Week 3 Challenge

ANSWERS: 

 

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

Ans :- The main load cases that must be taken into account for the design are the Dead, Live, Snow,         Seismic, and Wind loads are among them.

  

2. What is One – Way slab?

Ans :- 

When loads are distributed between two parallel walls or beams and a slab with a length to breadth ratio of two or higher, it only bends in one direction (the spanning direction). This is due to the geometry of the slab.

A one-way slab is, in essence, one with a length to breadth ratio of two or higher.

 

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

Ans :- Unit weight of Steel = 7850 Kg/m^3

     Unit weight of soil = 1800 Kg/m^3

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

Ans : Sections name that can be defined using Properties tab in STAAD Pro.

    Circle ,Rectangle , Trapeziodal , Tapered I , Tapered Tube ,Tee

 5. Define Primary Beams.

Ans : -The horizontal flexural (beam) structural member known as the primary beam is directly connected to the supporting compressive structure member, the column.

To the Columns, it transfers the burdens.

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

Ans :- Pz = 0.6 Vz^2

     Where; Pz = Design wind pressure
                 Vz = Design Speed 
                 Vz = Vb x K1 x K2 x K3
                 Vb = Wind Speed
                 K1 = Probability Factor
                 K2 = Terrain, Height and Structure size factor
                 K3 = Tropography Factor

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

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

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

Ans :- In one way slab, the ratio of longer span (l) to shorter span (b) is equal or greater than 2, 

The slabs are supported by the beams on the two opposite sides. 

The loads are carried along one direction.

In Two Slab, the ratio of longer span (l) to shorter span (b) is less than 2, 

In two way slab, as the loads are carried in both directions (longer and shorter directions)

Main reinforcement bars are provided in both directions.

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

Ans :-   a connecting horizontal beam (simply supported or shear connected) between primary beams.

The main beam will receive the load transfer.

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

Ans : Dead Load ( IS 875 Part 1)

 Dead Load ( Unit Weight) of following Materials are

Cement = 1440 Kg/m^3

Steel = = 7850 Kg/m^3

Sand = 1600 Kg/m^3

Concrete = 2400 Kg/m^3

RCC = 2500 Kg/m^3

Soil = 1800 Kg/m^3

Timber = 700 Kg/m^3

Factor for Dead Load is 1.

Live Load ( IS 875 Part 2)

Live loads on floors and roofs consists of all the loads which are temporarily placed on the structure,

All rooms and kitchens = 2 KN/m^2

Bathroom and Toilets = 2KN/m^2

Corridors / Stair case= 3 KN/m^3

Balcony = 3KN/m^3

Wind Load :

When designing a structure, one must take the horizontal component of wind's force into account. Wind loads are influenced by the building's shape, size, and wind speed. In IS 875 (Part-3):1987, the method for calculating wind loads on structures is described.

Snow Load : 

Buildings that are situated in areas with heavy snowfall must be built to withstand snow loads. The IS 875 (Part-4):1987 code addresses snow loads on building roofs.

Earthquake Load :

 The location of the building affects the building's earthquake loads. India has four seismic zones according to IS 1893-2016 (Part-I) (General Provisions for Buildings). The code makes suggestions for designing structures that can withstand earthquakes.

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

Ans :-

•  Go to Properties Tab
•  Click on Define Material
•  Choose Section and Material 
•  Click on Add and Close.
•  Now Click on Assign to view and assign the Property
•  Go to Support Tab 
•  Click on Create 
•  Choose Support and click on Add and close
•  Now Click on Use Cursor to Assign.

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

Ans :- 

Add node at the desired location of the beam. New Add beam between Nodes.

• Now go to the Specification
• Select Beam
• Click on Release
• Click on Start and Tick on Fx and Mz and click Add.
• Now Click on End and Tick Fx and Mz and click Add   
• Now Assign to the selected Beam

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

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)   (Categort- 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.