Seely F.B. Analytical Mechanics for Engineers

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CRANES

101

Solution.

Consider

first the

equilibrium

of the

entire

crane,

free-body

diagram

which

is shown

Fig.

107

(a).

By applying

the three

equations

equilibrium,

the

unknown

quantities

may

found.

Thus,

-C#

cos 30

.'.

Ai,

1000+1300

23001b.

Whence,

(750)

(2300)

2420

Ib.

and,

Consider

next the

equilibrium

of the

member

BH.

The forces

acting

BH are

shown

Fig.

107(6)

From the

equations

equilibrium

the

unknown

quantities may

found

follows

10.6

1000

-CD

cos

2830

.707

2000

-Hy+CD

cos 45-1000

-1000+2830X707

1000

Ib.

Whence,

V(2000)

+(1000)

2240

Ib.

and,

PROBLEMS

81.

the

crane

shown

Fig.

108

the

post

weighs

1600

Ib. and the

member

weighs

1200 Ib.

The

remaining

members

are cables the

weights

which

may

neglected.

The

member

passes

through

a slot in

the

post

AE. Determine

the

external

reactions,

and

the

tensile

stresses

the

members

BC,

CD,

and DH. Assume

the

reaction

to be

horizontal and

that there

no reaction

between CH

and

102

EQUILIBRIUM

FORCE

SYSTEMS

82. The crane in

Fig.

109 is in

equilibrium

under

the

action

the

two

loads shown

and

the

reactions at

and

the

reaction at

being

horizontal.

Find

the

reactions

at A and B

and the tension

in the

member

DF.

Find

also

the

pressure

the

at C on the

member

CF.

Ans.

52001b.

15400

Ib.

6000

Ib.

FIG.

108.

FIG.

109.

58. Flexible

Cables.

the

following

two

articles

the

equi-

librium of flexible

cables or cords will

discussed. A

cable

said

perfectly

flexible when it can offer no

resistance

bend-

ing.

A flexible

cable, then,

can

transmit

stress

only

along

its

axis,

that

is,

the

stress at

any point

of a flexible

cable is

tangent

the

curve

assumed

the

cable.

Although

physical

cables

and cords

are

not

perfectly

flexible the resistance

they

offer to

bending

generally

small

that

can be

neglected

without

serious

error.

the

discussion of cables it will be assumed

that

the cables

are

perfectly

flexible and

inextensible.

59. The Parabolic

Cable. If a

flexible

cable

suspended

from two

points

and carries a load

that

distributed

uniformly

horizontally

(Fig.

llOa),

the

curve

assumed

the

cable

is a

para-

bola,

will

presently

be shown. In the

present

discussion the

points

from which

the cable

suspended

will be assumed

to be in

the

same

horizontal

plane.

example

a cable

carrying

a load

which

closely

approximates

that above indicated

is the

cable of

suspension

bridge,

since the

weight

of the

roadway

uniformly

distributed

horizontally

and

the

weights

of the

cable and

hangers

are small

comparison

with the

weight

the

roadway

and

THE PARABOLIC

CABLE

103

therefore

may

neglected.

Another

example

that of^ar

tightly

stretched

cable

(that

is,

one

which

the

sag

small as

compared

with

the

span)

which

carries no

load

except

its

own

weight,

for

example,

the

cable of an electric transmission

line,

telegraph

wire,

etc.

In this case the load carried

the

cable

(its

weight)

is distributed

uniformly along

the curve assumed

the

cable,

but since

the

sag

small

the horizontal

projection

of an

arc

of the

curve

approximately

equal

to the

length

of the

arc,

and

hence

the load

distributed

approximately

uniformly

in the

horizontal

direction.

(6)

FIG. 110.

the

solution

problems

involving

the

parabolic

cable,

use

made

the

equation

the

curve

assumed

the

cable

(para-

bola)

and

equations

which

express

relations

between

the

span,

sag,

length

the

cable, tension,

etc.

order

determine

the

equation

the

parabola

portion

the

cable will

con-

sidered

as a

free

body

(Fig.

106)

. A

the

lowest

point

of the

cable,

will

be taken

the

origin

coordinates

and

the

tension

this

point

will

be denoted

The

tension

any point,

will

denoted

T. The

portion

cable

AB,

then,

equilibrium

under

the action

of the

three

forces

H, T,

and

the

vertical load

which

acts

through

the

point

midway

between

and

Since

these three

forces are in

equilibrium

they

must

concurrent,

104

EQUILIBRIUM

FORCE

SYSTEMS

and hence the action

line

passes

through

The

equa-

tions

equilibrium

are:

cos a-PI

(1)

=Tsma-wx

(2)

dividing

(2)

(1),

the

equation

obtained

is,

tan a

-7=-.

