Consider the bar in Figure 2, which has cross-sectional area A = 1-10-3 m², modulus of elasticity E1-10¹¹1 N/m², and length 1 m. The bar is fixed to a wall on its left hand side. Along the left half of the bar, x=[0 m, 0.5 m], there is a constant distributed force of 1(x) = 10 kN/m. Along the right half of the bar, x=[0.5 m, 1 m], there is a constant distributed force of 1(r) = 20 kN/m. 0m 1 AE L 10 kN/m Figure 2: Bar domain with varying distributed forces. a) Given the finite element mesh in Figure 3, consisting of two elements of equal size L = 0.5 m, sketch the associated basis functions of the mesh, and write the approximation of the displacement function u(x) as an expansion of these basis functions. Use u₁, u₂ and us to denote the expansion coefficients for each node. 1 0.5 m 0.5 m L 20 kN/m Figure 3: Mesh of 2 elements. Elements are numbered with underlines. 1m 0.5 m b) The general form of the element stiffness matrix system, with nodes indexed by i and j, is given as, FO [1]{"}-{ N₁(2)(x) dx} + {N(1) S(1) - N,(0) Fo} (2) where F0 and f(1) denote boundary forces at positions = 0 and 1, respectively. Form the two basis functions for element 2, and evaluate the right hand side vector of the matrix system 2 to form the local system of equations for element 2. Then use the local system for element 1 given by AE (3) = #1]{}-{25% to form and solve the global system of equations for u₁, U₂ and us. [2500 - FO 2500 J

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Consider the bar in Figure 2, which has cross-sectional area A = 1-10-³ m², modulus of elasticity
E = 1.10¹¹ N/m², and length 1m. The bar is fixed to a wall on its left hand side. Along the left half
of the bar, x=[0 m, 0.5 m], there is a constant distributed force of 1(x) = 10 kN/m. Along the right
half of the bar, x=[0.5 m, 1 m], there is a constant distributed force of 1(x) = 20 kN/m.
0 m
1
10 kN/m
1
0.5 m
Figure 2: Bar domain with varying distributed forces.
a) Given the finite element mesh in Figure 3, consisting of two elements of equal size L = 0.5 m,
sketch the associated basis functions of the mesh, and write the approximation of the displacement
function u(x) as an expansion of these basis functions. Use u₁, ₂ and us to denote the expansion
coefficients for each node.
0.5 m
2
20 kN/m
=
2
X
0.5 m
1m
Figure 3: Mesh of 2 elements. Elements are numbered with underlines.
b) The general form of the element stiffness matrix system, with nodes indexed by i and j, is given
as,
AE
4 [11]{}- [N₁(x)l(x)dx] [N(1)ƒ(1) - N(0) FO
+
N₁(x)l(x)dx) [N, (1)f(1) - N,(0) FO)*
(2)
where F0 and f(1) denote boundary forces at positions z = 0 and r = 1, respectively. Form the
two basis functions for element 2, and evaluate the right hand side vector of the matrix system
2 to form the local system of equations for element 2. Then use the local system for element 1
given by
AE
[u |
- FO
AF [17] {m} = {2505 F0},
(3)
to form and solve the global system of equations for u₁, ₂ and us.
Transcribed Image Text:Consider the bar in Figure 2, which has cross-sectional area A = 1-10-³ m², modulus of elasticity E = 1.10¹¹ N/m², and length 1m. The bar is fixed to a wall on its left hand side. Along the left half of the bar, x=[0 m, 0.5 m], there is a constant distributed force of 1(x) = 10 kN/m. Along the right half of the bar, x=[0.5 m, 1 m], there is a constant distributed force of 1(x) = 20 kN/m. 0 m 1 10 kN/m 1 0.5 m Figure 2: Bar domain with varying distributed forces. a) Given the finite element mesh in Figure 3, consisting of two elements of equal size L = 0.5 m, sketch the associated basis functions of the mesh, and write the approximation of the displacement function u(x) as an expansion of these basis functions. Use u₁, ₂ and us to denote the expansion coefficients for each node. 0.5 m 2 20 kN/m = 2 X 0.5 m 1m Figure 3: Mesh of 2 elements. Elements are numbered with underlines. b) The general form of the element stiffness matrix system, with nodes indexed by i and j, is given as, AE 4 [11]{}- [N₁(x)l(x)dx] [N(1)ƒ(1) - N(0) FO + N₁(x)l(x)dx) [N, (1)f(1) - N,(0) FO)* (2) where F0 and f(1) denote boundary forces at positions z = 0 and r = 1, respectively. Form the two basis functions for element 2, and evaluate the right hand side vector of the matrix system 2 to form the local system of equations for element 2. Then use the local system for element 1 given by AE [u | - FO AF [17] {m} = {2505 F0}, (3) to form and solve the global system of equations for u₁, ₂ and us.
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