Seely F.B. Analytical Mechanics for Engineers

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METHOD OF

SECTIONS

sider

the section

which cuts

the

members

DF, DG,

and

EG. The

free-body

diagram

the

portion

the truss

to the left of

this

section

is shown

Fig.

96(a).

This

part

the truss

is in

equi-

librium

under the

action of six

forces;

namely,

Ri,

PI,

and

which

are assumed

known,

and

the

unknown

stresses

in the

members

DF,

DG,

and

EG.

The stress

)

in the member

may

obtained

from the

equation

since

neither

of the

other two unknowns

will

appear

in this

equation.

Similarly

the

stress

(^5)

in the member

may

obtained

from

the

single

equation

and

the stress

(^7)

the

member

may

obtained from

the

single

equation

(a}

(b)

FIG. 96.

important

to note that the forces

^5, ^6,

and

F-?

are

regarded

internal forces

(stresses)

when

the

equilibrium

of the

truss as

whole

considered,

but

that

they

are

regarded

external forces when the

part

of the truss on one side of the section

is considered. In

considering

the

equilibrium

the

part

the

truss

on the left of the section

(Fig.

96a)

only

the

external forces

which

act on the left

part

are involved

and

hence the

stresses

the members

of the left

part

that

are

not

cut

the section

not

affect the

values of

FS, FQ,

and

F?.

Thus

the left

part

may

thought

as a solid block as

indicated in

Fig. 96(6)

which

the forces

FQ,

and

act

together

with

Ri, PI,

and

PI,

the six

forces

holding

the

block in

equilibrium.

ILLUSTRATIVE PROBLEM

71.

Determine

the stresses in

the

members of

Howe truss

when loaded

shown

Fig.

(a).

Solution.

The stress in

any

member

will

be denoted

the

same

letters

the member itself.

Thus,

the

stress

in member AB will

denoted

AB.

considering

the

equilibrium

the

truss

as a

whole,

the

reactions

EQUILIBRIUM

OF FORCE

SYSTEMS

40,000

lb.

20.W

20,000

lb.

20,000

lb.

(d)

FIG. 97.

METHOD OF SECTIONS

and

are

found

to be

40,000

Ib.

find the stresses

members

and

the

joint

method

is used.

free-body

diagram

joint

A is

shown

Fig.

97(6).

applying

the

equations

equilibrium

for

a concurrent

force

system,

the stresses

the two

members

are

found as follows:

-40,OOOX20+ACX30

.'.

i-!

26,667

Ib.

(tension)

VFy^-AB

cos

+40,000

4-0 000

~^-

=40,000

sec

48,000

Ib.

(compression)

find the stresses

in members

and

BD let

a section aa

passed

(Fig.

97c).

The forces

acting

on the

part

the truss

to the

left of

the section

form a

non-concurrent

force

system.

But since

the

stress

member

AC is

already

known

only

two

the three

equations

equilibrium

are

needed.

Thus,

40,000

+BDX

40,000X20 ,

26,667

Ib.

(compression)

40,000 -20,000-50

.*.

20,000

Ib.

(tension)

In a similar

manner

the stresses in CD

and

may

found

the

method

sections.

considering

the

part

of the truss

to the

left of

sec-

tion 66

(Fig.

97d),

the

stresses in the two

members

are

found

follows:

-40,

000X40

+20,

000

+CEX

CE=

iz?Wj9??

40,000

Ib.

(tension)

40,000

-20,

000

+CD

cos

0=0,

CD=

-20,000

sec

-24,000

Ib.

(compression)

will

noted that all of the stresses

except

(Fig.

97rf)

are

assumed

to act in the

correct

directions,

and

hence

are

found

positive

whether

they

are tensile

compressive.

The

stress in CD is

assumed to be

tension

and

hence the

negative sign

indicates

that it is

compression.

considering

the

equilibrium

the

forces

which

act

the

(Fig.

97e),

it is evident

that

the

stress in

zero,

and

that

equals

EG.

Fur-

ther,

since the truss

symmetrical

with

respect

to the

center

line

and

the

loads are also

symmetrical

with

respect

this

line,

obvious

that

the

stresses in

the

members

the

right

half of

the

truss

are

equal

the

stresses

in the

corresponding

members

of the

left

half.

