Week 3 Challenge

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:Part1).
• Live Load (IS 875:Part2).
• Wind Load (IS 875:Part3).
• Snow Load (IS 875:Part4).
• Seismic Load (IS 1893:2016).

2. What is a one-way slab?

Answer:

• A one-way slab is a uniformly loaded planar surface that flexes into a cylindrical shape, developing a bending force exclusively in one direction. One-way slabs are this kind of slab.
• A one-way slab is one where the larger span's ratio to its shorter span is greater than or equal to two.
• A one-way slab is one where the larger span's ratio to its shorter span is greater than or equal to two.

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/cu.m.
• Unit-weight of soil: 18KN/cu.m.

4. Name few section that can be defined using properties tab in Staad pro.

Answer: Few section that can be defined using propertise tab in Staad pro are listed below:

• Circular Sections.
• Rectangular Sections.
• I-Sections.
• T-Sections.
• Traped-I sections.
• Trapezoidal Sections.
• 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.

2.1 State the factors and parameters that influnece the design pressure intensity of a high rise building.

Answer:

The following section has covered the variables and elements that affect a high-rise building's design pressure intensity:

Therefore, the following formula can be used to govern the design speed:

Vz= VbXk1Xk2Xk3Xk4

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

k1= Risk Co-efficent factor.

k2= Terrain and height factor.

k3= Topography factor.

k4= Imfortance 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.

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

3.1 Differentiate one-way and two way salbs. Elaborate each slab type.

Answer:

One way slabs:

• One way slab is supported by a beam on two opposite sides only.
• In one way slab laod is carried in one direction perpendicular to the supported beam.
• The slab is regarded as a one-way slab if the L/b ratio is larger than or equal to 2.
• The one-way slab has a cylindrical deflected shape.
• There is less steel in a one-way slab.
• Only one direction of bending occurs in a one-way slab, and that is along the shorter span.

Two way slabs:

• A beam supports a two-way slab from all four sides.
• Additionally, load is carried in both directions in a two-way slab.
• The slab is referred to as a two-way slab if the L/b ratio is less than or equal to 2.
• The two-way slab has a saucer-like deflected shape.
• A two-way slab The amount of steel is greater than that of the on-way slab.
• In a two-way slab, bending occurs on both spans.

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

Answer:

• The secondary beams are essentially supported beams that are connected to the primary beams.
• Together with the major beams that rest on the slabs, secondary beams are given for a better distribution of slab loads.
• Therefore, the beams that are not attached to the columns directly are mostly known as secondary beams. The other beams, referred to as the primary beams, are directly supported by the column on which the secondary beams rest.
• Therefore, the secondary beams support a portion of the overall load and aid in the proper distribution of the stresses.
• Since it's crucial to release the end moments at the supports when building secondary beams, every secondary beam has a pinned connection.

3. Mention the promary load cases.

a. State the Indian Standards for each load case.

b. Details the parameters of each case.

c. Brief each load case.

Answer:

a. State the Indian Standards for each load case:

• Dead Load (IS 875:Part1).
• Live Load (IS 875:Part2).
• Wind Load (IS 875:Part3).
• Snow Load (IS 875:Part4).
• Seismic Load (IS 1893:2016).

b. Detail the parameters of each load case:

• Member loads.
• Concentration loads.
• Floor loads.
• Plate loads.

c. Brief each load case:

Dead load:

• The definition of a dead load is any continuous load in a structure, such as the weight of a machine or other item, that is brought about by the weight of members that are permanently attached to that particular structure.
• The dead load of the structure includes the loads on columns, beams, floors, walls, roofing, windows, and electrical fixtures, among other loads that are permanent on the structure.
• The following materials have some of their dead loads (unit weights):

a. Cement:1440 Kg/cu.m.

b. Steel: 7850 Kg/cu.m.

c. Sand: 1600 Kg/cu.m.

d. Concrete: 2400 Kg/cu.m.

e. RCC: 2500 Kg/cu.m.

d. Soil: 1800 Kg/cu.m.

Dead load factor is 1.

Live load:

• Live loads are those that fluctuate in full or in part over time on the structure.
• A structure's live loads fluctuate during the course of its lifetime.
• Live loads may fluctuate over time according on occupancy or usage.
• The live load may differ depending on the kind of construction, such as an industrial, commercial, or residential building.
• 3KN/sq.m Live Load is what we utilise for residential buildings.
• All loads temporarily positioned on the structure are considered live loads and are found on the floors and roofs. Among the unit weights (live loads) are:
  

a. All rooms and kitchen: 2KN/sq.m.

b. Bathrooms and Toilets: 2KN/sq.m.

c. Balcony: 3KN/cu.m.

d. Staircase/Corridors: 3KN/cu.m.

Wind load:

• A force acting on a structure as a result of the wind's impact is called wind load.
• One significant load scenario for the structure is wind load.
• We took into consideration this kind of load in cases when the structure is situated in hilly or mountainous regions where strong winds may have an impact on it. We must take the wind loads into account in those places.
• Our primary focus was determining the wind loads on the structure's roofs.

