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

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1 Solution Manual A First Course in the Finite Element Method 5th Edition Logan Instant download and all chapters Solution Manual A First Course in the Finite Element Method 5th Edition Logan Contents Chapter 1 1 Chapter 3 Chapter 3 3 Chapter 4 17 Chapter Chapter 6 81 Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter

2 Appendix A 550 Appendix B 555 Appendix D 561

3 Chapter A finite element is a small body or unit interconnected to other units to model a larger structure or system. 1.. Discretization means dividing the body (system) into an equivalent system of finite elements with associated nodes and elements The modern development of the finite element method began in 1941 with the wor of Hrennioff in the field of structural engineering The direct stiffness method was introduced in 1941 by Hrennioff. However, it was not commonly nown as the direct stiffness method until 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 As computer developed it made possible to solve thousands of equations in a matter of minutes The following are the general steps of the finite element method. Step 1 Step 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 wor 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 unnown 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 The displacement method assumes displacements of the nodes as the unnowns of the problem. The problem is formulated such that a set of simultaneous equations is solved for nodal displacements Four common types of elements are: simple line elements, simple two-dimensional elements, simple three-dimensional elements, and simple axisymmetric elements Three common methods used to derive the element stiffness matrix and equations are direct equilibrium method () wor or energy methods (3) methods of weighted residuals The term degrees of freedom refers to rotations and displacements that are associated with each node.

4 1.1. Five typical areas where the finite element is applied are as follows. Structural/stress analysis () Heat transfer analysis (3) Fluid flow analysis (4) Electric or magnetic potential distribution analysis (5) Biomechanical engineering Five advantages of the finite element method are the ability to Model irregularly shaped bodies quite easily () 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 inds of boundary conditions (5) Vary the size of the elements to mae it possible to use small elements where necessary

5 Chapter.1 (a) [ ] [ () ] [ 3 (3) ] [K] [ ] + [ () ] + [ (3) ] [K] (b) Nodes 1 and are fixed so u 1 0 and u 0 and [K] becomes [K] 1 3 {F} [K] {d} F 3 x 1 u 3 F 4 x 3 u u 3 3 u 4 {F} [K] {d} [K 1 ] {F} [K 1 ] [K] {d} [K 1 ] {F} {d} Using the adjoint method to find [K 1 ] C C 1 ( 1) 3 ( ) C 1 ( 1) 1 + ( ) C 1 +

[C] 3 and C T 3 1 1 det [K] [K] ( 1 + ) ( + 3 ) ( ) ( ) [K] ( 1 + ) ( + 3 ) [K 1 ] T [C ] det K 3 3 [K 1 ] 1 1 ( 1 )( 3 ) 1 1 3 3 3 0 u 3 u 4 1 1 1 3 3 u 3 u 4 1 1 3 3 ( ) 1 1 1 3 3 (c) In order to

6 [C] 3 and C T det [K] [K] ( 1 + ) ( + 3 ) ( ) ( ) [K] ( 1 + ) ( + 3 ) [K 1 ] T [C ] det K 3 3 [K 1 ] 1 1 ( 1 )( 3 ) u 3 u u 3 u ( ) (c) In order to find the reaction forces we go bac to the global matrix F [K] {d} F 1x u 1 F x u F 3 x u 3 F 4 x u 4 F 1x 1 u F 1x ( ) F x 3 u F x ( ) [ ] () () (3) ; [ () () ] () (3) By the method of superposition the global stiffness matrix is constructed.

7 () (3) 0 0 [K] () [K] Node 1 is fixed u 1 0 and u 3 δ {F} [K] {d} 0 (3) 0 F 1x? 0 u 1 0 F x 0 u? F 3x? 0 u 3 0 u 0 u F 3 x F 3x u u Internal forces Element f 1x 1 F 3x (0.5 ) + (1 ) u 0.5 F 3x ( 1000 ) (0.5 ) + (1000 ) (1 ) F 3x 500 s u 0 1 () f x u 0.5 1x f ( 1000 ) (0.5 ) f 1x 500 Element () f x (1000 ) (0.5 ) f x 500 f x () () u 0.5 f x 500 () f 3x u 3 1 () f 3x (a) [ ] [ () ] [ (3) ] [ (4) ] By the method of superposition we construct the global [K] and nowing {F} [K] {d} we have F 1x? u 1 0 F x u F 3 x 0 0 u 3 F 4 x u 4 F 5 x? u 5 0

