=103. Lateral Wind Pressure on Trusses.=—The duties of a bridge
structure are not confined entirely to the supporting of vertical
loads. There are some horizontal or lateral loads of considerable
magnitude which must be resisted; these are the wind loads resulting
from wind pressure against both structure and moving train. In order
to determine the magnitudes of these loads it is assumed in the first
place that the direction of the wind is practically or exactly at
right angles to the planes of the trusses and the sides of the cars.
This assumption is essentially correct. There is probably nothing
else so variable as both the direction and pressure of the wind.
These variations are not so apparent in the exposure of our bodies
to the wind, for the reason that we cannot readily appreciate even
considerable changes either in direction or pressure. As a matter of
fact suitable measuring apparatus shows that there is nothing steady
or continued in connection with the wind unless it be its incessant
variability. Its direction may be either horizontal or inclined, or
even vertical, while within a few seconds its pressure may vary between
wide limits. Under such circumstances the wind is as likely to blow
directly against both bridge and train as in any other direction, and
inasmuch as such a condition would subject the structure to its most
severe duty against lateral forces, it is only safe and proper that
the assumption should be made. The open work of bridge-trusses enables
the wind to exert practically its full pressure against both trusses
of a single-track bridge, or against even three trusses if they are
used for a double-track structure. Hence it is customary to take the
exposed surface of bridge-trusses as the total projected area on a
plane throughout the bridge axis of both trusses if there are two, or
of three trusses if there are three. Inasmuch as the floor of a bridge
from its lowest point to the top of the rails or other highest point
of the floor is practically closed against the passage of the wind,
all that surface between the lowest point and the top of the rail or
highest floor-member is considered area on which wind pressure may act.
Many experimental observations show that on large surfaces, greater
perhaps than 400 or 500 square feet in area, the pressure of the wind
seldom exceeds 20 or 25 pounds per square foot, while it may reach 80
or 90 pounds, or possibly more on small surfaces of from 2 to 40 or 50
square feet in area. This distinction between small and large exposed
areas in the treatment of wind pressures is fundamental and should
never be neglected.
This whole subject of wind pressures has not yet been brought into
a completely definite or well-defined condition through lack of
sufficient experimental observations, but in order to be at least
reasonably safe civil engineers frequently, and perhaps usually,
assume a wind pressure acting simultaneously on both bridge and train
at 30 pounds per square foot of exposed surface and 50 pounds per
square foot of the total exposed surface of a bridge structure which
carries no moving load. This distinction arises chiefly from the fact
that a wind pressure of 30 pounds per square foot on the side of many
railroad trains, particularly light ones, will overturn them, and it
would be useless to use a larger pressure for a loaded structure. There
have been wind pressures in this country so great as to blow unloaded
bridges off their piers; indeed in one case a locomotive was overturned
which must have resisted a wind pressure on its exposed surface of not
less than 90 pounds and possibly more than 100 pounds per square foot.
The consideration of wind pressure is of the greatest importance in
connection with the high trusses of long spans, as well as in long
suspension and cantilever bridges, and in the design of high viaducts,
all of which structures receive lateral wind pressures of great
magnitude.
Some engineers, instead of deducing the lateral wind loads from the
area of the projected truss surfaces, specify a certain amount for
each linear foot of span, as in ”The General Specifications for Steel
Railroad Bridges and Viaducts” by Mr. Theodore Cooper it is prescribed
that a lateral force of 150 pounds for each foot of span shall be taken
along the upper chords of through-bridges and the lower chords of
deck-bridges for all spans up to 300 feet in length; and that for the
same spans a lateral force of 450 pounds for each foot of span shall
be taken for the lower chords of through-spans and the upper chords of
deck-spans, 300 pounds of this to be treated as a moving load and as
acting on a train of cars at a line 8⁵/₁₀ feet above the base of rail.
When the span exceeds 300 feet in length each of the above amounts
of load per linear foot is to be increased by 10 pounds for each
additional 30 feet of span.
Special wind-loadings and conditions under which they are to be used
are also prescribed for viaducts.
These wind loads are resisted in the bridges on which they act by a
truss formed between each two upper chords for the upper portion of the
bridge, and between each two lower chords for the lower portion of the
structure.
[Illustration: FIG. 26.]