EQUILIBRIUM

OF FORCE

SYSTEMS

PROBLEMS

72.

Find the

stresses in the

members of the

Warren

truss

shown

Fig.

98.

Ans.

AB= -4620

Ib. DB=

-3460 Ib.

4000

Ib.

2000

Ib.

FIG.

98.

73.

Find

the stresses in

the members

of the

Pratt

truss

shown

Fig.

in which

the

loads

are each

10,000

Ib.

Ans.

-30,000

Ib.

+10,000

Ib.

-26,670

Ib.

FIG. 99.

74.

Find

the

stresses

in the

members of the Howe

truss

shown

Fig.

100.

Ans. AB=

-12.700 Ib. C=-30001b.

D=-90001b.

FIG.

100.

GRAPHICAL

ANALYSIS

OF TRUSSES

75.

In the

Fink

truss shown

Fig.

101 BC

perpendicular

to AD

and

is the

mid-point

of AD.

Find

the

stresses

in the members.

Ans.

C= -1730

Ib.

AB=

-6000 Ib.

12000 Ib.

2000

Ib. 2000

Ib.

FIG. 101.

56.

Graphical

Analysis

of Trusses.

The

graphical

method

analysis

framed

structures

sometimes

simpler

than the

algebraic

method.

This

statement

true

particularly

when

the

form

the structure

is such that

considerable amount of cal-

culation

necessary

to determine

the

directions

and moment-arms

the

forces

involved.

The

graphical

method

consists

essentially

constructing

the

force

polygons

for the

concurrent forces at

the

joints

and

superimposing

these

polygons.

The method

will

explained

in detail

with

reference to

the Pratt

truss

shown

Fig.

102

(a).

It will

assumed that the loads

carried

at the

upper

panel

points

are

equal.

It is convenient to use the Bow

system

notation

(Art. 7).

this

system

the

regions

on either

side of

the

action

line

of a force

(either

external or

internal)

are denoted

numbers

letters as

shown

Fig.

102

(a)

and the

force is

denoted

the

two

numbers

letters

adjacent

the action

line

the

force.

For

example,

the left reaction is

denoted

Xi-Y;

the

load

at the

upper

panel point

at the

left end of

the

truss

Xi-Xz;

the

stress

the

diagonal

the

second

panel

2-3;

the

stress

in the lower

cord

at the left

end

of the

truss

F-l;

etc.

The order

of the

numbers or

letters

denoting

external

force,

an internal

stress in

any

member,

may

reversed.

Thus

the stress

the

diagonal

the

second

panel

may

also

denoted

3-2.

The

force

polygon

shown in

Fig. 102(6)

con-

structed

follows:

The

load

line

XiX^X^X^X^X^

first

laid

off.

Since

the loads

are

symmetrical,

the

reactions

will

equal,

and hence

will

midway

between

the

points

and

XQ.

EQUILIBRIUM

FORCE

SYSTEMS

Consider

now

the

three

concurrent forces

at the

lower left-hand

panel point.

these three

forces the left

reaction, Y-X\,

com-

pletely

known and

the

directions of the other

,two,

X\-\

and

1-Y,

are

known. Since

these

three forces are

equilibrium,

their force

polygon

must close.

Hence,

determine the

mag-

nitudes

of the last two forces a line is

drawn in

Fig.

102(6)

through

parallel

Xi-1,

and a line is

drawn

through

par-

allel

1-Y.

These lines intersect in the

point 1,

and the

mag-

nitudes

the

stresses

are indicated

the

lengths

of the lines

Xi-1

and

1-Y

the force

polygon.

It will be noted

that

the

pressure

of the member

X\-l

the

at the lower end

down-

ward

the

left and

hence the member

is in

compression.

FIG. 102.

As soon

the stresses

the

various

members

are found

arrow-heads

should be

placed

on the

members

indicating

whether

the stresses

are

tensions or

compressions.

