The general form of the element stiffness matrix system, with nodes indexed by i and j, is given as, AE 4² [1₁ 7] {u} = {√2, N. (a)l(a)x) L to form (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 (3) + = N₁(x)l(x)dx] {N;(1)ƒ(1) – N₂(0)F0}' AE FO £2500 [11] {2}-{25 L -1 2500 le the global "
The general form of the element stiffness matrix system, with nodes indexed by i and j, is given as, AE 4² [1₁ 7] {u} = {√2, N. (a)l(a)x) L to form (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 (3) + = N₁(x)l(x)dx] {N;(1)ƒ(1) – N₂(0)F0}' AE FO £2500 [11] {2}-{25 L -1 2500 le the global "
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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can you help me answer this question using the figure provided? thank you.
![The general form of the element stiffness
as,
AE 1
L
-17 Ui
nu;
J S
AE
matrix system, with nodes indexed by i and j, is given
N ₁ (x)l(x) dx 】
N₁(x)l(x) dx)
(2)
xi
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
L
2500 - FOL
#[41]{}-{²
ՂԱԶ
2500
to form and solve the global system of equations for u₁, u2 and u3.
[N₂(1)ƒ(1) - N₂(0) FO
[N;(1)ƒ(1) — N¡(0) F0];
+
=
9
(3)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f7fd42a-4300-4da8-8f14-b18670105553%2Fc16df6f0-38dc-494b-bb79-a475983f4910%2Fgwztxkk_processed.png&w=3840&q=75)
Transcribed Image Text:The general form of the element stiffness
as,
AE 1
L
-17 Ui
nu;
J S
AE
matrix system, with nodes indexed by i and j, is given
N ₁ (x)l(x) dx 】
N₁(x)l(x) dx)
(2)
xi
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
L
2500 - FOL
#[41]{}-{²
ՂԱԶ
2500
to form and solve the global system of equations for u₁, u2 and u3.
[N₂(1)ƒ(1) - N₂(0) FO
[N;(1)ƒ(1) — N¡(0) F0];
+
=
9
(3)
![=
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 1(x) = 20 kN/m.
0m
10 kN/m
0.5 m
20 kN/m
X
1m
Figure 2: Bar domain with varying distributed forces.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f7fd42a-4300-4da8-8f14-b18670105553%2Fc16df6f0-38dc-494b-bb79-a475983f4910%2Fxdb4j5o_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 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 1(x) = 20 kN/m.
0m
10 kN/m
0.5 m
20 kN/m
X
1m
Figure 2: Bar domain with varying distributed forces.
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