Week 3 Challenge

Questions 1.

1. 1. State the primary load cases to be considered for design.
2. What is One – Way slab?
3. What is the value of unit weight of structural steel and soil?
4. Name few sections that can be defined using Properties tab in STAAD Pro.
5. Define Primary Beams.

Answers.

State the primary load cases to be considered for design.

The primary load cases that are to be considered for design are listed below:

• Dead Load (IS 875: Part I).
• Live Load (IS 875: Part II).
• Wind Load (IS 875: Part III).
• Snow Load (IS 875: Part IV).
• Seismic Load (IS 1893: 2016).

2. What is One – Way slab?

ONE-WAY SLAB:

• One-way slab is a plane surface uniformly loaded which deforms into a cylindrical shape and hence the bending moment develops only in one direction. This type of slab is called a one-way slab.
• One-way slab is a slab in which the ratio of the longer span to its shorter span is more than or equal to 2.
• One-way slab load distribution takes place along its shorter span. Hence bending also happens at the shorter span of the slab.
• What is the value of unit weight of structural steel and soil?

The values of the unit weight of structural steel and soil are given below:

• Unit-weight of Structural Steel: 7850 Kg/cu.m.
• Unit-weight of Soil: 18 KN/cu.m.

4. Name a few sections that can be defined using the Properties tab in STAAD Pro.

A few sections that can be defined using the Properties tab in STAAD Pro are listed below:

• Circular sections.
• Rectangular sections.
• I-Sections.
• T-Sections.
• Tapered-I sections.
• Trapezoidal sections.
• Custom Geometric Sections.

5. Define Primary Beams.

Primary Beams:

• The primary beams are the beams that are directly connected with the column on both ends.
• The function of the primary beams is to transfer the load of the slab directly to the column that they are resting on.

Questions 2.

1. State the factors and parameters that influence the design pressure intensity of a high-rise building.
2. Mention the load transfer process of a Reinforced Concrete Building.

Answers.

2. 1. State the factors and parameters that influence the design pressure intensity of a high-rise building.

• The factors and parameters that influence the design pressure intensity of a high-rise building have been discussed below:

So the design speed can be governed by the below given formula:

Vz = Vb X k1 X k2 X k3 X k4.

where, Vz = Basic wind speed, which is based on the location of the project in India.

k1 = Risk Co-efficient factor.

k2 = Terrain and height factor.

k3 = Topography factor.

k4 = Importance factor.

And then the design wind pressure is given by the below formula:

Pz = 0.6 X Vz^2.

2. Mention the load transfer process of a Reinforced Concrete Building.

• The load transfer procedure of a reinforced concrete building has been illustrated below in the form of a flowchart:

Questions 3.

1. Differentiate One-Way and Two-way slabs. Elaborate on each slab type.
   2. Explain the concept of secondary beams with a proper sketch.
   3. Mention the Primary load cases.
   a. State the Indian Standards for each load case
   b. Detail the parameters of each load case
   c. Brief each load case
   4. Detail the steps involved in assigning Properties and Support to a model using STAAD Pro sequentially.
   5. Brief the steps to be followed to create and assign secondary beams in STAAD Pro.

Answers.

3. 1. Differentiate One-Way and Two-way slabs. Elaborate on each slab type.

One-Way Slabs	Two-Way Slabs
a. One-way slab is supported by a beam on two opposite sides only.	a. Two-way slab is supported by a beam on all four sides.
b. In a one-way slab, the load is carried in one direction perpendicular to the supported beam.	b. And in a two-way slab load is carried in both directions.
c. If the L/b ratio is greater than or equal to 2, then the slab is considered as the One-way slab.	c. If the L/b ratio is less than or equal to 2, then the slab is considered as the Two-way slab.
d. The deflected shape of the one-way slab is cylindrical.	d. The deflected shape of the two-way slab is a saucer-like shape.
e. In a one-way slab, the quantity of steel is less.	e. In the two-way slab, the quantity of steel is more as compared to the one-way slab.
f. In a one-way slab, bending takes place only in one direction that is along the shorter span.	f. In a two-way slab, bending takes place along both spans.

2. Explain the concept of secondary beams with a proper sketch.

• The secondary beams are primarily connected to the primary beams mainly simply supported beams.
• The secondary beams are provided for the better distribution of slab loads along with the primary beams which rest on the slabs.
• So mainly the secondary beams are the beams that are not directly connected to the columns. The secondary beams rest on the other beams that are directly rest on the column and those beams are known as the primary beams.
• So the secondary beams carry a certain amount of total load and help them to distribute the loads respectively.
• In secondary beams, it is important to release the end moments at the supports, because of this all the secondary beams are provided with a pinned connection.

