Week 3 Challenge

QUESTION:-1.

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

ANSWER:

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?

ANSWER:

ONE-WAY SLAB:-

• One-way slab is a plane surface of uniformly loaded which deforms into a cylindrical shape and hence the bending moment develops only in one direction. These type of slab is called 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.
• one-way slabs are generally rested over only two beams.

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

ANSWER:

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

• Unit-weight of Structural Steel: 7850Kg/m^3
• Soil - 18 kn/m^3

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

ANSWER:

Few sections that can be defined using Properties tab in STAAD Pro are listed below:

1. Circular sections.
2. Rectangular sections.
3. I-Sections.
4. T-Sections.
5. Tapered-I sections.
6. Trapezoidal sections.
7. Custom Geometric Sections.

5. Define Primary Beams.

ANSWER:

Primary Beams:-

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

QUESTION:-2.

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

ANSWER:

Below are the factors governing the design pressure intensity of a high-rise building.

Design wind speed is given by the following formula:

Vz = Vb x k1 x k2 x k3 x k4

Here,

• Vb -> Basic wind speed based on where the project is located within India.
• K1 -> Risk coefficient factor based on the basic wind speed of the location. (Table 1 - IS875 PART3)
• K2 -> Terrain and height factor.
• K3 -> Topography factor.
• K4 -> Importance factor for cyclonic regions

Then finally, the design wind pressure (Pz) is given by:

Pz = 0.6 X (Vz)^2

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

ANSWER:

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

QUESTION:-3

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

ANSWER:-

Sr No.	One-Way Slab	Two-Way Slab
1	One-way slab is supported by a beam on two opposite sides only.	Two-way slab is supported by beam on all the four sides.
2	In one-way slab load is carried in one direction perpendicular to the supported beam.	And in two-way slab load is carried in both directions.
3	If L/b ratio is greater than or equal to 2,then the slab is considered as the One-way slab.	If L/b ratio is less than or equal to 2,then the slab is considered as the Two-way slab.
4	The deflected shape of the one-way slab is cylindrical.	The deflected shape of the two-way slab is saucer-like shape
5	In one-way slab the quantity of steel is less	In two-way slab the quantity of steel is more as compared to the onw-way slab.
6	In one-way slab,bending takes place only in one direction that is along the shorter span.	In two-way slab,bending takes place along the both span.

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

ANSWER:-

SECONDARY BEAM :-

• 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 thed 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

ANSWER:

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 machine etc and this load caused by the weight of the members that are pemanently 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.

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 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. 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 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.
• 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 the time history analysis.

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

ANSWER:

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 dialoige box and in that dialouge box there is an option called Define menu just click on that option.
• After clicking the define option there will be another dialouge box,in that dialouge 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 dialouge 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 opion 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 dialouge box and in that dialouge box there will be an option called Create New just click on that option.
• After clicking on that option there will be another dialouge box and in that dialouge 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 opion Assign to the selected view and then click on the assign option and then assign the property and done

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

ANSWER:

• Right click on your staad screen and make sure you have selected the beam cursor.
• Now select the primary beam and right click over the same.
• A list of various options will appear one of which will be add node at the mid point.
• So, click on the add a node at mid point and the same will be automatically added.
• Now click on add beam optoin and start the from the added node and select the opposite beam.
• The software automatically locates the mid point of the opposite beam and thus the beam is added.
• Now, you need to make the secondary beam simply supported.
• So, select the beam>go to the specifications>beams>releases>Fx>Mz>add>OK.
• Now select the added specification and assign it to the view by checking the radio button.

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

ANSWER:

AIM:

• To calculate the Wind pressure of a 20m Height Reinforced cement concrete building located in Darjeeling.

INTRODUCTION:

• I Calculated wind load by using IS875 part-3,where each and every formula and values are given with detailed explanation. Besides this all required load data for this calculation has already provided by skill-lynce side in a excel file.

PROCEDURE:

Given:

• Building dimensions:- 20m * 30m * 20m.
• Buidling usage:- Hospital Block.
• Location of the building:- Darjeeling.

STEP 1:-

Calculate Design Wind Speed:-

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

Vz=Vb*k1*k2*k3*k4

Where,

Vb=Basic wind speed(m/sec)

k1=Risk Co-efficent.

We can find the value of Risk Co-efficent,k1 from IS code 875:Part-3 2015,Table1

k2=Terrain Roughness and height Factor

k3 =Topography Factor.

We can find the value of topography factor from IS code 875:part 3 2015 Caluse 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

STEP-1

CALCULATION:

Vz=Vb*k1*k2*k3*k4

Vz=47.1*1.07*0.8*1*1

Vz=40.23m/s

STEP:-2

Calculation of Wind pressure

Pz=0.6*Vz^2

Pz=0.6*(40.23)^2

Pz=971.071N

Pz=0.97117kN

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

QUESTION:-2.

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

ANSWER:

AIM:

• To calculate the dead load and live load for the above building with reference to IS standards(Assume suitable sections).

INTRODUCTION:

• I calculated live by using IS-875 part-2,Where each values are given with detailed explanation Besides this dead load calculation done manually considering the thickness width and length of eah section.

PROCEDURE:

STEP-1

Dead Load due to Slabs

• Assuming 150mm thick slabs,
• Dead load due to one slab = 20 x 30 x 25 x 0.15 = 225 kn
• Now, as there are a total of five slabs in the building,
• Total dead load due to all the slabs = 225 x 5 = 1125 kn

STEP-2

Dead Load due to the beams

• Considering the beams of cross-section 350 x 450 mm.
• Dead load due to beams = 0.35 x 0.45 x 25 = 3.94 kn/m
• Rounding off the above value, we get,
• Dead load due to the beams = 4 KN/m

STEP-3

Superimposed loads due to the partition walls

• Assuming 150 mm thick brickwork.
• Height of the brickwork = floor height - beam depth = 3.75 - 0.45 = 3.3 meters

Therefore,

• Dead load due to the brick masonry partition = 3.3 x 0.15 x 18 = 8.91 KN/m

STEP-4

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:

Sr No	OCCUPANCY	LIVE LOAD
1	For ward rooms,dressing rooms,dormitories and lounges	2.0KNm2">2.0KNm22.0KNm2
2	Kitchen and Laboratory	3.0KNm2">3.0KNm23.0KNm2
3	Dining room and Cafeteria	3.0KNm2">3.0KNm23.0KNm2
4	Toilets and Bathrooms	2.0KNm2">2.0KNm22.0KNm2
5	X-ray rooms and Operating Rooms	3.0KNm2">3.0KNm23.0KNm2
6	Office and OPD Room	2.5KNm2">2.5KNm22.5KNm2
7	Corridor and Passages	4.0KNm2">4.0KNm2

RESULT:

• All the dead and live loads are shown as per given question.