chapter 4: equilibrium of rigid bodies...

School of Mechanical Engineering

Statics

Chapter 4: Equilibrium of Rigid Bodies (강체의평형)

최해진

[email protected]


Statics

4 - 2

Contents

Introduction

Free-Body Diagram

Reactions at Supports and Connections for a Two-Dimensional Structure

Equilibrium of a Rigid Body in Two Dimensions

Statically Indeterminate Reactions

Sample Problem 4.1

Sample Problem 4.3

Sample Problem 4.4

Equilibrium of a Two-Force Body

Equilibrium of a Three-Force Body

Sample Problem 4.6

Equilibrium of a Rigid Body in Three Dimensions

Reactions at Supports and Connections for a Three-Dimensional Structure

Sample Problem 4.8


Statics

4 - 3

4.1 Introduction

• The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces form a system equivalent to zero,

( )å å =å ´== 00 FrMF Orrrr

ååå

ååå

======

000000

zyx

zyx

MMMFFF

• Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium,

• For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.


Statics

4 - 4

4.1 Introduction

만약 물체에 작용하는 힘계의 합력이 0이라면, 물체는 평형(equilibrium)이라고 말한다. 평형은 합력과 합우력이 모두 0임을의미한다.

합력이 0 이라는 것

힘계가 초기에 정지하여 있는 물체에 작용할 때, 물체가이동하려는 경향이 없음을 의미한다.

동역학 평형이 아닌 힘계에 대한 물체의 응답에 관심

정역학 공학적 구조물의 평형상태를 유지하기 위한 힘의 상태에대한 필요충분조건을 제시


Statics

4 - 5

4.1 Introduction

* Equilibrium Æ

; necessary & sufficient

conditions for equilibrium

÷÷

ø

ö

ïî

ïíì

==

==

åå

0MM

0FR

i

i

평형상태 : ▷ 어떤 물체에 여러 개의 힘이 작용하여그 물체가 이동하거나 회전하지 않고 정지되어 있는 상태

평형해석의 첫 단계 물체에 작용하는 모든 힘들을 찾아내는 것, 이것은자유물체도에 의해 이루어진다.

ååå

ååå

======

000000

zyx

zyx

MMMFFF


Statics 4.2 Free-Body Diagram

물체의 자유물체도(Free-Body Diagram ; FBD)는 물체에작용하는 모든 힘을 나타내는 물체의 개략도이다.자유(free)라는 용어는 모든 지지부가 제거되고, 물체에작용하는 힘(반력)에 의해 대체되는 것을 의미.

물체에 작용하는 힘

1. 작용력 (applied force)

2. 반작용력 (reactive force)

반작용력 단순히 반력(reactions)이라고 하고 반력은 물체에결합되어 있는 지지부에 의해 물체에 작용하는 힘

작용력 지지부에 의해 주어지지 않고 물체에 작용하는 힘

4 - 6


Statics

4 - 7

4.2 Free-Body Diagram

First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram.

• Select the extent of the free-body and detach it from the ground and all other bodies.

• Include the dimensions necessary to compute the moments of the forces.

• Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body.

• Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.


Statics

4 - 8

4.2 Free-Body Diagram

1. 모든 지지부(접촉면, 지지용 케이블 등)가 제거된 것으로 가정한물체의 개략도를 그린다.

2. 개략도에 모든 작용력을 그리고, 기호로 표시한다. 물체의무게는 무게중심에 작용하는 힘으로 간주한다.

균질(homogenous)인 물체의 무게중심(center of gravity)은체적의 도심(centroid)와 일치한다.

3. 개략도에 각 지지부에 기인하는 반력을 그리고, 기호로표시한다. 만약 반력의 방향이 미지이면 가정되어야 한다. 이는구해진 해로부터 정확한 방향이 결정될 것이다. 양(+)의 결과는가정된 방향이 올바르다는 것을 나타내고, 반면에 (-)의 결과는정확한 방향이 가정된 방향과 반대라는 것을 의미한다.

4. 4. 개략도에 모든 관련된 각도와 차원을 나타낸다.

자유물체도를 구성하는 일반적인 과정


Statics

4 - 94.3 Reactions at Supports and Connections for a Two-Dimensional Structure

• Reactions equivalent to a force with known line of action.


Statics

4 - 104.3 Reactions at Supports and Connections for a Two-Dimensional Structure

• Reactions equivalent to a force of unknown direction and magnitude.

• Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude


Statics

4.4 Equilibrium of a Rigid Body in Two Dimensions

• For all forces and moments acting on a two-dimensional structure,

Ozyxz MMMMF ==== 00• Equations of equilibrium become

å å å === 000 Ayx MFF

where A is any point in the plane of the structure.

• The 3 equations can be solved for no more than 3 unknowns.

