Introduction To Finite Element Analysis And Design
Introduction To Finite Element Analysis And Design
2nd Edition
ISBN: 9781119078722
Author: Kim, Nam H., Sankar, Bhavani V., KUMAR, Ashok V., Author.
Publisher: John Wiley & Sons,
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Chapter 2, Problem 6E

A one-dimensional heat conduction problem can be expressed by the following differential equation: k d 2 T d x 2 + Q = 0 , 0 x L . where k is the thermal conductivity, T ( x ) is the temperature, and Q is heat generated per unit length. Q, the heat generated per unit length, is assumed constant. Two essential boundary conditions are given at both ends: T ( 0 ) = T ( L ) = 0 . Calculate the approximate temperature T ( x ) using the Galerkin method. Compare the approximate solution with the exact one. Hint: Start with an assumed solution in the following form: T ˜ ( x ) = c 0 + c 1 x + c 2 x 2 , and then make it satisfy the two essential boundary conditions.

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Suppose that we have a wire (or a thin metal rod) of length that is insulated except at the endpoints. Let L denote the position along the wire and let t denote time. The 1- D heat equation is * temperature u 3 insulation O (a^2 u)/(at^2)=c^2 (a^2 u)/(ax^2) (a^2 u)/(at^2 )=c^2 du/əx au/Ət=c^2 (a^2 u)/(ðx^2 ) au/at=c^2 au/Əx none
1. The general form of linear second-order differential equation can be written in the form: و بار / كلية الهندسة Q4)/ grap dy q(x)y = r(x) d'y +p(x) dx dy b. dx - F(x)y = F(x) x2 dy dx - xy = C. d. r2 d?y dx2 -f(x)y = F(x) 2431)(5-1) 3 (3-21)2 a. (221 -91i) / 169 b. (21 + 52i)/ 13 c. (-90+220i)/169 d. (-7+17i)/ 13 2. Simplify: الحدار المك المراغة 3. If the roots of second order differential equation is complex conjugate, then the gene contain: a. sinusoidal functions and exponentials b. constant and two exponentials c. two constants and two exponentials d. two constants and one exponential 5 4. The order and degree of the differential: 3(3 - + 4y = sinx* are: d²y a. First-order, First-degree- b. First-order, second-degree Second -order, First -degree d. Second -order, second-degree dx2 lo - 2i tisi. 8- 12i 5. The particular solution of (D² + 4)y = cos 2x is equal to: a. sin 2x b. cos 2x 13+159 C. 4 cos 2x d. 4 sin 2x 5-12 lo Best wishes الامتحانية د. مازن ياسین عبود رئيس القسم بن فاضل…
a. A slab of thermal insulator is 100 cm² in cross section and 2 cm thick. Its thermal conductivity is 2.4 x 10 -+ cal/(s cm C"). If the temperature difference between opposite faces is 180 F', how much heat flows through the slab in one day?

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