chapter 6: analysis of structures - cauisdl.cau.ac.kr/education.data/statics/ch6.pdf · 2010. 11....

School of Mechanical Engineering

Statics

Chapter 6: Analysis of Structures

최해진

[email protected]


Contents

6 - 2

• Introduction• Definition of a Truss• Simple Trusses• Analysis of Trusses by the Method

of Joints• Joints Under Special Loading

Conditions• Space Trusses• Sample Problem 6.1• Analysis of Trusses by the Method

of Sections

• Trusses Made of Several Simple Trusses

• Sample Problem 6.3• Analysis of Frames

• Frames Which Cease to be Rigid When Detached From Their Supports

• Sample Problem 6.4

• Machines


Introduction

6 - 3

• 구조물의평형을위해서는외부력(external forces) 뿐만이아니라내부력(internal forces)도고려해야한다.

• 연결부에서는뉴톤제3법칙인작용과반작용법칙(forces of action and reaction)이성립

• 구조물의분류 :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

6 - 4

• 트러스는절점(joints)에연결된직선부재(member) 로구성된다.

• 실제로볼트또는용접으로연결될지라도부재는핀으로연결된다고가정함.

• 부재의각단에작용하는힘은단일하중이되고, 우력이되지않는다.

• 부재에작용하는하중은바로부재의각단에서단일하중이다. 각부재는 2력(two-force members) 부재로취급한다.

• 모든하중은여러절점에작용해야하며, 부재그자체에작용해서는안된다.

• 각 2력부재는인장(tension)상태또는압축(compression)상태에놓인다.

트러스(truss) :삼각형의형태를이루도록서로결합된곧고, 가는봉으로구성된구조물


Definition of a Truss

6 - 5

부재들은세장부재이며, 횡하중(lateral loads)을지탱할수없다. 모든하중은여러절점에작용해야하며, 부재그자체에작용해서는안된다고가정한다.

분포하중이작용할때는하중이보강재(stinger 또는 floor beam)에의해절점에전달되도록상부(floor system)가제공되어야한다.


Types of a Truss

6 - 6


Simple Trusses

6 - 7

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

• A simple truss (단순트러스)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

트러스해석방법 : (1)절점법 (2)단면법


Analysis of Trusses by the Method of Joints

6 - 8

• Dismember the truss and create a freebody diagram 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


6 - 9


åå

=

=

0

0

y

x

F

F절점법 :각절점의연결핀에작용되는힘이평형상태에있어야한다.한점에작용하는힘의평형문제이므로두개의방정식만필요.

미지력이두개를초과하지않는A점에서부터시작….

åå

=

=

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


6 - 10


미지의힘(BF, EF)을결정기지 : AF

미지의 힘(BE, BC)을결정

기지 : L, AB, BF

미지의 힘

CD결정

미지

D점에서평형상태유지

해석의정확도를검증하게됨

2

3

4

5

6


6 - 11


Joints Under Special Loading Conditions

6 - 12

• 맞은편부재의힘들은같아야한다.• 2 개의맞은편부재의힘은같아야 하고, 나머지부재에서의힘은 P와같아야한다. P=0 이면부재AC 에서의힘은 0 이되고, 부재AC를무부하부재(zero-force member)라고한다.

• 절점A가평형이되자면두부재에서의힘들은반드시같아야한다. 그림6.13b 의두부재는무부하부재이어야한다.

• 특수하중하에있는절점에대한이해는해석에큰도움됨. 부재 BC, IJ, JK 는무부하

그림 6.13 a, b

그림 6.12 a, b

그림 6.11


Space Trusses

6 - 13

• 강성공간트러스는 6개부재들이연결되어4면체ABCD의변들을형성함

• 단순공간트러스(simple space truss)는 3개의새부재로형성되는하나의절점으로확정되어진다.

• 단순공간트러스에서부재수는, m = 3n – 6

• 각절점에관한평형조건식은 3개이다. 따라서n개의절점이있으면 3n개의평형식이있다.


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.

