Week 10 Challenge

Modeling , Analysis & Design of Industrial building (PEB/Truss) in STAAD pro.

1. 1.Purlin & Girts number determination problem

Aim: To Calculate total number of purlins and girts required for the building with following inputs

Given Data:

• Rafter span = 50 m
• Bay spacing = 8 m
• Roof slope = 1 in 10
• Therefore Roof height = 2.5 m
• Height of column = 12 m
• Brick wall height = 2 m
• Purlin spacing = 1.2 m
• Girt spacing = 1.5 m

Solution:

Step 1: Open STAAD Pro Software and draw the building using commands for the given inputs.

Step 2: No. of Horizontal member at Roof Level along single roof = (Length of Rafter / Spacing) + 1

Length of Rafter = √[(Half of Span)2 + (Height of Rafter)2]

=√(252 + 2.52)

Length of Rafter = 25.1247 m

No. of Horizontal members at roof level = (25.1247 / 1.2) + 1 = 20.94 + 1 = 21.94

Says = 22 Nos

In 22 Nos,

Top most one is Ridge and Last one is Eave Strut.

Therefore Remaining Members are Purlin.

No. of Purlin for entire span = 2 x 20 = 40 Nos.

Step 3: No. of Girts = Height of column / Spacing

Height of Column = 12 m

Brick wall height = 2 m

Hence girts arranged in remaining 10 m

No. of Girts on one side = 10 / 1.5 = 6.67

Says 7 Nos

Total number of Girts on both side = 2 x 7 = 14 Nos.

Result:

• Number of Purlins on Entire span = 40 Nos
• Number of Girts on both Side = 14 Nos
• Number of Ridge = 1 No
• Number of Eave Struts = 2 Nos

2. 2.Taper the beam according to BMD envelope

Aim: To Taper the Beams according to BMD envelope and check Utility ratio within limit for the given STAAD file.

Solution:

Open the given STAAD file in the STAAD Pro software.

• Go to Analysis Tab and check whether the Parameters applied.

• Then Click Run Analysis.
• Output shows 0 Errors and 0 Warnings.

Now select Utility Ratio, and see the Ratio for the beams.

• Then Enable Beam Label to identify the beams.

Member Name	Tag	Utility Ratio	Remarks
End Column	1	0.634	The column is utilized for 63.4% only. We have to utilize the member according to BMD envelope.
Middle Column	5	0.863	In this column no moment is to be transferred as moment is released so its capacity can be utilized further up to 100% as in future this member is not supposed to carry moment
End Column	2	0.598	The column is utilized for 59.8% only. We have to utilize the member according to BMD envelope.
One Rafter	3	0.866	The Rafter is utilized for 86.6% only. We have to utilize the member according to BMD envelope.
Another Rafter	4	0.866	The Rafter is utilized for 86.6% only. We have to utilize the member according to BMD envelope.

Now modify the Section properties.

• Split the beam where ever required.
• It is a Iterative process, modify the sections as utilize the section more effective.

Result: The given STAAD model was reviewed and modified to utilize the sections more effective for the given BMD envelope.

Utilization Ratio

Utilization Ratio in 3D view

Bending Moment Diagram

3. 3.Determining Wind load for certain category and condition as per IS 875 Part 3 :2015

Given: Dimension of Building = 10 m x 5 m, Height = 6m, Bay spacing = 2.5 m, Wall opening = 15%, Building is situated in Chennai

Assumptions:

1) Terrain Category 3

2) Design life of the structure 50 years

3) Topography: Plane with upwind slope Ɵ < 3°

4) Roof Angle = 5°

Aim: To calculate the wind pressure on the wall of building with the given inputs.

Solution:

Step 1:

Basic wind speed Vb = 50 m/s ( For Chennai as per IS 475 part-3 Fig. 1)

Design wind speed Vz = Vb x K1 x K2 x K3 x K4

Where K1­ is Probability factor (risk coefficient) taken from Table 1

Considering general building and structure and Vb =50 m/s

K1 = 1.0 for design life of structure 50 years & and for given Basic wind speed of 50 m/s

K2 is Terrain roughness & height factor taken from Table 2

K2 = 0.91

K3 = Topography factor for Upwind slope < 3° from Clause 6.3.3

K3 = 1.0

K4 = Importance factor for cyclonic region for Industrial Structure from Clause 6.3.4

K4 = 1.15

Therefore Vz = 1.0 x 0.91 x 1.0 x 1.15 x 50 = 52.325 m/s

Step 2:

Wind pressure Pz = 0.6 (Vz)2

Where,

Pz = wind pressure at height z in N/m2

Vz = design wind pressure at height z in m/s

Therefore

Pz = 0.6 x (52.325)2 = 1.643 kN/m2

Step 3:

Design wind pressure Pd = Kd x Ka x Kc x Pz

Where,

Kd = wind directionality factor 1.0 for cyclone affected region

Ka = Area averaging factor 0.9 for tributary area < 25 m2

(Tributary area = 6 x 2.5 = 15 m2)

Kc = combination factor 0.9

Therefore

Pd = 1.0 x 0.9 x 0.9 x 1.643 = 1.331 kN/m2

Step 4:

Wind load acting in a direction normal to the individual structural element or cladding unit

F = (Cpe – Cpi) x A x Pd

Where,

Cpi = internal pressure coefficient

Cpi = +0.5 and -0.5 for opening of 15% of wall area i.e. medium permeability (Clause 7.3.2.2 of IS :475, part 3)

External pressure coefficient:

