Week 3 Challenge

ANSWERS:

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

Ans: Dead Load ,Live Load ,Snow Load, Seismic Load, Wind Load

2. What is One – Way slab?

Ans: The one-way slab is a slab, which is supported by parallel walls or beams, and whose length to breadth ratio is equal to or greater than two and it bends in only one direction (spanning direction) while it is transferring the loads to the two supporting walls or beams, because of its geometry.

In short one-way slab is a slab whose length to breadth ratio is equal to or greater than two.

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 : Primary Beam is a horizontal flexural (beam) structure member which directly connected to supporting compressive structure member that is column.

It transfer the loads to the Columns

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 direction)

Main reinforcement bars are provided in both directions.

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

Ans : A horizontal beam connecting primary beams (simply supported or shear connected.)

It will transfer the load to the primary beam.

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 :The force exerted by the horizontal component of wind is to be considered in the design of building. Wind loads depends upon the velocity of wind, shape and size of the building. The method of calculating wind loads on structure is given in IS 875 (Part-3):1987.

Snow Load : The building which are located in the regions where snowfall is very common, are to be designed for snow loads. The code IS 875 (Part-4):1987 deals with snow loads on roofs of the building.

Earthquake Load : Earthquake loads depend upon the place where the building is located. As per IS 1893-2016 (Part-I) (General Provisions for Buildings), India is divided into four seismic zones. The code gives recommendations for earthquake resistant design of structures.

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.

AIM:

• To calculate the wind pressure for the given building.

PROCEDURE:

Given Data :

Building dimensions : 20m X 30m X 20m.

Building usage : Hospital Block.

Location : Darjeeling.

STEP 1: 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 Co-efficient.

We can find the value of risk co-efficient,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 cide 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 = Vb X k1 X k2 X k3 X k4.

= 47 X 1.07 X 0.8 X 1 X 1.

Vz = 40.23 m/s.

STEP 2: Calculation of Wind pressure:

Pz=0.6xVz^2

=0.6x(40.23)^2

= 971.071 N

Pz=0.97117KN

Similarly, we can calculate the for different heights, the calculation for different height has been done in the excel sheet so the excel sheet has been given below:

Question 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 the slab thickness of 150mm.

Dead Load due to the Slabs = Length X Breadth X Height X Slab Thickness.

Dead Load due to the Slabs = 20 X 30 X 25 X 0.15

Dead Load due to the Slabs = 225 KN.

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

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

Total dead load due to all the slabs = 5 X 225.

Total dead load due to all the slabs = 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 = 0.35 X 0.45 X 25.

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

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

Superimposed Loads due to the Partition walls:

Assuming 150mm thickness of brickwork.

Height of brickwork = Floor height - depth of the beam.

Height of the brickwork = 3.75 - 0.45.

Height of brickwork = 3.3 meters.

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

The Dead Load due to the brick masonary partition = 8.91 KN/m.

Live Load:

• We can calculate the live load of a given structure by using the IS 875:Part 2,which states different live loads values for different rooms.
• As our building is the Hospital building, the all the proposed live load values for hospital building according to the IS 875:Part 3 has been listed below:

OCCUPANCY	LIVE LOAD
1. For ward rooms, dressing rooms, dormitories and lounges	2.0 KN/sq.m.
2. Kitchen and Laboratory	3.0 KN/sq.m
3. Dining room and Cafeteria	3.0 KN/sq.m
4. Toilets and Bathrooms	2.0 KN/sq.m
5. X-ray rooms and Operating Rooms	3.0 KN/sq.m
6. Office and OPD Room	2.5 KN/sq.m
7. Corridor and Passages	4.0 KN/sq.m