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 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 l(x) = 10 kN/m. Along the right half of the bar, x=[0.5 m, 1 m], there is a constant distributed force of l(x) = 20 kN/m. 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 u3 to denote the expansion coefficients for each node. b) The general form of the element stiffness matrix system, with nodes indexed by i and j, is given as shown in the image of eqaution 2, where F0 and f(1) denote boundary forces at positions x = 0 and x = 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 equation 3 in the image.
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 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 l(x) = 10 kN/m. Along the right
half of the bar, x=[0.5 m, 1 m], there is a constant distributed force of l(x) = 20 kN/m.
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 u3 to denote the expansion
coefficients for each node.
b) The general form of the element stiffness matrix system, with nodes indexed by i and j, is given
as shown in the image of eqaution 2, where F0 and f(1) denote boundary forces at positions x = 0 and x = 1, respectively. Form the two basis functions for element 2, and evaluate the right hand side
given by equation 3 in the image.
![[ f* N₁(x)l(x)dx \ _ [N;(1)ƒ(1) – N;(0) F0]
{ [ª N₁(x)l(x)dx) [N₁(1)ƒ(1) – N;(0) Fo) '
Xi
AE
F[₁1] { ";} = +
L
AE
u1
AF [17] {} = {250
L
Աշ
[2500 - FO
2500
(3)
(2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a64e81e-0af4-4cb0-8221-b3f2646e3736%2Fb150bab7-6971-4b5a-b674-1027f990f1bf%2Ftg7ndcs_processed.png&w=3840&q=75)
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