Week 3 Challenge

WEEK3 CHALLENGE

1.

1. State the primary load cases to be considered for design.

A) The primary load cases to be considered for design are:

• 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?

A)  One – Way slab:

                   The one-way slab is a slab, which is supported by parallel walls or beams, and whose length to breadth ratio is equal to or greater than two and it bends in only one direction (spanning direction) while it is transferring the loads to the two supporting walls or beams, because of its geometry. Simply stating it spans and bends in only one direction.

3. What is the value of unit weight of structural steel and soil?

A)  Unit Weights:

• Unit weight of Structural Steel is 7850 kg/m³
• Unit weight of soil is 18 kN/m³

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

A)  The following are different sections that can be defined using properties tab in STAAD Pro:

• Circular section
• Rectangular section
• Trapezoidal section
• Tee section
• General section
• Tapered Tube

5. Define Primary Beams.

A) Primary Beams:

                 The beams that are connecting columns for transferring loads of a structure directly to the columns are known as primary beams. Usually, primary beams are shear connected or simply supported and they are provided in a regular building structure. The depth of the primary beams is always greater than secondary beams. Primary beam act as a medium between columns and secondary beams.

2.

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

A) The factors and parameters that influence the design pressure intensity of a high-rise building are

• K1 = Probability Factor (risk coefficient)
• K2 = Terrain , height and structure size
• K3 = Topography factor
• Vz = Vb X k1 X k2 X k3 X k4.
• Pz=Design wind pressure in N/m2 at height z
• Vz =Design wind speed at any height z in m/s.

Design wind pressure:

• The wind pressure at any height above mean ground level shall be obtained by the following relationship between wind pressure and wind speed.
• Pz = 0.6 Vz^2

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

A) The load transfer process of a Reinforced Concrete Building

3.

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

A) One way Slab:

• If L/B ratio is greater than or equal to 2 then it is considered as One Way slab
• In one way, slab bending moment only occurs in shorter span direction
• The main reinforcement is provided in short span and distribution reinforcement is provided in a long span
• The crank is provided in two directions
• It is supported by beams on two opposite sides only
• The load is carried in one direction perpendicular to the supporting beam
• Bending occurs only in one direction i.e. in a short span
• The deflected shape of the one way slab is cylindrical
• Chajja and Varandah are practical examples of one way slab
• The one way slab is economical upto 3.6 mts
• In one way slab quantity of steel is less
• While designing a one way slab we provide less steel hence the depth of the slab increases as a result the thickness of one way slab is more as compared to the two way slab.

Two way Slab:

• If L/B ratio is less than 2 then it is considered as Two Way slab
• In two way slab bending moment occurs in both directions long and shorter span
• The main reinforcement is provided in both directions
• The crank is provided in four directions
• It is supported by beams on all four sides
• The load is carried in both directions
• Bending occurs in both directions
• The deflected shape of the two way slab is dish or saucer-like shape
• Two way slabs are used in the constructive floors of the multistorey building
• The two way slab is economical upto 6m x 6m
• In two way slab quantity of steel is more compared to one way slab
• While designing a two way slab we provide more steel hence the depth of the slab decreases as a result the thickness of two way slab is less as compared to the one way slab.

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

A)  Secondary beams:

• The beams that are connecting primary beams for transferring loads of a structure to the primary beams are known as primary beams. These beams are provided for supporting and reducing the deflection of beams and slabs.
• 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 rests on then 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 carries a certain amount of total load and help them to distribute the loads in the respective manner.
• In secondary beams, it is important to release the end moments at the supports, because of this all the secondary beams are provided with the 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)  a. 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 machine etc and this load 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 may be the load of the electrical fixture. All the loads which is permanent on the structure is called as the dead load of the structure.

 Dead Loads(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 are fully or partially vary from time to time on the structure.
• Live loads are not constant loads throughout of the life span of a structure.
• Live loads may vary time to time according to its usage or the occupancy.
• Live load may vary depend on the type of the 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 the roofs consist of all loads which are temporarily placed on the structure.

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 is arising 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 are 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 of the high-rise structure of the building.
• Mainly we considered this type of loads where there are high chances of earthquakes or we can say we have to consider this load in the earthquake prone areas, where there is a high risk of earthquake.
• 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 the time history analysis.

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

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

Assigning Properties:

• 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.
• Now there will be another option called Add just click on that option and then click OK.
• 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.

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.
• 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.

5. Brief the steps to be followed to create and assign secondary beams in STAAD Pro

A) 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.
• 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.
• Now after creating the secondary beam we need to make the secondary beam the simply supported beam.
• 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.
• After this there will be another option called releases just select Fx and then Mz. 
• Next there will be an option called Add just click on that option and then click on the OK.
• 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
2. Calculate the Dead load and Live load for the above building with reference to IS standards ( assume suitable sections )

A)

1. 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 topography factor from IS cide 875:Part 3 2015 clause no 6.3.3.

           k4 = Importance factor.

Given Data form 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

RESULT:

The wind pressure for the given building is 0.97117KN

 2. Calculate the Dead load and Live load for the above building with reference to IS standards (Assume suitable sections).

A)

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 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 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 masonary partition = 3.3 X 0.15 X 18.

The Dead Load due to the brick masonary 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 loads values for different rooms.
• As our building is the Hospital building, the all the proposed live load values for hospital building according to the IS 875:Part 3 has been listed below:

1. For ward rooms, dressing rooms, dormitories and lounges- 2.0 KN/sq.m.

2. Kitchen and Laboratory -3.0 KN/sq.m

3. Dining room and Cafeteria- 3.0 KN/sq.m

4. Toilets and Bathrooms- 2.0 KN/sq.m.

5. X-ray rooms and Operating Rooms- 3.0 KN/sq.m

6. Office and OPD Room- 2.5 KN/sq.m.

7. Corridor and Passages- 4.0 KN/sq.m