A FIRST COURSE IN THE FINITE ELEMENT METHOD FIFTH EDITION PDF

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INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY A FIRST COURSE IN THE FINITE ELEMENT METHOD FIFTH EDITION DARYL L. 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 Chapt...

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INSTRUC TO R'S SO LUTIO NS M A NUA L TO A C C O M PA NY

A FIRST C O URSE IN THE

FINITE ELEM ENT M ETHO D FIFTH EDITIO N

DA RYL L. LO G A N

Contents Cha p te r 1

1

Cha p te r 2

3

Cha p te r 3

23

Cha p te r 4

127

Cha p te r 5

183

Cha p te r 6

281

Cha p te r 7

319

Cha p te r 8

338

Cha p te r 9

351

Cha p te r 10

371

Cha p te r 11

390

Cha p te r 12

414

Cha p te r 13

432

Cha p te r 14

473

Cha p te r 15

492

Cha p te r 16

518

Ap p e nd ix A

550

Ap p e nd ix B

555

Ap p e nd ix D

561

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

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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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Chapter 2 2.1 (a)

0 – k1 0 0 0 k1 0 0

[k(1)] =

Ë k1 Ì 0 Ì Ì – k1 Ì Í 0

[k(2)] =

Ë0 Ì 0 Ì Ì0 Ì Í0

0 0 0 0 0 k2 0 – k2

[k 3(3)] =

Ë0 Ì 0 Ì Ì0 Ì Í0

0 k3 0 – k3

0Û 0Ü Ü 0Ü Ü 0Ý 0 Û 0 Ü Ü – k2 Ü Ü k2 Ý

0 0 Û 0 – k3 Ü Ü 0 0 Ü Ü 0 k3 Ý

[K] = [k(1)] + [k(2)] + [k(3)]

[K] =

0 k3 0 – k3

Ë k1 Ì 0 Ì Ì – k1 Ì Í 0

– k1 0 k1  k2 – k2

0 Û – k3 Ü Ü – k2 Ü Ü k2  k3 Ý

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

Ëk1  k2 [K] = Ì Í – k2

– k2 Û k 2  k3 ÜÝ

{F} = [K] {d}

Ëk1  k2 Î F3 x Þ Ï ß= Ì F Í – k2 Ð 4x à

– k2 Û k 2  k3 ÜÝ

Î u3 Þ Ï ß Ðu4 à

Ë k1  k2 Î0Þ ⇒ Ï ß= Ì P Ð à Í – k2

– k2 Û k2  k3 ÝÜ

Îu3 Þ Ï ß Ðu4 à

{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 3

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Ë k  k3 [C] = Ì 2 Í k2

k2 Û Ë k  k3 Û and CT = Ì 2 Ü k1  k2 ÜÝ k1  k2 Ý Í k2 k2

det [K] = | [K] | = (k1 + k2) (k2 + k3) – ( – k2) (– k2) ⇒

| [K] | = (k1 + k2) (k2 + k3) – k22 [K –1] =

[C T ] det K

k2 Û k2 Û Ë k2  k3 Ëk 2  k3 Ì Ì k Ü k1  k2 ÜÝ k1  k2 Ý Í k2 Í 2 [K –1] = = k1 k2  k1 k3  k2 k3 ( k1  k2 ) ( k2  k3 ) – k2 2 k2 Û Î 0 Þ Ëk2  k3 Ï ß Ì k1  k2 ÜÝ Ð P à Îu3 Þ Í k2 Ï ß= k1 k2  k1 k3  k2 k3 Ðu4 à ⇒ u3 =

k2 P k1 k 2  k1 k3  k2 k3

⇒ u4 =

(k1  k2 ) P k1 k 2  k1 k3  k2 k3

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

Î F1x Þ Ë k1 ÑF Ñ Ì 0 Ñ 2x Ñ Ï ß= Ì Ì k1 Ñ F3 x Ñ Ì Ñ Í 0 Ð F4 x Ñ à

0 k3 0 k3

 k1 0 k1  k2  k2

F1x = – k1 u3 = – k1 ⇒ F1x =

k2 P k1 k2  k1 k3  k2 k3

k1 k2 P k1 k2  k1 k3  k2 k3

F2x = – k3 u4 = – k3 ⇒ F2x =

0 Û Îu1 Þ Ü Ñu Ñ  k3 Ü ÑÏ 2 Ñß  k2 Ü Ñu3 Ñ Ü k 2  k3 Ý Ñ Ðu4 Ñà

( k1  k2 ) P k1 k 2  k1 k3  k 2 k3

k3 (k1  k2 ) P k1 k 2  k1 k3  k 2 k3

2.2

k1 = k2 = k3 = 1000

lb in.

