basic truss matlab program for truss analysis -...

School of Mechanical EngineeringChung-Ang University

Basic Truss 이론 및MATLAB Program for Truss Analysis

김석민


Introduction

• 구조물의 평형을 위해서는 외부력(external forces) 뿐만이 아니라 내부력(internal forces)도 고려해야 한다.

• 연결부에서는 뉴톤 제3법칙인 작용과 반작용법칙(forces of action and reaction)이 성립

• 구조물의 분류 :a) Trusses: formed from two-force members (2력부재),

i.e., straight members with end point connectionsb) Frames: contain at least one multi-force member (다력부재),

i.e., member acted upon by 3 or more forces.c) Machines: structures containing moving parts designed to

transmit and modify forces.

구조물 : 하중을 지지하거나 전달하기 위하여 또는 작용하는 하중을안전하게 견디어 내기 위하여 만들어진 부재들의 조립체


Definition of a Truss

• 트러스는 절점(joints)에 연결된 직선 부재(member) 로 구성된다.

• 실제로 볼트 또는 용접으로 연결될지라도부재는 핀으로 연결된다고 가정함.

• 부재의 각 단에 작용하는 힘은 단일 하중이다.

• 부재에 작용하는 하중은 바로 부재의 각 단에서단일 하중이다. 각 부재는 2력(two-force members) 부재로 취급한다.

• 모든 하중은 여러 절점에 작용해야 하며, 부재그 자체에 작용해서는 안 된다.

• 각 2력 부재는 인장(tension)상태 또는압축(compression)상태에 놓인다.

트러스(truss) : 삼각형의 형태를 이루도록 서로 결합된 곧고, 가는 봉으로 구성된 구조물


Definition of a Truss

부재들은 세장 부재이며, 횡하중(lateral loads)을 지탱할 수 없다. 모든 하중은 여러 절점에 작용해야 하며, 부재 그 자체에 작용해서는안 된다고 가정한다.

분포하중이 작용할 때는 하중이 보강재(stinger 또는 floor beam)에의해 절점에 전달되도록 상부(floor system)가 제공되어야 한다.


Types of a Truss


Simple Trusses

• A rigid truss(강성 트러스)will not collapse under the application of a load.

• A simple truss (단순 트러스)is constructed by successively adding two members and one connection to the basic triangular truss.

• In a simple truss, m = 2n – 3 where m is the total number of members and n is the number of joints.

단순 트러스 해석법 순서 :

단계1. 전체 구조물에 대한 평형방정식에서 반력을 결정

단계2. 각 부재들의 내력을 결정

트러스 해석방법 : (1) 절점법 (2) 단면법


Analysis of Trusses by the Method of Joints

• Dismember the truss and create a freebodydiagram for each member and pin.

• The two forces exerted on each member are equal, have the same line of action, and opposite sense.

• Forces exerted by a member on the pins or joints at its ends are directed along the member and equal and opposite.

• Conditions of equilibrium on the pins provide 2n equations for 2n unknowns. For a simple truss, 2n = m + 3. May solve for m member forces and 3 reaction forces at the supports.

• Conditions for equilibrium for the entire truss provide 3 additional equations which are not independent of the pin equations.


CT



0

0

0

M

F

F

y

x절점법 : 각 절점의 연결핀에 작용되는 힘이 평형상태에 있어야 한다.한 점에 작용하는 힘의 평형문제이므로 두 개의 방정식만 필요.

미지력이 두 개를 초과하지 않는 A점에서부터 시작….

0

0

y

x

F

F

기지

미지

미지

단순 트러스 해석법 순서 :

단계1. 전체 구조물에 대한 평형방정식에서 반력을 결정

단계2. 각 부재들의 내력을 결정

1

2

3 4

5

6

1



미지의 힘(BF, EF)을 결정기지 : AF

미지의 힘(BE, BC)을결정

기지 : L, AB, BF

미지의 힘

CD 결정

미지

D점에서 평형상태유지

해석의 정확도를 검증하게 됨

2

3

4

5

6



Space Trusses

• 강성 공간트러스는 6개 부재들이 연결되어4면체 ABCD의 변들을 형성함

• 단순 공간트러스(simple space truss)는 3개의 새부재로 형성되는 하나의 절점으로확정되어진다.

• 단순 공간트러스에서 부재 수는, m = 3n – 6

• 각 절점에 관한 평형조건식은 3개이다. 따라서n 개의 절점이 있으면 3n 개의 평형식이 있다.


Sample Problem 6.1

Using the method of joints, determine the force in each member of the truss.

SOLUTION:

• Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C.

• Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements.

• In succession, determine unknown member forces at joints D, B, and E from joint equilibrium requirements.

