Q3 Figure Q3 shows an arrangement of four springs in series...

Q3 Figure Q3 shows an arrangement of four springs in series being depressed with a force 2000 kg. At equilibrium, force balance equations can be developed defining the interrelationships between the springs as: k2(x2 - x1) = k1x1 k3(x3 - x2) = k2(x2 - x1) k4(x4 - x3) = k3(x3 - x2) F = k4(x4 - x3) where k's are spring constants. (a) Can we solve the displacement of spring in these force balance equations using Thomas Method? Justify your answer. (5 marks) (b) Given k1 = k4 = C N/m, k2 = k3 = D N/m, where C is the last two digits of your matrix

Transcript

00:01 In this question, we have been given arrangement of four springs in series.

00:06 They are being depressed with a force of 2000 newton.

00:11 At equilibrium, force balances equation can be developed defining the interrelationship between the spring as follows.

00:20 So, this is the relationship.

00:22 So, k2 times x2 minus x1 equals to k1 x1.

00:27 Next, we have k3 times x3 minus x2 equals to k2 times x2 minus x1.

00:37 Here we have k4 times x4 minus x3 equals to k3 times x3 minus x2 and f we know it is equals to k4 x4 minus x3 where ki's are the spring constant.

00:57 In the first question, it is asked, can we solve the displacement of a spring in these forces using thomas method? so, yes, we can apply the thomas method.

01:10 Now, let me write down the reasoning why we can apply this thomas method because the above first of all, we'll try to form one matrix corresponding to this.

01:24 Okay, this system i'm writing in the form of matrix.

01:28 So, i will write down here i will get k1 plus k2 then i get minus k2.

01:36 If i open the bracket and solve, so i will get the coefficient of x1 as k1 plus k2.

01:42 This is the coefficient of x2 and x3 x4 the coefficient will be zero.

01:47 Next, in the second equation, i get coefficient of x1 is minus k2 then for x2 it is k2 plus k3.

01:57 Here it is minus k3 and zero then zero minus k3 and then here i get k3 plus k4 and then i get minus k4 here.

02:14 Next, we have zero zero minus k4 and this is k4.

02:20 So, this is the coefficient matrix that i am getting and in the system form the variable are x1, x2, x3, x4.

02:31 This is equal to the zero matrix.

02:34 So, this is converted into this system.

02:38 Now, here we can apply the thomas method because the matrix that we obtain here because the matrix is diagonal domino matrix.

02:50 It is a diagonal domino matrix and it is also a symmetric matrix and we know that thomas method applies on this matrix.

03:13 So, hence now i will write down the value of k1, k4.

03:18 I am taking it to be equals to 3 and k2, k4, sorry k2, k3 value i am taking it to be 9.

03:35 Force we already have been given 2000 newton.

03:38 We are taking gravity as 10 meter per second square to reduce the calculation.

03:47 So, what we get? we get the system as 12 minus 9 0 0 corresponding value of k1, k2, k3, k4.

03:56 If i put i get these things.

03:58 Here i get minus 9 18.

04:00 This i got using thomas method minus 9 18 minus 9 0 0 minus 9 12 minus 3 and this is 0 0 minus 3 and 3 times here the variables x1, x2, x3 and x4.

04:20 This is equals to the zero matrix 0 0 0.

04:26 Now from this i get the value of v1 will be equals to 12.

04:35 D1 value is 0.

04:37 Then i write down v2 value.

04:39 This is now i am using the thomas method.

04:43 So, i am getting these things.

04:45 I am not writing the formula that you can refer minus 9 times minus 9 over 2.

04:52 It comes out to be 11 .25...

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