=104. Upper and Lower Lateral Bracing.=—Fig. 26 shows what are called
the upper and lower lateral bracing for such trusses as are shown in
the preceding figures. The wind is supposed to act in the direction
shown by the arrow. _DERA_ and _KLBC_ are the two portals at the ends
of the structure, braced so as to resist the lateral wind pressures.
It will be observed that the systems of bracing between the chords
make an ordinary truss, but in a horizontal plane, except in the
case of inclined chords like that of Fig. 24. In the latter case the
lateral trusses are obviously not in horizontal planes, but they may
be considered in computations precisely as if they were. These lateral
trusses are then treated with their horizontal panel wind loads just
as the vertical trusses are treated for their corresponding vertical
loads, and the resulting stresses are employed in designing web and
chord members precisely as in vertical trusses. The wind stresses in
the chords, in some cases, are to be added to those due to vertical
loading, and in some cases subtracted. In other words, the resultant
stresses are recognized and the chord members are so designed as
properly to resist them. At the present time it is the tendency
in the best structural work to make all the web members of these
lateral trusses of such section that they can resist both tension
and compression, as this contributes to the general stiffness of the
structure. On account of the great variability of the wind pressures
and the liability of the blows of greatest intensity to vary suddenly,
some engineers regard all the wind load on structure or train as a
moving load and make their computations accordingly. It is an excellent
practice and is probably at least as close an approximation to actual
wind effects as the assumption of a uniform wind pressure on a
structure.
Both the lateral and transverse wind bracing of railroad bridges have
other essential duties to perform than the resistance of lateral wind
pressures. Rapidly moving railroad trains produce a swaying effect on
a bridge, in consequence of unavoidable unevenness of tracks, lack of
balance of locomotive driving-wheels, and other similar influences.
These must be resisted wholly by the lateral and transverse bracing,
and these results constitute an important part of the duties of that
bracing. These peculiar demands, in connection with the lateral
stability of bridges, make it the more desirable that the lateral and
transverse bracing should be as stiff as practicable.
=105. Bridge Plans and Shopwork.=—After the computations for a bridge
design are completed in a civil engineer’s office they are placed in
the drawing-room, where the most detailed and exact plans of every
piece which enters the bridge are made. The numerical computations
connected with this part of bridge construction are of a laborious
nature and must be made with absolute accuracy, otherwise it would be
quite impossible to put the bridge together in the field. The various
quantities of bars, plates, angles, and other shapes required are then
ordered from the rolling-mill by means of these plans or drawings. On
receipt of the material at the shop the shopwork of manufacture is
begun, and it involves a great variety of operations. The bridge-shop
is filled with tools and engines of the heaviest description. Punches,
lathes, planers, riveters, forges, boring and other machines of the
largest dimensions are all brought to bear in the manufacture of the
completed bridge.
=106. Erection of Bridges.=—When the shop operations are completed
the bridge members are shipped to the site where the bridge is to be
erected or put in place for final use. A timber staging, frequently
of the heaviest timbers for large spans, called false works, is first
erected in a temporary but very substantial manner. The top of this
false work, or timber staging, is of such height that it will receive
the steelwork of the bridge at exactly the right elevation. The bridge
members are then brought onto the staging and each put in place and
joined with pins and rivets. If the shopwork has not been done with
mathematical accuracy, the bridge will not go together. On the accuracy
of the shopwork, therefore, depends the possibility of properly fitting
and joining the structure in its final position. The operations of the
shop are so nicely disposed and so accurately performed that it is not
an exaggeration to state that the serious misfit of a bridge member
in American engineering practice at the present time is practically
impossible. This leads to rapid erection so that the steelwork of a
pin-connected railroad bridge 500 feet long can be put in place on the
timber staging, or false works, and made safe in less than four days,
although such a feat would have been considered impossible twenty years
ago.
[Illustration: FIG. 27.]
