Replace the given loads with an equivalent force couple...

Replace the given loads with an equivalent force-couple system at O by specifying the resultant R and couple $M_o$. Next, determine the equation for the line of action of the stand-alone resultant force R. 1.3 m 1.3 m 0.25 m 0.6 5.3 kN m 1.7 m 65° 19 kN-m 9.7 kN 7.4 kN 0.25 m Answers: Equivalent force-couple system at O. R = ( 6.89 i + 3.21 j) kN Q Search

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00:01 In this question we have been given uniform varying load.

00:04 So here at this point we have 8 kn per meter and then it goes up to certain distance and then from here it is constant and then again it starts decreasing up to this point.

00:19 So here initially it was 8 kn per meter.

00:22 Then at this point it is 5 kn per meter.

00:27 And then here this point this is same 5 kilo newton per meter and this distance is given as 1 .5 meter and then this distance is 0 .75 meter and this is also 0 .75 meter so first of all we will find the force acting as a point load so we we will consider this as f1.

00:56 So this is the force acting due to this uniform varying load in this section.

01:02 Similarly we will find this f2 and then we will find f3.

01:10 So here we can write this force f1 will be equal to the area of this trapezoid and this force f1 will be equal to half of base that is 8 plus 5 and then height is given as 1 .5 meter so this will be of 9 .75 kilonewton then similarly we will find f2 and this f2 will be the force due to this rectangle so this will be 5 times 0 .75 and this will be also in now kilo newton so here this was in kilo newton per meter and if we solve this we will get 3 .75 kilo newton and now similarly we will find f3 and this will be the area under the triangle which is half of base times height and this will be again in kilo newton which is equal to 1 .875 kilo newton so we have calculated the total force and if you want to calculate total force so this will be 1 f2 and f3 and if we add all these three so we will get 15 .375 kilo newton now total force is known to us now we need to find the moment about this point let us consider this point is o.

02:44 So first of all we will find this distance where f1 is acting we will take this as x1 then we will find the distance x2 where f2 is acting so this is x2 and then this is x3 so here can write this x1 will be equal to 2a plus b divided by a plus b multiplied by h divided by 3 so this will be 2 times 5 plus 8 divided by 5 plus 8 multiplied by 1 .5 divided by 3 and this is equal to 0 .6923 meter similarly we will find for x2 and this x2 will be 1 .5 meter that is the total distance so let us consider this as a, b and c so oa is 1 .5 then plus now this is a rectangle so we need to take the half of this distance which is 0 .75 divided by 2 and this will be equal to 1 .875 meter...

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