Week 10 Challenge

1. AIM:

Purlin & Girts number determination problem

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

Create nodes as shown

Add beams to complete the structure

Insert nodes for purlins

To include the brick wall height of 2m , insert a node at 2m in the columns

Add nodes for the girts

2. AIM:

Taper the beam according to BMD envelope

For the STAAD file attached. For given loading taper the beam for the BMD envelope and check utility ratio within limit. (Design Parameter is already applied)

Procedure:

Download and open the attached file in StaadPro

Check the loadings and parameters applied

Now run the analysis

Check the utilization ratio

Now check the BMD for envelope load. Enable beam numbers in Label settings

Member Name	Tag	Utility Ratio	Remarks
End Column	1	0.634	The column is utilized for 63.4% only.
Middle Column	5	0.863	The column is utilized for 86.3% only. In this column no moment is to be transferred as moment is released.
End Column	2	0.598	The column is utilized for 59.8% only.
Rafter 1	3	0.866	The Rafter is utilized for 86.6% only.
Rafter 2	4	0.866	The Rafter is utilized for 86.6% only.

Now go to Analytical modeling and check the properties table

Check the node to node distance under Utilities - Node tools

Now modify the tapered section for the end columns to match with the BMD

Run the analysis to check the utilization ratio

We can see that the utilization ratio is very small. So we can decrease the tapered member size

So decrease it further more

Hence end columns are OK

Now add nodes to the beams

Assumption from gravity analysis standpoint is that the point of contraflexure in beams is at 0.2L from the column = 0.2 * 25.028 = 5m from either end of the nodes

The following are the tapered sections assumed for the members

Now check for the utilization ratio

3. AIM:

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

An Industrial building of plan dimension 10m x 5m to be constructed (as in Figure below). Building is situated in Chennai. Find the wind pressure on the wall of the building. Assume wall opening as 15 %. Bay spacing = 2.5m.

Procedure:

Step 1: Design Wind speed Vz

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

Design wind speed V_z = V_b x K₁ x K₂ x K₃ x K₄

Where K_1­ is Probability factor (risk coefficient) taken from Table 1

Considering general building and structure and V_b =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

For the height of 6m and Terrain category 3,

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

Step 2:-

Wind pressure Pz = 0.6 (V_z)²

Where,

P_z = wind pressure at height z in N/m²

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

Therefore

P_z = 0.6 x (52.325)² = 1.643 kN/m²

Step 3:-

Design wind pressure P_d = K_d x K_a x K_c x P_z

Where,

K_d = wind directionality factor 1.0 for cyclone affected region

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

(Tributary area = 6 x 2.5 = 15 m²)

K_c = combination factor 0.9

for framed structure

Therefore

P_d = 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 = (C_pe – C_pi) x A x P_d

Where,

C_pi = internal pressure coefficient

C_pi = +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

On roof:

4. AIM:

Find support reaction for given load. Static check problem

Mono sloped structure (20 degree) of width 10 m & length 30 m with bay spacing of 6 m having no wall sheeting (fully open all around) has following loads. Model the structure in STAAD. Apply Dead load, Live Load & Wind load on the structure and find the support reaction. Also perform static check if the total load applied on the structure is captured in support reaction for each load case.

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

Procedure:

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/m

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

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

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

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

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

• Open STAAD Software and create a New File.

Add beams to connect the nodes

Use translational repeat to create a length of 30m

Under loading, add DL, LL and WL cases

Under DL,

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

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

Now add LL

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

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

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

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

^{Now start assigning each load to the rafters}

^DL:

^LL:

WL:

Add parameters and perform Static Check

From this we could understand that the applied load is equal to the total reaction load

Hence static check is done