6 - 14


Sample Problem 6.1

SOLUTION:

• 전체트러스의 FBD : 점 E , C에서 3개평형식

( )( ) ( )( ) ( )m 1.8m 3.6kN 4.5m 7.2kN 9.00

EMC

-+==å

= kN 45E

å == xx CF 0 0=xC

å ++--== yy CF kN 45 kN 4.5 kN 0.90¯= kN 5.31yC

kN 9

6 - 15


Sample Problem 6.1

6 - 16

• 절점A 에서의 FBD

534kN 0.9 ADAB FF

==kN 25.11

kN 75.6==

AD

AB

FF

( ) DADE

DADB

FF

FF

532=

=

kN 5.13kN 25.11

==

DE

DB

FF

• 절점A 에서의 FBD미지의힘, FAD , FAB

미지의힘, FDE , FDB


Sample Problem 6.16 - 17

( )lb 3750

25.115.40 54

54

-=

---==å

BE

BEy

FFF

kN 9.16=BEF

( ) ( )kN63.23

9.169.1675.60 53

53

+=

---==å

BC

BCx

FFF

kN 63.23=BCF

( )kN 4.39

9.65.130 53

53

-=

++==å

EC

ECx

FFF

kN 4.39=ECF

• 절점 E 에서의 FBD

• 절점 B 에서의 FBD

미지의힘, FBC , FBE

미지의힘, FEC

6 - 17


Statics Sample Problem 6.16 - 18

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

( ) ( )( ) ( )checks 04.39kN 5.31

checks 04.39kN 63.23

54

53

=+-=

=+-=

å

å

y

x

FF

6 - 18


Analysis of Trusses by the Method of Sections

• When the force in only one member or the forces in a very few members are desired, the method of sections works well.

• To determine the force in member BD, pass a section through the truss as shown and create a free body diagram for the left side ABC.

• With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including FBD.

6 - 19


Trusses Made of Several Simple Trusses

6 - 20

• Compound trusses are statically determinant, rigid, and completely constrained.

32 -= nm

• Truss contains a redundant memberand is statically indeterminate.

32 -> nm

• Necessary but insufficient condition for a compound truss to be statically determinant, rigid, and completely constrained,

nrm 2=+

non-rigid rigid32 -< nm

• Additional reaction forces may be necessary for a rigid truss.

42 -= nm부재m=26개, 미지반력 4개, 절점 n=15개

부재m=26개, 미지반력 3개, 절점 n=15개


Sample Problem 6.3

6 - 21

Determine the force in members FH, GH, and GI.

SOLUTION:

• Take the entire truss as a free body. Apply the conditions for static equilib-rium to solve for the reactions at A and L.

• Pass a section through members FH, GH, and GI and take the right-hand section as a free body.

• Apply the conditions for static equilibrium to determine the desired member forces.


Sample Problem 6.3

6 - 22

SOLUTION:

• Take the entire truss as a free body. Apply the conditions for static equilib-rium to solve for the reactions at A and L.

( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )

=

++-==

=

+--

---==

å

å

kN 5.12

kN 200 kN 5.7

m 30kN 1m 25kN 1m 20kN 6m 15kN 6m 10kN 6m 50

A

ALFL

LM

y

A


Sample Problem 6.3

6 - 23

• Pass a section through members FH, GH, and GIand take the right-hand section as a free body.

( )( ) ( )( ) ( )kN 13.13

0m 33.5m 5kN 1m 10kN 7.500

+==--

=å

GI

GI

H

FF

M

• Apply the conditions for static equilibrium to determine the desired member forces.

TFGI kN 13.13=


Sample Problem 6.3

6 - 24

( )( ) ( )( ) ( )( )( )( )

kN 82.130m 8cos

m 5kN 1m 10kN 1m 15kN 7.50

07.285333.0m 15m 8tan

-==+

--=

°====

å

FH

FH

G

FF

MGLFG

a

aa

CFFH kN 82.13=

( )

( )( ) ( )( ) ( )( )kN 371.1

0m 15cosm 5kN 1m 10kN 10

15.439375.0m 8

m 5tan32

-==++

=

°====

å

GH

GH

L

FF

MHIGI

b

bb

CFGH kN 371.1=


Analysis of Frames

6 - 25

• Frames and machines are structures with at least one multiforcemember. Frames are designed to support loads and are usually stationary. Machines contain moving parts and are designed to transmit and modify forces.

