EML 5526 Finite Element Analysis and Design Exam-I

Last Name: Mandal
First Name: Anirudha


EML 5526 Finite Element Analysis and Design
Exam-I
50 minutes
1 3 2
Problem 1: The following equation is to be solved using a single 3-node 1D
isoparametric element. x1
x2
d 2
+  = 0 , 0  x  L , where  = constant
dx 2
a) Derive the weak form for this differential equation. (10pts)
d 2
Here the governing equation is given as + = 0
dx 2
To convert the above function to the weak form we integrate it with a weighted function 𝛿𝛹𝑋
𝐿
𝑑 𝑑𝜓
∫ [ ( ) + 𝜎] 𝛿𝜓𝑋 𝑑𝑥
𝑑𝑥 𝑑𝑥
0
Here we perform integration by parts to find the weak form hence we get.
𝑳
𝑳 𝑳
𝒅 𝒅𝝍 𝒅𝝍
−∫ ( ) 𝜹𝝍𝑿 . 𝒅𝒙 + [ . 𝜹𝝍𝑿 ] + ∫ 𝝈. 𝜹𝝍𝑿 . 𝒅𝒙
𝟎 𝒅𝒙 𝒅𝒙 𝒅𝒙 𝟎
𝟎




b) Will you get the exact solution to this problem using just one such elements? Explain. (10 pts)

There is no fixed way to denote weather we will get an exact solution using just one element but we can say
that by increasing the number of elements reduces the number of errors each time and that results in a close
approximation of an exact solution. It is difficult to predict the accuracy of a given problem, since FEA is a
study of approximate solution.



c) What is the mapping x(r) between the real coordinate x and the parametric coordinates r, for the element if the
nodal coordinates are x1 = 0 , x2 = 1 and the third node is at the mid-point of the element? (10 pts)

We know that for three node elements the values at the node are for a parametric coordinate (r)
1
𝑁1 (𝑟) = − 𝑟(1 − 𝑟)
2
1
𝑁2 (𝑟) = 𝑟(1 + 𝑟)
2
𝑁3 (𝑟) = 1 − 𝑟 2

Performing Mapping for the three node elements we get
𝑋(𝑟) = 𝑥1 𝑁1 (𝑟) + 𝑥2 𝑁2 (𝑟) + 𝑥3 𝑁3 (𝑟)

1 1
𝑥(𝑟) = 𝑥1 (− 𝑟(1 − 𝑟)) + 𝑥2 ( 𝑟(1 + 𝑟)) + 𝑥3 (1 − 𝑟 2 )
2 2
𝐻𝑒𝑟𝑒 𝑖𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 , 𝑥1 = 0, 𝑥2 = 1, 𝑥3 = 0.5
We get ,

1 1
𝑥(𝑟) = 0 (− 𝑟(1 − 𝑟)) + 1 ( 𝑟(1 + 𝑟)) + 0.5(1 − 𝑟 2 )
2 2


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, Last Name: Mandal
First Name: Anirudha
1
𝑥(𝑟) = ( 𝑟(1 + 𝑟)) + 0.5(1 − 𝑟 2 )
2
1 1
𝑥(𝑟) = (𝑟 + 𝑟 2 ) + (1 − 𝑟 2 )
2 2
𝑟 𝑟2 1 𝑟2
𝑥(𝑟) = + + −
2 2 2 2
𝟏
𝒙(𝒓) = (𝒓 + 𝟏)
𝟐

d
d) If the nodal values of  are computed to be 1 = 10 ,  2 = 30 , and  3 = 40 , determine  ( r ) and within
dx
the element. (10 pts)
We get the interpolating equation as
𝜓(𝑟) = 𝜓1 𝑁1 (𝑟) + 𝜓2 𝑁2 (𝑟) + 𝜓3 𝑁3 (𝑟)

1 1
∴ 𝜓(𝑟) = 𝜓1 (− 𝑟(1 − 𝑟)) + 𝜓2 ( 𝑟(1 + 𝑟)) + 𝜓3 (1 − 𝑟 2 )
2 2
1 1
∴ 𝜓(𝑟) = 10 (− 𝑟(1 − 𝑟)) + 30 ( 𝑟(1 + 𝑟)) + 40(1 − 𝑟 2 )
2 2
∴ 𝜓(𝑟) = (5 + 15 − 40)𝑟 2 + (−5 + 15)𝑟 + 40
∴ 𝜓(𝑟) = −20𝑟 2 + 10𝑟 + 40
∴ 𝝍(𝒓) = −𝟐𝒓𝟐 + 𝒓 + 𝟒
Differentiating w.r.t r,
∴ 𝜓(𝑟) = −2𝑟 2 + 𝑟 + 4
We get ,
𝑑𝜓(𝑟)
=𝑟+1
𝑑𝑟
Differentiating w.r.t r,
1
𝑥(𝑟) = (𝑟 + 1)
2
We get ,

𝑑𝑥(𝑟) 1
=
𝑑𝑟 2

𝒅𝝍
= 𝟐(𝒓 + 𝟏)
𝒅𝒙

e) Derive the contribution {Fe } to the right-hand side of the equilibrium equations from one element. (15 pts)

R.H.S,
𝐿

∫ 𝜎. 𝛿𝜓𝑋 . 𝑑𝑥
0
𝐻𝑒𝑟𝑒 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝜎 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 , 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝐿

𝜎. ∫ 𝛿𝜓𝑋 . 𝑑𝑥
0
We know that , 𝛿𝜓𝑋 = {𝑁1 , 𝑁2 , 𝑁3 } . 𝛿𝜓𝑋

𝐿 𝑁
1
= 𝜎. 𝛿𝜓𝑋 ∫ {𝑁2 } . 𝑑𝑥
0 𝑁3




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