=107. Statically Determinate Trusses.=—The bridge structures which
have been treated require but the simplest analysis, based only on
statical equations of equilibrium of forces acting in one plane, i.e.,
the plane of the truss. It is known from the science of mechanics that
the number of those equations is at most but three for any system of
forces or loads, viz., two equations of forces and one of moments. This
may be simply illustrated by the system of forces _F₁_, _F₂_, etc., in
Fig. 27. Let each force be resolved into its vertical and horizontal
components _V_ and _H_. Also let _l₁_, _l₂_, etc. (not shown in the
figure), be the normals or lever-arms dropped from any point _A_ on the
lines of action of the forces _F₁_, _F₂_, etc., so that the moments of
the forces about that point will be _F₁l₁_, _F₂l₂_, etc. The conditions
of purely statical equilibrium are expressed by the three general
equations
_H₁ + H₂_ + etc. = _F₁_ cos _a₁_ + _F₂_ cos _a₂_ + etc. = 0; (35)
_V₁ + V₂_ + etc. = _F₁_ sin _a₁_ + _F₂_ sin _a₂_ + etc. = 0; (36)
_Fl = F₁l₁ + F₂l₂_ + etc. = 0. (37)
If all the forces except three are known, obviously those three can
be found by the three preceding equations; but if more than three are
unknown, those three equations are not sufficient to find them. Other
equations must be available or the unknown forces cannot be found. In
modern methods of stress determinations those other needed equations
express known elastic relations or values, such as deflections or the
work performed in stressing the different members of structures under
loads. A few fundamental equations of these methods will be given.
In Figs. 19, 20, and 21 let the truss be cut or divided by the
imaginary sections _QS_. Each section cuts but three members, and as
the loads and reactions are known, the stresses in the cut members
will yield but three unknown forces, which may be found by the three
equations of equilibrium (35), (36), (37). If more than three members
are cut, however, as in the section _TV_ of Figs. 22 and 23, making
more than three unknown equations to be found, other equations than
the three of statical equilibrium must be available. Hence the general
principle that _if it is possible to cut not more than three members
by a section through the truss, it is statically determinate_, but _if
it is not possible to cut less than four or more, the stresses are
statically indeterminate_.
At each joint in the truss the stresses in the members meeting there
constitute, with the external forces or loads acting at the same point,
a system in equilibrium represented by the two equations (35) and (36).
If there are _m_ such joints in the entire structure, there will be
2_m_ such equations by which the same number of unknown quantities
may be found. Since equilibrium exists at every joint in the truss,
the entire truss will be in equilibrium, and that is equivalent to
the equilibrium of all the external forces acting on it. This latter
condition is expressed by the three equations (35), (36), and (37),
and they are essentially included in the number 2_m_. Hence there
will remain but 2_m_ - 3 equations available for the determination of
unknown stresses or external forces.
If, therefore, all the external forces (loads and reactions) are
known, the 2_m_ - 3 equations of static equilibrium can be applied
to the determination of stresses in the bars of the truss or other
structure. It follows, therefore, that the greatest number of bars that
a statically determinate truss can have is
_n_ = 2_m_ - 3. (38)
In Fig. 19 there are twelve joints and twenty-one members, omitting
counter web members and the verticals _ab_ and _fl_, which are,
statically speaking, either superfluous or not really bars of the
truss. Hence
_m_ = 12 and 2_m_ - 3 = 21. (39)
Again, in Fig. 21 there are fifteen joints. Hence
_m_ = 15, 2_m_ - 3 = 27,
and there are twenty-seven bars or members of the truss. The number of
joints and bars in actual, statically determinate trusses, therefore,
confirm the results.
=108. Continuous Beams and Trusses—Theorem of Three Moments.=—These
considerations find direct application to what are known as ”continuous
beams,” i.e., beams (or trusses) which reach continuously over two or
more spans, as shown in Fig. 28.
[Illustration: FIG. 28.]
The beam shown is continuous over three spans, but a beam or truss may
be continuous over any number of spans. In general the ends of the beam
or girder may be fixed or held at the ends _A_ and _D_, so that bending
moments _M_ and _M₃_ at the same points may have value. The bending
moments at the other points of support are represented by _M₁_, _M₂_,
etc. The points of support may or may not be at the same elevation, but
they are usually assumed to be so in engineering practice. Finally,
it is ordinarily assumed that the continuous structure is straight
before being loaded, and that in that condition it simply touches the
points of support. Whether the preceding assumptions are made or not,
a perfectly general equation can be written expressing the relation
between the bending moments over each set of three consecutive points
of support, as _M_, _M₁_, and _M₂_, or _M₁_, M₂, and M₃. Such an
equation expresses what is called the ”Theorem of Three Moments.” It is
not necessary to give the most general form of this theorem, as that
which is ordinarily used embodies the simplifying assumptions already
described. This simplified form of the ”Theorem of Three Moments”
applied to the case of Fig. 28 will yield the following two equations:
1 ₁
_Ml₁ + 2M₁_(_l₁ + l₂_) + _M₂l₂_ + ---- ∑ _W_(_l₁² - z²_)_z_
_l₁_
1 ₂
+ ----- ∑ _W_(_l₂² - z²__)z_ = 0. (40)
_l₂_
1 ₂
_M₁l₂ + 2M₂_(_l₂ + l₃_) + _M₃l₃_ + ------ ∑ _W_(_l₂² - z²_)
_l₂_
1 ₃
+ ---- ∑ _W_(_l₃² - z²_)_z_ = 0. (41)
_l₃_
The figure over the sign of summation shows the span to which the
summation belongs. If there is but one weight or load _W_ in each span,
the sign of summation is to be omitted. In an ordinary bridge structure
or beam the ends are simply supported and _M = M₃ = 0_. In any case if
the number of supports be _n_, there will be _n_ - 2 equations like the
preceding.
