=88. The Truss Element or Triangle of Bracing.=—A number of the
preceding formulæ find their applications to bridge-trusses, as well
as to beams; hence it is necessary to give attention at least to some
simple forms of those trusses.
[Illustration: FIG. 18.]
[Illustration: FIG. 18_a_.]
The skeleton of every bridge-truss properly designed to carry its load
is an assemblage of triangles. In other words, the truss element, i.e.,
the simplest possible truss, is the triangular frame, such as is shown
in skeleton in Figs. 18 and 18_a_. These simple triangular frames are
sometimes called the King-post Truss. The action of such a triangular
frame in carrying a vertical load is extremely simple. In Fig. 18 let
the weight _W_ be suspended from the apex _C_ of the triangle. The
line _CF_ represents that weight, and if the latter be resolved into
its two components parallel to the two upper members of the triangular
frame, the two component forces _CG_ and _CD_ will result. If from _D_
and _G_ the horizontal lines _DH_ and _GO_ be drawn, those two lines
will represent the horizontal components of the forces or stresses in
the two bars _CA_ and _CB_. The force _HD_ will act to the left at the
point _A_, and the force _CG_ will act to the right at _B_, and as
these two forces are equal and opposite to each other, equilibrium will
result. Either of the horizontal forces will represent the magnitude
of the tension in _AB_. Both _AC_ and _CB_ will be in compression,
the former being compressed by the force _CD_, and the latter by the
force _CG_. The manner of drawing a parallelogram of forces makes the
triangle _COG_ similar to _CNB_, and _CHD_ similar to _CNA_; hence
_HW_ divided by _CH_ will be equal to _AN_ divided by _NB_. But _HW_
is the vertical component of the stress in _CB_, while _CH_ is the
vertical component of the stress in _AC_, the latter being represented
by the reaction _R_ and the former by the reaction _R′_. It is seen,
therefore, that the weight _W_ is carried by the frame to the two
abutment supports _A_ and _B_, precisely as if it were a solid beam. In
other words, the important principle is established that when weights
rest upon a simple truss supported at each end they will produce
reactions at the ends in accordance with the principle of the lever,
precisely as in the case of a solid beam. In engineering parlance it
is stated that the weight _W_ is divided according to the principle of
the lever, and that each portion travels to its proper abutment through
the members of the triangular frame. If the two inclined members of the
triangular frame are equally inclined to a vertical, the case of Fig.
18_a_ results, in which one half of the weight goes to each abutment.
The triangular frame, with equally inclined sides, shown in Fig. 18_a_,
is evidently the simplest form of roof-truss, constituting two equally
inclined members with a horizontal tie.
=89. Simple Trusses.=—The simplest forms of trussing used for bridge
purposes are those shown in Figs. 19, 20, and 21. There are many other
forms which are exhibited in complete treatises on bridge structures,
but these three are as simple as any, and they have been far more used
than any other types. The horizontal members _af_ and _AB_ are called
the “chords,” the former being the upper chord and the latter the lower
chord. The vertical and inclined members connecting the two chords are
called the web members or braces. When a bridge is loaded, either by
its own weight only, or by its own weight added to that of a moving
train of cars, the upper chord will evidently be in compression, while
the lower chord is in tension. A portion, which may be called a half,
of the web members will be in tension and the other portion, or half,
will be in compression.
The function of the upper and lower chords is to take up or resist
the horizontal tension and compression which correspond to the direct
stresses of tension and compression existing in the longitudinal
fibres of a loaded solid or flanged beam. The metal designed to take
these so-called direct stresses is concentrated in the chords of
trusses, whereas it is distributed throughout the entire section of a
beam, whether that beam be solid or flanged. The function of the web
members of a truss is to resist the transverse or vertical shear which
is represented by the algebraic sum of the reactions and loads. The
total section of a solid beam resists these vertical shears, while
the web only of a flanged beam is estimated to perform that duty. The
horizontal shears, which have already been recognized as existing along
the horizontal planes in a bent beam, are resisted by the inclined
web members of a truss, the horizontal stress components being the
horizontal shears, whereas the vertical shears are resisted by the
vertical web members of a truss. If the web members are all inclined,
as shown in Fig. 21, each web member resists both horizontal and
vertical shear. It is thus seen that the members of a truss perform
precisely the same duties as the various portions of either solid or
flanged beams. Inasmuch as the chords of bridge-trusses resist the
direct or horizontal stresses of tension and compression produced
by the bending in the truss, it is obvious that the greatest chord
stresses will be found at the centre of the span, and that they will
be the smallest at the ends of the span. In the web members, on the
contrary, since the vertical shear is the greatest at the ends of the
span and equal to the reactions at those points, decreasing towards
the centre precisely as in solid beams, the greatest web stresses
will be found at the ends of the span and the least near the centre.
