For the system subject to a varying distributed load as...

For the system subject to a varying distributed load as shown before, define the distributed loading function, w(x), then determine expressions for the effective applied force, effective moment, and location of the effective force as a function of q1, q2, a, and L. Finally, then determine the reactions at both supports given a = 1m, L = 2m, q1 = 10$\frac{N}{m}$, and q2 = 20$\frac{N}{m}$. Note that support B is located a distance "a" from the left hand side. ($f_{eff}$ = 60 N, $M_{eff}$ = 133 N-m 91 A B a 92 D C La Figure 2: Trapezoidal distributed loaded beam

Transcript

00:01 In this problem we are given this beam.

00:06 We have a pin support here at a and we have a roller support at b.

00:14 And on top of this we have some external load.

00:24 So we have this constant portion and then linearly increasing portion.

00:30 We have w1 here, w2, so this is w1 as well.

00:35 And as for the dimensions, we have d1 here, d2 here.

00:44 Now we are given that w1 is 3 kilonitons per meter, w2 is 8 kilonitons per meter, d1 is 4 meters, d2 is 4 .5 meters.

01:08 And with that, we have three questions.

01:14 First we are going to replace this force system with an equivalent force f sub r.

01:22 Okay, now we are going to essentially compute this area under the given load functions.

01:32 So we will do it in two parts.

01:36 First, we have this rectangular area over here, and then we have this triangular area over here.

01:47 So corresponding to this rectangular area, let's say we have r1 acting at half of this point d1 plus d2 over 2.

02:04 And corresponding to this triangular part, let's say we have r2.

02:10 So this is 1 third of the base.

02:16 So this is d1 plus d2 minus 1 third of d2.

02:24 Okay, now r1 is that rectangular area.

02:34 We have base d1 plus d2 times height, w1.

02:40 If we plug in the numbers, you are going to obtain 25 .5 kiloons.

02:48 For r2, we have the area of this triangular region.

02:54 We have one half times base times height.

02:58 The height is w2 minus w1 and if we plug in the numbers we are going to obtain 11 .25 kiloons.

03:11 If we add them up, we are going to obtain the single resultant force, which is equal to 36 .75 kilonitons.

03:28 Okay, the problem tells us to run our final answers to three significant figures.

03:34 And we are going to do that at the very end...

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