Introduction of Reinforced Concrete Design

Introduction of Reinforced Concrte Design, Advantages and DIsadvantages of RC, & Sample Computation of Loads

1. REINFORCED CONCRETE DESIGN Concrete: Concrete is a stone like substance obtained by permitting a carefully proportioned mixture of cement, sand and gravel or other aggregate and water to harden in forms of the shape and of dimensions of the desired structure . Reinforced cement concrete: Since concrete is a brittle material and is strong in compression. It is weak in tension, so steel is used inside concrete for strengthening and reinforcing the tensile strength of concrete. When completely surrounded by the hardened concrete mass it forms an integral part of the two materials, known as " Reinforced Concrete ".

2. Advantages and Disadvantages of Reinforced Concrete Flexural Strength of Concrete Advantages of reinforced concrete It has relatively high compressive strength It has better resistance to fire than steel It has long service life with low maintenance cost In some types of structures, such as dams, piers and footings, it is most economical structural material It can be cast to take the shape required , making it widely used in pre-cast structural components It yields rigid members with minimum apparent deflection Yield strength of steel is about 15 times the compressive strength of structural concrete and well over 100 times its tensile strength By using steel, cross sectional dimensions of structural members can be reduced. Disadvantages of reinforced concrete It needs mixing, casting and curing, all of which affect the final strength of concrete The cost of the forms used to cast concrete is relatively high It has low compressive strength as compared to steel (the ratio is about 1:10 depending on material) which leads to large sections in columns/beams of multi-storey buildings Cracks develop in concrete due to shrinkage and the application of live loads

3. “ The design of a structure may be regarded as the process of selecting proper materials and proportioned elements of the structure, according to the art, engineering science and technology. In order to fulfil its purpose, the structure must meet its conditions of safety, serviceability, economy and functionality .” Reinforced Cement Concrete Design Philosophy and Concepts

4. Design Methods Strength Design Method Working Stress Design Limit State Design

5. Loads -Forces for which a structure should be proportioned. Loads that act on structure can be divided into three categories. Dead Loads: Dead loads are those that are constant in magnitude and fixed in location throughout the lifetime of the structure. It includes the weight of the structure and any permanent material placed on the structure, such as roofing, tiles, walls etc. They can be determined with a high degree of accuracy from the dimensions of the elements and the unit weight of the material. Live loads: Live loads are those that may vary in magnitude and may also change in location. Live loads consists chiefly occupancy loads in buildings and traffic loads in bridges. Live loads at any given time are uncertain, both in magnitude and distribution. Environmental loads: Consists mainly of snow loads, wind pressure and suction, earthquake loads. Soil pressure on subsurface portion of structures, loads from possible pounding of rainwater on flat surfaces and forces caused by temperature differences. Like live loads, environmental loads at any given time are uncertain both in magnitude and distribution.

6. 1.Required Strength (Factored Load) U 1.1 To resist dead load & live load: U=1.4DL + 1.7LL 1.2 If resistance to structural effects of specific wind load U= 0.75(1.4DL+1.7LL+1.7W) U=0.9DL+1.3W not less than 1.4DL+1.7LL 1.3 If resistance to specified earthquake loads U= 0.75(1.4DL+1.7LL+1.87E) U=0.9DL+1.43E not less than 1.4DL+1.7LL 1.4 If resistance to specified earth pressure U= 1.4DL+1.7LL+1.7H U=0.9DL not less than 1.4DL+1.7LL 1.5 Where structural effects T of differential settlement, creep, shrinkage or temperature change are significant. U= 0.75(1.4DL+1.4T+1.7LL) not less than 1.4(DL+T)

7. Sample Problem 1. CE BOARD 2012 A singly reinforced T-beam with a cross sectional area shown in the figure, span of 5m is subjected to support dead load and live load of 10KN/m and 28KN/m respectively. Find the factored load required in ff; using DL and LL only. With additional wind pressure, w=5KN/m. b.1. use liveload b.2. No liveload With additional Earthquake load, E=40KN/m c.1. use liveload c.2. No liveload

8. Solution: Area of the T-Beam = (0.3x0.12)+(.220x.150) = .0069 Deadload of T-Beam = 25KN/m 3 x .0069 = 1.725 KN/m DL total = DL + DL T-Beam = 10 + 1.725 = 11.725KN/m - Combination of DL & LL a.) U=1.4DL + 1.7LL = 1.4(11.725) + 1.7(28) = 21.175KN/m - with Wind Load = 5KN/m, b.1) U= 0.75(1.4DL+1.7LL+1.7W) = 0.75(1.4(11.725)+1.7(28)+1.7(5)) = 54.39KN/m b.2) U=0.9DL+1.3W = 0.9(11.725)+1.3(5) = 17.05, Use: = 21.175KN/m with Earthquake Load = 40KN/m, c.1) U= 0.75(1.4DL+1.7LL+1.87E) = 0.75(1.4(11.725)+1.7(28)+1.87(40)) = 104KN/m c.2) U=0.9DL+1.43E = 0.9(11.725)+1.43(40) = 67.75 KN/m

9. 2. Determine the Axial stress acting on a circular column shown in the figure. Solution: U = 0.75(1.4DL+1.7LL+1.87E) = 121.725N Stress = Force/Area = (121.725)/ π (0.125) 2 = 2,479.76N/m2 Or 2.48KPa DL = 20N LL = 35N D = 250 E = 40N

10. Thank you!