23.12.2014 Views

OCTOBER 19-20, 2012 - YMCA University of Science & Technology

OCTOBER 19-20, 2012 - YMCA University of Science & Technology

OCTOBER 19-20, 2012 - YMCA University of Science & Technology

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Proceedings <strong>of</strong> the National Conference on<br />

Trends and Advances in Mechanical Engineering,<br />

<strong>YMCA</strong> <strong>University</strong> <strong>of</strong> <strong>Science</strong> & <strong>Technology</strong>, Faridabad, Haryana, Oct <strong>19</strong>-<strong>20</strong>, <strong>20</strong>12<br />

Consistent mass<br />

The consistent mass matrix for a beam element is a filled matrix. The filled matrix can be combined with other<br />

consistent mass matrices <strong>of</strong> other elements <strong>of</strong> the structure, in the same manner as the element stiffness matrices<br />

are combined, to yield the final global mass matrix. The element consistent mass matrix for a prismatic beam is,<br />

with mass per unit length m and length l.<br />

m<br />

e<br />

=<br />

ml<br />

4<strong>20</strong><br />

⎡ 156<br />

⎢<br />

221<br />

⎢<br />

⎢ 54<br />

⎢<br />

⎣−131<br />

221<br />

41<br />

2<br />

131<br />

2<br />

− 31<br />

54<br />

131<br />

156<br />

− 221<br />

−131⎤<br />

2<br />

− 31<br />

⎥<br />

⎥<br />

− 221⎥<br />

2 ⎥<br />

41 ⎦<br />

Assuming the two elements have the same properties and lengths, the global mass matrix becomes:<br />

2<br />

2<br />

⎡ 156m1 22m1 54m1 −13m1<br />

0 0 ⎤<br />

⎢ 2<br />

3<br />

2<br />

3<br />

⎥<br />

⎢ 22m1 4m1 13m1 − 3m1 0 0 ⎥<br />

⎢<br />

2<br />

2<br />

1 54m1 13m1 312m1 0 54m1 −13m1<br />

⎥<br />

mg<br />

= ⎢<br />

2<br />

3<br />

3<br />

2<br />

3<br />

⎥<br />

4<strong>20</strong> ⎢−13m1<br />

− 3m1 0 8m1 13m1 − 3m1 ⎥<br />

⎢<br />

2<br />

2<br />

0 0 54m1 13m1 156m1 − 22m1 ⎥<br />

⎢<br />

⎥<br />

2<br />

3<br />

2<br />

3<br />

⎢⎣<br />

0 0 −13m1<br />

− 3m1 − 22m1 4m1 ⎥⎦<br />

Taking into account the two constrained degrees <strong>of</strong> freedom at the built in end, we can eliminate the first two<br />

rows and columns:<br />

m<br />

g<br />

=<br />

1<br />

4<strong>20</strong><br />

⎡ 312m1<br />

⎢<br />

⎢<br />

⎢<br />

⎢<br />

⎣<br />

0<br />

54m1<br />

2<br />

−13m1<br />

0<br />

8m1<br />

3<br />

13m1<br />

2<br />

3<br />

− 3m1<br />

54m1<br />

13m1<br />

2<br />

156m1<br />

2<br />

− 22m1<br />

−13m1<br />

− 3m1<br />

− 22m1<br />

4m1<br />

3<br />

3<br />

2<br />

2<br />

⎤<br />

⎥<br />

⎥<br />

⎥<br />

⎥<br />

⎦<br />

Having the mass and stiffness matrices allows us to solve the eigenvalue problem for the homogeneous<br />

equations <strong>of</strong> motion:<br />

mg ž + kgz<br />

= [ 0]<br />

It is better to reduce the 4x4 problem down to 2x2 size. Therefore, the Guyan reduction [15] will be used to<br />

reduce the size <strong>of</strong> the problem.<br />

Two element cantilever eigen value closed form solution using guyan reduction<br />

Repeating the rearranged global stiffness matrix from the static run,<br />

⎡ 24<br />

– 12<br />

⎢<br />

0<br />

3<br />

3<br />

1<br />

1<br />

⎢<br />

8 – 6<br />

⎢ 0<br />

2<br />

k = ⎢<br />

1 1<br />

g<br />

EI<br />

⎢ – 12 – 6 12<br />

⎢ 3<br />

2<br />

3<br />

1 1 1<br />

⎢ 6 2 – 6<br />

⎢<br />

2<br />

2<br />

⎣ 1 1 1<br />

6<br />

2<br />

1<br />

2<br />

1<br />

– 6<br />

2<br />

1<br />

4<br />

1<br />

⎤<br />

⎥<br />

⎥<br />

⎥<br />

⎥<br />

⎥<br />

⎥<br />

⎥<br />

⎥<br />

⎦<br />

Equation <strong>of</strong> motion is: mss<br />

x + kssx = [0]<br />

392

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!