But,

tana

Hence,

_wx

TT'

That

is,

The

curve,

then,

is a

parabola

with

its vertex

and

its axis

vertical.

squaring

and

adding (1)

and

(2)

and

extracting

the

square

root

each

side of the

resulting

equation,

the

following

expression

for

T is

found,

(4)

applying

the

above

equations

are

concerned with

the tension

at the

point

support

since at this

point

the

tension is a

maximum.

Hence,

if the

span

be denoted

a and the

maximum

value of

(that

is,

the

sag)

equations

(3)

and

(4)

reduce

(5)

and

which

represents

the

tension

the

points

support.

The

length

the cable will

now

determined

terms of the

span

and

sag.

any

curve

the

length

arc is obtained from

the

equation

From

equation

(3) ,

-7-

-77-.

THE

PARABOLIC

CABLE 105

Hence,

if the

length

of the

cable

be denoted

we have

Substituting

for

H from

equation

(5),

this

equation

becomes

The

exact

expression

for

obtained from

this

integral,

involves

logarithmic

function

and

difficult

apply.

simpler

expres-

sion

for

may

be obtained

expanding

the

expression

under

the

integral

into a series

and

integrating

the series term

term.

This method

leads to

the

following equation,

[S/A2

^9//\4

l(a

)

' '

.....

(7)

Since the

sag

ratio

f/a

generally

small,

the series

converges

rapidly

and

is sufficient

most

practical computations

to use

only

the

first two

three terms of the

series to obtain a close

approximation

the value

ILLUSTRATIVE

PROBLEM

83. The

horizontal

load carried

each cable of a

suspension

bridge

1000

Ib.

per

ft.

The

span

of the

bridge

is 800

ft. and

the

sag

is 50

ft.

Deter-

mine

the tensions

the ends and at

the

middle of

the cable and

also

find the

length

of the

cable.

Solution.

From

equations (5)

and

(6),

1000X(800)

8X50

600

and,

^XlOOOX800^/l+^5^1_

)650>

o001b.

The

length

of the cable

may

be determined

using

equation

(7)

Thus,

2 QO

/ ^0 \

-f]

=808

24ft

106

EQUILIBRIUM

OF FORCE

SYSTEMS

PROBLEMS

84.

telegraph

wire

weighing

0.1 Ib.

per

foot

stretched

between

two

poles

150

ft.

apart.

The tension in the

wire at

the

insulators

(which

are

in the same

horizontal

plane)

is 500

Ib. Find the

sag,

assuming

that

the

weight

of the wire

uniformly

distributed

horizontally.

Find also

the

length

of the wire.

Ans.

/=0.562

ft.

150.005

ft.

85. Each

cable of a

suspension bridge

carries a load of 1200

Ib.

per

foot

uniformly

distributed

along

the horizontal.

The

span

1000

ft. and

the

sag

is 50 ft.

Find

the

maximum

stress in the cable and the

length

of the

cable.

86.

A cable 100

ft.

length

suspended

from

two

points

in a horizontal

plane

which are 99 ft.

apart.

the cable

carries a load

that is

uniformly

distributed

along

the

horizontal

what

is the

sag

of the

cable?

Ans.

6.10

ft.

60.

The

Catenary.

The curve

assumed

a flexible cable of

uniform

cross-section which is

suspended

from

two

points,

and

which

carries no

load

except

its own

weight (Fig.

Ill

a),

called

FIG. 111.

catenary.

The

load which causes a cable to assume

the form of

catenary,

then,

differs

from that which causes

the form of

parabola

in that the

load

is distributed

uniformly

along

the

cable

the former

case,

whereas in the latter case the

load is distrib-

uted

uniformly

horizontally.

The

discussion

the

catenary

is of

practical

importance

only

THE CATENARY

107

for

cables in

which the

sag

ratio is

large,

since for a small

~so,g

ratio the

curve assumed

cable

may

regarded

with

small

error as

being

parabola,

as discussed

in the

preceding

article.

order to

determine the

equation

the

catenary

and also

to derive

certain

important

relations between

such

quantities

the

sag,

span, length

cable,

tension,

etc.,

the

equilibrium

portion, OA,

of the

cable

(Fig.

1116)

will

considered, being

the

lowest

point

of the

cable and

any

other

point.

The

point

will be taken

as the

origin

coordinates,

the

weight

of the

cable

per

unit of

length

will

denoted

and the

length

the

arc

will

denoted

s. The

portion, OA,

of the

cable

equilibrium

under

the influence

of three

forces, namely,

the

ten-

sion H

at the

point

the tension T at the

point A,

and

the

weight

ws. The

angle

which T

makes

with the horizontal will

denoted

6. The

equations

equilibrium

for the

concurrent

force

system

are,

From

(1)

and

(2)

have,

, H

tan 6

77-

where

constant)

He w

Hence,

or s

c-r

(3)

This

equation

is the intrinsic

equation

of the

catenary.