3. Mention the Primary load cases.
   a. State the Indian Standards for each load case
   b. Detail the parameters of each load case
   c. Brief each load case

a.State the Indian Standards for each load case:

• Dead Load (IS 875:Part 1).
• Live Load (IS 875:Part 2).
• Wind Load (IS 875:Part 3).
• Snow Load (IS 875:Part 4).
• Earthquake Load (IS1893:2016).

b.Detail the parameters of each load case:

• Member Loads.
• Concentration Loads.
• Floor Loads.
• Plate Loads.

c.Brief each load case:

Dead Load:

• The dead load is defined as the continuous load in a structure i.e. the load of a machine etc and this load is caused by the weight of the members that are permanently attached to the particular structure.
• The dead load of the structure comprised of the load of the column, load of the beam, loads of the floor slab, roofing, walls, the load of the windows or maybe the load of the electrical fixture. All the loads which are permanent on the structure are called the dead load of the structure.

Some of the Dead Load(Unit Weight) of the following materials:

• Cement : 1440 Kg/cu.m.
• Steel: 7850 Kg/cu.m.
• Sand: 1600 Kg/cu.m.
• Concrete : 2400 Kg/cu.m.
• RCC: 2500 Kg/cu.m.
• Soil: 1800 Kg/cu.m.
• Dead Load factor is 1.

Live Load:

• Live loads are the loads that fully or partially vary from time to time on the structure.
• Live loads are not constant loads throughout the life span of a structure.
• Live loads may vary from time to time according to their usage or occupancy.
• Live load may vary depending on the type of structure such as the residential building, commercial building, industrial building etc.
• For residential building we use 3KN/sq.m Live Load.
• Live loads on floors and roofs consist of all loads which are temporarily placed on the structure.

Some of the live loads (Unit Weight) are:

• All rooms and kitchen: 2 KN/sq.m.
• Bathrooms and Toilets: 2 KN/sq.m.
• Balcony : 3 KN/cu.m.
• Staircase/Corridors : 3 KN/cu.m.

Wind Load:

• Wind Load is a force on a structure that arises from the impact of the wind.
• Wind load is an important load case for the structure.
• We considered this type of load where the structure is situated or located in the hilly areas of the mountain areas where we can expect high speed of the winds that can affect the structure.
• In those areas we need to consider the wind loads.
• Mainly we calculated the wind loads for the roofs of the structure.

Earthquake Load:

• Earthquake loads or seismic loads are also an important load case that needs to be considered while designing and analysing the high-rise structure of the building.
• Mainly we considered this type of load where there are high chances of earthquakes or we can say we have to consider this load in earthquake-prone areas, where there is a high risk of earthquakes.
• And according to the codal provision, the dynamic seismic analysis can be carried out via any of the two methods and these two methods are spectrum analysis and time history analysis.

• Question 4.

• Detail the steps involved in assigning Properties and Support to a model using STAAD Pro sequentially.

  Assigning Properties:

  The steps that are involved in assigning the properties and support to a model using the STAAD Pro software have been discussed below:

  • The first step in order to assigning the properties to a model in a model using STAAD Pro software, go to the Properties tab and then there will be a dialogue box and in that dialogue box there is an option called Define menu just click on that option.
  • After clicking the define option there will be another dialogue box, in that dialogue box there will be a list of existing sectional properties, from that list choose the desired sectional property like Circular section, Rectangular section, Trapezoidal section etc.
  • After choosing the section properties specify the dimensions in that dialogue box.
  • And now there will be another option called Add just click on that option and then click OK.
  • And now assign the property to the model or structure and to assign the property select the section that we have just created from the created section and click on the opinion Assign to the selected view and then click on the assign option and then assign the property and done.

  Assigning Support:

  • In order to assign the Support to the model using the STAAD Pro software, go to the Section tab and then there will be another dialogue box and in that dialogue box there will be an option called Create New just click on that option.
  • After clicking on that option there will be another dialogue box and in that dialogue box there will be another options like Fixed, Pinned, Fixed but etc.
  • Click on the desired end conditions, then there will be an option called Add just click on that option.
  • And now assign the Support to the model or structure and to assign the support select the support that we have just created from the created section and click on the opinion Assign to the selected view and then click on the assign option and then assign the property and done.

  • Question 5.