It is obvious that the areas of cross-sections of either chords or
web members must be proportioned to the stresses which they carry.
Hence the distribution of stresses just described tends to a uniform
distribution of the truss weights over the span.
=90. The Pratt Truss Type.=—In the discussion of these three simple
types of trusses, the simplest possible loading of a perfectly uniform
train will be assumed. The portions into which the trusses are divided
by the vertical or inclined bracing are called panels. In Fig. 19, for
instance, the points 1, 2, 3, 4, 5, and 6 of the lower chord and _a_,
_b_, _c_, _d_, _e_, and _f_ of the upper chord are called panel-points.
The distance between each consecutive two of these points is called a
panel length. The uniform train-load which is to be assumed will be
represented by the weight _W_ at each panel-point. This is called the
“moving load” or “live load.” The own weight of the structure is called
the “dead load” or the “fixed load.” The dead load per upper-chord
panel will be taken as _Wʹ_, and _W_₁ for the lower chord. The loads to
be used will, therefore, be as follows:
Panel moving load = _W_;
Upper-chord panel dead load = _Wʹ_;
Lower ” ” ” ” = _W_₁.
There will also be used the length of panel and depth of truss as
follows:
Panel length = _p_;
Depth of truss = _d_.
In these simple trusses with horizontal upper and lower chords the
stress in any inclined web members is equal to the shear multiplied by
the secant of the inclination of the members to a vertical line. Also,
at each panel-point every inclined web member, in passing from the end
to the centre of the span, adds to either chord stress at that point
an amount represented by the horizontal component of the stress which
it carries; or, what is the same thing, an amount equal to the shear
at the panel in question multiplied by the tangent of its angle of
inclination to a vertical line.
It has already been shown in discussing solid beams that the greatest
shear at any section will be found when the uniform moving load covers
one of the segments of the span. This principle holds equally true for
trusses carrying uniform panel-loads like those under consideration. In
determining the stresses in these trusses, therefore, the inclined web
members will take their greatest stresses when the moving train or load
extends from the farthest end of the span up to the foot of the member
in question. In this connection it is to be observed also that any two
web members meeting in the chord which does not carry the moving load
take their greatest stresses for the same position of the latter. The
so-called “counter web members” take no stresses from the dead load.
Inasmuch as every load placed upon a truss will produce compression in
the upper chord and tension in the lower, the greatest chord stresses
will obviously exist when the moving load covers the entire span,
and that condition of loading is to be used for the stresses in the
following cases.
Bearing these general observations in mind, the ordinary simple method
of truss analysis yields the tabulated statement of stresses given
below for the three types selected for consideration. The first case to
be treated is that of Fig. 19, which represents the Pratt truss type.
The moving load is supposed to pass across the bridge from right to
left. The plus sign indicates tension and the minus sign compression.
[Illustration: FIG. 19.]
_Stress in c₁_ = + (¹/₇ + ²/₇) _W_ sec _a_ = ³/₇ _W_ sec _a_.
_Stress in T₄_ = + (¹/₇ + ²/₇ + ³/₇) _W_ sec _a_ = ⁶/₇ _W_ sec _a_;
” ” _T₃_ = + [(¹/₇ + ²/₇ + ³/₇ + ⁴/₇) _W_ + _Wʹ_ + _W₁_] sec _a_
= (¹⁰/₇ _W_ + _Wʹ_ + _W₁_) sec _a_;
” ” _T₂_ = + [(¹/₇ + ²/₇ + ³/₇ + ⁴/₇ + ⁵/₇) _W_ + _2wʹ_ + _2w₁_] sec _a_
= (¹⁵/₇ W + 2wʹ + 2w₁) sec _a_;
” ” _T₁_ = + (_W_ + _W₁_).
