Solution Manual a First Course in the Finite Element Method 5th Edition Logan

Contents

Chapter 1

1

Chapter 2

3

Chapter 3

23

Chapter 4

127

Chapter 5

183

Chapter 6

281

Chapter 7

319

Chapter 8

338

Chapter 9

351

Chapter 10

371

Chapter 11

390

Chapter 12

414

Chapter 13

432

Chapter 14

473

Chapter 15

492

Chapter 16

518

Solution Manual A First Course in the Finite Element Method 5th Edition Logan

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Appendix A

550

Appendix B

555

Chapter 1

1.1. A finite element is a small body or unit interconnected to other units to model a larger

structure or system.

1.2. Discretization means dividing the body (system) into an equivalent system of finite

elements with associated nodes and elements.

1.3. The modern development of the finite element method began in 1941 with the work of

Hrennikoff in the field of structural engineering.

1.4. The direct stiffness method was introduced in 1941 by Hrennikoff. However, it was not

commonly known as the direct stiffness method until 1956.

1.5. A matrix is a rectangular array of quantities arranged in rows and columns that is often

used to aid in expressing and solving a system of algebraic equations.

1.6. As computer developed it made possible to solve thousands of equations in a matter of

minutes.

1.7. The following are the general steps of the finite element method.

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8

Divide the body into an equivalent system of finite elements with associated nodes and choose the most appropriate element type.

Choose a displacement function within each element.

Relate the stresses to the strains through the stress/strain law—generally called the constitutive law.

Derive the element stiffness matrix and equations. Use the direct equilibrium method, a work or energy method, or a method of weighted residuals to relate the nodal forces to nodal displacements.

Assemble the element equations to obtain the global or total equations and introduce boundary conditions.

Solve for the unknown degrees of freedom (or generalized displacements). Solve for the element strains and stresses.

Interpret and analyze the results for use in the design/analysis process.

1.8. The displacement method assumes displacements of the nodes as the unknowns of the

problem. The problem is formulated such that a set of simultaneous equations is solved for nodal displacements.

1.9. Four common types of elements are: simple line elements, simple two-dimensional

elements, simple three-dimensional elements, and simple axisymmetric elements.

1.10 Three common methods used to derive the element stiffness matrix and equations are

(1) direct equilibrium method (2) work or energy methods (3) methods of weighted residuals

1.11. The term ‘degrees of freedom’ refers to rotations and displacements that are associated

1.12. Five typical areas where the finite element is applied are as follows.

(1) Structural/stress analysis (2) Heat transfer analysis (3) Fluid flow analysis

(4) Electric or magnetic potential distribution analysis (5) Biomechanical engineering

1.13. Five advantages of the finite element method are the ability to

(1) Model irregularly shaped bodies quite easily (2) Handle general load conditions without difficulty

(3) Model bodies composed of several different materials because element equations are evaluated individually

(4) Handle unlimited numbers and kinds of boundary conditions

(5) Vary the size of the elements to make it possible to use small elements where necessary

Chapter 2

0 0 0 0 0 0 0 0 2.1 (a) k1 0 – k1 0 [k(1)] = 0 0 0 0 – k1 0 k1 0 0 0 0 0 [k(2)] = 0 0 k2 – k2 0 0 – k2 k2 0 0 0 0 0 k 0 – k [k 3(3)] = 3 3 0 0 0 0 0 – k3 0 k3 [K] = [k(1)] + [k(2)] + [k(3)] k1 0 – k1 0 0 k₃ [K] = 0 – k3 – k₁ 0 k₁ k₂ – k₂ 0 – k₃ – k₂ k₂ k₃

(b) Nodes 1 and 2 are fixed so u1 = 0 and u2 = 0 and [K] becomes

[K] = k1 k2 – k2 – k₂ k₂ k₃ {F} = [K] {d} F3 x k1 k2 – k2 u3 = F4 x – k2 k2 k3 u4 0 k k – k u ⇒ = 1 2 2 3 P – k2 k2 k3 u4 {F} = [K] {d} ⇒[K –1] {F} = [K –1] [K] {d} ⇒ [K –1 ] {F} = {d} Using the adjoint method to find [K –1]

C11 = k2 + k3 C21 = (– 1) 3 (– k2) C12 = (– 1) 1 + 2 (– k2) = k2 C22 = k1 + k2

2 T 2 2 [C] = k2 k3 k 2 and CT = k2 k3 k2 k₂ k₁ k₂ k₂ k₁ k₂ det [K] = | [K] | = (k1 + k2) (k2 + k3) – ( – k2) (– k2) ⇒ | [K] | = (k1 + k2) (k2 + k3) – k 2 [K –1] = [C ] det K k2 k3 k2 k2 k3 k2 [K –1] = k2 k1 k2 = k2 k1 k2 (k1 k2 ) (k2 k3 ) – k2 k1 k2 k1 k3 k2 k3 k2 k3 k2 0 u3 = u4 k2 k1 k2 P k₁ k₂ k₁ k₃ k₂ k₃ k P ⇒ u3 = 2 k₁ k₂ k₁ k₃ k₂ k₃ (k k ) P ⇒ u4 = 1 2 k k k k k k_{1 2 1 3 2 3}