  • Brief the steps to be followed to create and assign secondary beams in STAAD Pro.
  • The steps that are involved in creating and assigning the secondary beams in STAAD Pro are discussed below:
  • In order to assign the secondary beam in the model, first we need to right-click on the screen and select the cursor as the beam cursor.
  • After that click the primary beam and right-click on the beam ,after that there will be a list of various options appears on the screen and in that list there will be an option called add node at the mid-point just click on that option.
  • After clicking on that option the node at the mid-point of the primary beam added automatically.
  • After added the node at the mid-point of the primary beam, now we have to draw the beam, and to add beam we need to go to the option Add beam and start adding the beam from the created node and create beam from the node that we have just created up to the opposite primary beam.
  • And the thing is that the software automatically locates the mid-point at the another beam, and that is how the beam has been added on the nodes. And this beam is called as the secondary beam.
  • And now after creating the secondary beam we need to make the secondary beam the simply supported beam.
  • And in order to do that first select the beam after that go to the specification tab then there will be an option called beams just click on that option.
  • And after this there will be another option called releases just select Fx and then Mz.
  • After this there will be an option called Add just click on that option and then click on the OK.
  • And now we have to assign the support to the model or structure, and to assign the support select the support that we have just created from the created section, and click on the opinion Assign to the selected view and then click on the assign option and then assign the property and done.

  Practical
  1. Calculate the wind pressure for the given building
  a. Building Dimension : 20m x 30m x 20m height RCC
  b. Building usage : Hospital block in city centre
  c. Location : Darjeeling.

AIM:

• To calculate the wind pressure for the given building.

PROCEDURE:

Given Data :

Building dimensions: 20m X 30m X 20m.

Building usage: Hospital Block.

Location: Darjeeling.

STEP 1: Calculate Design Wind Speed:

We know the formula to calculate the design wind speed, that is:

Vz=Vbxk1xk2xk3xk4

where, Vb = Basic Wind Speed (m/sec).

k1 = Risk Co-efficient.

We can find the value of risk co-efficient,k1 from IS code 875:Part 3 2015, Table 1).

k2 = Terrain Roughness and Height Factor.

k3 = Topography Factor.

We can find the value of the topography factor from IS cide 875:Part 3 2015 clause no 6.3.3.

k4 = Importance factor.

Given Data from the IS code:

• Vb = 47 m/s.
• k1 = 1.07.
• k2 = 0.8.
• k3 = 1.
• k4 = 1.

CALCULATION:

Vz = Vb X k1 X k2 X k3 X k4.

= 47 X 1.07 X 0.8 X 1 X 1.

Vz = 40.23 m/s.

STEP 2: Calculation of Wind pressure:

Pz=0.6xVz^2

=0.6x(40.23)^2

= 971.071 N

Pz=0.97117KN

Similarly, we can calculate the for different heights, the calculation for different heights has been done in the Excel sheet so excel sheet given below:

Question 2. Calculate the Dead load and Live load for the above building concerning IS standards (Assume suitable sections).

Answer.

Dead Load due to Slabs:

Assuming the slab thickness of 150mm.

Dead Load due to the Slabs = Length X Breadth X Height X Slab Thickness.

Dead Load due to the Slabs = 20 X 30 X 25 X 0.15

Dead Load due to the Slabs = 225 KN.

In this project we have a total of five-floor slabs, therefore:

Total dead load due to all the slabs = 5 X Dead Load due to the Slabs.

Total dead load due to all the slabs = 5 X 225.

Total dead load due to all the slabs = 1125 KN.

Dead Load due to the beams:

Assume the beams of cross-section 350X450mm.

Dead load due to the beams = Size of the beam X Height.

Dead load due to the beams = 0.35 X 0.45 X 25.

Dead load due to the beams = 3.94 KN/m.

Dead load due to the beams = 4 KN/m.

Superimposed Loads due to the Partition walls:

Assuming a 150mm thickness of brickwork.

Height of brickwork = Floor height - depth of the beam.

Height of the brickwork = 3.75 - 0.45.

Height of brickwork = 3.3 meters.

Therefore, the dead load due to the brick masonry partition = 3.3 X 0.15 X 18.

The Dead Load due to the brick masonry partition = 8.91 KN/m.

Live Load:

• We can calculate the live load of a given structure by using the IS 875:Part 2, which states different live load values for different rooms.
• As our building is the Hospital building, all the proposed live load values for the hospital building according to the IS 875:Part 3 have been listed below:

LIVE LOAD

2.0 KN/sq.m.

3.0 KN/sq.m

3.0 KN/sq.m

2.0 KN/sq.m

3.0 KN/sq.m

2.5 KN/sq.m

4.0 KN/sq.m