(1) (2) Ë k k Û (1) ; [k(2)] = [k(1)] = Ì Í k k ÜÝ (2)

(2) (3) Ë k k Û (2)

ÌÍ k

k ÜÝ (3)

By the method of superposition the global stiffness matrix is constructed. 4 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(1)

(2)

(3)

0Û (1) Ë k k Ì [K] = k k  k  k Ü (2) ⇒ [K] = Ì Ü k ÜÝ (3) ÌÍ 0  k

0Û Ë k k Ì k 2k  k Ü Ì Ü k ÜÝ ÌÍ 0 k

Node 1 is fixed ⇒ u1 = 0 and u3 = δ {F} = [K] {d}

Î F1x ? Þ Ñ Ñ Ï F2 x 0ß = ÑF Ñ Ð 3x ? à

Ë k Ì k Ì ÌÍ 0

Î 0 Þ Ë 2k ⇒ Ï ß = Ì Í k Ð F3 x à

⇒

u2 =

k 2k k

0Û Îu1 0 Þ Ñ Ü Ñu Ï 2 ?ß

k

Ü

k ÜÝ Ñ Ðu3

 k Û Îu2 Þ Î 0 Ï ßÀ Ï k ÜÝ Ð E à Ð F3 x

E Ñà

2k u2  kE Þ ß  k u2  kE à

kE E 1 in. = = ⇒ u2 = 0.5″ 2 2k 2

F3x = – k (0.5″) + k (1″) F3x = (– 1000

lb lb ) (0.5″) + (1000 ) (1″) in. in.

F3x = 500 lbs Internal forces Element (1) (1) Î Ë k Ñ f1x Þ Ñ Ï (2) ß = Ì Í k Ñ f2 x à Ñ Ð

⇒

k Û Îu1 0 Þ Ï ß k ÝÜ Ðu2 0.5” à

f1x (1) = (– 1000 f 2x (1) = (1000

lb ) (0.5″) ⇒ f1x (1) = – 500 lb in.

lb ) (0.5″) ⇒ f 2x (1) = 500 lb in.

Element (2) (2) Î Ñ f2 x Þ Ñ Ë k Ï (2) ß = Ì Í k Ñ f 3x à Ñ Ð

f 2 x (2) – 500 lb  k Û Îu2 0.5”Þ ⇒ Ü Ïu 1” ß k ÝÐ à f 3 x (2) 500 lb 3

2.3

Ë k k Û (a) [k(1)] = [k(2)] = [k(3)] = [k(4)] = Ì Í k k ÜÝ By the method of superposition we construct the global [K] and knowing {F} = [K] {d} we have Î F1x ? Þ Ë k  k 0 0 0Û Îu1 0 Þ ÑF Ñ Ì Ñ ÜÑ ÑÑ 2 x 0 ÑÑ Ì k 2k  k 0 0Ü ÑÑu2 ÑÑ 0Ü Ïu3 Ï F3 x P ß = Ì 0  k 2k k ß Ì 0 0  k 2k  k Ü Ñ ÑF Ñ Ñ 0 Ü Ñu4 Ñ Ñ 4x Ñ Ì ÑÐ F5 x ? Ñà ÌÍ 0 0 0 k k ÜÝ ÑÐu5 0Ñà 5 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Î0 Þ Ë 2 k Ñ Ñ (b) Ï P ß = Ì k Ì Ñ0 Ñ Ì 0 Ð à Í ⇒

