Week 3 Challenge

AIM:-

To explain the answer for each and every question from 1 to 3

INTRODUCTION:-

In this section of question from 1 to 3, I used staad.pro software to answer these question. Besides this, I have

used some reference from recorded video session of skill-lync.

PROCEDURE:-

Here first I will mention question one by one then answer then in the same.

Q1.

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.

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:

* The unit weight of a structural steel is 7850 kg/m^3

* The unit weight of soil is 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:

* Circular sections.

* Rectangular sections.

* I-Sections.

* T-Sections.

* Tapered-I sections.

* Trapezoidal sections.

* Custom Geometric Sections.

Circular section

Rectangular Sections

Tee Section

Tapered I section

Trapezoidal section

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.

Q2.

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

ANSWER:

The factors and Parameters influence the design pressure Intensity as per IS 875 part 3

1) Design wind speed

Vz= Vb K1 K2 K3 K4

where ,

Vb=Basic wind speed (m/sec)

K1=Risk coefficient (Table 1 of IS 875 (Part 3)-2015)

K2 = Terrain Roughness and height factor

K3=Topography factor (Clause 6.3.3 of IS 875 (Part 3)-2015)

K4 =Importance factor for Cyclic Region

2) Design wind pressure (pd)

pd= Kd Ka Kc pz

where,

Kd= Wind directionality factor

Ka=area averaging factor

Kc = Combination factoras

3) The wind pressure pz at any height z is given by

Pz= 0.6 Vz^2

where,

Pz= Design Wind speed in N/m2 at height z

Vz= Design wind speed in m/s.

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

ANSWER:-

The load transfer process of a reinforced concrete building is the following

> Slab - The load will be transferred from Slab to Primary Beams

> Secondary beam - If Secondary beams are provided its load transferred to the Primary beams

>imary beam - The loads from Primary beams to Columns

> Columns - The loads from Column to the Foundation

> Foundation - Foundation transferred the load to the soil beneath the footing.

Q3.

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

ANSWER:-

One-Way Slab	Two-Way Slab
* 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.
* 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.
* 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.
* The deflected shape of the one-way slab is cylindrical.	* The deflected shape of the two-way slab is saucer-like shape
* 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.
* 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:-

STEP-1:-

First open the staad.pro software and fill all the project details and select unit system as Metric. Then select the model as analytical.

STEP-2:-

Go to geometry add node then go to node cursor, Then create a cube of (X=5m, y=6, Z=8m)

STEP-3:-

STEP-6:-

Go to material tab and assign material as concrete.

RESULT:-

All the qusestions from 1 to 3 are answered with detailed elaboration.

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 hight Reinforced cement concrete building located in Darjeeling

INTRODUCTION:-

I calculted wind load by using IS875 part 3, where each and every formula and values are given with detailed explanation. Besides

this, all required data for this calculation has already provided by skill-lync side in a excel file.

PROCEDURE:-

I will mention each and every steps-

DETAILED PROCEDURE FOR

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

STEP-1:-

Calculattion for Design Wind Speed:-

We know

Vz = Vb x k1 x k2 x k3 x k4

where ,

Vb=Basic wind speed (m/sec)

K1=Risk coefficient (Table 1 of IS 875 (Part 3)-2015)

K2 = Terrain Roughness and height factor

K3=Topography factor (Clause 6.3.3 of IS 875 (Part 3)-2015)

K4 =Importance factor for Cyclic Region

AS Vb = 47 m/s ( Darjeeling ) Is 875 Annexure A

K1 = 1.07 (given in Table 1 of IS-875 part 3)

K2 = 0.80 ( Ht for 20m) (Category 4 as building is located in city center)

K3 = 1 [Topography factor (Clause 6.3.3 of IS 875 (Part 3)-2015)]

K4 = 1 [Clause 6.3.4 of IS 875 (Part 3)-2015] As this zone does not falls under cyclone zone