L = 10 m; w = 5 m; h = 6 m

h/w = 6 / 5 = 1.2

l/w = 10 / 5 = 2

Thus as per table 5 of clause 7.3.3.1 of IS:475, Part 3

For 1/2 < h/w < 3/2 and 3/2 ≤ l/w < 4

Wind angle Ɵ	Cpe
	Wall A	Wall B	Wall C	Wall D
0°	+0.7	-0.3	-0.7	-0.7
90°	-0.5	-0.5	+0.7	-0.1

Case-I Ɵ = 0°

Cpnet = Cpe - Cpi

For Wall A

Cpnet = +0.7 – (+0.5) = +0.2 (Pressure)

Cpnet = +0.7 – (-0.5) = +1.2 (Pressure)

Design Wind Force for Wall A

F = Cpnet x A x Pd

FA = +0.2 x 6 x 2.5 x 1.331 = +3.993 kN (Pressure)

FA = +1.2 x 6 x 2.5 x 1.331 = +23.958 kN (Pressure)

For Wall B

Cpnet = -0.3 – (+0.5) = -0.8 (Suction)

Cpnet = -0.3 – (-0.5) = +0.2 (Pressure)

Design Wind Force for Wall B

F = Cpnet x A x Pd

FB = -0.8 x 6 x 2.5 x 1.331 = -15.972 kN (Suction)

FB = +0.2 x 6 x 2.5 x 1.331 = +3.993 kN (Pressure)

For Wall C & D

Cpnet = -0.7 – (+0.5) = -1.2 (Suction)

Cpnet = -0.7 – (-0.5) = -0.2 (Suction)

Design Wind Force for Wall C & D

F = Cpnet x A x Pd

FC&D = -1.2 x 6 x 5 x 1.331 = -47.916 kN (Suction)

FC&D = -0.2 x 6 x 5 x 1.331 = -7.986 kN (Suction)

Case-II Ɵ = 90°

Cpnet = Cpe - Cpi

For Wall A & B

Cpnet = -0.5 – (+0.5) = -1.0 (Suction)

Cpnet = -0.5 – (-0.5) = 0.0

Design Wind Force for Wall A & B

F = Cpnet x A x Pd

FA&B = -1.0 x 6 x 2.5 x 1.331 = -19.965 kN (Pressure)

FA&B = 0.0 x 6 x 2.5 x 1.331 = 0.000 kN

For Wall C

Cpnet = +0.7 – (+0.5) = +0.2 (Pressure)

Cpnet = +0.7 – (-0.5) = +1.2 (Pressure)

Design Wind Force for Wall C

F = Cpnet x A x Pd

FC = +0.2 x 6 x 5 x 1.331 = +7.986 kN (Pressure)

FC = +1.2 x 6 x 5 x 1.331 = +47.916 kN (Pressure)

For Wall D

Cpnet = -0.1 – (+0.5) = -0.6 (Suction)

Cpnet = -0.1 – (-0.5) = +0.4 (Pressure)

Design Wind Force for Wall D

F = Cpnet x A x Pd

FD = -0.6 x 6 x 5 x 1.331 = -23.958 kN (Suction)

FD = +0.4 x 6 x 5 x 1.331 = +15.972 kN (Pressure)

Result: Maximum design Pressure on the Wall = 47.916 kN

4. 4. Find support reaction for given load. Static check problem

Given:

Self Weight factor – 1.1

Roof Sheet loading = 0.5 kN/m2

Purlin load = 6 kN/m2

Live Load on Roof = 1.5 kN/m2

Wind Load on rafter (Pressure) = 2 kN/m2

Assume Frame height as 5m

Support as Fixed.

Total Dead Load = 0.5 + 6 = 6.5 kN/m2

Dead Load on Main frame = 6.5 x 6 = 39 kN/m2

Dead Load on Gable frame = 6.5 x 6 / 2 = 19.5 kN/m2

Live Load on Main frame = 1.5 x 6 = 9 kN/m2

Live Load on Gable frame = 1.5 x 6 / 2 = 4.5 kN/m2

Wind Load on Main frame = 2 x 6 = 12 kN/m2

Wind Load on Gable frame = 2 x 6 / 2 = 6 kN/m2

Aim: To apply Dead load, Live Load & Wind Load on the structure and find the support reaction.

Solution:

• Open STAAD Software and create a New File.
• Using Node, Beams and Translational repeat command create a Model as per given input.
• Then create and assign Fixed support to all columns.

• Click Load Cases details and add Load cases Dead Load as DL, Live Load as LL and Wind Load as WL.

• Then add Dead load cases, Selfweight of factor 1.1, Member load for Main frame (39 kN/m2) and Member load for Gable frame (19.5 kN/m2)
• Then add Live load cases, Member load for Main frame (9 kN/m2) and Member load for Gable frame (4.5 kN/m2)
• Then add Wind load cases, Member load for Main frame (12 kN/m2) and Member load for Gable frame (6 kN/m2)

• Assign all load cases for members.

• Go to Analysis tab, click define command and define Perform Analysis as Static Check.

• Then click Run Analysis.
• Output shows 0 Errors.

• Open Output file and Go to Applied and Reaction load page.
• We can see the Reaction forces at columns.

Result:

• A Mono Sloped structure modelled and the given load cases are assigned.
• Also Found the Support reactions for columns.

For Load Case 1 (Dead Load)

• Applied Force at Y Direction = -2152.24 kN
• Summation Force at Y Direction = 2152.24 kN

For Load Case 2 (Live Load)

• Applied Force at Y Direction = -478 kN
• Summation Force at Y Direction = 478 kN
• For Load Case 3 (Wind Load)
• Applied Force at Y Direction = -638.51 kN
• Summation Force at Y Direction = -638.51 kN
• It is concluded that all applied forces are equal to Reaction forces. Hence the static equilibrium is maintained.