equilibrium of a rigid body - civil engineering...

Engineering Mechanics: Statics

Equilibrium of a Rigid Body

Chapter Objectives

• Revising equations of equilibrium of a rigid

body in 2D and 3D for the general case.

• To introduce the concept of the free-body

diagram FBD for a rigid body.

• Showing how to solve rigid-body equilibrium

problems.

THREE DIMENSIONAL FORCE SYSTEMS

• Remember M = r (vector) X F (vector)

• Find the length ‘r’ Vectorially for each Force ‘F’

• Also if not given, find the vectorial representation of ‘F’

Finding the length ‘r’

• Read the coordinates of the two ends of the line

• Get the difference of the coordinates from the tail of the force coordinate to the

point of interest.

Moment of a 3D Force Review

zyx

zyx

FFF

rrr

kji

rM

== F X

In determinant form:

Sarrus‟ Rule


zyx

zyx

zyx

zyx

FFF

rrr

kji

FFF

rrr

kji

F Xr =

+ + + - - -

= + (ryFz –rzFy) i + (rzFx -rxFz) j + (rxFy – ryFx) k


A man exerts a 485 N pull on the

rope which is looped around the

branch at point A. Determine the

moment about point C of this force

that exerts on the branch of the tree

and state the magnitude of this

moment.

Example:

Moment of a Force AT A POINT ‘3D’

CAr

Example:

ABT


Example:

Three forces act on the rod. Determine the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.


Solution Position vectors are directed from point

O to each force

rOA = {5j} m and

rOB = {4i + 5j - 2k} m

For resultant moment about O,

Ro A 1 B 2 C 3M (r X F) r X F + r X F + r X F

i j k i j k i j k

0 5 0 0 5 0 4 5 2 30i 40 j 60k N.m

60 40 20 0 50 0 80 40 30


OA OA OB

Solution

For magnitude

For unit vector defining the direction of moment axis,

2 2 2

Ro

Ro

Ro

M (30) ( 40) (60) 78.103 N.m

M 30i 40 j 60ku

78.103M

0.3841i 0.5121j 0.7682k


λ

Solution

For the coordinate angles of the moment axis,

cos 0.3841; 67.4121

cos 0.5121; 120.8038

cos 0.7682; 39.806


Example (T):

a) Determine the moment of the force F about the point P:

b) Find the perpendicular distance D acting on the line of action of

the given force that causes the same moment.

The vector from P to the point of application of F is:

r = (12 3) i + (6 4) j + (5 1) k

= 9 i + 2 j 6 k (m)

N

m

m m

m


The moment is:

The magnitude of MP :

m)-(N 288738

744

629 kji

kji

FrM +==×=P

m-N 980.98288738222PM

Example (T):


The perpendicular distance D acting on the line of action of the given force that causes the same moment is

The direction of MP tells us both the orientation of the plane of the plane containing D and F and the direction of the moment.

m 998.10N 9

m-N 980.98===

F

MPD

Example (T):


Example (T):

The cable tension has a magnitude of

34.185 kN. Calculate the moment of this

cable T, which produces about the base

‘O’ of the construction crane.


ABT

EXAMPLE (T):

A force is applied to the tool, to open a gas valve. Find the magnitude of the moment

of this force about the point A.

A

B


A

B

rAB = {0.25 sin 30° i + 0.25 cos30° j} m

= {0.125 i + 0.2165 j} m

F = {-60 i + 20 j + 15 k} N

Mz = (rAB F)

0 0 1 0.125 0.2165 0 -60 20 15

Mz =

= 1{0.125(20) – 0.2165(-60)} N·m

= 15.490 N·m


i j k

Example (T): 3 cables are attached to a bracket . Replace the forces exerted by the cables with:

an equivalent resultant force and couple moment acting at point A.

Moment of a Force About A LINE ‘3D’

Further Reduction of Forces and Couples System

Solution: General system of forces and moments

Further Reduction of Forces and Couples System

z

x

When an object acted upon by a system of forces & moments is in equilibrium, the following conditions are satisfied:

1. The sum of the forces is zero:

2. The sum of the moments about any point is zero:

∑ 0=F

0∑ int =poanyM

o 1

F

d1

d2 d3

2F

3F

1M

2M

Equilibrium Conditions Coplanar (2D) General case

Equilibrium Conditions Coplanar (2D)General case

o 1

F

d1

d2 d3

2F

3F

1M

2M

Body in Equilibrium

Fundamental Equations/Conditions for Statics !

