Something of Everything

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

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!

# For wind case : permissible max. Horizontal deflection is H/500. # For EQ case : permissible max. Horizontal deflection is H/250. # For EQ case you have to satisfy story drift criteria of 0.004H also. All deflection worked out shall be for service loads only i.e. for unfactored combinations.

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

1)       IS 1904 - Code of practice for design and construction of foundations in soils : general requirements. 2)       IS 1080 - Code of practice for design and construction of shallow foundations in soils (other than raft, ring and shell ) 3)       IS 2950 - Code of practice for design and construction of raft foundations. 4)       IS 111089 - Code of practice for design and construction of ring foundation. 5)       IS 2911 - Code of practice for design and construction pile foundations. 6)       Code of practice for design and construction machine foundations.   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.

Basic Properties of steel Unit mass of steel = 7850 kg/m3 Modulus of elasticity = 2.0 x 10^5 N/mm2 Poisson ratio = 0.3 Modulusof rigidity G = 0.769 x 10^5 N/mm2 Co-efficient of thermal expansion = 12 x 10^-5 /degree celcius