A crank shaft is operated as shown in the figure. A load F1...

A crank shaft is operated as shown in the figure. A load F1 = 9.89 [N] is applied and the operator applies a load P to counter. The system is in equilibrium and all of the bearings are perfectly aligned such that they do not produce moments on the rotating crank. The geometry is given by, w = 8.44 [m], d1 = 9.09 [m], d2 = 2.15 [m], d3 = 5.92 [m], d4 = 1.77 [m], d5 = 4.69 [m], and h1 = 8.21 [m], find the following: Part 1. Express the reaction forces exerted on the bar by the journal bearing at point B. Use the coordinate system shown in the diagram. \cdot 5.87 \hat{i} + 0 \hat{j} + 0 \hat{k} Part 2. Express the reaction forces exerted on the bar by the journal bearing at Point A. Use the coordinate system shown in the diagram.

Transcript

00:01 Maximum torsional stress occurs at point c and is given by tau max the maximum torsional stress tau max is equals to tc divided by jc.

00:14 So, here tc is the torque acting on section c and jc is the polar moment of inertia of section c ok.

00:22 So, the maximum bending stress occur at point b we will give it by sigma max is equals to mc.

00:32 So, mc here is the bending moment acting on section b and divide by ic.

00:37 So, ic here is the moment of inertia of section b.

00:41 Now, we have given here the quantities.

00:44 So, let us see it the given quantities first i am giving it equation 1 and this equation 2.

00:53 Now, ab given to us 125 mm in the figure we can see tc given to us 100 mm cf given to us 38 mm and force given to us here 1 .3 kilo newton.

01:12 Now, we can calculate here the tc easily.

01:17 So, the formula for tc is f multiply with cf.

01:25 So, when we will put the values of f and cf we will get the value of tc 49 .4 newton meter.

01:33 Now, for jc formula is pi divide by 2 and multiply with 32 divide by d to the power 4 minus small d to the power 4.

01:49 So, we have given the values of capital d and small d here.

01:53 So, when we will put this pi the value of pi we know 3 .14 divide by 2 multiply with 32 and divide by the value of capital d is 0 .02 and to the power 4 minus the value of small d is 0 .018 and to the power 4.

02:17 So, when we will solve this the value of jc will be 1 .57 and multiply with 10 to the power minus 7 meter to the power 4.

02:34 So, from here we can easily calculate the tau maximum.

02:37 So, the formula for tau maximum in equation 1 we will put the value of tc and jc here.

02:43 So, the value of tc we have calculated earlier 49 .4 and divide by value of jc is 1 .57 multiply with 10 to the power minus 7.

02:59 When we will solve this tau maximum will be 3 .14 multiply with 10 to the power 11 and pascal.

03:14 Now, we will calculate the mc.

03:18 So, it is f multiply by ab when we will put the value of force and ab here plus tc multiply by bc.

03:33 So, when we will put its value of all this the value of tc here is 49 .4 which we have calculated earlier.

03:45 So, from here when we put all the values and calculate it this is 6 .67 multiply with 10 to the power 3 newton mm and now we will calculate the ic.

04:03 So, it is 5 divide by 64 and multiply with d to the power 4 minus small d to the power 4.

04:12 When we know here now the value of capital d and small d we are directly putting the values here.

04:17 So, this will be 2 .47 and multiply with 10 to the power minus 8 meter to the power 4.

04:26 So, here we can calculate the sigma maximum.

04:31 So, it is mc divided by ic...

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