(c) In order to find the reaction forces we go back to the global matrix F = [K] {d}

F1x k1 0 k1 0 u1 F2 x 0 = k3 0 k3 u F3 x _k1 0 k1 k2 k2 u3 F_{4 x} 0 k₃ k₂ k₂ k_{3 u4} k P F1x = – k1 u3 = – k1 2 k₁ k₂ k₁ k₃ k₂ k₃ k k P ⇒ F1x = 1 2 k₁ k₂ k₁ k₃ k₂ k₃ (k k ) P F2x = – k3 u4 = – k3 1 2 k₁ k₂ k₁ k₃ k₂ k₃ k (k k ) P 2.2 ⇒ F2x = 3 1 2 k₁ k₂ k₁ k₃ k₂ k₃ k1 = k2 = k3 = 1000 lb in. (1) (2) (2) (3) [k(1)] = k k (1) k ; [k(2)] = k (2) k k (2) k k (3)

⇒ f (1) (2) (3) k k 0 (1) k k 0 [K] = k k k k (2) ⇒ [K] = k 2k k 0 k k (3) 0 k k Node 1 is fixed ⇒ u1 = 0 and u3 = δ

{F} = [K] {d} F1x ? k k 0 u1 0 F2 x 0 = k 2k k u2 ? F3x ? 0 k k u3 0 2k k u 0 2k u k ⇒ = 2 2 F3 x k k F3x k u2 k ⇒ u2 = k ₌ 2k 2 = 1 in. 2 ⇒ u2 = 0.5″ F3x = – k (0.5″) + k (1″) lb lb F3x = (– 1000 F3x = 500 lbs Internal forces Element (1) in. ) (0.5″) + (1000 in. ) (1″) f1x(1) k k u 0 = 1 f2 x (2) k k u2 0.5 (1) 1x = (– 1000 lb ) (0.5″) ⇒ in. f1x (1) = – 500 lb f_{2 x} Element (2) (1) = (1000 lb ) (0.5″) ⇒ in. f2 x (1) = 500 lb f2 x(2) k k u2 0.5 f2 x(2) – 500 lb = ⇒ 2.3 f3x (2) _{k k u} 3 1 f3x (2) 500 lb (a) [k(1)] = [k(2)] = [k(3)] = [k(4)] = k k k k

By the method of superposition we construct the global [K] and knowing {F} = [K] {d} we have F1x ? k k 0 0 0 u1 0 F_{2 x} 0 k 2k k 0 0 u2 F3 x P = 0 k 2k k 0 u3 F4 x 0 0 0 k 2k k u4 F5 x ? 0 0 0 k k u5 0

P P 0 2k k 0 u2 0 2ku2 ku3 (b) P = k 2k k u3 P ku2 2ku3 ku4 0 0 k u 2k u4 u 0 ku₃ 2ku₄ ⇒ u2 = 3 2 ; u4 = 3 2

Substituting in the equation in the middle

P = – k u2 + 2k u3 – k u4 u ⇒ P = – k 3 + 2k u – k u 3 2 3 2 ⇒ P = k u3 P ⇒ u3 = u2 = k P P ; u4 = 2k 2k

(c) In order to find the reactions at the fixed nodes 1 and 5 we go back to the global equation {F} = [K] {d} Check P F1x = – k u2 = – k 2k P F5x = – k u4 = – k 2k ⇒ F1x = 2 ⇒ F5x = 2 ΣFx = 0 ⇒ F1x + F5x + P = 0 P P 2.4 ⇒ 2 + ⇒ 0 = 0 2 + P = 0 (a) [k(1)] = [k(2)] = [k(3)] = [k(4)] = k _{k k} k

By the method of superposition the global [K] is constructed. Also {F} = [K] {d} and u1 = 0 and u5 = δ

F1x ? k k 0 0 0 u1 0 F2 x 0 k 2k k 0 0 u2 ? F3x 0 = 0 k 2k k 0 u3 ? F_{4 x} 0 0 0 k 2k k u4 ? F5 x ? 0 0 0 k k u5

2 (b) 0 = 2k u2 – k u3 (1) 0 = – ku2 + 2k u3 – k u4 (2) 0 = – k u3 + 2k u4 – k δ (3) From (2) u3 = 2 u2 From (3) 2 u u4 = 2 2 Substituting in Equation (2) 2 ⇒ – k (u2) + 2k (2 u2) – k 2 x ⇒ – u2 + 4 u2 – u2 – 2 = 0 ⇒ u2 = 4 ⇒ u3 = 2 4 ⇒ u3 = 2 ⇒ u4 = 2