• All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.


Sample Problem 6.1

SOLUTION:

• 전체 트러스의 FBD : 점 E , C에서 3개 평형식

m 1.8m 3.6kN 4.5m 7.2kN 9.00

EM C

kN45E

xx CF 0 0xC

yy CF kN 45 kN 4.5 kN 0.90 kN 5.31yC

kN 9


Sample Problem 6.1

• 절점 A 에서의 FBD

534kN0.9 ADAB FF

kN 25.11

kN75.6

AD

AB

FF

DADE

DADB

FF

FF

532

kN 5.13kN25.11

DE

DB

FF

• 절점D에서의 FBD미지의 힘, FAD , FAB

미지의 힘, FDE , FDB


Sample Problem 6.1

BEy FF 54

54 25.115.40 kN 875.16BEF

kN625.23

875.1625.1175.60 53

53

BC

BCx

FFF

kN625.23BCF

kN 375.39

875.165.130 53

53

EC

ECx

FFF

kN 375.39ECF

• 절점 E 에서의 FBD

• 절점 B 에서의 FBD

미지의 힘, FBC , FBE

미지의 힘, FEC


Sample Problem 6.16 - 16

• All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results.

checks 0375.39kN 5.31

checks 0375.39kN625.23

54

53

y

x

F

F


MATLAB Program for Truss Analysis- Input parameters (1)

nnp (number of nodal point)5 nel (number of element)7nnl (number of nodal points per element)2ndf (number of DOF)3ndm (number of spatial dimension)3nmt (number of material)1

1

1

2

3

4

52

3

4

5

6

7



xyz (각 노드의 좌표 data)xyz (1,:) = [ 0,0,0]xyz (2,:) = [ 1.8,-2.4,0]xyz (3,:) = [ 3.6,0,0] …

lc (Element connectivity)lc (1,:) = [1 2]lc (2,:) = [1 3] …

lm (Element material)lm(1) = 1lm(2) = 2 …

mat (material information)mat(1, :) = [ 2.0e11 , 0.01 ] (E, A 값) …

1

1

2

3

4

52

3

4

5

6

7


nbc (number of boundary condition)4

bcc (boundary condition code)bcc(1,:) = [ 0 0 1 ]bcc(3,:) = [ 0 0 1 ]bcc(5,:) = [ 1 1 1 ]bcc(4,:) = [ 1 0 1 ]

bcv (boundary condition value)bcc(1,:) = [ -9000 0 0 ]bcc(3,:) = [ -4500 0 0 ]bcc(5,:) = [ 0 0 0 ]bcc(4,:) = [ 0 0 0 ]


1

1

2

3

4

52

3

4

5

6

7


Tensile Test

Most common test for studying stress-strain relationship, especially metals In the test, a force pulls the material,

elongating it and reducing its diameter

Figure 3.1 Tensile test: (a) tensile force applied in (1) and (2) resulting elongation of material

- 20 -


Tensile Test Specimen

ASTM (American Society for Testing and Materials) specifies preparation of test specimen

- 21 -


Figure 3.2 Typical progress of a tensile test: (1) beginning of test, no load; (2) uniform elongation and reduction of cross-sectional area; (3) continued elongation, maximum load reached; (4) necking begins, load begins to decrease; and (5) fracture. If pieces are put back together as in (6), final length can be measured.

Tensile Test Sequence

- 22 -


Engineering Stress & Strain

Engineering Strain- Defined as force divided by original area:

oe A

F

where e = engineering stress, F = applied force, and Ao = original area of test specimen

Engineering Strain - Defined at any point in the test as

where e = engineering strain; L = length at any point during elongation; and Lo = original gage length

o

oL

LLe

- 23 -


Typical Engineering Stress-Strain Plot

- 24 -


Elastic Region in Stress-Strain Curve

Relationship between stress and strain is linear Material returns to its original length when stress is

removed

Hooke's Law: e = E e

where E = modulus of elasticity

E is a measure of the inherent stiffness of a material Its value differs for different materials

- 25 -


Yield Point in Stress-Strain Curve As stress increases, a point in the linear relationship is

finally reached when the material begins to yield Yield point Y can be identified by the change in slope at the

upper end of the linear region Y = a strength property Other names for yield point

- yield strength- yield stress- elastic limit

- 26 -


Plastic Region in Stress-Strain Curve

Yield point marks the beginning of plastic deformation The stress-strain relationship

is no longer guided by Hooke's Law

As load is increased beyond Y, elongation proceeds at a much faster rate than before, causing the slope of the curve to change dramatically

- 27 -


Tensile Strength in Stress-Strain Curve

Elongation is accompanied by a uniform reduction in cross-sectional area, consistent with maintaining constant volume