If the end moments _M_ and _M₃_ are not zero, they will be determinable
by the local conditions in each instance. In any event, therefore, they
will be known, and there will be but _n_ - 2 unknown moments to be
found by the same number of equations. When the moments are known the
reactions follow from simple formulæ.
=109. Application to Draw- or Swing-bridges.=—In general the
reactions or supporting forces of the beams and trusses of ordinary
civil-engineering practice are vertical, and all their points
of application are known. Hence there are but two equations of
equilibrium, equations (36) and (37), for external forces. These two
equations for the external forces and the _n_ - 2 equations derived
from the theorem of three moments are therefore always sufficient to
determine the _n_ reactions. After the reactions are known all the
stresses in the bars or members of the trusses can at once be found.
The preceding equations and methods as described are constantly
employed in the design and construction of swing- or drawbridges.
=110. Special Method for Deflection of Trusses.=—The method of finding
the elastic deflections produced by the bending of solid beams has
already been shown, but it is frequently necessary to determine the
elastic deflections of bridge-trusses or other jointed or so-called
articulate frames or structures. It is not practicable to use the
same formulæ for the latter class of structures as for the former.
The elastic deflection of a bridge- or roof-truss depends upon the
stretching or compressions of its various members in consequence of the
tensile or compressive forces to which they are subjected. Any method
by which the deflection is found, therefore, must involve these elastic
changes of length. There are a number of methods which give the desired
expressions, but probably the simplest as well as the most elegant
procedure is that which reaches the desired expression through the
consideration of the work performed in the truss members in producing
their elastic lengthenings and shortenings.
The general features of this method can readily be shown by reference
to Fig. 29. It may be supposed that it is desired to find the
deflection of any point, as _J_, of the lower chord produced both by
the dead and live load which it carries. It is known from what has
preceded that every member of the upper chord will be shortened and
that every member of the lower chord will be lengthened; and also that
generally the vertical web members will be shortened and the inclined
web members lengthened. If there can be obtained an expression giving
that part of the deflection of _J_ which is due to the change of
length of any one member of the truss independently of the others,
then that expression may be applied to every other member in the
entire truss, and by taking the sum of all those effects the desired
deflection will at once result. While this expression will be found for
some one particular truss member, it will be of such a general form
that it may be used for any truss member whatever; it will be written
for the upper-chord member _BC_ in Fig. 29.
[Illustration: FIG. 29.]
The general problem is to determine the deflection of the point _J_
when the bridge carries both dead and moving load over the entire span,
as shown in Fig. 29. The general plan of procedure is first to find
the stresses due to this combined load in every member of the truss,
so that the corresponding lengthening or shortening is at once shown.
The effect of this lengthening and shortening for any single member
_BC_ in producing deflection at _J_ is then determined; the sum of all
such effects for every member of the truss is next taken, and that sum
is the deflection sought. In this case the vertical deflection will be
found, because that is the deflection generally desired in connection
with bridge structures, but precisely the same method and essentially
the same formulæ are used to find the deflection in any direction
whatever. The following notation will be employed:
Let _w_ = deflection in inches at any panel-point or joint
of the truss;
” _P_ = any arbitrary load or weight supposed to be hung at
the point where the deflection is desired and
acting as if gradually applied. This may be taken
as unity;
” _Z_ = stress produced in any member of truss by _P_;
” _S_ = stress produced in any member of truss by the combined
dead and moving loads;
Let _l_ = length in inches of any member of the truss in which
_Z_ or _S_ is found;
” _A_ = area of cross-section of same member in square inches;
” _E_ = coefficient of elasticity.
