Beam practice question 500 lbft e 16 x 106 psi i 100 in4 20...

Name: beam practice question 500 lbft e 16 x 106 psi i 100 in4 20 ft 20 ft k 4000 lbin determine a the displacements and slopes at each node b the local nodal forces and moments in each element 18395
Uploaded: 2024-07-29T20:45:25-08:00
Duration: 10 min 59 s
Channel: Adi S
Description: beam practice question 500 lbft e 16 x 106 psi i 100 in4 20 ft 20 ft k 4000 lbin determine a the displacements and slopes at each node b the local nodal forces and moments in each element 18395

Beam- Practice question 500 lb/ft E = 1.6 x 10⁶ psi I = 100 in⁴ 20 ft 20 ft k = 4000 lb/in. Determine: (a) the displacements and slopes at each node; (b) the local nodal forces and moments in each element.

Transcript

00:01 In this question solution for the problem will be like this and here it's a given.

00:06 What is the given in the question? here it's e is given which is 230 gpa.

00:14 For this it's actually gpa and here for this i is 1 .4 into 10 to 4 minus 4.

00:24 M4 and here for this q is 7 low newton per meter, l is equal to 1 .4.

00:30 6 meter let v1 b to b b b b 3 and theta 1 theta 2 theta 3 are the diffraction and slopes of node 1 to 3 so here k e stiffness matrix we will like this as a e i upon l q will be something like this here it will be 1 2 and here it's a 6l here it's a minus 1 2 here it's a 6 l here it's a minus 1 2 here it's a 6 l and here it's a 6 l and here it's a 6 l here it's a 4l square, minus 6 of l, 2l squared.

01:06 And here, when we write this as here it's a minus 12, minus 6 l, 12, and here it's a 6 minus 6 of l, and here it's a 6 of l, here it's 2l square, here it's a minus 6 of l, and here it's 4l square.

01:24 Okay, so here it's the stiffness matrix.

01:27 Now as both beam element have same length and ei so we can write this as a k1 is equal to k2 is equal to 230 into 10 to power 9 into 1 .4 into 10 to or minus 4 divided by 6 to power 3 and here the matrix will be something like this 12 6 into 6 minus 12 6 into 6 yet the value of l we put as 6 6 6 6 6 6 2 x 2 x squared and here minus 6 into 6 and here 2 into 6 square we can write this as a minus 12 minus 6 into 6 and here 12 and here it will be minus 6 into 6 here it will be 6 into 6 here it will be 2 into 6 square minus 6 into 6 4 into 6 is square for this we will make the matrix like this okay now for the k1 is equal to k2 is equal to this we can sort the values it's easy to find okay k 3 will be k minus k here it's a minus k here it's a k now the value of k is something like 6 into 10 to bar 5 to 10 to bar 5 then we can common in every time say like minus 6 6 and here it's minus 6 here it will be 6 okay now the global stiffness matrix will be something like this i'm making the global stiffness matrix okay so it's quite long actually so it takes some more space k c will be equal to 10 to r 5 okay and here it will be like it's long in terms like 13 .888 and here it's like 53 .64 minus 17 .88.

03:39 53 .64 here it's a 0 -00.

03:43 Okay and here it's 53 .64 214 .214.

03:49 Minus 53 .64107 .28 and here at 00000.

03:57 Here it's like minus 17 .88.

04:00 Here it's a minus 53 .64.

04:04 Here it's 41 .76.

04:06 Here it's 0 and here is minus 17 .88.

04:09 Here is 53 .64.

04:12 Here it's a minus 6.

04:13 Now we can add this as a 53 .64.

04:17 107 .28.

04:19 Here it's 0...