Necessary

Sufficient

∑∑∑ 0,0,0 === oyx MFF

∑∑∑ 0,0,0 === oyx MFF+ +

+

1 unknown

2 unknowns

2 unknowns

3 unknowns

STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS

EQ

UILIB

RIU

M

EQ

UA

TIO

NS

CO

ND

ITIO

NS

OF

EQ

UILIB

RIU

M

3 unknowns

5 unknowns

3 unknowns

6 unknowns

There are 2 sets of Independent Equilibrium Equations

Set 1: Equilibrium of Forces (3 components)

Set 2: Equilibrium of Moments (3 components)

Equilibrium Conditions 3D General case

∑∑∑

∑∑∑

0 ,0 ,0

0 ,0 ,0

===

===

zyx

Zyx

MMM

FFF+ i + j

+ i + j + k

The Free-Body Diagram „FBD‟

It is the sketch of the body that shows all the applied known and unknown forces and moments acting on this body.

This sketch, shows the particle “free” (isolated) from its surroundings with all the external forces acting on it.

Note that; while showing the externally applied forces and moments, their directions should be indicated.

If the directions are not given, ASSUME:

Force: +ve for all directions (x, y and z)

(i.e. Same direction with the defined axis directions)

Moments: +ve for i, j, k components.

The Free-Body Diagram FBD

To apply equilibrium equations we must account for all known and unknown forces and moments acting on the object by drawing a free-body diagram FBD of the body.

Force Types Active Forces : tend to set the body in motion. Reactive Forces : result from constraints or supports and tend to prevent motion

Procedure for Analysis Equilibrium Problems

1) Draw Free-Body Diagram FBD

Establish the x, y (for 2D) axes and x, y and z (for 3D)

axes in any suitable orientation.

Label all known and unknown force and moment

magnitudes and directions on the FBD.

The sense of an unknown force and moment may be

assume.

2) Apply equations of equilibrium.

Components of the force and moments are positive if

directed along a positive axis and negative if directed

along a negative axis.

3) If solution yields a negative result the force or the

moment is in the opposite sense of that was assumed

and showed on the FBD.

Supports

Forces & couples exerted on an object by its

supports are called reactions.

E.g. a bridge is held up by the reactions exerted

by its supports.

support

support

Support Reactions

Reactive Forces General Rule:

If a support prevents the translation of a

body in a given direction, then a force

(REACTION) is developed on the body in that

direction.

Likewise, if the rotation is prevented, a

couple moment (REACTION) is exerted on

the body.

2 D SUPPORTS

• Pin Support:

– Figure a: pin support

a bracket to which an object is attached by a smooth

pin that passes through the bracket & the object

– Figure b: side view

The arrows indicate the directions of

the reactions Ax and Ay in 2D.

SYMBOL USED IN NOTEBOOK

Pin Support Details:

The arrow indicate the directions of the reaction A


• Roller Support: It can move freely in the direction parallel to the surface on which it rolls, it can’t exert a force parallel to the surface but can exert a force normal (perpendicular) to this surface.

Fixed Support

– The fixed support shows the supported object

literally built into a wall (built-in).


A fixed support can exert

2 components of force

AX AY

and a couple

MA

Different Support Types in 2D

Different

Unknow

n is d

irecte

d

alo

ng

th

e a

xis

of th

e

sh

ort

lin

k

Different Support Types in 2 D


Different Support Types

(a) Pin in a slot (b) Slider in a slot (c) Slider on a shaft


Different 2D Support Types in Practice

Example:

For the given supports and loadings of the body draw its FBD?

Free Body Diagram

FBD

2D

Example:

Free Body Diagram

FBD

2D

Example: Free Body Diagram

FBD

2D

Example:

Free Body Diagram

FBD 2D

30°

Free Body Diagram FBD 2D


PULLEYS

When a cable (cord) wraps over a frictionless pulley, definitely

each and every portion of this cables (cords) is under

TENSION. To satisfy the static equilibrium, the magnitudes of

the tension T1 = T2 should be the SAME.

r =15 cm

7.8 kN 7.8 kN 7.8 kN


PULLEYS

Example:

• The object in this figure has a fixed support at the left end

(point A).

• A cable passing over a pulley is attached to the object at 2

points.