u2 =

k

2 ku2  ku3 ku2  2ku3  ku4 ku3  2ku4

0 Û Îu2 Þ 0 Ñ Ñ k ÜÜ Ïu3 ß À P 2k ÜÝ ÑÐu4 Ñà 0

2k

k

u3 u ; u4 = 3 2 2

Substituting in the equation in the middle P = – k u2 + 2k u3 – k u4

u3 Ø È u3 Ø Ù + 2k u3 – k É Ù Ú Ê 2Ú 2

⇒

P = – k ÈÉ

⇒

P = k u3

⇒

u3 =

P k

u2 =

P P ; u4 = 2k 2k

Ê

(c) In order to find the reactions at the fixed nodes 1 and 5 we go back to the global equation {F} = [K] {d} F1x = – k u2 = – k

P P ⇒ F1x =  2k 2

F5x = – k u4 = – k

P P ⇒ F5x =  2k 2

Check ΣFx = 0 ⇒ F1x + F5x + P = 0 ⇒ 

P + 2

È É Ê

PØ Ù +P=0 2Ú

⇒0=0 2.4

k Û k ÜÝ

Ë k (a) [k(1)] = [k(2)] = [k(3)] = [k(4)] = Ì Í k

By the method of superposition the global [K] is constructed. Also

Î F1x ÑF 2x Ñ Ñ F Ï 3x ÑF Ñ 4x Ñ F5 x Ð

{F} = [K] {d} and u1 = 0 and u 5 = δ

?Þ Ë k  k 0 0 0Û Îu1 0 Þ Ñ Ì  k 2k  k 0 0 Ü Ñu ? Ñ 0 Ñ Ì Ü ÑÑ 2 ÑÑ Ñ 0 ß = Ì 0  k 2k k 0Ü Ïu3 ? ß Ì 0 0  k 2k  k Ü Ñ Ñ 0Ñ Ì Ü Ñu4 ? Ñ Ñ ÌÍ 0 0 0 k k ÜÝ ÐÑu5 E àÑ ? àÑ

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(b) 0 = 2k u2 – k u3

(1)

0 = – ku2 + 2k u3 – k u 4

(2)

0 = – k u3 + 2k u4 – k δ

(3)

From (2) u3 = 2 u2 From (3) u4 =

E  2 u2 2

Substituting in Equation (2) ⇒ – k (u2) + 2k (2 u2) – k ⇒ – u2 + 4 u2 – u2 – ⇒ u3 = 2 ⇒ u4 =

ÈE Ê

 2 E2 x Ø Ú

2

E E = 0 ⇒ u2 = 2 4

E E ⇒ u3 = 4 2

E  2  E4 3E ⇒ u4 = 2 4

(c) Going back to the global equation {F} = [K] {d} F1x = – k u2 = k

kE E ⇒ F1x =  4 4 3EØ Ù +kδ Ê 4 Ú

F5x = – k u4 + k δ = – k ÈÉ ⇒ F5x =

kE 4

2.5 2

1 kip 1 in.

4

d1

5

4 5

4 kip in.

d2

kip in.

3

3

2

1

kip in. 2

d2

kip in.

3

x

d4

Ë 2  2Û Ë 1 1Û [k (1)] = Ì ; [k (2)] = Ì Ü 2 ÜÝ Í 2 Í 1 1Ý d2

d4

d2

d4

Ë 4  4Û Ë 3  3Û [k (3)] = Ì ; [k (4)] = Ì Ü Í  4 4ÜÝ Í  3 3Ý d4

d3

Ë 5  5Û [k (5)] = Ì Í  5 5ÜÝ 7 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Assembling global [K] using direct stiffness method

0 0 1 Ë1 Û Ì 1 1  2  3  4 0  2  3  4 Ü Ü [K] = Ì 0 5 Ì0 5 Ü Ì 0  2  3  4  5 2  3  4  5Ü Í Ý Simplifying

0 Û Ë 1 1 0 Ì  1 10 0  9 Ü kip Ü [K] = Ì Ì 0 0 5  5 Ü in. Ì 0  9  5 14 Ü Í Ý 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