_S_ or _Z_ may be either tension or compression, and the formulæ will
be so expressed that tension will be made positive and compression
negative.
The change of length of the chord member _BC_ produced by a stress
gradually increasing from zero to _S_ is
_S_
----_l_.
_AE_
If it be supposed that _BC_ is a spring of such stiffness that it will
be compressed by the gradual application of _Z_ exactly as much as
the shortening of the actual member by the stress _S_, the deflection
of the point 4 with the weight _P_ hung from it, and due to that
compression alone, will be precisely the same as that due to the actual
shortening of _BC_ by the combined dead and moving loads.
It is known by one of the elementary principles of mechanics that,
since _P_ acts along the direction of the vertical deflection _w_, the
work performed by the weight _P_ over that deflection is equal to the
work performed by _Z_ over the change of length _l_. Hence
_l l Sl_
--- _Pw_ = --- _Z_ ----, or
2 2 _AE_
_Z Sl_
_w_ = --- ----. (42)
_P AE_
The quantity _Z÷P_ is the stress produced in the member by a unit load
applied at the joint or point where the deflection is desired. Again,
_S÷A_ is the stress per unit of area, i.e., intensity of stress, in the
member considered by the actual dead and moving loads. For brevity let
these be written
_Z S_
--- = _z_ and --- = _s_;
_P A_
then
_zsl_
_w_ = ------. (43)
_E_
If the influence of every member of the truss is similarly expressed,
the value of the total deflection produced by the dead and moving loads
will be
_zsl_
_w_ = ∑ ------. (44)
_E_
The sign of summation ∑ indicates that the summation is to extend over
all the web and chord members of the truss.
=111. Application of Method for Deflection to Triangular Frame.=—Before
applying those equations to the case of Fig. 29 it is best to consider
a simpler case, i.e., that of the triangular frame shown in Fig. 18.
The reactions are
_l₂ l₁_
_R_ = --- _W_ and _Rʹ_ = --- _W_. (45)
_l l_
The stresses in the various members are:
_l₁_
In _CB_, _S_ = ---- _W_ sec α.
_l_
_l₂_
” _CA_, _S_ = ---- _W_ sec β.
_l_
_l₂ l₁_
” _AB_, _S_ = --- _W_ tan β = --- _W_ tan α.
_l l_
Also: _CB = h_ sec α; area of section = _A₁_.
_CA = h_ sec β; ” ” ” = _A₂_.
_AB = l_; ” ” ” = _A₃_.
In this instance it is simplest to take _P = W_. Equation (44) then
gives
( _l₁² h_ sec³ α _l₂² h_ sec³ _β_
_w_ = (----- ---------- + ----- -------------
( _l² A₁ l² A₂_
_l₂² l_ tan² _β_ ) _W_
+ ---- ------------) ------. (46)
_l² A₃_ ) _E_
Let it be supposed that
_l_ = 25 feet = 300 inches;
_h_ = 8 feet 4 inches = 100 inches;
_l₂_ = 16 feet 8 inches = 200 inches and _l₁_ = 100 inches;
tan β = 1; sec β = 1.414;
sec α = 2.24;
_W_ = 10,000 pounds.
If the bars are all supposed to be of yellow-pine timber, there may be
taken
_E_ = 1,000,000 pounds;
_A₁_ = 10″ × 12″ = 120 square inches;
_A₂_ = 10″ × 10″ = 100 square inches;
_A₃_ = 10″ × 12″ = 120 square inches.
The insertion of these quantities in equation (46) gives the deflection
_w_ = .01042 + .01253 + .01111 = 0.034. (47)
Equation (47) is so written as to show the portion of the deflection
due to each member of the frame.
In applying either equation (43) or equation (44) care must be taken
to give each stress and its corresponding strain (lengthening or
shortening) the proper sign. As the formulæ have been written and
used, a tensile stress and its resulting stretch must each be written
positive, while a compressive stress must be written negative. This
holds true for both the stresses _Z_ and _S_ (or _z_ and _s_). The
magnitude of the assumed load _P_ is a matter of indifference, since
the stress _Z_ will always be proportional to it and the ratio _P ÷ Z_
will therefore be constant. _P_ is frequently taken as unity; or, as in
the case just given, it may have any value that the conditions of the
problem make most convenient.