Earthquake load:-

• Earthquake loads, also known as seismic loads, are a significant load scenario that must be taken into account when planning and assessing the building's high-rise structure.
• This type of load was mostly taken into consideration in places with a high probability of earthquakes, or, to put it another way, in areas where there is a high risk of earthquakes.
• Additionally, the codal provision states that any of the two methods—spectrum analysis and time history analysis—may be used to conduct a dynamic seismic study.

4. Detail the steps involved in assigning properties and support to a model using Staad Pro sequentially.

Answer:

Assigning properties:

The following is a discussion of the stages involved in giving properties and support to a model using the STAAD Pro software:

• In order to assign properties to a model in STAAD Pro software, the first step is to navigate to the Properties tab. From there, a dialogue box will appear. Click the Define menu option.
• Select the desired sectional property, such as a circular section, rectangular section, trapezoidal section, etc., from the list of existing sectional properties that appears in the dialogue box that appears after clicking the "define" option.
• Once the section characteristics have been selected, enter the dimensions in that dialogue box.
• After selecting the Add option, which is the next one to appear, click OK.
• Next, assign the property to the model or structure. To do this, choose the newly generated section from the created section, click Assign to the selected view, select the assign option, assign the property, and you're done.
  

Assigning support:

• Click the "Create New" option in the dialogue box that appears after selecting the "Section" tab in the STAAD Pro software. This will allow you to assign the Support to the model electronically.
• A new dialogue box will appear after selecting that option, with other options such as Fixed, Pinned, Fixed But, etc.
• Click the desired end conditions. A "Add" option will then appear; simply click it.
• Finally, assign the support to the model or structure by selecting the newly generated support from the created section, clicking the Assign to the selected view option, clicking the assign option, assigning the property, and you're done.

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

Answer:

The following is a discussion of the processes needed to create and assign secondary beams in STAAD Pro:

• To designate the secondary beam in the model, we must first do a right-click on the screen and choose the pointer to serve as the beam cursor.
• Next, click the main beam and do a right-click on it. A selection of options will then display on the screen; select the "add node" option located at the halfway point.
• Upon selecting that option, the primary beam's midpoint node was immediately added.
• Once the node has been added at the middle of the primary beam, we must draw the beam. To do this, select the Add Beam option, begin adding the beam from the created node, and build the beam from the newly produced node all the way to the opposite primary beam.
• The reason the beam has been added to the nodes is that the software finds the mid-point at another beam automatically. And we refer to this beam as the secondary beam.
• We now need to turn the secondary beam into a simply supported beam after producing it.
• To accomplish that, first choose the beam. Then, select the specification tab. There, you'll see the beams option; simply click on it.
• Following this, there will be a release choice; choose Fx first, then Mz.
• A menu item named "Add" will appear after this; choose it, and then select "OK."
• Now that the support needs to be assigned, choose the newly formed support from the created section, click Assign to the selected view, click the assign option, assign the property, and you're done. This will assign the support to the model or structure.

Practical:

1. Calculate the wind pressure for the given building.

a. Building Dimension: 20mX30mX20m Height RCC.

b. Building usage: Hospital block in city centre.

c. Location: Darjeeling.

Aim: To calculate the wind pressure for the given building.

Procrdure:

Given Data:

Building dimensions: 20mX30mX20m

Building usage: Hospital block in city centre.

Location: Darjeeling.

Step1: 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 coefficient.

We can find the value of risk coefficient, 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 code 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= VbXk1Xk2Xk3Xk4.

Vz= 47X1.07X0.8X1X1.

Vz= 40.23 m/s.

Step2: Calculation of Wind pressure:

Pz= 0.6 X Vz^2

Pz= 0.6 X 40.23^2

Pz= 971.071N

Pz= 0.97117KN

In a similar manner, we may compute for other heights. This computation has been completed in an Excel file, which is shown below.

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

Answer:

• Dead load due to slad thickness:

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 slab= 20X30X25X0.15

Dead load due to the slab= 225 KN.

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

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

Total dead load due to the slab= 5X225

Total dead ;oad due to the slab= 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= 0.32X0.45X25

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

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

• Superimposed loads due to the Partition wllas:

Assuming 150mm thickness of brickwork.

Height of brickwork= Floor height - depth of the beam

Height of brickwork= 3.75 - 0.45

Height of brickwork= 3.3m

Therefore the dead load due to the brick masonary partition= 3.3X 0.15X 18

= 8.91 Kn/m.

Live load:

A specific structure's live load can be determined by utilising IS 875:Part 2, which specifies various live load values for various rooms.

Since our building is a hospital, the following is a list of all the suggested live load values for hospital buildings as per IS 875:Part 3:

OCCUPANCY LIVE LOAD

For ward rooms, dressing rooms, dormitories 2.0 KN?sq.m.

and lounges.

Kitchen and Laboratry 3.0 KN/sq.m.

Dining room and Cafeteria 3.0 KN/sq.m.

Toilets and Bathrooms 2.0 KN/ m.

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

Office and OPD Room 2.5 KN/sq.m.

Corridor and Passages 4.0 KN/sq.m.