0 Û Îu1 0 Þ Î F1 Þ Ë 1 1 0 Ñ Ñ2 Ñ Ì  1 10 0  9 Ü ÑÑu Ì Ü Ï 2 Ñß = ÑÏ Ñß Ì 0 0 5  5 Ü Ñu3 0Ñ Ñ F3 Ñ Ì 0  9  5 14 Ü Ñ Í Ý Ðu4 Ñà ÑÐ0 Ñà

(A)

where u 1 = 0, u 3 = 0 as nodes 1 and 3 are fixed. Using Equations (1) and (3) of (A)

Ë 10  9 Û Îu2 Þ Î 2Þ ÌÍ  9 14ÜÝ Ïu ß = Ï 0 ß Ð 4à Ð à Solving u2 = 0.475 in., u4 = 0.305 in. 2.7

conv.

f1x = C, f2x = – C f = – kδ = – k(u2 – u1) ∴

f1x = – k(u2 – u1) f2x = – (– k) (u2 – u1) Î f1x Þ Î k Ï ß = Ï f Ð –k Ð 2x à

∴

Î k [K] = Ï Ð –k

–k Þ ß kà

Îu1 Þ Ï ß Ðu2 à

–k Þ same as for ß k à tensile element 8

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2.8 lb k = 500 in.

Ë 1 k1 = 500 Ì Í 1

lb k = 500 in.

1 Û Ë 1 ; k2 = 500 Ì Ü 1Ý Í 1

500 lb

1 Û 1 ÜÝ

So

Ë 1 [K] = 500 Ì 1 Ì ÍÌ 0

1

0Û 1 Ü Ü 1 ÝÜ

2 1

{F} = [K] {d}

Ë F1 ? Û Ë 1 Ì Ü Ì ⇒ F2 0 Ì Ü = 500 Ì 1 Ì 0 Í ÍÌ F3 1000ÝÜ ⇒

1 2 1

Îu1 0 Þ Ñ Ñ Ïu2 ?ß Ü 1ÝÜ ÑÐu3 ? Ñà

0Û 1Ü

0 = 1000 u2 – 500 u3

(1)

500 = – 500 u2 + 500 u3

(2)

From (1) u2 =

500 u3 ⇒ u2 = 0.5 u3 1000

(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 (1) Î Ë 1 Ñ f1x Þ Ñ Ï (1) ß = 500 Ì Í 1 Ñ f2 x à Ñ Ð

1 Û Î0 Ï 1 ÜÝ Ð1

f1 x (1) in.Þ À ß in.à f 2 x (1)

 500lb 500lb

Element 2–3

Î f 2 x (2) Ñ Þ Ë 1 Ñ Ï (2) ß = 500 Ì Í 1 Ñ f 3x à Ñ Ð

1 Û Î1 in. Þ f 2 x (2)  500 lb Ï ßÀ 1 ÜÝ Ð2 in.à f 3 x (2) 500 lb

F1x = 500 [1 – 1 0]

Ë0 Ì 1 Ì Ì Í2

Û in.Ü À F1x Ü in.ÜÝ

 500 lb

2.9 lb in.

lb in.

lb in.

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(1) (2) 1000  1000 Ë Û [k(1)] = Ì Í 1000 1000ÜÝ (2) 1000 Ë [k(2)] = Ì Í 1000 (3) Ë 1000 [k(3)] = Ì Í  1000 (1) 1000 Ë Ì 1000 [K] = Ì Ì 0 Ì 0 Í

Î F1x ? Þ ÑF Ñ Ñ 2 x  1000Ñ Ï ß= Ñ F3x 0 Ñ ÑÐ F4 x 4000 Ñà ⇒

u1 = u2 = u3 = u4 =

(3)

1000Û 1000ÜÝ (4)  1000Û

1000ÜÝ (2) (3) (4) 0 0 Û 1000 2000 1000 0 Ü Ü 1000 2000 1000Ü 0  1000 1000 ÜÝ

0 Ë 1000  1000 0 Û Ì 1000 2000 1000 0 Ü Ì Ü Ì 0 1000 2000 1000Ü Ì 0 0 1000 1000 ÜÝ Í

Îu1 0Þ Ñu Ñ Ñ 2 Ñ Ï ß Ñu3 Ñ ÑÐu4 Ñà

0 in. 3 in. 7 in. 11 in.