• The 3 equations can be replacedå å å === 000 BAx MMF

ååå

ååå

======

000000

zyx

zyx

MMMFFF

4 - 11


Statics 4.5 Statically Indeterminate Reactions

• More unknowns than equations

• Fewer unknowns than equations, partially constrained

• Equal number unknowns and equations but improperly constrained

평형방정식의수 < 미지수의수 평형방정식의수 > 미지수의수 평형방정식의수 = 미지수의수

4 - 12


Statics

4 - 13

4.5 Statically Indeterminate Reactions

평형을 유지하는 데 필요한 구속만을 가지고 또한미지의 지지력(unknown reaction forces)이 독립된평형방정식에 의해서 완전히 결정될 수 있는 정정(statically determinate) 인 물체에 국한하여 취급함

구속과 정정( Constraints and statically determinate )

1. 독립된 평형방정식의 수 = 미지수의 수

정역학적으로 풀 수 있고, 정정(statically determinate) 문제이다. 정정문제는 평형해석 자체만으로 해결될 수 있다.

2. 독립된 평형방정식의 수 < 미지수의 수

정역학적으로 풀 수 없고, 부정정 (statically indeterminate) 문제이다.


Statics Sample Problem 4.14 - 14

A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G.

Determine the components of the reactions at A and B.

SOLUTION:

• Create a free-body diagram for the crane.

• Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A.

• Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components.

• Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.



1. Create the free-body diagram.

4. Check the values obtained. 점 B 에 대한모멘트가 제로이면확인됨

2. Determine B by solving the equation for the sum of the moments of all forces about A.

( ) ( ) ( ) 0m6kN5.23m2kN81.9m5.1:0

=--+

=åBM A

kN1.107+=B

3. Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces.

0:0 =+=å BAF xx

kN1.107-=xA

0kN5.23kN81.9:0 =--=å yy AF

kN 3.33+=yA


Statics Sample Problem 4.3

A loading car is at rest on an inclined track. The gross weight of the car and its load is 25 kN, and it is applied at atG. The cart is held in position by the cable.

Determine the tension in the cable and the reaction at each pair of wheels.

SOLUTION:

• Create a free-body diagram for the car with the coordinate system aligned with the track.

• Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle.

• Determine the cable tension by solving the equation for the sum of force components parallel to the track.

• Check the values obtained by verifying that the sum of force components perpendicular to the track are zero.

4 - 16



1. Create a free-body diagram

( )

( )kN 0.51–

25sinkN 25

kN 65.2225coskN 25

=

-=

+=+=

o

o

y

x

W

W

2. Determine the reactions at the wheels.

( ) ( ) ( ) 0mm 1250mm 150kN 65.22–mm625kN10–:0

2 =+

=åR

M A

kN 82 =R

( ) ( ) ( ) 0mm0125–mm150kN 65.22–mm625kN 5.10:0

1 =+

=åR

M B

kN 50.21 =R

3. Determine the cable tension.0–kN 65.22:0 =+=å TFx

kN 65.22+=T

4. Check the values obtained.y-방향의 힘의합이 제로이거나, 점A 또는 B 이외의한점에 대한모멘트가 제로이면 확인됨


Statics

4 - 18

Sample Problem 4.4

The frame supports part of the roof of a small building. The tension in the cable is 150 kN.

Determine the reaction at the fixed end E.

SOLUTION:

• Create a free-body diagram for the frame and cable.

• Solve 3 equilibrium equations for the reaction force components and couple at E.


Statics

4 - 19

Sample Problem 4.4

1. Create a free-body diagram for the frame and cable.

2. Solve 3 equilibrium equations for the reaction force components and couple.

( ) 0kN1505.75.4:0 =+=å xx EF

kN 0.90-=xE

( ) ( ) 0kN1505.7

6kN204:0 =--=å yy EF

kN 200+=yE

å = :0EM ( ) ( )( ) ( )

( ) 0m5.4kN1505.7

6

m8.1kN20m6.3kN20

m4.5kN20m7.2kN20

=+-

++

++

EM

mkN0.180 ×=EM


Statics

4 - 20

Sample Problem 4.5

연습문제 #1, #3, #5

•

AB

k =45 N/mm

O

W =1800 N

l = 200 mm

q

r = 75 mm

C

• when q =0 ,spring is unstretched.Equilibrium position = ?

SOLUTION:

• Create a free-body diagram for the frame and cable.

°=°==×´-´=

=×-=

=Þ=-=

-=Þ=+=

=

ååå

3.80,00mm)75(mm/N45sin mm200N1800

0sin

0

0

2

qqqq

qq

qqq

rkrWlM

WRWRFkrRkrRF

krF

O

yyy

xxx •

A

O

W = 1800 N

q

l sin q

RxRy

F = ks

W

r

s

Undeformedposition


Statics

4 - 21

4.6 Equilibrium of a Two-Force Body

• Consider a plate subjected to two forces F1 and F2

• For static equilibrium, the sum of moments about Amust be zero. The moment of F2 must be zero. It follows that the line of action of F2 must pass through A.

