Finite Element Analysis
Assignment 3 Answers
Year 2025
0027 63 985 5033
,
,
, QUESTION 1
(a)
𝛽𝑖 = 𝑦𝑗 − 𝑦𝑚 = 5 − 15 = −10
𝛽𝑗 = 𝑦𝑚 − 𝑦𝑖 = 15 − 15 = 0
𝛽𝑚 = 𝑦𝑖 − 𝑦𝑗 = 15 − 5 = 10
𝛾𝑖 = 𝑥𝑚 − 𝑥𝑗 = 25 − 15 = 10
𝛾𝑗 = 𝑥𝑖 − 𝑥𝑚 = 5 − 25 = −20
𝛾𝑚 = 𝑥𝑗 − 𝑥𝑖 = 15 − 5 = 10
1 𝑥𝑖 𝑦𝑖 1 5 15
2𝐴 = |1 𝑥𝑗 𝑦𝑗 | = |1 15 5 |
1 𝑥𝑚 𝑦𝑚 1 25 15
2𝐴 = [(15)(15) − (25)(5)] − 1[(15)(5) − (25)(15)] + 1[(5)(5) − (15)(15)]
2𝐴 = 100 + 300 − 200 = 200
𝛽𝑖 0 𝛽𝑗 0 𝛽𝑚 0
1
[𝐵] = [ 0 𝛾𝑖 0 𝛾𝑗 0 𝛾𝑚 ]
2𝐴
𝛾𝑖 𝛽𝑖 𝛾𝑗 𝛽𝑗 𝛾𝑚 𝛽𝑚
1 −10 0 0 0 10 0
[𝐵] = [ 0 10 0 −20 0 10 ]
200
10 −10 −20 0 10 10
Assignment 3 Answers
Year 2025
0027 63 985 5033
,
,
, QUESTION 1
(a)
𝛽𝑖 = 𝑦𝑗 − 𝑦𝑚 = 5 − 15 = −10
𝛽𝑗 = 𝑦𝑚 − 𝑦𝑖 = 15 − 15 = 0
𝛽𝑚 = 𝑦𝑖 − 𝑦𝑗 = 15 − 5 = 10
𝛾𝑖 = 𝑥𝑚 − 𝑥𝑗 = 25 − 15 = 10
𝛾𝑗 = 𝑥𝑖 − 𝑥𝑚 = 5 − 25 = −20
𝛾𝑚 = 𝑥𝑗 − 𝑥𝑖 = 15 − 5 = 10
1 𝑥𝑖 𝑦𝑖 1 5 15
2𝐴 = |1 𝑥𝑗 𝑦𝑗 | = |1 15 5 |
1 𝑥𝑚 𝑦𝑚 1 25 15
2𝐴 = [(15)(15) − (25)(5)] − 1[(15)(5) − (25)(15)] + 1[(5)(5) − (15)(15)]
2𝐴 = 100 + 300 − 200 = 200
𝛽𝑖 0 𝛽𝑗 0 𝛽𝑚 0
1
[𝐵] = [ 0 𝛾𝑖 0 𝛾𝑗 0 𝛾𝑚 ]
2𝐴
𝛾𝑖 𝛽𝑖 𝛾𝑗 𝛽𝑗 𝛾𝑚 𝛽𝑚
1 −10 0 0 0 10 0
[𝐵] = [ 0 10 0 −20 0 10 ]
200
10 −10 −20 0 10 10
