Week 3 Challenge

1.

1.

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.

• The bending force only develops in one direction in a one-way slab because it is a planar surface that is uniformly loaded and deforms into a cylindrical shape. The one-way slab is the term given to this kind of 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.
• Along its shorter span, the slab load is distributed in a single direction. As a result, bending also occurs at the slab's shorter span.

3.

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

• Unit-weight of Structural Steel: 7850 Kg/cu.m.
• Unit-weight of Soil: 18 KN/cu.m.

4.

A few sections that can be defined using the Properties tab in STAAD Pro are listed below:

• Circular sections.
• Rectangular sections.
• I-Sections.
• T-Sections.
• Tapered-I sections.
• Trapezoidal sections.
• Custom Geometric Sections.

5.

• The beams that are immediately attached to the column on both ends are known as main beams.
• The primary beam's job is to convey the slab's weight straight to the column they are supported.

2.

1.

• The factors and parameters that influence the design pressure intensity of a high-rise building have been discussed below:

So the design speed can be governed by the below-given formula:

Vz = Vb X k1 X k2 X k3 X k4.

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

k1 = Risk Co-efficient factor.

k2 = Terrain and height factor.

k3 = Topography factor.

k4 = Importance factor.

And then the design wind pressure is given by the below formula:

Pz = 0.6 X Vz^2.

2. Gravity load (Dead + Live) on the floor and roof slabs are transferred to the columns, down to the foundations, and then to the supporting soil beneath.

3.

1. Differentiate One-Way and Two-way slabs.

One-Way Slabs	Two-Way Slabs
a. The one-way slab is supported by a beam on two opposite sides only.	a. Two-way slab is supported by a beam on all four sides.
b. in a one-way slab load is carried in one direction perpendicular to the supported beam.	b. And in a two-way slab load is carried in both directions.
c. If the L/b ratio is greater than or equal to 2, then the slab is considered a One-way slab.	c. If the L/b ratio is less than or equal to 2, then the slab is considered as the Two-way slab.
d. The deflected shape of the one-way slab is cylindrical.	d. The deflected shape of the two-way slab is saucer-like shape.
e. In a one-way slab, the quantity of steel is less.	e. In a two-way slab, the quantity of steel is more as compared to the one-way slab.
f. In a one-way slab, bending takes place only in one direction that is along the shorter span.	f. In a two-way slab, bending takes place along both spans.

2.

• The primary beams and subsidiary beams are generally joined by simply supported beams.
• Along with the major beams that rest on the slabs, supplemental beams are offered for better distribution of slab loads.
• Therefore, the beams that are mostly secondary beams are those that are not directly attached to the columns.
• The primary beams are the other beams that directly support the column, on top of which the secondary beams rest.
• Therefore, the secondary beams carry a portion of the overall weight and aid in the proper load distribution.
• It is crucial for secondary beams to release end moments at the supports, which is why all secondary beams have pinned connections.

3. 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 continuous load in a structure, such as the weight of a machine or other object, and this load induced by the weight of the members that are permanently linked to the specific structure are both referred to as dead loads.
• The dead load of the building is made up of loads of the beam, the column, the floor slab, the roof, the walls, the windows, and sometimes the electrical fixtures. The term "dead load" refers to all loads that are placed permanently on a 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 those placed on a structure that occasionally change in full or part.
• Live loads are not consistent loads over the course of a structure's life.
• Live loads may occasionally change depending on how it is used or how full the space is.
• Depending on the kind of structure, such as a residential building, a commercial building, an industrial building, etc., the live load may vary.
• 3KN/sq.m Live Load is what we utilize for residential buildings.
• Live loads are all loads that are temporarily imposed on the structure, and they include loads on floors and roofs.

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

• A force on a building known as wind load results from the wind's influence.
• For the structure, wind load is a significant load condition.
• We took this sort of load into account in locations where the structure is positioned or located that are hilly or mountainous where we may anticipate strong winds that could have an impact on the building.
• The wind loads in certain places must be taken into account.
• We primarily calculated the structure's roofs' wind loads.