• Similarly, the line of action of F1 must pass through B for the sum of moments about B to be zero.

• Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense.


Statics

4 - 22

4.7 Equilibrium of a Three-Force Body

• Consider a rigid body subjected to forces acting at only 3 points.

• Assuming that their lines of action intersect, the moment of F1 and F2 about the point of intersection represented by D is zero.

• Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about any axis must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D.

• The lines of action of the three forces must be concurrent or parallel.

세힘이 작용하는 물체가 평형을 이루자면세힘의 작용선은 동일점에 작용하거나, 평행하여야 함


Statics

4 - 23

Sample Problem 4.6

A man raises a 10 kg joist, of length 4 m, by pulling on a rope.

Find the tension in the rope and the reaction at A.

SOLUTION:

• Create a free-body diagram of the joist. Note that the joist is a 3 force body acted upon by the rope, its weight, and the reaction at A.

• The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R.

• Utilize a force triangle to determine the magnitude of the reaction force R.


Statics

4 - 24

Sample Problem 4.6

• Create a free-body diagram of the joist.

• Determine the direction of the reaction force R.

( )

( )( )

636.1414.1313.2tan

m 2.313m 515.0828.2m 515.020tanm 414.1

m 414.1m828.245cosm445cos

21

===

=-=-==°=

===

=°=°=

AECEBDBFCE

BDAFAECD

ABAF

a

o6.58=a


Statics

4 - 25

Sample Problem 4.6

• Determine the magnitude of the reaction force R.

ooo 38.6sinN 1.98

110sin4.31sin==

RT

N 8.147

N9.81

=

=

R

T

연습문제 #67, (#73, #75 )

A

CB


Statics

4 - 264.8 Equilibrium of a Rigid Body in Three Dimensions

• Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three dimensional case.

å =å =å =å =å =å =

000000

zyx

zyxMMMFFF

• These equations can be solved for no more than 6 unknowns which generally represent reactions at supports or connections.

• The scalar equations are conveniently obtained by applying the vector forms of the conditions for equilibrium,

( )å å å =´== 0FrM0F O

rrrr,

3차원강체의평형조건


Statics

4 - 274.9 Reactions at Supports and Connections for a Three-Dimensional Structure


Statics

4 - 284.9 Reactions at Supports and Connections for a Three-Dimensional Structure



A sign of uniform density weighs 1200 N and is supported by a ball-and-socket joint at A and by two cables.

Determine the tension in each cable and the reaction at A.

SOLUTION:

• Create a free-body diagram for the sign.

• Apply the conditions for static equilibrium to develop equations for the unknown reactions.



• Create a free-body diagram for the sign.

Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading.

( )

( )kjiT

kjiT

rrrrTT

kjiT

kjiT

rrrrTT

EC

EC

EC

ECECEC

BD

BD

BD

BDBDBD

rrr

rrr

rrrrr

rrr

rrr

rrrrr

72

73

76

32

31

32

7236

12848

++-=

++-=

--

=

-+-=

-+-=

--

=

평형방정식의수 > 미지수의수

부분적인구속



• Apply the conditions for static equilibrium to develop equations for the unknown reactions.

( )

( ) ( )

0mN144077.08.0:

0514.06.1:

0N 1200m .21

0:

0N 1200:

0:

0N 1200

72

32

73

31

76

32

=×-+

=-

=-´+´+´=

=+-

=-++

=--

=-++=

å

å

ECBD

ECBD

ECEBDBA

ECBDz

ECBDy

ECBDx

ECBD

TTk

TTj

jiTrTrM

TTAk

TTAj

TTAi

jTTAF

rr

rrrrrrr

r

r

r

rrrrr

( ) ( ) ( )kjiA

TT ECBD rrvrN 100.1N 450.1N 1503

N 1401N 6.450

-+=

==

Solve the 5 equations for the 5 unknowns,

연습문제 #96



• Determine (a) when G should be located if the tension in the cable is to be minimum

• (b) the corresponding minimum value of the tension

2000 N

3.6 m

1.8 m

3.6 m1.8 m 1.8 m

3.6 m

3.6 m

W=2000 N3.6 m

1.8 m

1.8 m


Statics 4장 EX. 1 4 - 33


Statics EX. 1 4 - 34



교정 : 원본에서도 1.05 m 임, 실제 105mm

세힘이 작용하는 물체가 평형을 이루자면세힘의 작용선은 동일점에 작용하거나, 평행하여야 함

chapter 4: equilibrium of rigid bodies...

Documents