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Problem 1-37. So the statement of problem is the plate has a width of 0.5 meter. If stress distribution at the support varies as shown, determine the load P applied to the plate and distance D to the plate. where it is applied.
So you can see the figure. This is a plate and there is a load that is varying and the stress is given as 15 into X power one by two mega Pascal. And this load is varying and applied to the fixed fixed support at the bottom so you have to find this force as well as the location of this force from this end which means you have to find D here so it is also given that width of the plate I will draw the side view the width of the plate is 0.5 mm so this width is 0.5 meter not mm this is meter so first we will draw the free body diagram so i will draw it first and then we will solve this so this is the plate clear there is a load p that is acting at this point so this is load p that is acting at a distance of d from this end Also, if this is the X direction, clear and the load is given as varying load, clear. So I will draw it like this.
This is a varying load and this load. is given as Sigma is equal to 15 X power 1 by 2 mega Pascal so if I take a small element of this force for example if I take the force small element so this is the small element and this force is known is this force is that this force is df that is acting on this plate and this force is at a distance of x x is any distance from this side clear so this is a small element and the force acting on a small element is df and this the thickness of this small element is dx this thickness is dx so what we will do is that this is origin point so we will solve this in order to find the value of p so the resultant force df of the bearing pressure acting on the plate of area da so this is small element so its area will be equal to d in this area will be equal to width of this plate which is 0.5 and thickness, which is the X. So it will be equal to be into DX. This small area is equal to be into DX clear.
And that if you put the value of BB 0.5, so it will be equal to 0.5 DX. So this force will become equal to, as we know that stress is equal to force per unit area. So here we have differential force and differential area.
So therefore we will have this D F will be equal to bearing stress into differential area. F will be here and sigma into A will give you the differential force. Now you have the value of stress which is this one.
So I will put the values 15 into X power 1 over 2. and that is Mega Pascal so I will convert into Pascal so it is 10 to the power 6 and DA is equal to 0.5 into DX so again if you solve it further so DF will become equal to 7.5 into 10 to the power 6 X power 1 by 2 into DX So this is the value of differential force that is acting at a distance of x from this point O over a differential area dA. Now we will apply equilibrium condition in order to find the value of P. So for that we will apply the first equilibrium condition that sum of All force along y direction is equal to zero in taking upward force as positive.
So here you can see there are two forces acting vertically. One is P and the second one is this force. So I will write P is down.
so it will be negative plus this force is upward so this is a differential force so in order to find the net force we will integrate this differential force so it will be equal to d integral of df here some must be equal to 0. Okay. So you have to put minus P plus integral of, you have to put the value of D F, which is equal to 7.5 into 10 to the power 6 X. This is 6x1 by 2 into dx is equal to 0. Again, since it is differential, so you have to take limit from x is equal to 0 till x is equal to 4 meter. Because the force is acting on a length starting from 0 to 4. So 0 to 4 limit.
So when you solve this integral, this 7.5 into 10 raised to the power is constant, it will be outside and integral of 1 over 2 will give you x 1 over 2 plus 1 divided by 1 plus 2 plus 1. So when you solve this is very easy, you can do it. So when you solve it, so you will get p is equal to 40 into 10 raised to the power 6 Newton. or you can say that it is 40 mega Newton. So this is the force that can be applied to this plate.
Now we have solved this. So what will about the position of this force or location of this force from this X is equal to zero. So for that we have to find D.
So for equilibrium condition, equilibrium in order to bring this plate in equilibrium. We will apply the another equation of equilibrium that sum of all movements about point O is equal to zero. So let's apply its point O is equal to zero and taking the counterclockwise movement as positive.
So you can see one force is producing clockwise movement that is P and P is acting at a distance of D from this O. So this is one moment. So P into D is the first movement and it is clockwise so it will be negative. The second movement that is produced by this differential force about point O is DF.
So DF into perpendicular distance is X. So as we are going to find it the whole effect of this force. So we will integrate it.
So F DF into perpendicular distance is X. Clear. I will show you. This is DF and this is X and it is producing counterclockwise movement.
Again, we will take limit from 0 to 4 because this force is wearing force and it starts from 0 to X is equal to 4. Their sum must be equal to 0. Now you will have the value of DF which is this one. You can put it minus PP minus P is P is 40 mega Newton. So it is 40 into 10 raised to power 6 multiplied by this D plus integral of 0 to 4 DF is equal to 7.5 into 10 raised to power 6 X power 1 over 2 into DX. Clear?
And perpendicular distance is X. Okay. So this is force and this is this.
X. The sum must be equal to zero. When you solve this, when you integrate this term, it's very easy.
You can do it. There is a D and when you solve it, you will get D is equal to 2.40. So we have find out the location of point P to bring this plate in equilibrium.
So this D must be equal to 2.40. It means that P must be applied at a distance of 2.40 meter from this end. This was a very easy problem.
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