Week 3 Challenge

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

ANSWER:

. 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: One way slab is the slab in which the ratio of its longer span to the shorter span is more than or equal to 2.

. The load distribution in a one-way slab takes place along its shorter span.

. Consequently, the bending also happens along the shorter span of the one-way slab.

. As a result, one-way slabs are generally rested over only two beams as opposed to the two-way slabs that are rested over four beams along their periphery.

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

ANSWER: Structural Steel - 78 kn/m^3

. Soil - 18 kn/m^3

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

ANSWER:

. Circular

. Rectangular

. T - Sections

. I - sections

. Trapezoidal

. Tapered - I

. Custom Geometric Sections

5. Define Primary Beams.

ANSWER: Primary beams are then ones that are connected with columns on both ends.

. Primary beams are supposed to transfer loads of slabs directly to the column they are resting on.

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:

                        

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

 ANSWER:

ONE-WAY SLABS	TWO-WAY SLABS
L/B ration is more than or equal to 2.	L/B ratio is less than 2.
Bending takes place only in one direction that is along the shorter span.	Bending takes place along both spans.
One-way slabs are supported by beams in only two directions.	Two-way slabs are supported by beams on all four sides.
Main steel is only provided in one direction that is along the direction of bending.	Main steel is provided in both directions which are X and Y.

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

 ANSWER: Secondary Beams

Secondary beams are provided for better distribution of various slab loads along the primary beams on which the slab rests. These are the beams that are not directly connected to the columns and rather rest on other beams that are directly resting on columns also known as the primary beams. Secondary beams are supposed to carry a certain amount of the total load and help distribute the same in an improved fashion. Also, it is important to release end moments at the support for secondary beams. And for the same purpose, all the secondary beams are provided with pinned connections.

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: Dead Load (IS 875 - Part I)

Dead loads are the gravity loads generated due to the self-weight of the structure. The name is dead because these loads are static and do not change throughout the service period of the structure.  As we know, all the structural elements such as slabs, beams, and columns are heavy and possess a significant amount of load that needs to be accounted for. Hence, part one of the IS 875 specifically deals with the density of structural materials using which we can obtain the dead load caused based on the size of the section.

. Live Load (IS 875 - Part II)

 Live loads are loads that are not constant and may vary during the functioning of a building. Live loads are different for different types of structures. For example, the live load for an industrial building with heavy machinery is certainly going to be more than that of a residential structure. Similarly, it also varies with the type of room and its occupancy. Therefore the part of IS 875 provides us with the guidance on how to consider live loads for different structures and different parts of the same building. Accordingly, the live load for all rooms, bathrooms, and kitchens is taken as 2kn/m^2. On the other hand, it is taken as 3kn/m^2 for stairs and hall areas where people are supposed to gather.

. Wind Load (IS 875 - Part III)

Wind load is also an important load case that is dealt with in part three of the Indian standard 875. The formula and all the governing factors are briefly explained in this code so as to obtain the total design wind pressure. 

. Snow Load (IS 875 - Part IV)

 Snow load is a governing load case for structures that are located in the mountain and hilly areas that encounter regular snow rains. The loads are generally calculated for the building roof.

. Earthquake Load (IS 1893 - 2016)

Earthquake loads or seismic loads are also a primary type of load case considered for the design and analysis of high-rise buildings. And as per the codal provisions, the dynamic seismic analysis can be conducted via any one of the two methods namely response spectrum analysis and time history analysis.

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

 ANSWER: Go to the properties tab and click on the define menu.

. A list of existing sectional properties will be generated.

. Choose the desired section from the list is it circular, rectangular, trapezoidal, T-section, etc.

. Specify the section dimensions in the dialogue box.

. Click on add and then ok.

. Now, to assign, select the same section from the cerated section list and check the radio button assign selected to the view and done.

. Similarly to assign end conditions, go to supports, create new, specify fixed, pinned, or other.

. Click on add>then select the added and assign it to the view.

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 the city center
c. Location: Darjeeling

 ANSWER:

*Screenshot of an excel spreadsheet that was created to calculate the design wind load for the given building has been attached 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 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

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

. 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

. Live Load

As the proposed project is a hospital building, the IS 875 PART III states different live load values for different rooms within a hospital that are listed below:

OCCUPANCY	LIVE LOAD
For ward rooms, dressing rooms, dormitories, and lounges	2.0 KN/m^2
Kitchen and laboratory	3.0 KN/m^2
Dining room, cafeteria	3.0 KN/m^2
Toilets and bathrooms	2.0 KN/m^2
X-ray room and operating theatre	3.0 KN/m^2
Office and OPD room	2.5 KN/m^2
Corridor, passages	4.0 KN/m^2