Reactions

Îu1 Ñu Ñ 2 F1x = [1000 – 1000 0 0] Ï Ñu3 ÑÐu4 Element forces Element (1)

0Þ 3 ÑÑ ß ⇒ F1x = – 3000 lb 7Ñ 11Ñà

(1) Î Ñ f1x Þ Ñ Ë 1000 Ï (1) ß = Ì Í 1000 Ñ f2 x à Ñ Ð Element (2)

f1x (1) 3000 lb 1000 Û Î0Þ ⇒ Ï ß 1000 ÜÝÐ3 à f 2 x (1) 3000 lb

(2) Î Ñ f2 x Þ Ñ Ë 1000 Ï (2) ß = Ì Í 1000 Ñ f 3x à Ñ Ð Element (3)

f 2 x (2) 1000 Û Î3 Þ ⇒ Ï ß 1000 ÜÝÐ7à f 3 x (2)

ÎÑ f 3 x (3) ÞÑ Ë 1000 Ï (3) ß = Ì Í 1000 ÐÑ f 4 x àÑ

 4000lb 4000 lb

f 3 x (3)  4000 lb 1000 Û Î7 Þ ⇒ Ï ß 1000 ÜÝ Ð11à f 4 x (3) 4000 lb

2.10 lb in. lb in.

lb in.

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Ë 1000 [k(1)] = Ì Í 1000

1000 Û 1000 ÜÝ

Ë 500 [k(2)] = Ì Í  500

 500 Û 500 ÜÝ

Ë 500 [k(3)] = Ì Í  500

 500 Û 500 ÜÝ

{F} = [K] {d}

Î F1x ÑF Ñ 2x Ï Ñ F3 x Ñ Ð F4 x

0 0 Û Îu1 0 Þ Ë 1000  1000 Þ Ì Ñ  1000 2000  500  500Ü ÑÑu2 ? ÑÑ – 4000Ñ ÜÏ ß ß= Ì  ? 0 500 500 0 Ì Ü Ñu3 0 Ñ Ñ Ì 0 Ü Ñ ?  500 0 500Ý ÑÐu4 0Ñà Í à

?

⇒

u2 =

 4000 2000

= – 2 in.

Reactions

0 0 Û Î0 Þ Î F1x Þ Ë 1000  1000 ÑF Ñ Ì 1000 2000 500 500 Ü ÑÑ2 ÑÑ Ñ 2x Ñ ÜÏ ß Ï ß= Ì 500 0 Ü Ñ0 Ñ Ì 0 500 Ñ F3 x Ñ Ì 0 Ü Ñ 500 0 500Ý ÑÐ0 àÑ Í Ð F4 x Ñ à

Î F1x Þ Î 2000 Þ ÑF Ñ Ñ 4000Ñ Ñ Ñ Ñ 2x Ñ ⇒ Ï ß= Ï ß lb 1000 F Ñ 3x Ñ Ñ Ñ Ñ Ñ Ð 1000 Ñà Ð F4 x Ñ à Element (1)

Î f1x (1) Ñ Þ Ñ Ë 1000 Ï (1) ß = Ì Í 1000 Ñ f2 x à Ñ Ð

1000 Û Î 0 Þ ÎÑ f1x (1) ÑÞ Î 2000 Þ ß lb Ï ß ⇒ Ï (1) ß = Ï 1000 ÜÝ Ð – 2à ÐÑ f 2 x àÑ Ð 2000à

Element (2)

Î f 2 x (2) Ñ Þ Ë 500 Ñ Ï (2) ß = Ì Í 500 Ñ Ð f 3x Ñ à

500 Û Î  2 Þ Ï ß⇒ 500 ÜÝÐ 0 à

(2) ÑÎ f 2 x ÑÞ Î –1000 Þ ß lb Ï (2) ß = Ï ÑÐ f 3 x Ñà Ð 1000à

500 Û Î  2 Þ Ï ß⇒ 500 ÜÝÐ 0 à

(3) ÑÎ f 2 x ÑÞ Î  1000 Þ ß lb Ï (3) ß Ï ÑÐ f 4 x Ñà Ð 1000 à

Element (3)

