b) The general form of the element stiffness matrix system, with nodes indexed by i and j, is given as, - N₂(0) FO Uj AE 1 L [ ₁ ] {";} = {√ N₂(2)(z)dx} + {N;(1)ƒ(1) – N;(0) F0] 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 4F[17]{4} = {25 ¹ to form and solve the global system of equations for u₁, ₂ and uz. AE L U₂ £2500 - FO 2500 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 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 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 uz to denote the expansion coefficients for each node. 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, - N₂(0) FO Uj AE 1 L [ ₁ ] {";} = {√ N₂(2)(z)dx} + {N;(1)ƒ(1) – N;(0) F0] 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 4F[17]{4} = {25 ¹ to form and solve the global system of equations for u₁, ₂ and uz. AE L U₂ £2500 - FO 2500 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 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 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 uz to denote the expansion coefficients for each node. 2 20 kN/m 2 X 0.5 m 1m Figure 3: Mesh of 2 elements. Elements are numbered with underlines.
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
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Chapter2: Axially Loaded Members
Section: Chapter Questions
Problem 2.7.12P: A bungee cord that behaves linearly elastically has an unstressed length L0= 760 mm and a stiffness...
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![b) The general form of the element stiffness matrix system, with nodes indexed by i and j, is given
as,
- N₂(0) FO
Uj
AE 1
L
[ ₁ ] {";} = {√ N₂(2)(z)dx} + {N;(1)ƒ(1) – N;(0) F0]
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
4F[17]{4} = {25
¹
to form and solve the global system of equations for u₁, ₂ and uz.
AE
L
U₂
£2500 - FO
2500](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F507f99da-9967-4c1f-a8fd-4e55d78bcba3%2Fd8caad66-ae04-43c5-be1e-e35c41db988f%2Fd10sugd_processed.png&w=3840&q=75)
Transcribed Image Text:b) The general form of the element stiffness matrix system, with nodes indexed by i and j, is given
as,
- N₂(0) FO
Uj
AE 1
L
[ ₁ ] {";} = {√ N₂(2)(z)dx} + {N;(1)ƒ(1) – N;(0) F0]
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
4F[17]{4} = {25
¹
to form and solve the global system of equations for u₁, ₂ and uz.
AE
L
U₂
£2500 - FO
2500
![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 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
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 uz to denote the expansion
coefficients for each node.
2
20 kN/m
2
X
0.5 m
1m
Figure 3: Mesh of 2 elements. Elements are numbered with underlines.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F507f99da-9967-4c1f-a8fd-4e55d78bcba3%2Fd8caad66-ae04-43c5-be1e-e35c41db988f%2F4xddeo_processed.png&w=3840&q=75)
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 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(x) = 20 kN/m.
0 m
1
10 kN/m
1
0.5 m
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 uz to denote the expansion
coefficients for each node.
2
20 kN/m
2
X
0.5 m
1m
Figure 3: Mesh of 2 elements. Elements are numbered with underlines.
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