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Solution Manual for A First Course in the Finite Element Method, 6th Edition

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    An ’s Solutions Manual to Accompany A FIRST COURSE IN THE FINITE ELEMENT METHOD, 6 TH E DITION D ARYL L. L OGAN

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    Contents Chapter 1:............................................................................................................................ 1 Chapter 2:............................................................................................................................ 3 Chapter 3:.......................................................................................................................... 25 Chapter 4:........................................................................................................................ 137 Chapter 5:........................................................................................................................ 203 Chapter 6:........................................................................................................................ 315 Chapter 7:........................................................................................................................ 363 Chapter 8:........................................................................................................................ 383 Chapter 9:........................................................................................................................ 397 Chapter 10:...................................................................................................................... 423 Chapter 11:...................................................................................................................... 449 Chapter 12:...................................................................................................................... 477 Chapter 13:...................................................................................................................... 499 Chapter 14:...................................................................................................................... 539 Chapter 15:...................................................................................................................... 561 Chapter 16:...................................................................................................................... 591 Appendix A: ..................................................................................................................... 629 Appendix B: ..................................................................................................................... 635 Appendix D: ..................................................................................................................... 641

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    1 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 Divide the body into an equivalent system of finite elements with associated nodes and choose the most appropriate element type. Step 2 Choose a displacement function within each element. Step 3 Relate the stresses to the strains through the stress/strain law—generally called the constitutive law. Step 4 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. Step 5 Assemble the element equations to obtain the global or total equations and introduce boundary conditions. Step 6 Solve for the unknown degrees of freedom (or generalized displacements). Step 7 Solve for the element strains and stresses. Step 8 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

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    2 (3) methods of weighted residuals 1.11. The term ‘degrees of freedom’ refers to rotations and displacements that are associated with each node. 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

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    3 Chapter 2 2.1 (a) [ k (1) ] = 1 1 1 1 0 – 0 0 0 0 0 – 0 0 0 0 0 0 k k k k [ k (2) ] = 2 2 2 2 0 0 0 0 0 0 0 0 0 0 – 0 0 – k k k k [ k 3 (3) ] = 3 3 3 3 0 0 0 0 0 0 – 0 0 0 0 0 – 0 k k k k [ K ] = [ k (1) ] + [ k (2) ] + [ k (3) ] [ K ] = 1 1 3 3 1 1 2 2 3 2 2 3 0 – 0 0 0 – – 0 – 0 – – k k k k k k k k k k k k (b) Nodes 1 and 2 are fixed so u 1 = 0 and u 2 = 0 and [ K ] becomes [ K ] = 1 2 2 2 2 3 – – k k k k k k { F } = [ K ] { d } 3 4 x x F F = 1 2 2 2 2 3 – – k k k k k k 3 4 u u  0 P = 1 2 2 2 2 3 – – k k k k k k 3 4 u u { F } = [ K ] { d }  [ K ] – 1 { F } = [ K ] – 1 [ K ] { d }

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    4   K ] – 1 { F } = { d } Using the adjoint method to find [ K – 1 ] C 11 = k 2 + k 3 C 21 = ( – 1) 3 ( – k 2 ) C 12 = ( – 1) 1 + 2 ( – k 2 ) = k 2 C 22 = k 1 + k 2 [ C ] = 2 3 2 2 1 2 k k k k k k and C T = 2 3 2 2 1 2 k k k k k k det [ K ] = | [ K ] | = ( k 1 + k 2 ) ( k 2 + k 3 ) – ( – k 2 ) ( – k 2 )  | [ K ] | = ( k 1 + k 2 ) ( k 2 + k 3 ) – k 2 2 [ K – 1 ] = [ ] det T C K [ K – 1 ] = 2 3 2 2 1 2 2 1 2 2 3 2 ( ) ( ) – k k k k k k k k k k k = 2 3 2 2 1 2 1 2 1 3 2 3 k k k k k k k k k k k k 3 4 u u = 2 3 2 2 1 2 1 2 1 3 2 3 0 k k k k k k P k k k k k k  u 3 = 2 1 2 1 3 2 3 k P k k k k k k   u 4 = 1 2 1 2 1 3 2 3 ( ) k k P k k k k k k (c) In order to find the reaction forces we go back to the global matrix F = [ K ]  { d } 1 2 3 4 x x x x F F F F = 1 1 1 2 3 3 1 1 2 2 3 3 2 2 3 4 0 0 0 0 0 0 u k k u k k k k k k u k k k k u F 1 x = – k 1 u 3 = – k 1 2 1 2 1 3 2 3 k P k k k k k k  F 1 x = 1 2 1 2 1 3 2 3 k k P k k k k k k F 2 x = – k 3 u 4 = – k 3  1 2 1 2 1 3 2 3 ( ) k k P k k k k k k  F 2 x = 3 1 2 1 2 1 3 2 3 ( ) k k k P k k k k k k 2.2