Therefore ,

Vz = 47 x 1.07 x 0.80 x 1 x 1

= 40.23 m/s

ii) Wind Pressure = Pz

Pz = 0.6 x Vz ^2

= 0.6 x 40.23 x 40.23

= 971.168 N

= 0.97 KN

RESULT:-

Similarly we can calculate for different heights

Height (m)	k2	Vz (m/s)	Pz	Pz in kN
3.5	0.80	40.232	971.17	0.97
6.5	0.80	40.232	971.17	0.97
9.5	0.80	40.232	971.17	0.97
12.5	0.80	40.232	971.17	0.97
15.5	0.80	40.232	971.17	0.97
18.5	0.80	40.232	971.17	0.97
20	0.80	40.232	971.17	0.97

Practical

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 of a 20m x 30m x 20m hight Reinforced cement concrete building located in Darjeeling.

INTRODUCTION:-

I calculted live by using IS875 part 2, where each values are given with detailed explanation. Besides this, dead load calculation

done manually considering the thickness, width and length of each section.

PROCEDURE:-

I will mention each and every steps-

DETAILED PROCEDURE FOR

1. Calculate the Dead load and Live load-
a. Building Dimension : 20m x 30m x 20m height RCC
b. Building usage : Hospital block in city centre
c. Location : Darjeeling

STEP-1:-

To Calculate Dead Load of Column , Beam , Wall and Slab are as follows -

A) For Column -

i) As we have assume Column Size as (0.5 x 0.5 )

ii) Length of column = 20m (upto roof level)

Therefore

Dead Load = L X B X H x Unit weight of Concrete

= 20 x 0.5 x 0.5 x 25

= 125 KN

So Dead Load of Each Column is 125 KN

2) For Beams -

i) As we have assume Beams Size as ( 0.300 x 0.450 m)

Therefore-

Dead Load = B X H x Unit weight of concrete

= 0.3 x 0.45 x 25

= 3.375 KN/m

So Dead Load of Beam is 3.375 KN/m

3) For wall - As it has Block

i) Assume the Height of wall 2.8 m

ii) Thickness of wall = 0.230 m

iii) Unit weight of Blocks = 20KN/m^3

Therefore

Dead Load = B X H x unit weight of block

= 0.230 x 2.8 x 20

= 12.88 KN / m

So Dead Load of wall is 12.88 KN /m

4) For Slab

i) Assume the thickness of slab = 200 mm

Therefore

Dead Load = thickness X Unit weight of concrete

= 0.200 x 25

= 5 KN/ m^2

Dead Load of Slab = 5 KN/m^2

2) For Live Load - As per IS 875 Part 2

Sl. No.	Description	UDL Load (kN/m2)	Concentrated Load (kN)
1.	Bathrooms and toilets in all types of buildings	2	1.8
3	Office rooms in
	(i)Hostel, hotels, hospitals and business building with separate store	2.5	2.7
	(ii)In assembly buildings	3	4.5
4.	Kitchens in
	(i) Dwelling houses	2	1.8
	(ii) Hostels, hotels and hospitals	3	4.5
5.	Banking halls, class rooms, X-ray rooms, operation rooms	3	4.5
6.	Dining rooms in
	(i)Educational buildings, institutional and mercantile buildings	3	2.7
	(ii)Hostels and hotels	4	2.7
7.	Corridors, passages, stair cases in
	(i) Dwelling houses, hotels, and hostels	3	4.5
	(ii)Educational, institutional and assembly buildings	4	4.5
	(iii) Mercantile buildings	5	4.5
8.	Reading rooms in libraries
	(i)With separate storage	3	4.5
	(ii)Without separate storage	4	4.5
9.	Assembly areas in assembly buildings
	(i)With fixed seats	5
	(ii)Without fixed seats	5	3.6
10.	Store rooms in educational buildings	5	4.5
11.	Store room in libraries	6 for a height of 2.24 2kN/m2 for every 1m additional height	4.5
12.	Boiler rooms and plant rooms in
	(i)Hostels, hotels and hospitals, mercantile and industrial buildings	5	4.5
	(ii)Assembly and storage buildings	7.5	4.5

RESULT:-

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