The

cartesian

equation

will now be found.

any curve,

Hence,

from

(3),

the

following

equat'on

obtained,

_Vs

+c?

~r~ \i

Therefore,

, sds

Integrating,

108

EQUILIBRIUM

OF FORCE SYSTEMS

now

the

origin

is transferred

0',

where 00'

then

when

and hence

A=0. The last

equation,

therefore, becomes,

y=Vs

(4)

Eliminating y

from

(3)

and

(4),

, cds

Integrating

this

equation,

x+B

log

(s+Vs

Since

when

log

c and hence the last

equation

becomes,

s+Vs

y-\-s

X=--C\Og

C\Og

...

(5)

Equation

(5)

can also be written

in the

form,

(6)

inverting

each side of

(6)

and

rationalizing

the

denominator of

the

left

side,

the

following equation

obtained,

(7)

Adding

(6)

and

(7)

and

using (4),

- --

y=-~(e

)=ccosh-

(8)

This

the

cartesian

equation

of the

catenary.

Subtracting

(7)

from

(6)

have,

s=^(e~

-e~~

)=csmh-

(9)

Squaring

and

adding

(1)

and

(2)

have,

Hence,

T=wy

(10)

THE CATENARY

109

Summarizing, then,

the

following important properties

of the

catenary

may

stated

(1)

The

horizontal

component

of the stress at

any

point

is con-

stant

and

equal

to we.

(2)

The

vertical

component

of the stress

any point

equal

ws.

(3)

The

total

stress

T at

any

point

equal

wy.

engineering

problems

which

involve

the

catenary

we are

concerned

particularly

with

the tension

at the

points

support,

since

these

points

the

tension is

a maximum.

Hence,

the

above

formula,

T will

regarded

the tension

the

points

support

and

the values

x, y,

and

will

regarded

the

values

the

variables

these

points.

Therefore,

the

length

the

cable

denoted

by I,

the

span

and

the

sag

by /,

then

the

values

and s

the

above

equations

become

/+c,

and

respectively.

may

be noted

that

when

the

sag

of the

catenary

is small

the

curve

very

closely

approximates

parabola,

since the

load

approximately

uniform

horizontally.

The

formulas

of Art.

are

generally

used when

the

sag

small,

since

they

are much easier

apply

and the

results

obtained

are

sufficiently

accurate

for

prac-

tical

purposes.

When

the

sag

large

compared

with

the

span,

however,

the

above

formulas

should

be used.

Since

the

relations

between

the

quantities

expressed

the

above

equations

are

complicated, many

of the

problems

which

involve

the

catenary

can

solved

only

trial.

ILLUSTRATIVE

PROBLEM

87.

cable

weighing

Ib.

per

foot is stretched

between two

points

in the

same

horizontal

plane.

The

length

the

cable

600 ft.

and

the

tension at

the

points

support

2000

Ib.

Find the

sag

and also

the

distance

between

the

points

support.

Solution.

From

equation

(10),

T 2000

?/=

500

ft.

From

equation

(4),

V^^=

V(500)

(300)

=400

ft.

Hence,

/=j/-c

500-400

100ft.

110

EQUILIBRIUM

OF FORCE

SYSTEMS

From

equation

(5),

400

log

Hence,

PROBLEMS

88.

A cable

100

ft.

long

suspended

between

two

points

which are in

the

same

horizontal

plane

and 80 ft.

apart.

What

is the

sag

at the

mid-point

the cable?

Ans.

26.54

ft.

89.

cable

weighing

Ib.

per

foot is stretched between two

points

in the

same

horizontal

plane

which

are 150 ft.

apart.

If the

sag

5 ft. what

is the

length

the cable

and the tension

at the

points

support.

Ans.

150.44 ft.

T^llSOlb.

7. CONCURRENT

FORCES IN SPACE

61.

Equations

Equilibrium.

non-coplanar,

concurrent

system

forces

is in

equilibrium

if the

algebraic

sums of

the com-

ponents

of the forces

along any

three

non-coplanar

lines

through

the

point

of concurrence

of the forces

are

equal

to zero.

As a

matter of convenience

the three

lines will be

taken

as a set of

rectangular

axes

through

the

point

concurrence,

which

case

the

equations

equilibrium

may

written

Proof.

The resultant

of a concurrent

system

forces in

space,

not in

equilibrium,

is a force

(Art.

32).

order

sat-

isfy

the

equation 2/^=0,

the

resultant,

if there

one,

must

lie in

the

2/z-plane.

In order

satisfy

the

equation

the

result-

ant

must lie

the

zz-plane,

and

in order to

satisfy

the

equa-

tion

2^2

the resultant must

lie

the

:n/-plane.

It is

impossible

for a force to lie in the

three

planes

simultaneously

and

hence,

the forces

of the

system

satisfy

the

above

equations,

the

resultant

cannot

be a force

and,

therefore,

the

system

must be

equilibrium.