=112. Application of Method for Deflection to Truss.=—In making
application of the deflection formulæ to any steel railroad truss
similar to that shown in Fig. 29, it will first be necessary to
determine the stresses in all its members due to the dead and moving
loads, since the deflection under the moving load is sought. These
loads will be considered uniform, and that is sufficiently accurate for
any railroad bridge. The moving train-load will be taken as covering
the entire span, assumed, for a single-track railroad, 240 feet in
length between centres of end pins. There are eight panels of 30
feet each, and the depth of truss at centre is 40 feet. Other truss
dimensions are as shown in Fig. 29. The dead loads, or own weight, are
taken at 400 pounds per linear foot of span for the rails and other
pieces that constitute the track; at 400 pounds per linear foot for
the steel floor-beams and stringers, and 1600 pounds per linear foot
for the weight of trusses and bracing. The moving train-load will be
taken at 4000 pounds per linear foot. This will make the panel-loads
for each truss as follows:
Lower-chord dead load, 30 × 800 = 24,000 pounds per panel.
Lower-chord moving load, 30 × 2000 = 60,000 ” ” ”
------
Total load on lower chord = 84,000 ” ” ”
Upper-chord dead load, 30 × 400 = 12,000 ” ” ”
The structure is a “through” bridge, hence all moving loads rest on the
lower chord.
[Illustration: FIG. 30.]
The stresses in the truss members due to the combined uniform dead and
moving load are best found by the graphical method. One diagram only is
needed to determine all the stresses, and it is shown in Fig. 30. This
diagram is drawn accurately to scale, and the stresses measured from it
are shown in the table on page 136.
The stresses in all the truss members due to the unit load hung at
_J_ are readily found by the single diagram shown in Fig. 31, also
carefully drawn to scale. These stresses measured from the diagram
are given in the table as indicated by the column _z_; they are also
represented in equation (44) by the letter _z_. The quantity _s_ in
equation (44) is the intensity of the stress (pounds per square inch of
cross-section of member) produced by the combined dead and moving loads
in each member. As shown, these stresses are least in the web members
near the centre of the span, and greatest in the chord members. The
lengths in inches of the truss members are shown in the proper column
of the table. It will be observed that all counter web members are
omitted, as they are not needed for the uniform load. The coefficient
of elasticity (_E_) is taken at 28,000,000 pounds. The quantities
represented by the second member of equation (44) are computed from
these data, and they appear in the last column of the table, the sum
of which gives the desired deflection in inches. The elements of the
table show how much of the deflection is due to the chords and to the
web members, and they show that disregarding the latter would lead to a
considerable error.
+-------+----------+---------+--------+-----+---------+
| | _S_ | _s_ | _z_ | _l_ | _w_ |
+-------+----------+---------+--------+-----+---------+
| _L₁_ | +373,300 | +12,000 | +.555 | 360 | +.08563 |
| _L₂_ | +373,300 | +12,000 | +.555 | 360 | +.08563 |
| _L₃_ | +480,000 | +12,000 | +.833 | 360 | +.1284 |
| _L₄_ | +540,000 | +12,000 | +1.125 | 360 | +.1736 |
| _P₁_ | -502,300 | -9,000 | -.748 | 472 | +.1132 |
| _U₁_ | -501,000 | -9,500 | -.870 | 376 | +.1108 |
| _U₂_ | -544,800 | -10,000 | -1.135 | 363 | +.1472 |
| _U₃_ | -576,000 | -10,000 | -1.50 | 360 | +.1928 |
| _T₁_ | +84,000 | +9,000 | 0 | 324 | -- |
| _T₂_ | +143,500 | +10,000 | +.3738| 472 | +.0629 |
| _P₂_ | -12,000 | -1,000 | -.250 | 432 | +.00386 |
| _T₃_ | +93,720 | +7,400 | +.456 | 562 | +.0677 |
| _P₃_ | +12,000 | +1,000 | -.35 | 480 | -.0060 |
| _T₄_ | +60,000 | +4,800 | +.625 | 600 | +.0643 |
| _P₄_ | -12,000 | -1,000 | 0 | 480 | -- |
+-------+----------+---------+--------+-----+---------+
Deflection for ½ truss members = 1.2300 inches.
Deflection at _J_ = 2 × 1.2300 = 2.4600 inches.
[Illustration: FIG. 31.]
As the deflection is usually desired in inches, the lengths of members
must be taken in the same unit.