Finally, the applied load F reaches a maximum value, and engineering stress at this point is called the tensile strength TS(ultimate tensile strength)

TS = oA

Fmax

- 28 -


Ductility in Tensile Test Ability of a material to plastically strain without fracture Ductility measure = elongation EL

where EL = elongation; Lf = specimen length at fractureLo = original specimen lengthLf is measured as the distance between gage marks after two pieces of specimen are put back together

o

ofL

LLEL

- 29 -


Input file for example Visual C++ code

Run FEM_solver.exe using test_001


Nano-molding&Optica

Patch test #1 ( Model )

노드번호 X좌표 Y좌표 Z좌표

1 0 0 0

2 0 -10 0

3 10 -10 0

요소번호 구성 노드

1 1 , 2

2 2 , 3

노드번호

BCC

X

BCC

Y

BCC

Z

BCV

X

BCV

Y

BCV

Z

1 1 1 1 0 0 0

2 0 0 1 0 -50000

0

3 1 1 1 0 0 0

1

2 3

(1)

(2)Y

X 50000N

E = 200GPa = 2.0 E11 Pa

A = 0.01m^2


노드 번호 x 변위 y 변위 z변위

[1] 0.000000e+000 0.000000e+000 0.000000e+000

[2] 0.000000e+000 -2.500000e-004 0.000000e+000

[3] 0.000000e+000 0.000000e+000 0.000000e+000

요소번호 구성 노드 Strain Stress Force

[1] 1 2 2.500000e-005 5.000000e+006 5.000000e+004

[2] 2 3 0.000000e+000 0.000000e+000 0.000000e+000

노드번호 X 반력 Y 반력 Z 반력

[1] 0.000000e+000 -5.000000e+004 0.000000e+000

[2] 0.000000e+000 -5.000000e+004 0.000000e+000

[3] 0.000000e+000 0.000000e+000 0.000000e+000

Patch test #1 ( Result )

요소별 Force를 계산한 후각 노드의 BCC 조건이 1 인 경우

요소 Force를 노드 반력에더하는 계산 수행


Patch test #2 ( Model )

E = 200GPa = 2.0 E11 Pa

A = 0.01m^2

1 6

72

49

3 8

5

d = 1.0

d = 1.0

d = 1.0

d = 1.0

요소번호

구성노드

요소번호

구성노드

요소번호

구성노드

1 1,2 9 6,7 17 6,5

2 2,3 10 7,8 18 7,5

3 3,4 11 8,9 19 8,5

4 4,1 12 9,6 20 9,5

5 1,6 13 1,5

6 4,9 14 2,5

7 3,8 15 3,5

8 2,7 16 4,5

노드번호 좌표 노드번호 좌표

1 0,0,0 6 10,0,0

2 0,10,0 7 10,10,0

3 0,10,10 8 10,10,10

4 0,0,10 9 10,0,10

5 5,5,5

노드번호 구속방향

1,2,3,4 X

5 Y,Z


Patch test #2 ( Result )