_Stress in P₃_ = -(⁶/₇ _W_ + _Wʹ_);
” ” _P₂_ = -(¹⁰/₇ _W_ + 2_Wʹ_ + _W₁_);
” ” _P₁_ = -3(_W_ + _Wʹ_ + _W₁_) sec _a_.
_Stress in L₁_ =_Stress in L₂_ = + 3(_W_ + _Wʹ_ + _W₁_) tan _a_;
” ” _L₃_ = ” ” _L₂_ + 2(_W_ + _Wʹ_ + _W₁_)tan _a_
+ + 5(_W_ + _Wʹ_ + _W₁_) tan _a_;
” ” _L₄_ = ” ” _L₃_ + (_W_ + _Wʹ_ + _W₁_) tan _a_
+ 6(_W_ + _Wʹ_ + _W₁_) tan _a_.
_Stress in U₁_ = _-Stress in L₃_ = -5(_W_ + _Wʹ_ + _W₁_) tan α;
” ” _U₂_ = - ” ” _L₄_ = -6(_W_ + _Wʹ_ + _W₁_) tan α;
” ” _U₃_ = ” ” _U₂_ = -6(_W_ + _Wʹ_ + _W₁_) tan α.
It is easy to check any of the chord stresses by the method of moments.
As an example, let moments first be taken about the panel-point 5
in the lower chord, and then about the panel-point _c_ in the upper
chord. The following expressions for the chord members _U₁_ and _L₄_
will be found, and it will be noticed that they are identical with
the stresses for the same members given in the preceding tabulation,
the counter-members, shown in broken lines, being omitted from
consideration as they are not needed.
_R.2p_ - (_W + Wʹ + W₁_)_p_
_Stress in U₁_ = ------------------------------
_d_
_p_
= 5(_W + Wʹ + W₁_) ---- = 5(_W + Wʹ + W₁_) tan α. (29)
_d_
_R.3p_ - 2(_W + Wʹ + W₁_) . 1½_p_
_Stress in L₄_ = ----------------------------------
_d_
= 6(_W + Wʹ + W₁_) tan α. (30)
[Illustration: FIG. 20.]
=91. The Howe Truss Type.=—The truss shown in Fig. 20 is the skeleton
of the Howe truss, to which reference has already been made. The
inclined web members are all in compression, while the vertical web
members are all in tension. In the Howe truss all compression members
are composed of timber. It has the disadvantage of subjecting the
longest web members to compression. It thus makes the truss, if built
all in iron or steel, heavier and more expensive than the trusses of
the Pratt type. As in the preceding case, the moving train or load is
supposed to pass across the bridge from _B_ to _A_. Also, as before,
the + sign indicates tension and the - sign compression. The greatest
stresses, given in the tabulated statement below, can be computed or
checked by the method of moments in this case, precisely as in the
preceding.
_Stress in c₁_ = -(¹/₇ + ²/₇) _W_ sec _a_ = -³/₇ _W_ sec _a_.
_Stress in P₄_ = -(¹/₇ + ²/₇ + ³/₇) _W_ sec _a_ = -⁶/₇ _W_ sec _a_;
” ” P₃_ = -(_¹⁰/₇ W + Wʹ + W₁_) sec _a_;
” ” P₂_ = -(_¹⁵/₇ W + 2Wʹ + 2W₁_) sec _a_;
” ” P₁_ = -3(_W + Wʹ + W₁_) sec _a_.
_Stress in T₃_ = + (_¹⁰/₇ W + W₁_) sec _a_;
” ” T₂_ = + (_¹⁵/₇ W + Wʹ + 2W₁_) sec _a_;
” ” T₁_ = + (_3W + 2Wʹ + 3W₁_) sec _a_.
_Stress in L₁_ = + 3(_W + Wʹ +W₁_) tan _a_;
” ” L₂_ = + 3(_W + Wʹ + W₁_) tan _a_+2(_W + Wʹ + W₁_) tan _a_
= + 5(_W + Wʹ + W₁_) tan _a_;
” ” L₃_ = + 5(_W + Wʹ + W₁_) tan _a_+(_W + Wʹ + W₁_) tan _a_
= + 6(_W + Wʹ + W₁_) tan _a_;
” ” L₄_ = _Stress in L₃._
_Stress in U₁_ = _- Stress in L₁_
” ” U₂_ = _- ” ” L₂_;
” ” U₃_ = _- ” ” L₃_.
