written 7.2 years ago by | modified 3.2 years ago by |
Subject: Finite Element Analysis
Topic: One Dimensional Problems
Difficulty: Medium
written 7.2 years ago by | modified 3.2 years ago by |
Subject: Finite Element Analysis
Topic: One Dimensional Problems
Difficulty: Medium
written 7.0 years ago by |
a) Number of elements and nodes,
b) Element matrix Equation is given by,
[K]e=AEhe[1−1−11]
Forelement1,A=250mm2,he=300mm
AEhe=250∗2∗105300=167.67∗103
[K]1=103[167.67−167.67−167.67167.67]
Forelement2,A=400mm2,he=200mm
AEhe=400∗2∗105200=400∗103
[K]2=103[400−400−400400]
Forelement3,A=400mm2,he=200mm
AEhe=400∗2∗105200=400∗103
[K]3=103[400−400−400400]
c) Elemental matrix
[K]=103[167.67−167.6700−167.67167.67+400−40000−400400+400−40000−400400]
d) Global Matrix Equation,
[K]{u}={P}
103[167.67−167.6700−167.67567.67−40000−400800−40000−400400]{u1u2u3u4}={P1P2P3P4}
e) Imposing boundary condition,
u1=0,u4=3.5mm,P3=75∗103N
P2=0 -for balancing,
Frame the equation,
103(−167.67∗u2)P1 -(1)
103(567.67u2−400u3)=0 -(2)
103(−400u2+800u3−400∗3.5)=75∗103 -(3)
103(−400u3+400∗3.5)=P4 -(4)
Solving equation (2) and (3) we get,
∴u2=0.10389mmu3=0.1474mm
Substituting in equation (1) and (4), we get,
P1=−17.41kNP4=−57.56
∑Px=P1+P2+P3+P+4=−17.41+0+75−57.56=0 -verified