The second lower

panel

point

should next be taken. The stress

in the member

Y 1 is

now known and in order that the

polygon

for this

joint

shall close

evident that the

point

must

coincide

with

Therefore

the

stress in 12

zero

and the stress

2 Y is

the

same as the

stress

in Y-l. The first

upper panel point

can now

be taken

and

the stresses

.XV-3

and

3-2

may

be found.

The forces

at the

remaining

joints

may

be found

similar

manner.

The student

should

through

the

construction

(Fig.

1026)

and build

the

complete

polygon.

GRAPHICAL

ANALYSIS

OF TRUSSES

ILLUSTRATIVE

PROBLEM

76. Find the stresses

in the

members of the truss

shown

Fig.

103

(a).

The members

X^-l

and 2-3 are

parallel

and

the members

Xi~2

and

7-1

are

parallel.

30000

lb.

^^\

EQUILIBRIUM

OF FORCE SYSTEMS

78.

Find,

the

graphical

method,

the

stresses in the members of

the

truss

shown in

Fig. 105,

the

value

being

10,000

Ib.

FIG. 105.

57. Cranes.

It was

stated

in Art.

53 that

the

analysis

the

forces

which

act

on the

members

of a crane

balanced

coplanar

force

systems

are involved.

The loads

cranes,

unlike

those on

trusses,

are

not

applied

at the

pins

only.

That

is,

assumptions

(2) , (3)

and

(4)

of Art.

may

not

always

made in

the anal-

ysis

of the forces

acting

cranes.

The force

system

acting

member

of a

crane,

therefore,

includes forces

other than the

two

pressures

the ends of

the

member,

and

hence the stress

a crane

member,

general,

is not a direct axial

tension or com-

pression

is the case

a truss.

In order to calculate

the stress

a crane member a

knowledge

of the

subject

Strength

Materials is

needed,

and

consequently

the stresses in

crane

mem-

bers will

not,

general,

here be

found. The main

problem

here

considered in connection with

cranes, therefore,

that

of deter-

mining

only

the forces

acting

on the various

members

the

crane.

Since the stresses

the

members

of a

crane

are not

single

forces

directed

along

the

axes

the

members,

the method

sections

as used in the

preceding

articles

cannot be

employed.

The

general

method

procedure

the

analysis

of the forces

acting

crane

members will be

first to consider the

equilibrium

the

structure as a

whole

and

determine the external

reactions,

and then

to consider

the

equilibrium

the various

members

separately

and

determine the

forces

(pin

pressures,

etc.)

which

act

the

separate

members.

CRANES

ILLUSTRATIVE PROBLEM

79.

the crane shown

Fig.

106 the

post

weighs

1200

Ib. and the

boom

weighs

1000

Ib. The member

CE is

a cable

the

weight

of which

will

neglected.

The

reaction at

will be assumed

to be

horizontal. Find

the

reactions

A and

the

reaction

and

the tension in

the cable

CE.

6000

Ib.

FIG.

106.

Solution.

First consider the

equilibrium

of the entire

crane,

free-body

diagram

which is

shown

Fig.

106(a).

The

reaction at

will be

denoted

The reaction

at the

A will be

denoted

and,

for

convenience,

will be

resolved into its

horizontal and

vertical

components

which

will be

denoted

and A

The

unknown

reactions are obtained

applying

the

equations

equilibrium

follows:

20D

1000

6000

1200

1000

6000

EQUILIBRIUM

FORCE

SYSTEMS

V(3900)2+(8200)

9075 Ib.

35'.

100

Whence,

and,

consider

the

equilibrium

the member

BE. The

forces

acting

this

member are

four

number; namely,

the tension in

CE which

will be

denoted

CE,

the

load

6000

Ib.,

the

weight

of the

member,

and the

pressure

the

on the

member,

as shown

the

free-body

diagram (Fig.

1066)

. The

last

force

resolved into its

horizontal and

vertical

components

as shown

the

free-body diagram. By applying

the

equations

equilibrium

the

un-

known

quantities may

found as follows:

XC#-6X1000

-12X6000=0,

*ZF

-CE=Q,

1000

6000

Whence,

and,

80. Determine the

forces

acting

on the members

of the crane

shown

Fig.

107

(a).

Neglect

the

weights

of the members.

1000

Ib.

~~*

(6)

1000

Ib.

Fio. 107.