5번 노드의 변위가 다른 변위의 반값이며대칭 변위가 나타남

노드 번호 x 변위 y 변위 z변위

[1] 0.000000e+000 1.250000e-001 1.396315e-002

[2] 0.000000e+000 -1.396315e-002 1.250000e-001

[3] 0.000000e+000 -1.250000e-001 -1.396315e-002

[4] 0.000000e+000 1.396315e-002 -1.250000e-001

[5] 5.000000e-001 0.000000e+000 0.000000e+000

[6] 1.000000e+000 1.666667e-001 -2.770352e-002

[7] 1.000000e+000 2.770352e-002 1.666667e-001

[8] 1.000000e+000 -1.666667e-001 2.770352e-002

[9] 1.000000e+000 -2.770352e-002 -1.666667e-001


Patch test #2 ( Result )요소번호 구성 노드 Strain Stress Force

[1] 1 2 -1.389631e-002 -2.779263e+009 -2.779263e+007 [2] 2 3 -1.389631e-002 -2.779263e+009 -2.779263e+007 [3] 3 4 -1.389631e-002 -2.779263e+009 -2.779263e+007 [4] 4 1 -1.389631e-002 -2.779263e+009 -2.779263e+007 [5] 1 6 1.000000e-001 2.000000e+010 2.000000e+008 [6] 4 9 1.000000e-001 2.000000e+010 2.000000e+008 [7] 3 8 1.000000e-001 2.000000e+010 2.000000e+008 [8] 2 7 1.000000e-001 2.000000e+010 2.000000e+008 [9] 6 7 -1.389631e-002 -2.779263e+009 -2.779263e+007 [10] 7 8 -1.389631e-002 -2.779263e+009 -2.779263e+007 [11] 8 9 -1.389631e-002 -2.779263e+009 -2.779263e+007 [12] 9 6 -1.389631e-002 -2.779263e+009 -2.779263e+007 [13] 1 5 2.406912e-002 4.813825e+009 4.813825e+007 [14] 2 5 2.406912e-002 4.813825e+009 4.813825e+007 [15] 3 5 2.406912e-002 4.813825e+009 4.813825e+007 [16] 4 5 2.406912e-002 4.813825e+009 4.813825e+007 [17] 6 5 2.406912e-002 4.813825e+009 4.813825e+007 [18] 7 5 2.406912e-002 4.813825e+009 4.813825e+007 [19] 8 5 2.406912e-002 4.813825e+009 4.813825e+007 [20] 9 5 2.406912e-002 4.813825e+009 4.813825e+007


Practical Test #1 ( Model )

1KN

1KN

1KN

1KN

1KN

1 2 3 4 5

6 7

8

(1) (2) (3) (4)

(5) (6)(7)

(12)

(8)(9)

(10)

(11) (13)

노드번호 좌표 노드번호 좌표

1 0,0,0 6 2,1,0

2 2,0,0 7 6,1,0

3 4,0,0 8 4,2,10

4 6,0,0

5 8,0,0

E = 200GPa = 2.0 E11 Pa

A = 0.01m^2<Engineering Mechanics –STATICS 185p 4/19>

Solution(5),(10) : 3.35KN 압축(11),(13) : 2.24KN 압축(6),(9) : 0KN (7),(8) : 1.12KN 압축(1),(2),(3),(4) : 3KN 인장 (X방향 반력이 없다고 가정할 경우)


요소번호 구성 노드 Strain Stress Force[1] 1 2 -1.323489e-021 -2.646978e-010 -2.646978e-012 [2] 2 3 -1.323489e-021 -2.646978e-010 -2.646978e-012 [3] 3 4 1.270549e-021 2.541099e-010 2.541099e-012 [4] 4 5 1.376429e-021 2.752857e-010 2.752857e-012 [5] 6 1 -1.677051e-006 -3.354102e+005 -3.354102e+003 [6] 6 2 0.000000e+000 0.000000e+000 0.000000e+000 [7] 6 3 -5.590170e-007 -1.118034e+005 -1.118034e+003 [8] 7 3 -5.590170e-007 -1.118034e+005 -1.118034e+003 [9] 7 4 0.000000e+000 0.000000e+000 0.000000e+000 [10] 7 5 -1.677051e-006 -3.354102e+005 -3.354102e+003 [11] 8 6 -1.118034e-006 -2.236068e+005 -2.236068e+003 [12] 8 3 5.000000e-007 1.000000e+005 1.000000e+003 [13] 8 7 -1.118034e-006 -2.236068e+005 -2.236068e+003

노드번호 X 반력 Y 반력 Z 반력

[1] 3.000000e+003 1.500000e+003 0.000000e+000 [2] 0.000000e+000 0.000000e+000 0.000000e+000 [3] 0.000000e+000 0.000000e+000 0.000000e+000 [4] 0.000000e+000 0.000000e+000 0.000000e+000 [5] -3.000000e+003 1.500000e+003 0.000000e+000 [6] 0.000000e+000 -1.000000e+003 0.000000e+000 [7] 0.000000e+000 -1.000000e+003 0.000000e+000 [8] 0.000000e+000 -1.000000e+003 0.000000e+000

Solution 과 결과값이일치함

Practical Test #1 ( Result )


노드번호 좌표

1 0,0,0

2 3,3,0

3 0,6,0

4 1,3,6

E = 200GPa = 2.0 E11 Pa

A = 0.01m^2

<Engineering Mechanics –STATICS 203p 4/53 참고>

Solution(4),(5) : -26.4KN(3) : -2.46KN

Practical Test #2 ( Model )7KN

1(1)

2

3

4

(2)

X

Y

Z

(4)

(3)

(5)


요소번호 구성 노드 Strain Stress Force

[1] 1 2 0.000000e+000 0.000000e+000 0.000000e+000

[2] 2 3 0.000000e+000 0.000000e+000 0.000000e+000

[3] 4 2 -1.229775e-006 -2.459549e+005 -2.459549e+003

[4] 4 3 -1.318786e-006 -2.637573e+005 -2.637573e+003

[5] 4 1 -1.318786e-006 -2.637573e+005 -2.637573e+003

Solution 과 결과값이일치함

Practical Test #2 ( Result )

basic truss matlab program for truss analysis -...

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