=113. Method of Least Work.=—The so-called theorem or principle of
“Least Work” is closely related to the subject of elastic deflections
just considered in its availability for furnishing equations of
condition in addition to those of a purely statical character in cases
where indetermination would result without them. This principle of
least work is expressed in the simple statement that when any structure
supports external loading the work performed in producing elastic
deformation of all the members will be the least possible. Although
this principle may not be susceptible of a complete and general
demonstration, it may be shown to hold true in many cases if not all.
The hypothesis is most reasonable and furnishes elegant solutions in
many useful problems.
The application of this principle requires the determination of
expressions for the work performed in the elastic lengthening and
shortening of pieces subjected either to tension or compression, and
for the work performed in the elastic bending of beams carrying loads
at right angles to their axes. Both of these expressions can be very
simply found.
Let it be supposed that a piece of material whose length is _L_ and the
area of whose cross-section is _A_ is either stretched or compressed by
the weight or load _S_ applied so as to increase gradually from zero to
its full value. The elastic change of length will be _SL/AE_, _E_ being
the coefficient of elasticity. The average force acting will be ½_S_,
hence the work performed in producing the strain will be
1 _S²L_
--- -----. (48)
2 _AE_
It will generally be best, although not necessary, to take _L_ in
inches. The expression (48) applies either to tension or compression
precisely as it stands.
To obtain the expression for the work performed by the stresses in a
beam bent by loads acting at right angles to its axis, a differential
length (_dL_) of the beam is considered at any normal section in
which the bending moment is _M_, the total length being _L_. Let _I_
be the moment of inertia of the normal section, _A_, about an axis
passing through the centre of gravity of the latter, and let _k_ be the
intensity of stress (usually the stress per square inch) at any point
distant _d_ from the axis about which _I_ is taken. The elastic change
produced in the indefinitely short length _dL_ when the intensity _k_
exists is (_k/E_)_dL_. If _dA_ is an indefinitely small portion of
the normal section, the average force or stress, either of tension or
compression, acting through the small elastic change of length just
given, can be written by the aid of equation (5) as
_Md_
½_k.dA_ = ----- ._dA._ (49)
2_I_
Hence the work performed in any normal section of the member, for which
_M_ remains unchanged, will be, since ∫_k.dA.d_ = _M_,
⌠ _M M²_
⌡ ------ _kd.dA.dL_ = ----- _dL._ (50)
2_IE_ 2_IE_
The work performed throughout the entire piece will then be
⌠ _M²_
⌡ ------ _dL._ (51)
2_IE_
Each of the expressions (48) and (51) belongs to a single piece or
member of the structure. The total work performed in all the pieces
subjected either to direct stress or to bending, and which, according
to the principle of least work, must be a minimum, is found by taking
the summation of the two preceding expressions:
1 ⎲ _S²L_ 1 ⎲ ⌠ _M²_
_e_ = ----⎳ ------ + ----- ⎳ ⌡ ----- _dL_ = minimum. (52)
2_E_ _A_ 2_E_ _I_
In making an application of equation (52) it is to be remembered that
_S_ is the direct stress of tension or compression in any member, and
that _M_ is the general value of the bending moment in any bent member
expressed in terms of the length _L_.
=114. Application of Method of Least Work to General Problem.=—The
problem which generally presents itself in the use of equation (52)
is the finding of an equation which expresses the condition that the
work expended in producing elastic deformation shall be a minimum, some
particular stress in the structure or some external load or force being
the variable. If _t_ represent that variable, then the desired equation
of condition will be found simply by placing the first differential
coefficient of _e_ in equation (52) equal to zero:
_de_ 1 (⎲ _S dS_ ⎲ ⌠ _M dM_ )
----- = ----(⎳ --- --- _dL_ + ⎳ ⌡ ---- --- _dL_) = 0. (53)
_dt_ _E_( _A dt_ _I dt_ )
The solution of equation (53) will give a value of _t_ which will make
the work performed as expressed in equation (52) a minimum. This method
is not a difficult one to employ in such cases as those of drawbridges
and stiffened suspension bridges. In the latter case particularly it is
of great practical value.
=115. Application of Method of Least Work to Trussed Beam.=—The method
of least work may be illustrated by the application of the preceding
equations to the simple truss shown in Fig. 32. The pieces _BC_ and
_GD_ are supposed to be of yellow-pine timber, the former 10 inches by
14 inches (vertical) in section and the latter 8 inches by 10 inches,
while each of the pieces _BD_ and _DC_ are two 1⅝-inch round steel
bars. The coefficient of elasticity _E_ will be taken at 1,000,000
pounds for the timber and 28,000,000 for the steel. The length of _BC_
is 360 inches; _GD_ 96 inches; _BD_ = 96 × 2.13 = 204.5 inches.
tan α = 1.875 and sec α = 2.13.