It will be noticed in the cases of Figs. 19 and 20 that upper and lower
chord panels in the same lozenge or oblique panel have identically
the same stresses, but with opposite signs. For instance, in Fig. 20
the stress in _U₂_ is equal in amount to that in _L₂_; and the same
observation can be made in reference to the stresses in _U₂_ and _L₄_
of Fig. 19. This must necessarily always be the case in trusses having
vertical web members.
In making computations for these forms of trusses it is very essential
to observe where the first counter-member, as _c₁_, must be used. These
counter-members may be omitted if the proper main web members near the
centre of the span are designed to take both tension and compression.
=92. The Simple Triangular Truss.=—The truss shown in Fig. 21, in which
all the web members have equal inclination to a vertical line, is
sometimes called the Warren Truss, although that term has also been
applied specially to this type of truss so proportioned as to make the
depth just equal to the panel length. As before, the moving train is
supposed to pass over the bridge from _B_ toward _A_, while the + sign
represents tension and the - sign compression. The greatest stresses
are the following.
[Illustration: FIG. 21.]
{ -(_⁶/₇W + ½Wʹ_) sec _a_, or
_Stress in P₄_ = { +(_⁶/₇W - ½W′_) sec _a_;
” ” P₃_ = { -(_¹⁰/₇W + 1½W′ + W₁_) sec _a_, or
{ +(_³/₇ W - 1½Wʹ - W₁_) sec _a_;
” ” P₂_ = -(_¹⁵/₇ W + 2½Wʹ + 2W₁_) sec _a_;
” ” P₁_ = -(_3W + 3½ Wʹ + 3W₁_) sec _a_.
{ +(_¹⁰/₇W + ½Wʹ + W₁_) sec _a_, or
_Stress in T₃_ = { -(_³/₇W - ½Wʹ - W₁_) sec _a_;
” ” T₂_ = +(_¹⁵/₇W + 1½W′ + 2W₁_) sec _a_;
” ” T₁_ = +(_3W + 2½Wʹ + 3W₁_) sec _a_.
_Stress in L₁_ = + 3(_W + Wʹ + W₁_) tan _a + ½W′_ tan _a_;
” ” L₂_ = _Stress in L₁_ + (_5W + 5Wʹ + 5W₁_) tan _a_
= + 8(_W + Wʹ + W₁_) tan _a + ½Wʹ_ tan _a_;
” ” L₃_ = _Stress in L₂_ + 3(_W + Wʹ + W₁_) tan _a_
= + 11 (_W + Wʹ + W₁_) tan _a + ½Wʹ_ tan _a_;
” ” L₄_ = _Stress in L₃_ + (_W + Wʹ + W₁_) tan _a_
= + 12 (_W + Wʹ + W₁_) tan _a + ½Wʹ_ tan _a_.
_Stress in U₁_ = -6(_W + Wʹ + W₁_) tan _a_;
” ” U₂_ = -6(_W + Wʹ + W₁_)tan _a_ - 4(_W + Wʹ + W₁_) tan _a_
= -10(_W + Wʹ + W₁_) tan _a_;
” ” U₃_ = -10(_W+ Wʹ+ W₁_) tan _a_ - 2(_W + Wʹ + W₁_) tan _a_
= -12(_W + Wʹ + W₁_) tan _a_.
The chord stresses may be checked or found by the method of moments,
precisely as in the case of Fig. 19. If, for instance, it is desired
to determine the stresses in the upper chord member _U₂_, moments must
be taken about the lower-chord panel-point 5, and about the upper-chord
panel-point _d_ for the lower-chord stress in _L₄_. Taking moments
about those points, results given in equations (31) and (32) will at
once follow, which it will be observed are identical with the values
previously found for the same members.
(_3W + 3½ W′ + 3W₁_)._2p_ - 2_W′p_ - (_W + W₁_)_p_
_Stress in U₂_ = - -------------------------------------------------
_d_
= -10(_W + W′ + W₁_) tan _a_. (31)
(_3W + 3½ W′ + 3W₁_)._3½ p_ - 3(_W + W₁_)._1½p - 3W′.2p_
_Stress in L₄_ = + -------------------------------------------------------
_d_
= + 12(_W + W₁ + W′_) tan _a_ + ½_W′_ tan _a_. (32)
=93. Through- and Deck-Bridges.=—These simple trusses have all been
taken as belonging to the “through” type, i.e., the moving load passes
along their lower chords. It is quite common to have the moving load
pass along the upper chords, in which cases the bridges are said to be
“deck” structures. The general methods of computation are precisely
the same whether the trusses be deck or through. It is only necessary
carefully to observe that the application of the methods of analysis
depends upon the position of each panel-load as it passes across the
structure.