• A free body diagram of the complete frame is used to determine the external forces acting on the frame.

• Internal forces are determined by dismembering the frame and creating free-body diagrams for each component.

• Forces between connected components are equal, have the same line of action, and opposite sense.

부재AD(혹 부재CF) 의 자유물체도에서 미지력구하면

나머지부재에서 확인할수있음.

• Forces on two force members have known lines of action but unknown magnitude and sense. 부재 BE : 압축

• Forces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components. 부재AD(혹 부재CF)


Frames Which Cease To Be Rigid When Detached From Their Supports

6 - 26

• Some frames may collapse if removed from their supports. Such frames can not be treated as rigid bodies.

• A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium conditions.

• The frame must be considered as two distinct, but related, rigid bodies.

• With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components.

• Equilibrium requirements for the two rigid bodies yield 6 independent equations.


Sample Problem 6.4

6 - 27

Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force in link DE and the components of the force exerted at Con member BCD.

SOLUTION:

• Create a free-body diagram for the complete frame and solve for the support reactions.

• Define a free-body diagram for member BCD. The force exerted by the link DEhas a known line of action but unknown magnitude. It is determined by summing moments about C.

• With the force on the link DE known, the sum of forces in the x and y directions may be used to find the force components at C.

• With member ACE as a free-body, check the solution by summing moments about A.


Sample Problem 6.4

6 - 28

SOLUTION:

• Create a free-body diagram for the complete frame and solve for the support reactions.

N 4800 -==å yy AF = N 480yA

( )( ) ( )mm 160mm 100N 4800 BM A +-==å®= N 300B

xx ABF +==å 0 ¬-= N 300xA

°== - 07.28tan 150801a

Note:


Sample Problem 6.4

6 - 29

• Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C.

( )( ) ( )( ) ( )( )N 561

mm 100N 480mm 08N 300mm 250sin0-=

++==åDE

DEC

FFM a

CFDE N 561=

• Sum of forces in the x and y directions may be used to find the force components at C.

( ) N 300cosN 561 0 N 300cos0

+--=+-==å

aa

x

DExxC

FCFN 795-=xC

( ) N 480sinN 5610

N 480sin0

---=

--==åa

a

y

DEyy

C

FCF

N 216=yC


Sample Problem 6.4

6 - 30

• With member ACE as a free-body, check the solution by summing moments about A.

( )( ) ( )( ) ( )( )( ) ( )( ) ( )( ) 0mm 220795mm 100sin561mm 300cos561

mm 220mm 100sinmm 300cos=---+-=

-+=åaaaa xDEDEA CFFM

(checks)


Machines

6 - 31

• Machines are structures designed to transmit and modify forces. Their main purpose is to transform input forces into output forces.

• Given the magnitude of P, determine the magnitude of Q.

• Create a free-body diagram of the complete machine, including the reaction that the wire exerts.

• The machine is a nonrigid structure. Use one of the components as a free-body.

• Taking moments about A,

PbaQbQaPM A =-==å 0


※ number of joint(관절, 절점개수 n), members(부재개수 m)

n=2 , m=1 n=3 , m=3 n=4 , m=5

내적, 외적 잉여상태(internal and external redundancy)

외적잉여상태 :평형상태유지하기에필요한수보다많은지지조건을갖는경우, 부정정내적잉여상태 :부재를제거해도구조물의붕괴가없는부재가있을경우, 부정정

edeterminatstaticallytrusssimple:23 nm =+

unstable:23 nm <+

ateindeterminstatically:23 nm >+

number of joint(n), members(m), 3=미지의반력수

부재개수가부족è붕괴

n 개의절점 : 2n개의평형방정식수가생기고, m 개의부재 : m 개의미지의힘이생김.

정적 정정문제

내적잉여상태, 정적 부정정문제

참고: Simple Trusses

6 - 32

chapter 6: analysis of structures - cauisdl.cau.ac.kr/education.data/statics/ch6.pdf · 2010. 11....

Documents