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Basic wind speed for wind load calculations of important Indian cities

Basic wind speed at 10m height for some important cities of India (as per IS 875 part 3 - 2015 latest amendment) is given as below: Basic wind speed of  Agra  = 47 m/s Basic wind speed of  Ahmedabad = 39 m/s Basic wind speed of  Ajmer = 47 m/s Basic wind speed of  Almora = 47 m/s Basic wind speed of  Amritsar = 47 m/s Basic wind speed of  Asansol = 47 m/s Basic wind speed of  Aurangabad = 39 m/s Basic wind speed of  Bahraich = 47 m/s Basic wind speed of  Bengaluru = 33 m/s Basic wind speed of  Baurauni = 47 m/s Basic wind speed of  Bareilly = 47 m/s Basic wind speed of  Bhatinda = 47 m/s Basic wind speed of  Bhilai = 39 m/s Basic wind speed of  Bhopal = 39 m/s Basic wind speed of  Bhubneshwar = 50 m/s Basic wind speed of  Bhuj = 50 m/s Basic wind speed of  Bikanre = 47 m/s Basic wind speed of  Bokaro = 47 m/s Basic wind speed of  Chandigarh = 50 m/s Basic wind speed of  Chennai = 50 m/s Basic wind speed of  Coimbatore = 39 m/s Basic wind speed of  Cuttack = 50 m/s Basic wind speed of 

Difference between Lap Length and Development Length

Difference between Lap Length and Development Length  is as below Lap Length - Lap length is provided to safely transfer load of one reinforcement to another. Lap length is provided such that load is taken by the lap sufficiently enough that no bars lapped fails.  As reinforcement is available in max. length of approx. 11m, there comes the need of lapping of  reinforcement in structural members.  Lap length is to be given as per SP-16. Lap length is different for tension bars as well as compression bard.  Development Length - Development Length or anchorage length is provided to transfer the load from reinforcement to concrete safely.  As per IS 456:2000, the calculated tension or compression in any reinforcement at any section needs to be safely transferred to concrete by anchorage So that the member should not be come away from the support. Anchorage length needs to be provided at all the supports. 

Chequered plate weight calculation guide

Chequered plates are steel flooring used in industrial structures. Normally the maximum spacing of chequered plates are between 1 metre to 1.2 metre.  Chequered plate thickness is measured excluding the chequeres. That is to say 6mm thick chequred plate means 6 thick plate with chequres above it. Hence the weight of chequred plate comes higher as compared to plain plates. Weight of chequeres are approximately 6.1 kg/m² over and above the plate weight.  Dead weight of chequered plates are as below.  5mm thick chequred plate weight = 45.35 kg/m² 6mm thick chequred plate weight = 53.2 kg/m² 7mm thick chequred plate weight = 61.05 kg/m² 8mm thick chequred plate weight = 68.9 kg/m² 10mm thick chequred plate weight = 84.6 kg/m² 12mm thick chequred plate weight = 100.3 kg/m² Weight of chequered plate (kg/m²)= thickness (m) x 7.85 (kg/m³) + 6.1 If you want to find weight of total plate to be used in project, then Total weight of chequered plate (kg) = Length (m) x Width (m) x (Thickness (m) x

How to calculate water tank capacity in liters?

Let us suppose your tank dimensions are  L = 8ft B = 6 ft D/H = 5 ft Then your volume of tank will be  8 x 6 x 5 = 240 ft³ Now 1ft³ = 28.31 liter, hence, total volume of tank in liter shall be 240 x 28.31 = 6794.4 liter.  That means almost 6800 liter of water can be stored in the tank.  Here is a quick video to show you  How to calculate water tank capacity in liters?

A guide to structural load calculations - 1: Dead Load

How a structure is designed? Loads are assessed: basic loads are defined as which loads will be coming on the structure. Structure is analyzed upon applying these loads and combination of it. Structure is designed for the results of analysis. Structure is designed to withstand the stresses caused by these forces and its combinations. Structure is designed for the shear force and bending moments found through analysis done based on these load cases and combination. In this article, let us learn the first part of structure design, how to assess the load. How to know as to which load will be coming on the structure in its life span. Load: every object in universe have some mass, this mass when multiplied with gravity exerts some weight. This weights are creating pressure on the structure they are resting on. These are called loads. Types of loads: Dead Load: Self weight of the structure. It can be find by density x volume of the member. Every country has defined densities of

What are the development lengths of bars for different grade of reinforcement and concrete?

Here are the development lengths of different grade reinforcement bars for different concrete grades. Development lengths are in mm. 1. Development lengths for plain bars of grade Fe250. 2. Development lengths for HYSD bars of grade Fe415. 3. Development lengths for HYSD bars of grade Fe500. Hope you find it useful. Comment if you have doubts or suggestions. Like, tweet and + post if you find it useful. follow us to get more. Build Your Own Website 4000+ templates, customize and go live!

Angles for different known slopes and slopes for different angles.

Here are some known slopes with their angle and vice versa. Slope Angle 1 in 1 45° 1 in 2 26.56° 1 in 3 18.43° 1 in 4 14.04° 1 in 5 11.31° 1 in 6 9.46° 1 in 7 8.13° 1 in 8 7.13° 1 in 9 6.34° 1 in 10 5.71° 1 in 25 2.29° 1 in 50 1.15° 1 in 100 0.57° 1 in 200 0.29° 1 in 500 0.11 Angle Slope 5° 1 in 11.43 10° 1 in 5.67 12° 1 in 4.70 15° 1 in 3.73 18° 1 in 3.08 30° 1 in 1.73 Enjoy and share with friends.  Hope you find it useful. Comment if you have doubts or suggestions. Like, tweet and + post if you find it useful. follow us to get more.

How to find the mass of reinforcing bar quickly?

Simple way to find the weight of reinforcing bar: The density of steel is 7850 kg/m³. Now let say you want weight per m. then what will be the formulation? Check this out. 7850(kg/m³) x [pi()/4 x d² ](m²)= mass of the bar per m.        (eq. 1) Now resolving this we get 7850 x pi()/4 = 6165.375 right?   Putting this into eq. 1                          6165.375 (kg/m³) x d² (m²) = mass per m.                                          (eq. 2)   What can be bar diameter? Let say 10mm, 12mm. . . we are used to mm format bar diameters, right? So, let us make formulation to eq. 2 for inputting bar dia as mm.              1m = 1000mm. => 6165.375 x 10^-6 =   1/162.196 Hence, mass per m of bar = d² (mm²)/162.196. Now if you want to get mass per feet, then use this relation 1m = 3.281 feet  Therefore mass of bar per feet = d² (mm²)/(162.196 x 3.281) = d² (mm²)/532.139. So now onwards, please don’t bother to use d²/162.2 for finding mass of reinforcement. Enjoy

Quantities for unit volume.

Quantities for unit volume (m 3 ) Work Proportion Cement (bags) Sand      (cu m) Agg.       (cu m) Bricks (no.) Rubble (cu m) Mason (no.) Labour (no.) Excavation 0.00 0.00 0.00 0.00 0.00 0.18 0.15 Concreting PCC   1:4:8 3.00 0.46 0.84 0.00 0.00 0.09 1.77   1:3:6 4.19 0.44 0.88 0.00 0.00 0.09 1.77   1:2:4 6.43 0.42 0.90 0.00 0.00 0.09 1.77 Concreting RCC   1:4:8 3.00 0.42 0.84 0.00 0.00 0.18 2.12   1:3:6 4.19 0.44 0.88 0.00 0.00 0.18 2.12   1:2:4 6.43 0.42 0.90 0.00 0.00 0.18 2.12 Brick Masonary