Introduction to the calculation of the loads | Building Construction

Students find it difficult to understand the concept of loads although it is a very simple concept. We are going to write a series of articles on “Load Calculations” and help you all in understanding different types of loads that are to be considered for structural designing and also how to calculate them.

An object is subject to mainly two types of forces:

Basically, an object subject to any type of force which could be gravitational force (weight), pressure or anything affects the object is called a load.

This concept is used in Mechanical and structural engineering. Let’s take in terms of Structural Engineering. Whenever a structure is designed, these concepts are taken into consideration because real world objects are analyzed in order to design the structure. This is very important in terms of structural stability.

As the name itself suggests, dead loads could be termed as self weight of the non-living objects. It could be the weight of the materials, equipments or any other components in the structure that will remain permanent throughout the life of the structure.

Dead load has to be considered in order to make the structural design accordingly. Dead loads vary from structure to structure. Every building is unique and has different considerations.

An additional load is considered in case additional forces build up in a structure in case of settlement or due to secondary effects of pre-stress construction or due to shrinkage of concrete.

• Columns
• Beams
• Footings
• Lintels
• Furniture
• Machinery and other equipment
• Walls
• Floors
• Roofs
• Ceilings
• Stairways
• Built-in partitions
• Finishes (POP – Plaster of Paris)
• Cladding (Use of various materials which increase the self weight of the structure) etc.

Basically, all the permanent loads are to be considered.

As the name suggests, live load is the load of human beings living in the building. Their movement is not fixed. The number of people at a time in a structure can also vary.

For example:

A person lives in a 4BHK apartment with his wife and two kids. If he happens to throw a party for 50 persons, the live load on the structure increases considerably for that period of time.

As soon as the guests leave, the number of persons reduces from 50 to 4.

So, here’s what I mean by variable force.

Live load to be considered while designing a staircase:

• Pressure of the feet
• Wind load on the stair in case the staircase is located outside the house

Live load to be considered while designing the roof:

Movement of workers on the roof during construction, maintenance along with their materials and equipments

Also, if the owner of the house plans to make a terrace garden on the roof, that adds additional load to it.

For dwelling houses to a 10KN/m2. In any building project, slabs are assumed to be 100m thick from stiffness/deflection consideration.

Beams are taken separately and the self-weight is calculated and added separately on the frame. The net weight of the above load is multiplied by a load of 1.5 for concrete.

Also, check out:

Bending moment and Fixed moment calculations

18 thoughts on “Introduction to the calculation of the loads | Building Construction”

1. hello
is there service for structural drawing for residential building with you
regards

• Dear Sathya,

Yes, we do provide architectural design and civil engineering services. To further discuss please email benruk@gmail.com.

2. Hello.
Can you pls send also the step by step methods of load calculation in a building and what is KN/m2

3. could you please send me the load calculations while designing the structures for the building in step by step.. please i want to know this.. Thanks the lord.

4. HI
COULD YOU PLEASE SEND ME STEP BY STEP WAY OF CALCULATING LOAD IN A BUILDING.
THANKS.

5. Plz help me to solve this problem

6. Solve this numerical
1.A column 4m long has to ssupport a factored load of 3500KN column is hinged at both ends. Design a suitable column member.Fe410 steel is used.

7. Do u organise training programmes.

8. hello all,
i am civil engineer and i have a doubt regarding a girder beam which is of L-shape consisting of conected 2 parts one is 5mts x 2.5mts and pouring is 1.7mtrs and other is 10mtrs x 5mtrs and same thickness of 1.7mtrs steel reinforcement consists of 32 mm dia bars 20 Mt and stirups are of 16mm dia 19 mt. the girder beams are located on the 2 nd podium and it ll take load of 40 flrs above it and new columns are planted on the beam. so can u please give me a short explaination regarding types of loads and bending moment calculations and how to place props for proper support and distribution while concreting.

regards

9. my building is around 864 Sqft tolal piller 16 nos in 3 rows. the all piller size is 6″ X 6″ and 4 rod in all piller 2 nos of 10mm and 2 Nos of 8mm. my foundation basement is 2′ X 2′ now my question it is possible to construct RCC roof /Slabs on this structure?

10. Hi.. First of all i really appreciate what you are doing on this site.
It’s best for all who are like me,
So., Assume that i have to design 6 storey residential building. then what’s the first to last (inception to execution) procedure to design.? i am that type of engineer, who have just Degree.!

11. Dear Writer

The main object for such examples is to help out the engineers in calculating the fixed moment reactions etc as how the engineers are to calculate the nos. of reactions for the simply supported beams, over hanging beams and for the fixed supports in order to calculate the static determinacy for the particular beams etc for which i mean is to have a detailed explanation for the topics to be covered.

12. The way of ur explaination is very easy to understand.tnx….keep it up.

13. Can you please tell me how to calculate self weight of a rcc beam( ie whether in kq/sq.m or kg/cu.m ) while designing RCC Beam

14. After calculating the total load of the building on the footings, then how to get the load on each footing? Do we have to divide the total load equally as per the area around the footing?

15. THE DIFFFRENT EXAMPLES U GAVE…..R ABLE TO EXPLAIN VERRY WELL.

• The examples given are understandable by the engineers who have covered all these topics like us as we are the structural engineers but for those new comers its very hard to understand as we have to give them step by step calculations as for example the reaction at support A is zero and we will be calculating Rb etc.

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