly we have to create Live Load by following same procedure.
• Loading – Live Load – Add – Member Load – uniform force – W1 -4 Kn/m – add – Now select the Rafter by using group option – Assign to selected member.

Apply member load of 7.5 kN/m on the secondary beams

• Follow Same procedure
• Loading – Live Load – Add – Member Load – uniform force – W1 -7.5 Kn/m – add – Now select the Secondary Beam by using group option – Assign to selected member

Form load combination 1.5DL+1.5LL

• Go to Loading – Load Case Details – Add – Define Combinations – Name and Factor 1.5 – Add

2. Find Maximum Force and Minimum Force for member and present the STAAD result.

Continue previous challenge, Run the analysis and find following results on the member mentioned

• Run the analysis
• Find the Maximum Moment in the Secondary beam for 1.5DL+1.5LL case
• Find the Maximum Axial force on the inclined bracing and identify the member having the maximum force for 1.5DL+1.5LL case
• Identify Bottom chord of Truss having Maximum shear Force & Maximum moment for 1.5DL+1.5LL case and present the values.

Run the Analysis

• Go to Menu bar – Analysis and Design – Run Analysis.

Find the Maximum Moment in the Secondary beam for 1.5DL+1.5LL case

• Go to Post Processing – Results – Select Load Case – Load Case – 1.5 DL+ 1.5 LL – Range Tab – Group – Secondary Beams – OK.

Find the Maximum Axial force on the inclined bracing and identify the member having the maximum force for 1.5DL+1.5LL case

• Post Processing – Results – Select Load Case – 1.5DL+1.5LL – Range Tab – Group – Top Chord – Ok.

Identify Bottom chord of Truss having Maximum shear Force & Maximum moment for 1.5DL+1.5LL case and present the values.

• Post Processing – Results – Select Load Case – 1.5DL+1.5LL – Range Tab – Group – Bottom Chord – Ok.

3. View and Rectify the Warning, instabilities in the structure

• For the attached STAAD file identify the Warnings and fix the same and run the analysis.
• For the attached STAAD file identify the Warnings and fix the same and run the analysis.
• Go to Utility – Analysis Output – Warnings

The warning zero stiffness is due to the inclined member connecting to bracing middle are assigned as a truss member

• To avoid this warning, the truss member specification to be removed for the bottom inclined member as shown below.

• From this type of assigning, the bottom inclined member will be made to take up the moments.

Result:- Given model analyzed and all warnings rectified.

4. Design Parameter exercise

In continuation to challenge 1, Apply Design parameter for structure modelled for code IS800:2007.

• Effective Length Lx, Ly, Lz for beams, Truss elements
• Effective length factors Ky, Kz for Columns
• Apply other design parameters for respective members

Procedure: -

• Go to analysis and design tab
• Now go to Design Steel
• Select the IS Code 800 2007
• Now define the parameters for ly, lz for beams an truss ky,kz for columns
• Run the analysis to check errors and warnings.

Result: - All parameters has been applied successfully.