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    5 k 1 = k 2 = k 3 = 1000 lb in. (1) (2) (2) (3) [ k (1) ] = (1) (2) k k k k ; [ k (2) ] = (2) (3) k k k k By the method of superposition the global stiffness matrix is constructed. (1) (2) (3) [ K ] = 0 (1) (2) 0 (3) k k k k k k k k  [ K ] = 0 2 0 k k k k k k k Node 1 is fixed  u 1 = 0 and u 3 =  { F } = [ K ] { d } 1 2 3 ? 0 ? x x x F F F = 0 2 0 k k k k k k k 1 2 3 0 ? u u u                3 0 x F = 2 2 3 2 0 2 2 x k u k k k u F k u k k k   u 2 = 2k k = 2 = 1 in. 2  u 2 = 0.5  F 3 x = – k (0.5  ) + k (1  ) F 3 x = ( – 1000 lb in. ) (0.5  ) + (1000 lb in. ) (1  ) F 3 x = 500 lbs Internal forces Element (1) (1) 1 (2) 2 x x f f = 1 2 0 0.5 u k k k k u  (1) 1 x f = ( – 1000 lb in. ) (0.5  )  (1) 1 x f = – 500 lb (1) 2 x f = (1000 lb in. ) (0.5  )  (1) 2 x f = 500 lb Element (2)

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    6 (2) 2 (2) 3 x x f f = 2 2 3 3 0.5 0.5 1 1 u u k k k k u u k k k k  (2) 2 (2) 3 – 500 lb 500 lb x x f f 2.3 (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 1 2 3 4 5 ? 0 0 ? x x x x x F F F P F F = 0 0 0 2 0 0 0 2 0 0 0 2 0 0 0 k k k k k k k k k k k k k 1 2 3 4 5 0 0 u u u u u (b) 0 0 P = 2 2 3 3 2 3 4 3 4 4 2 0 0 2 2 2 0 2 0 2 u k k ku ku k k k u P ku ku ku k k ku ku u  u 2 = 3 2 u ; u 4 = 3 2 u Substituting in the second equation above P = – k u 2 + 2 k u 3 – k u 4  P = – k 3 2 u + 2 k u 3 – k 3 2 u  P = k  u 3  u 3 = P k u 2 = 2 P k ; u 4 = 2 P k (c) In order to find the reactions at the fixed nodes 1 and 5 we go back to the global equation { F } = [ K ] { d } F 1 x = – k  u 2 = – 2 P k k  F 1 x = 2 P F 5 x = – k  u 4 = – 2 P k k  F 5 x = 2 P Check  F x = 0  F 1 x + F 5 x + P = 0 (1) (2) (3)

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    7  2 P + 2 P + P = 0  0 = 0 2.4 (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 u 1 = 0 and u 5 =  1 2 3 4 5 ? 0 0 0 ? x x x x x F F F F F = 0 0 0 2 0 0 0 2 0 0 0 2 0 0 0 k k k k k k k k k k k k k 1 2 3 4 5 0 ? ? ? u u u u u (b) 0 = 2 k u 2 – k u 3 (1) 0 = – ku 2 + 2 k u 3 – k u 4 (2) 0 = – k u 3 + 2 k u 4 – k  (3) From (2) u 3 = 2 u 2 From (3) u 4 = 2 2 2 u Substituting in Equation (2)  – k ( u 2 ) + 2 k (2  u 2 ) – k 2 2 2 u          – u 2 + 4 u 2 – u 2 – 2 = 0  u 2 = 4  u 3 = 2 4  u 3 = 2  u 4 = 4 2 2  u 4 = 3 4 (c) Going back to the global equation

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    8 { F } = [ K ]  { d } F 1 x = – k  u 2 = 4 k  F 1 x = 4 k F 5 x = – k u 4 + k  = – k 3 4 + k   F 5 x = 4 k 2.5 u 1 u 2 u 2 u 4 [ k (1) ] = 1 1 1 1 ; [ k (2) ] = 2 2 2 2 u 2 u 4 u 2 u 4 [ k (3) ] = 3 3 3 3 ; [ k (4) ] = 4 4 4 4 u 4 u 3 [ k (5) ] = 5 5 5 5 Assembling global [ K ] using direct stiffness method [ K ] = 1 1 0 0 1 1 2 3 4 0 2 3 4 0 0 5 5 0 2 3 4 5 2 3 4 5 Simplifying [ K ] = 1 1 0 0 1 10 0 9 kip 0 0 5 5 in. 0 9 5 14 2.6 Now apply + 3 kip at node 2 in spring assemblage of P 2.5.  F 2 x = 3 kip [ K ]{ d } = { F }

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    9 [ K ] from P 2.5 1 1 0 0 1 10 0 9 0 0 5 5 0 9 5 14 1 2 3 4 0 0 u u u u = 1 3 3 0 F F               (A) where u 1 = 0, u 3 = 0 as nodes 1 and 3 are fixed. Using Equations (1) and (3) of (A) 2 4 10 9 9 14 u u = 3 0       Solving u 2 = 0.712 in., u 4 = 0.458 in. 2.7 f 1 x = C , f 2 x = – C f = – k  = – k ( u 2 – u 1 )  f 1 x = – k ( u 2 – u 1 ) f 2 x = – ( – k ) ( u 2 – u 1 ) 1 2 x x f f = k –k –k k 1 2 u u  [ K ] = k –k –k k same as for tensile element 2.8 k 1 = 1000 1 1 1 1 ; k 2 = 1000 1 1 1 1 So

