lecture equivalent system [compatibility mode]

7/29/2019 Lecture Equivalent System [Compatibility Mode]

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System of Forces: Reduction to a Force and Couple

A system of forces may be replaced by a collection of force-couple systems

acting at a given point O

The force and couple vectors may be combined into a

resultant force vector and a resultant couple vector,

CFrMFRR

O

The force-couple system at O may be moved to O

with the addition of the moment ofR about O ,

RsMMR

O

R

O

'

Engineers Mechanics- Equivalent Force-Couple System

bodyrigidtheinpresentanyifmomentcoupleappliedadd ,,

Further Reduction of a Force System to a wrench


The pushing-turning action associated with

The tightening of a screw illustrates

collinear lines of action of the force and

couple vector that constitute a wrench


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Further Reduction of a System of Forces

If the resultant force and couple at O are mutually

perpendicular, they can be replaced by a single force actingalong a new line of action.

The resultant force-couple system for a system of forces

will be mutually perpendicular if:

1) the forces are coplanar, or

2) the forces are parallel.


coplanar

parallel

Reduction of a Coplanar Force System to a Single Force

System can be reduced to a single forceby moving the line of action of until

its moment about Obecomes RO

M

R

In terms of rectangular coordinates,

R

Oxy MyRxR


System of coplanar forces is reduced to a

force-couple system that is

mutually perpendicular.

R

OMR

and

FrMFR RO


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Example of a Coplanar Force System


3 tugboats moving USS John F. Kennedy


Equivalent Force Systems

Four tugboats are used to bring a ship to its pier.

Each tugboat exerts 100kN push in the direction

shown. Determinethepoint on hull (body) where a

single, more powerful tugboat should push to

produce the same effect as the original four

tugboats. Also determine the total push and its

directionto beexertedbythesingletugboat.

o

Solution Key:

Find the resultant force

and moment about O

Find the line of action

of resultant force

Find its intersectionwith the (hull) body of

the ship


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R = ( -100j ) + ( 0.8i - 0.6 j ) 100 + ( -100j ) + ( 0.6i + 0.8j ) 100

= T1 + T2 + T3 + T4= 100 ( 1.4i 1.8j ) kN



MoR= ri x Fi

= [ ( -100 x 24 ) + ( -100 x 0.6 x 144 - 100 x 0.8 x 24 ) + ( -100 x 224 )

+ ( 100 x 0.8 x184 + 100 x 0.6 x 24 ) ]= - 19200k kNm

SOLUTION

Y

XO

Prob.1

Equivalent Force SystemsExample of a Coplanar Force System

Equation of hull:y = (24/64)x & y = 24.

XO

Y

Try with y = (24/64)x

x = 19200/(180+140*24/64)

= 82.58 m > 64 m

x-intercept y-intercept

y = 24

(64, 24)

Y

X

Y

X

(0, 137.14)

(106.67, 0)

o

o

88

o


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Reduction of a Parallel Force System to a Single Force


X

Y

Z

X

Y

Z

X

Y

Z

O

G

In other words:Use

appropriate

signs

A concrete foundation mat in the shape

of a regular hexagon with 3-m sides

supports four column loads as shown.

Determine the magnitude and the point

of application of the resultant four

loads.


SOLUTION KEY

o Determine the magnitudeof the

equivalent force by just adding the

loads.

o Do the same for z axis, get x

co-ordinate.

o Equate the moment about the x axis

for the individual forces with that of

the equivalent force, get z co-ordinate.



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: A C D E F F F F F R

80 kN 40 kN 100 kN 60 kN R j j j j

280 kN j

:x A A C C D D E E GM F z F z F z F z R z

GZkN28060sinm3kN60

80 kN 0 40 kN 3 m sin 60 60 kN 0

0.185577 mGZ

:z A A C C D D E E GM F x F x F x F x R x

80 kN 3 m cos 60 1.5 m 40 kN 1.5 m 60 kN 1.5 m


100 kN 3 m cos60 1.5 m 280 kN Gx


Distributed Force-Reduction to a Single Force

A distributed load is represented by plotting the load

per unit length, w (N/m) . The total load is equal tothe area under the load curve.

AdAdxwW

L

0

wdxxdWxWx A distributed load can be replaced by a concentratedload with a magnitude equal to the area under the

load curve and a line of action passing through the

area centroid.


dAxAx

Total load (resultant)

Line of action

or


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Prob.3

Find the simplest resultant

for the forces acting on the

simply supported beam .


SOLUTION KEY

o Get the equivalent concentrated

load of the distribute loading.

O X

10 N/m

300 N-m

30N/m

3m 3m 3m 3m

100 N

M

Example:

simply supported steel beam for

which self weight is acting as

u.d.l.

o Do the moment balance of all

external forces and moments

about o.

Example of Distributed Force

Prob.3

Distributed load :

R1 = ( 10 x 12 ) + ( ) ( 6 ) ( 20 )

= 180 N

= 7.33 m

SOLUTION

O

300 N-m

20N/m

3m 3m 3m 3m

6 m

10 m

10N/m

100 N

100N

180N

300 N-m

3m7.33m

4.714m

280N

Equivalent Force SystemsExample of Distributed Force


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1 -17

Water

Y

X

Z

Dam

xddx

x


R

An automatic valve consists

of a 225 x 225 mm square

plate of uniform thickness

weighing 200 N (total). The

valve is pivoted about a

horizontal axis through A

located at a distance h = 100

mm above the lower edge.

Determine the depth of waterd for which the valve will

open.

b = 225mm

s=225mm

h = 100mm

d


Example of Hydrostatic Pressure


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Prob.4

o Calculate the total moment (M) of the thrust about the bottom edge of the plate.

SOLUTION KEY

R

o At the verge of

opening the valve, the

centre of pressure

(C.O.P. ) of the

hydrostatic thrust will

pass through hinge

A.

d

b = 225

o Find the C.O.P when water level is at the top edge of the plate.

o Decide whether the required water level (d) is below/up of the plate top edge.

o Calculate the total hydrostatic force (R) on the plate for the depth of water d.



Prob.4

SOLUTION

Rx

b = 225

dx

R

d

b = 225 mm

x

h = 100mm


Example of a Hydrostatic Pressure


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Prob.4

Rx

b = 225

dx

For any depth > 450 mm the C.O.P. will shift above hinge A and valve will open.

For d < 450 mm, the reaction at B prevents valve opening.

Have:M = h*R

For valve opening:

=h=100



1 -22

Redo the problem after

replacing the square plate

with an isosceles triangle

of width 225 mm at the top

and height of 225 mm.

SOLUTION KEY

The steps are same to the previous problem.

Find the C.O.P when the water level is at

the top edge of the plate:

SOLUTION

R

225mmx

b


225 mm

225 mm

dx



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Prob.5

R225mmx

b

225 mm

dx

*Note: for parabolic line loading the position of C.G. is L/2 (in this case 225/2)



Prob.5

225mm

xb=xR

At the verge of opening the valve, the centre of pressure (C.O.P. ) of the

hydrostatic thrust will pass through the hinge A.



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