₄

2 ⇒ u4 = 3 4 (c) Going back to the global equation

{F} = [K] {d} F1x = – k u2 = k 4 k ⇒ F1x = 4 3 F5x = – k u4 + k δ = – k 4 k + k δ 2.5 ⇒ F5x = 4 1 1 kip 2 in. 2 2 3 4 kip 3 in. 5 3 x kip 1 in. 4 kip 4 in. kip 5 in. d1 d2 d2 d4 [k (1)] = 1 1 2 ; [k (2)] = 2 1 1 2 2 d2 d4 d2 d4 [k (3)] = 3 3 4 ; [k (4)] = 4 3 3 d4 d3 4 4 [k (5)] = 5 5 5 5

0 0 5 5 0 9 5 14

Assembling global [K] using direct stiffness method 1 1 0 0 1 1 2 3 4 0 2 3 4 [K] = 0 0 5 5 0 2 3 4 5 2 3 4 5 Simplifying 1 1 0 0 1 10 0 9 kip [K] = 0 0 5 5 in. 0 9 5 14

2.6 Now apply + 2 kip at node 2 in spring assemblage of P 2.5.

∴ F2x = 2 kip [K]{d} = {F} [K] from P 2.5 1 1 0 0 u1 0 F1 1 10 0 9 u 2 2 = (A) u3 0 F3 u₄ 0

where u1 = 0, u3 = 0 as nodes 1 and 3 are fixed.

Using Equations (1) and (3) of (A) 10 9 u2 2 = 9 14 u4 Solving 0 2.7 u2 = 0.475 in., u4 = 0.305 in. conv. f1x = C, f2x = – C f = – kδ = – k(u2 – u1) ∴ f1x = – k(u2 – u1) f2x = – (– k) (u2 – u1) f1x k –k u1 = f2 x ∴ [K] = –k k u2 k –k same as for

in. in. 1 1x 2.8 k = 500 lb k = 500 lb 500 lb 1 1 1 1 k1 = 500 ; k2 = 500 1 1 So 1 1 1 1 0 [K] = 500 1 2 1 0 1 1 {F} = [K] {d} F1 ? 1 1 0 u1 0 ⇒ F2 0 = 500 1 2 1 u2 ? F3 1000 0 1 1 u3 ? ⇒ 0 = 1000 u2 – 500 u3 (1) 500 = – 500 u2 + 500 u3 (2) From (1) u2 = 500 1000 u3 ⇒ u2 = 0.5 u3 (3) Substituting (3) into (2) ⇒ 500 = – 500 (0.5 u3) + 500 u3 ⇒ 500 = 250 u3 ⇒ u3 = 2 in. ⇒ u2 = (0.5) (2 in.) ⇒ u2 = 1 in. Element 1–2 f_1x(1) 1 1 0 in. f (1) 500 lb = 500 1x f2 x (1) 1 1 1 in. f2 x (1) 500 lb Element 2–3 f_{2 x}(2) _{1 1 1 in.} f2 x (2) 500 lb (2) = 500 ₍₂₎ f3x 1 1 2 in. f3x 500 lb 0 2.9 F1x = 500 [1 – 1 0] in. F 500 lb 2 in. lb in. lb in. lb in.

(1) (2) [k(1)] = 1000 1000 1000 1000 (2) (3) [k(2)] = 1000 1000 1000 1000 (3) (4) 1000 1000 (1) (2) (3) (4) 1000 1000 0 0 1000 2000 1000 0 [K] = 0 1000 2000 1000 0 0 1000 1000 F1x ? 1000 1000 0 0 u1 0 F2 x 1000 1000 2000 1000 0 u2 = F3x 0 0 1000 2000 1000 u3 F4 x 4000 0 0 1000 1000 u4 ⇒ u1 = 0 in. u2 = 3 in. u3 = 7 in. u4 = 11 in. Reactions u1 0 u 3 F1x = [1000 – 1000 0 0] 2 u3 ⇒ F 1x = – 3000 lb 7 Element forces Element (1) u₄ 11 f1x (1) 1000 1000 0 f (1) 3000 lb = ⇒ 1x f_{2 x} (1) 1000 1000 3 f_{2 x} (1) 3000 lb Element (2) f2 x (2) 1000 1000 3 f (2) 4000 lb = ⇒ 2 x f3x (2) 1000 1000 7 f3x (2) 4000 lb Element (3) f_3x(3) ₁₀₀₀ 1000 7 f3x (3) 4000 lb = ⇒ 2.10 f_{4 x} (3) 1000 1000 11 f4 x (3) 4000 lb lb in. lb in. lb in.