The weight _W_ resting at _G_ is 20,000 pounds. A part of this weight
is carried by _BC_ as a simple timber beam, while the remainder of the
load will be carried on the triangular frame _BCD_ acting as a truss,
the elastic deflection of the latter throwing a part of the load on
_BC_ acting as a beam. According to the principle of least work the
division of the load will be such as to make the work performed in
straining the different members of the system a minimum.
That part of _W_ which rests on _BC_ as a simple beam may be
represented by _W₁_, while _W₂_ represents the remaining portion
carried by the triangular frame. As _G_ is at the centre of the span,
the beam reaction at either _B_ or _C_ is ½_W₁_. Hence the general
value of the bending moment in either half of the beam at any distance
_x_ from either _B_ or _C_ is
_M_ = ½_W₁x._ Hence _M²dL_ = ¼_W₁²x²dx_.
As there is but one member acting as a beam, whose moment of inertia
_I_ is constant, the second term of the second member of equation (52)
becomes, by the aid of the preceding equation,
1 1 ¹/₂ 1 _W₁²l³_
----- ⌠ _M²dL_ = ---- ⌠ _¼W₁²x²dx_ = ---- --------. (54)
2_EI_ ⌡ _EI_ ⌡₀ _EI_ 96
[Illustration: FIG. 32.]
The numerical elements of the expression for the work done in the
members of the triangular frame are:
Member. Stress. Length. Area of Section.
_BC_ ½_W₂_ tan _α_ 360 inches = _l_ 140 square inches
_DC_ ½_W₂_ sec _α_ 204.5 ” 4.14 ” ”
_DG_ _W₂_ 96 ” 80 ” ”
10 × 14³ 27440
_I_ = -------- = ------- = 2286.7.
12 12
The substitution of those quantities in the first term of the second
member of equation (52) will give
1 ⎲ _S²L_ 1 ( _W₂²_ tan² _α_.360 _W₂²_.96 )
-----⎳ ----- = ---------- (------------------ + ----------)
2_E A_ 2,000,000 ( 4 × 140 80 )
2 _W₂²_ sec² _α_ .204.5
+ ----------- --------------------- = .000,003,73 _W₂²_.
56,000,000 4 × 4.14
The substitution of numerical quantities in equation (54) gives
1 _W₁²l³_
---- -------- = .000,213_W₁²_.
_EI_ 96
Or, since _W - W₂ = W₁_,
_e_ = .000,003,73_W₂²_ + .000,213(_W - W₂)²_. (55)
Hence
_de_
---- = .000,007,46_W₂_ - .000,426(_W - W₂_) = 0. (56)
_dW₂_
The solution of this equation gives
_W₂_ = .893_W_ = 19,660 pounds.
_W₁_ = 340 ”
It is interesting to observe that the first term of the second member
of equation (56) is the deflection of the point of application of _W₂_
as a point in the frame, while the second term is the deflection of
the point of application of _W₁_ considered as a point of the beam.
In other words, the condition resulting from the application of the
principle of least work is equivalent to making the elastic deflections
by _W₁_ and _W₂_ equal. Indeed equation (53) expresses the equivalence
of deflections whenever the features of the problem are such as to
involve concurrent deflections of two different parts of the structure.
=116. Removal of Indetermination by Methods of Least Work and
Deflection.=—The indetermination existing in connection with the
computations for such trusses as those shown in Fig. 22 and Fig.
23 can be removed by finding equations of condition by the aid of
the method of least work or of deflections. It is evident that the
component systems of bracing of which such trusses are composed must
all deflect equally. Hence expressions may be found for the deflections
of those component trusses, each under its own load. Since these
deflections must be equal, equations of condition at once result. A
sufficient number of such equations, taken with those required by
statical equilibrium, can be found to solve completely the problem.
Such methods, however, are laborious, and the ordinary assumption of
each system carrying wholly the loads resting at its panel-points is
sufficiently near for all ordinary purposes.
The method of least work can be very conveniently used for the solution
of a great number of simple problems, like that which requires the
determination of the four reactions under the four legs of a table,
carrying a single weight or a number of weights, and many others of the
same character.