00:01
Hello everyone.
00:03
So the figure is shown on the screen.
00:05
The question reads that 1 .35 meter concrete post that is the value of l is given 1 .35 meter is reinforced with a 6 steel bars each with a 28 millimeter diameter diameter that is the value of ds is given 28 millimeter now further knowing that es modulus of e represents modulus of rigidity so modulus of rigidity in case of steel is 200 whereas in case of concrete it is 29 now further it is given that we have to determine the normal stresses in the steel and in the concrete when the axial centric force p is equal to 1560 kilo.
01:20
Also from the figure we see that diameter for concrete is equal to 0 .45 meter.
01:32
Right? so these are the only available of available given data.
01:39
Now let pc and ps be the portion of axial force or directly writing a portion of p carried by concrete and six steel rods respectively.
02:23
So we have the form delta is equals to pl divided by e multiplied by a where the symbols have the usual meaning.
02:35
So using this we can write pc is equal to e c multiplied by ac multiplied by l.
02:45
Divided by l where c represents the concrete and ps that is for steel is es multiplied by as multiplied by delta divided by l right so we find there's some so there's some is nothing but equals to total that is p total of excess centric force so this is pc plus p s or we can write this as ec ac plus e s as e s as multiplied by delta divided by l right or we can rearrange it and write it as delta divided by l is nothing but equals to p divided by e c ac ac plus e s as where e represents the modulus of of rigidity for concrete and steel respectively denoting cnss symbol and a represents the area, surface area, right? so this is equals to epsilon.
04:13
Now we find the values of ac and as directly and hence the value of epsilon.
04:22
So epsilon becomes p, that is 1 .560 kilo newton.
04:31
We'll be taking it in negative this is kilo newton, capital n.
04:38
10 raised to the power 3...