Î f 2 x (3) Ñ Þ Ë 500 Ñ Ï (3) ß = Ì Í 500 Ñ Ð f4 x Ñ à 2.11

N m

N m

d = 20 mm

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2000 Û Ë 2000 ; [k(2)] = Ì Ü 2000 Ý Í 2000

Ë 2000 [k(1)] = Ì Í 2000

2000 Û 2000 ÜÝ

{F} = [K] {d}

Î F1x ? Þ Ë 2000 2000 0 Û Ñ Ñ Ï F2 x 0 ß = ÌÌ 2000 4000 2000ÜÜ ÑF Ñ 2000 2000 ÜÝ Ð 3x ? à ÌÍ 0 ⇒

Ëu1 0 Û Ìu ? Ü Ì 2 Ü ÌÍu3 0.02 m ÜÝ

u2 = 0.01 m

Reactions F1x = (– 2000) (0.01) ⇒ F1x = – 20 N Element (1)

Î ÎÑ fˆ1x ÞÑ Î20Þ Ñ fˆ1x Þ Ñ Ë 2000 2000Û Î 0 Þ ⇒ = ψ ß Ïˆ ß = Ï ßN ÌÍ2000 2000 ÜÝ ÏÐ0.01ßà Ñ f2x à Ñ Ð ÐÑ f 2 x Ñà Ð 20à Element (2)

Î fˆ2 x Ñ Þ Ñ Ïˆ ß = Ñ Ð f3x Ñ à

–2000Û Î 0.01Þ ÑÎ fˆ2 x ÑÞ Î20 Þ ⇒ Ï ß ßN ψ ß = Ï 2000 ÜÝ Ð0.02 à ÑÐ f3 x Ñà Ð 20 à

Ë 2000 Ì –2000 Í

2.12 N m

N m

N m

1

2

3

Î 1 1Þ ß Ð1 1à

[k(1)] = [k(3)] = 10000 Ï

Î 2 2Þ ß Ð2 2à

[k(2)] = 10000 Ï {F} = [K] {d}

ÎF1x ? Þ Ë 1 1 0 0Û Îu1 0 Þ Ì 1 3 2 0Ü ÑÑu ? ÑÑ ÑF Ñ Ñ 2 x 4500 N Ñ Ì Ü Ï 2 ß = 10000 Ï ß F 0 0 2 3 1 Ì   Ü Ñu3 ? Ñ Ñ 3x Ñ Ì Ü Ñ Ñ Í 0 0 1 1Ý ÑÐu4 0Ñà ÐF4 x ? à 0 = – 2 u2 + 3 u3 ⇒ u2 =

3 u 3 ⇒ u2 = 1.5 u3 2

450 N = 30000 (1.5 u3) – 20000 u3 ⇒ 450 N = (25000 ⇒

N ) u3 ⇒ u 3 = 1.8 × 10–2 m m

u2 = 1.5 (1.8 × 10–2) ⇒ u 2 = 2.7 × 10–2 m

Element (1)

0 Î fˆ1x (1)  270 N Þ Ñ fˆ1x Þ Ñ Ë 1 1Û Î ⇒ = 10000 Ï ß ÌÍ 1 1ÜÝ Ï2.7 – 102 ß fˆ2 x (1) 270 N Ñ fˆ2 x à Ñ Ð à Ð 12 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Element (2) 2 Î fˆ2 x (2) Ñ fˆ2 x Þ Ñ Ë 1 1Û ÎÑ2.7 – 10 ÞÑ ⇒ = 20000 Ï ß ß ÌÍ 1 1ÜÝ Ï fˆ3 x (2) ÑÐ1.8 – 102 àÑ Ñ fˆ3 x à Ñ Ð

180 N

180 N

Element (3)

Î fˆ3 x (3) Ñ fˆ3 x Þ Ñ Ë 1 1Û Î1.8 – 102 Þ ⇒ = 10000 ψ ß ß ÌÍ 1 1ÜÝ Ï 0 fˆ4 x (3) Ñ f4 x à Ñ Ð à Ð

180 N

180 N

Reactions