0 0 u 0 u u 3 (b) u 3 u u 3 u 4 0 0 u 4 0 u 3 u 4 u3 u u ; u 4 Substituting in the equation in the middle 3 u + u 3 u 4 u 3 + u 3 u 3 u 3 u 3 u ; u 4 (c) In order to find the reactions at the fixed

8 0 0 u 0 u u 3 (b) u 3 u u 3 u u 4 0 u 3 u 4 u3 u u ; u 4 Substituting in the equation in the middle 3 u + u 3 u 4 u 3 + u 3 u 3 u 3 u 3 u ; u 4 (c) In order to find the reactions at the fixed nodes 1 and 5 we go bac to the global equation {F} [K] {d} Chec F 1x u F 5x u 4 F 1x F 5x ΣF x 0 F 1x + F 5x (a) [ ] [ () ] [ (3) ] [ (4) ] By the method of superposition the global [K] is constructed. Also {F} [K] {d} and u 1 0 and u 5 δ F 1x? u 1 0 F x u? F 3x u 3? F 4 x u 4? F 5 x? u 5

9 (b) 0 u u 3 0 u + u 3 u 4 () 0 u 3 + u 4 δ (3) From () u 3 u From (3) u 4 u Substituting in Equation () (u ) + ( u ) x u + 4 u u 0 u 4 u 3 4 u 3 u 4 4 u (c) Going bac to the global equation {F} [K] {d} F 1x u F 1x 4 4 F 5x u 4 + δ δ F 5x ip 3 4 ip 4 ip 4 ip 3 5 ip 5 3 x [ ] [ (3) ] [ (5) ] d 1 d d d ; [ () ] 1 1 d d 4 d d ; [ (4) 4 ] d 4 d

10 Assembling global [K] using direct stiffness method Simplifying [K] [K] ip Now apply + ip at node in spring assemblage of.5. [K] from.5 F x ip [K]{d} {F} u 1 0 F u u 3 0 F 3 u 4 0 (A).7 where u 1 0, u 3 0 as nodes 1 and 3 are fixed. Using Equations and (3) of (A) 10 9 u 9 14 u 4 0 Solving u 0.475, u conv. f 1x C, f x C f δ (u u 1 ) f 1x (u u 1 ) f x ( ) (u u 1 ) f 1x u 1 f x [K] u same as for

11 tensile element

12 So ; [K] {F} [K] {d} F 1? u 1 0 F u? From F u 3? u 500 u u u 3 () u Substituting (3) into () u 3 u 0.5 u 3 (3) (0.5 u 3 ) u u 3 Element 1 f 1x Element 3 u 3 u (0.5) ( ) u f 500 1x 500 f x f x 500 f x () f x () () 500 () f 3x 1 1 f 3x x F 1x 500 [1 1 0] F 500.9

13 [ ] [ () ] () () (3) (3) (4) [K] () (3) (4) F 1x? u 1 0 F x u F 3x u 3.10 F 4 x u 4 Reactions Element forces Element u 1 0 u 3 u 3 7 u 4 11 F 1x [ ] u 1 0 u 3 F 1x 3000 u 3 7 u 4 11 f 1x f x f x f x 3000 Element () f () () x f x 4000 () f 3x () f 3x 4000 Element (3) (3) (3) f 3x f 3x (3) f (3) 4 x f 4 x

A FIRST COURSE IN THE FINITE ELEMENT METHOD

A FIRST COURSE IN THE FINITE ELEMENT METHOD INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMANY A IRST COURS IN TH INIT LMNT MTHOD ITH DITION DARYL L. LOGAN Contents Chapter 1 1 Chapter 3 Chapter 3 3 Chapter 17 Chapter 5 183 Chapter 6 81 Chapter 7 319 Chapter