5 3 m 15 m 4 4m 16 m 1 35 pada gambar 1 ditunjukkan sebuah...

5 @3 m = 15 m 4 @4m = 16 m 1. [35%] Pada Gambar 1 ditunjukkan sebuah struktur truss yang diberi beban W = 150 N. A B M a. Dengan inspeksi, tentukan batang manakah yang merupakan batang dengan gaya nol? C N D E O b. Hitunglah reaksi pada tumpuan L dan V! F G Q P c. Hitunglah gaya batang AB, AC, serta DE, dan tentukan apakah batang tarik atau tekan! H S R 70° J K W U T L Figure 1 Truss Structure V 1. [35%] Fig. 1 shows a truss structure loaded by a weight W = 150 N. a. By inspection, determine which member is a zero force member? b. Calculate the reaction on the supports L and V! c. Calculate the forces on members AB, AC, and DE, and determine whether it is tension or compression!

Transcript

00:01 In this problem, we have a crane that's supporting the load of 2 ,000 newtons, and the purpose of the problem is to find the tension in this cable, and also the force, magnitude direction at the hinge at point a.

00:18 Now, right off the bat, we know the tension in this cable, the one directly attached to the load, it's going to be 2 ,000 neutins, equilibrium.

00:25 2 ,000 newtons up, 2 ,000 dunez down, equilibrium.

00:28 So that's taken care of it.

00:30 But that's not the tension in this cable.

00:32 It's a separate cable.

00:34 So first thing we need to do, it's a free -by diagram.

00:37 Remember, tensions always pull.

00:40 So this one here will be tl, l for load.

00:48 We have the weight, it's uniform rods.

00:50 It'll be right dead center of that rod, wb.

00:54 We have the tension in this cable.

00:56 One we're looking for, t -c.

00:58 We have the force at point a, f -y, f -x, we'll find out what direction they are from our equations.

01:10 Now before we go on, we have our axes, traditional axes.

01:13 We do need this one angle here.

01:17 This angle here will be 30 degrees.

01:19 Think about it.

01:21 If you look at this picture here, this is 30 degrees.

01:28 That makes this 150, but we know the whole thing has to be 180, so it makes this 30.

01:39 So that's where that comes from.

01:42 So now, equilibrium problem.

01:46 F net x equals zero, f net y equals zero, net torque equals zero.

01:52 So all we've got to add up now is all our x components.

01:55 Fx plus tc cosine 30 degrees equals zero.

02:03 That gives me that fx is going to be minus tc, cosine, cosine 30 degrees, and that's equation one.

02:12 That's just the side of the triangle here.

02:15 This is tcx, it says tcy.

02:19 That's cosine because it's the adjacent side.

02:23 Now, f net y, fy minus tl.

02:29 So we're only in negative y direction, minus wb, minus tc sine 30 degrees, is equal to zero, implies fy is equal to tl plus wb plus tc, sign 30 degrees and that's equation two.

02:55 There's nothing we can do with them.

02:56 We need tc.

02:59 The net torque is our third equation equal to zero.

03:04 But to get the net torque, we need moment arms and that's from, and we gotta choose a point.

03:12 We're gonna use our rotation axis through a.

03:15 Why? because fx, fy have no rotation effect.

03:20 Their moment arms are zero.

03:22 What is the moment arm? shortest distance from the line of action in the force, to the rotation axis.

03:31 If you're at the rotation axis, here's your line of actions, an infinite line through a force.

03:37 If you're on the rotation axis, how far was the shortest distance to walk to the dash line? it's under your feet, zero.

03:44 So there's no torque due to fx, likewise fy.

03:48 But we do need it from the other three.