Earthquake Load

• Seismic loads, often known as earthquake loads, are significant load situations that must be taken into account while planning and analyzing the high-rise structure of the building.
• We mainly took into account this sort of load in earthquake-prone locations where there is a significant probability of earthquakes, or you might say we had to take into account this load there.
• The codal requirement further states that any of the two methods—spectrum analysis or time history analysis—can be used to conduct dynamic seismic analysis.

Q4.

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 assigning properties to a model using the STAAD Pro program is to choose the Properties tab. There will then be a dialogue box; just click the Define menu item in the dialogue box.
• After selecting the define option, a new dialogue box will appear with a list of the sectional properties that are currently in use.
• From this list, select the required sectional property, such as a circular section, a rectangular section, a trapezoidal section, etc.
• In that dialogue box, after selecting the section attributes, enter the dimensions.
• There will then be a new option labeled Add; simply choose it before selecting OK.
• And now assign the property to the model or structure 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 and done.

Assigning Support:

• When using the STAAD Pro program, choose the Section tab to bring up a new dialogue box with the opportunity to create a new model; just click that option to assign Support to the model.
• Following the selection of that choice, a new dialogue box will appear with additional options such as Fixed, Pinned, Fixed but, etc.
• After selecting the required end conditions, a button labeled "Add" will appear; simply choose that button.

Q5.

The procedures for designing and allocating secondary beams in STAAD Pro are described below:

• Right-clicking on the screen and choosing the cursor as the beam cursor are the initial steps needed to allocate the secondary beam in the model.
• Then, after selecting the principal beam with the right mouse button, a list of options will show on the screen.
• At the midpoint of that list, click the option labeled "Add Node" to proceed.
• After selecting that choice, a node was automatically placed at the major beam's midpoint.
• After putting the node at the main beam's midpoint, we must now draw the beam.
• To add the beam, select the Add beam option, begin adding the beam from the newly generated node, and then build a beam from the newly created node up to the opposite primary beam.
• The program automatically determines the mid-point at a different beam, which is how the beam was added to the nodes. And we refer to this beam as the secondary beam.
• We must now turn the second beam into a simply supported beam after producing it.
• To achieve that, first choose the beam, then pick the specification tab. There, you should find an option labeled "beams"; simply select it.
• Following this, choose Releases from the menu, then Fx, and finally Mz.
• Once this is done, select the Add option by clicking on it and then select OK.
• And now we must assign support to the model or structure.
• To accomplish this, choose the support from the newly formed section, pick the opinion Assign to the selected view option, select the assign option, assign the property, and then click the Finish 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

Ans: For Wind Pressure (Pz) = 0.6 Vz^2

Where Vz = Vb x k1 x k2 x k3

The Structure in category - 4 and Class B and designed for 100 years.

From IS 875 Part 2

Vb = 47 m/s
K1 = 1.07 (Table 1, Clause 5.3.1)
K2 = 0.76 (Table 2, Clause 5.3.2.2) (Category- 4 ; Class - B)
k3 = 1 (Clause 5.3.3.1)

Hence; Vz = 47 x 1.07 x 0.76 x 1
= 38.22 m/s

And Pz = 0.6 (38.22)^2
= 0.87 N/m^2

Height	K2	Vz	Pz	Pz in KN
3	0.76	38.22	876.47	0.87
6	0.76	38.22	876.47	0.87
9	0.76	38.22	876.47	0.87
12	0.76	38.22	876.47	0.87
15	0.76	38.22	876.47	0.87
18	0.76	38.22	876.47	0.87
20	0.76	38.22	876.47	0.87

2.

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

Ans: Self Weight of Concrete = 25 KN/m^2
Self Weight of PCC = 22 KN/mm^2
Self Weight of Brick Wall = 18 KN/m^2

Dead load = (Weight of one Slab x 7) + Weight of Wall
(25 x 0.150 x 30x20) + ( 18 x 3 x 30 x 0.115) + ( 18 x 3 x 20 x 0.115)
=16060 KN

Assume live load = 3KN/m^2
Live Load = 20 x 30 x 3 = 1800 KN

Total Live Load = 1800 x no of floors = 1800 x 6= 10800 KN