• Isolate it from its supports & complete the free-body by

showing the reactions at the fixed support & the forces

exerted by the cable.


Don’t forget the couple at the fixed support,

Since the tension in the cable is assumed, on both

sides of the pulley, the 2 same forces exerted by

the cable having the magnitude T.

Once the free-body diagram of an object is obtained, by

identifying the loads & reactions acting on it, for

equilibrium, the equilibrium equations can be applied.


Ay

Ax


Example:


Example:

x

y

y



Example:

Solving rigid-body equilibrium problems

2D CASES

∑∑ 00 int == poanyforcesofdirection MF+ +

Only TWO unknowns can be found

G H Find the tensions in cables BG and DH

if P = 18.75 kN and Q = 12.5 kN ?

x

y

Example:

Only THREE unknowns can be found

∑∑∑ 0,0,0 === oyx MFF+ +

+

Example:

Find the reactions at the supports A and D

ϴ = 57 o

Note that the support at A is PIN type

and the support at D is ROLLER type

Example:

Find the reactions at the supports, if P = 2500 N

25 kNm

Example:

Example:

25 kNm

30o

Example:

25 kNm

6 sin 30 kN

6 cos 30 kN

0.8 cos 60 m

0.8 sin 60 m

Bx

By

Ay 1.2 sin 60 m

0.6+ 0.6+ 1.2 cos 60 m

EXAMPLE: For the given compound beam find the reactions at supports A

and B (ignore the size of the supports).

EXAMPLE: For the given compound beam find the reactions at supports A

and B (ignore the size of the supports).

Note: Seperate the given system into 2 parts as system CD and system AB.

Find the reactions at Supports C and D.

For system AB apply EQUAL MAGNITUDE BUT OPPOSITE DIRECTION

of the reaction values obtained for Support C (due to action-reaction).

Example:

For the given system, determine the reactions at A, B, C and D

(ignore the size of the supports)?

100 N

A SIMPLE PULLEY

100 N 100 N

100 N 100 N

MULTIPLE PULLIES

C

B

A

T = P

2T=P

4T=P

P /4 P

P /2

MULTIPLE PULLIES

Example: Calculate the tension T in the cable which supports 50 kN with the

given pulley arregement shown. Note that each pulley is free to rotate about its

bearing and has negligible weight compared with the carried load. Ignore the

weight of ther cable as well.

C

B

A

Solution:

C

B

A

25 kN

25 kN

12.5 kN 12.5 kN

= 12.5 kN

Example:

Relationship between pulled length ‘s’ of the rope and the

elevated height ‘h’ due to different pulley combinations.

Neglecting friction and the radius of the pulley, determine:

a) The tension in cable ADB

b) Reaction at C.

80 mm 80 mm 200 mm

150 mm

A B C

D

120 N

Example:

80 mm 80 mm 200 mm

150 mm

A B C

120 N

Example:

x

y

R1

R2 T T

FBD of the beam ABC

The mass of 700 kg is suspended from a trolley which moves along the

crane rail d = 3.5 m. Determine the force along short link BC and the

magnitude of the reaction force at pin A.

Example (T):


1 unknown

1 unknown

1 unknown

3 unknowns

4 unknowns


5 unknowns

5 unknowns

5 unknowns

5 unknowns

6 unknowns

All 2D SUPPORTS

Some 3D SUPPORTS

Example: Ball-and-socket joint at A.


Smooth journal bearing at B.

Roller support at C.

Example:


Ball-and-socket joint at A.

BD and BE are cables.

Example: Free Body Diagram FBD

Free Body Diagram FBD

Solving

3D

rigid-body

equilibrium problems

Determine the reactions due to force F = 300 N, at the fixed support O.

Example:

Example:

Solution:

FBD

Rod AB subjected to 200N force. Determine the reactions at the ball-and-socket

joint A and the tension in cables BD and BE. Note that point C is acting at the

middle of the rod AB.

Example:

x

y

z

Solution:

FBD

At point A, ball and socket reaction means only reaction forces may occur along

3 axes (no moment). Cable BD under tension along y-axis only and cable BE

under tension along x-axis only. Note that point C is at the middle of the two

points A and B.

Determine the support reaction at C (ball and socket type) if the applied force at

B is F = 3.6 kN.

Example T:

equilibrium of a rigid body - civil engineering...

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