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    10 [ K ] = 1000 1 1 0 1 2 1 0 1 1 { F } = [ K ] { d }   1 2 3 ? 0 500 F F F              = 1 1 0 10 1 2 1 00 0 1 1              1 2 3 0 ? ? u u u  0 = 2000 u 2 – 1000 u 3 (1) 500 = – 1000 u 2 + 1000 u 3 (2) From (1) u 2 = 1000 2000 u 3  u 2 = 0.5 u 3 (3) Substituting (3) into (2)  500 = – 1000 (0.5 u 3 ) + 1000 u 3  500 = 500 u 3  u 3 = 1 in.  u 2 = (0.5) (1 in.)  u 2 = 0.5 in. Element 1 – 2 (1) 1 (1) 2 x x f f = 1000 (1) 1 (1) 2 500lb 1 1 0 in. 1 1 0.5 in. 500lb x x f f                   Element 2 – 3 (2) 2 (2) 3 x x f f = 1000 (2) 2 (2) 3 500 lb 1 1 0.5 in. 1 1 1 in. 500 lb x x f f                   F 1 x = 500 [1 – 1 0] 1 0 0.5 in. 500 lb 1 in. x F              2.9 (1) (2) [ k (1) ] = 5000 5000 5000 5000         (2) (3)

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    11 [ k (2) ] = 5000 5000 5000 5000         (3) (4) [ k (3) ] = 5000 5000 5000 5000         (1) (2) (3) (4) [ K ] = 5000 5000 0 0 5000 10000 5000 0 0 5000 10000 5000 0 0 5000 5000                   1 2 3 4 ? 1000 0 4000 x x x x F F F F = 5000 5000 0 0 5000 10000 5000 0 0 5000 10000 5000 0 0 5000 5000                   1 2 3 4 0 u u u u  u 1 = 0 in. u 2 = 0.6 in. u 3 = 1.4 in. u 4 = 2.2 in. Reactions F 1 x = [5000 – 5000 0 0] 1 2 3 4 0 0.6 1.4 2.2 u u u u                    F 1 x = – 3000 lb Element forces Element (1) (1) 1 (1) 2 x x f f = 5000 5000 0 5000 5000 0.6                (1) 1 (1) 2 3000lb 3000lb x x f f Element (2) (2) 2 (2) 3 x x f f = 5000 5000 0.6 5000 5000 1.4                (2) 2 (2) 3 4000lb 4000lb x x f f Element (3) (3) (3) 3 3 (3) (3) 4 4 x x x x f f f f = 5000 5000 1.4 5000 5000 2.2                (3) 3 (3) 4 4000lb 4000lb x x f f 2.10

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    12 [ k (1) ] = 1000 1000 1000 1000 [ k (2) ] = 500 500 500 500 [ k (3) ] = 500 500 500 500 { F } = [ K ] { d } 1 2 3 4 ? – 8000 ? ? x x x x F F F F                   = 1 2 3 4 0 1000 1000 0 0 ? 1000 2000 500 500 0 0 500 500 0 0 500 0 500 0 u u u u  u 2 = 8000 2000  = – 4 in. Reactions 1 2 3 4 x x x x F F F F = 0 1000 1000 0 0 1000 2000 500 500 4 0 500 500 0 0 0 500 0 500 0                                   1 2 3 4 x x x x F F F F = 4000 8000 2000 2000                lb Element (1) (1) 1 (1) 2 x x f f = 1000 1000 0 1000 1000 – 4                (1) 1 (1) 2 x x f f = 4000 4000        lb Element (2) (2) 2 (2) 3 x x f f = 500 500 4 500 500 0                 (2) 2 (2) 3 x x f f = – 2000 2000       lb Element (3)

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    13 (3) 2 (3) 4 x x f f = 500 500 4 500 500 0                 (3) 2 (3) 4 x x f f 2000 2000        lb 2.11 [ k (1) ] = 1000 1000 1000 1000         ; [ k (2) ] = 3000 3000 3000 3000         { F } = [ K ] { d } 1 2 3 ? 0 ? x x x F F F = 1000 1000 0 1000 4000 3000 0 3000 3000               1 2 3 0 ? 0.02 m u u u  u 2 = 0.015 m Reactions F 1 x = ( – 1000) (0.015)  F 1 x = – 15 N Element (1) 1 2 x x f f       = 1000 1000 1000 1000         0 0.015        1 2 x x f f        15 15         N Element (2) 2 3 x x f f       = 3000 –3000 –3000 3000       0.015 0.02        2 3 x x f f       = 15 15         N 2.12 [ k (1) ] = [ k (3) ] = 10000 1 1 1 1 [ k (2) ] = 10000 3 3 3 3         { F } = [ K ] { d }

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