Why shrinkage is more in the Singly Reinforced section as compare to doubly Reinforced Section in RCC? Please give a detailed reason for what happens due to the uneven Distribution of Steel bars in the RCC section.
Mostly the ratio of 1:1.5:3 is used for M20 grade of concrete So, Volume of dry concrete = 1.54 to 1.57 times volume of wet concrete Assume, 1cum of concrete work ratio sum = 1+1.5+3 = 5.5 shrinkage or safety factor = 1.57 Total volume of wet concrete required is = 1.57cum Volume of broken stone reqRead more
Mostly the ratio of 1:1.5:3 is used for M20 grade of concrete
So, Volume of dry concrete = 1.54 to 1.57 times volume of wet concrete
Assume, 1cum of concrete work ratio sum = 1+1.5+3 = 5.5
shrinkage or safety factor = 1.57
Total volume of wet concrete required is = 1.57cum
Volume of broken stone require = (3/5.5) * 1.57 = 0.856 m3
Volume of sand require = (1.5/5.5) * 1.57 = 0.471 m3
Volume of cement = (1/5.5) * 1.57 = 0.285 m3
=0.285*1440 =411 kg
For 1 m3 of M20 (1:1.5:3)
Broken stone = 0.856 m3
Sand = 0.472 m3
Cement = 8.22 bag
Therefore,
8 bag of cement is required for 1cum of concrete work in M20
aviratdhodare
Analyzing the problem from a theoretical point of view, “Shrinkage” essentially means that the concrete is trying to shrink which is on account of the chemical interaction between various constituents of concrete like cement, water and aggregates. Now providing reinforcement steel in a way holds theRead more
Analyzing the problem from a theoretical point of view, “Shrinkage” essentially means that the concrete is trying to shrink which is on account of the chemical interaction between various constituents of concrete like cement, water and aggregates. Now providing reinforcement steel in a way holds the concrete together by way of restraining it and hence reduces the shrinkage.
Also, Shrinkage and temperature cracks have reasonably comparable behaviour.
Let us take a formula based approach to understand the relationship between shrinkage or temperature or shrinkage cracks and the reinforcement provided.
As per Cl. A-1.4, IS 3370 Part-II,
Smax = (Fct/Fb) x D / (2 x SR)
And
Wmax = Smax x Alpha x (T1+T2) x 0.5
Wmax is the crack width, which is quantity of interest here
From the above, we can deduce that all other variables being made constant valued,
Wmax is directly proportional to Smax.
Smax is inversely proportional to SR
It implies Wmax is inversely proportional to SR
So, if we want lesser shrinkage/temperature cracks, we have to increase this variable SR.
Now, this SR = Steel ratio = Percentage of steel provided / 100
Therefore, it is clear that, if we provide greater reinforcement, SR increases, Smax increases and Wmax decreases.
Since singly reinforced section has less reinforcement steel than a doubly reinforced section, shrinkage/temperature cracking is more in a singly reinforced section.
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