[Illustration: Fig. 22.]
=94. Multiple Systems of Triangulation.=—Figs. 19, 20, and 21 exhibit
what are called single systems of triangulation or single systems
of bracing, but in each of those types the system of web members
may be double or triple; in other words, they may be manifold.
There have been many bridges built in which two or more systems
of bracing are employed. Fig. 22 represents a truss with a double
system of triangulation, known at one time as the Whipple truss.
Fig. 23, again, exhibits a quadruple system of triangulation with all
inclined web members. The method of computation for such manifold
systems is precisely the same as for a single system, each system in
the compound truss being treated as carrying those loads only which
rest at its panel-points. This procedure is not quite accurate. The
complete consideration of an exact method of computation would take
the treatment into a region of rather complicated analysis beyond the
purposes of these lectures, but its outlines will be set forth on a
later page. The exact method of treatment of two or more web systems
involves the elastic properties of the material of which the trusses
are composed. In the best modern bridge practice engineers prefer to
design trusses of all lengths with single web systems, although the
panels are frequently subdivided to avoid stringers and floor-beams of
too great weight.
[Illustration: Fig. 23.]
=95. Influence of Mill and Shop Capacity on Length of Span.=—In
the early years of iron and steel bridge-building the sizes of
individual members were limited by the shop capacity for handling
and manufacturing, and by the relatively small dimensions of bars of
various shapes, and of plates which could be produced by rolling-mills.
As both mill and shop processes have advanced and their capacities
increased, corresponding progress has been made in bridge design. Civil
engineers have availed themselves of those advances, so that at the
present time single system trusses with depths as great as 85 feet or
more and spans of over 550 feet are not considered specially remarkable.
=96. Trusses with Broken or Inclined Chords.=—As the lengths of spans
have increased certain substantial advantages have been gained in
design by no longer making the upper chords horizontal in the case of
long through-spans, or indeed in the cases of through-spans of moderate
length. The greatest bending moments and the greatest chord stresses
have been shown to exist at the centre of the span, while the greatest
web stresses are found near the ends. Trusses may be lightened in view
of those considerations by making their depths less at the ends than at
the centre. This not only decreases the sectional areas of the heaviest
web members near the ends of the truss, but also shortens them. It
adds somewhat to the sectional area of the end upper-chord members,
but the resultant effect is a decrease in total weight of material and
increased stability against wind pressure by the decreased height and
less exposure near the ends. It has therefore come to be the ruling
practice at the present time to make through-trusses with inclined
upper chords for practically all spans from about 200 feet upward. A
skeleton diagram of such a truss is given in Fig. 24.
[Illustration: Fig. 24.]
=97. Position of any Moving Load for Greatest Web Stress.=—In the
preceding treatment of bridge-trusses with parallel and horizontal
chords a moving or live load has been taken as a series of uniform
weights concentrated at the panel-points. This simple procedure
was formerly generally used, and at the present time it is
occasionally employed, but it is now almost universal practice to
assume for railroad bridges a moving load consisting of a series of
concentrations, which represent both in amount and distribution the
weights on the axles of an actual railroad train. If a bridge is
supposed to be traversed by such a train, it becomes necessary to
determine a method for ascertaining the positions of the train causing
the greatest stresses in the various members of the bridge-truss. The
mathematical demonstration of the formulæ determining those positions
of loading need not be given here, but it can be found in almost any
standard work on bridges.
In order to show concisely the results of such a demonstration let it
be desired to find the position of a moving load which will give the
greatest stress to any web member, as _S_ in Fig. 24. Let the point of
intersection of _GK_ and _DC_ be found in the point _O_, then let _CK_
be extended, and on its extension let the perpendicular _h_ be dropped
from _O_. The distance of the point _O_ from _A_, the end of the span,
is _i_, while _m_ is the distance _AD_. Using the same notation which
has been employed in the discussion of beams, together with that shown
in Fig. 24, equation (33) expresses the condition to be fulfilled by
the train-loads in order that _S_ shall have its greatest stress. The
first parenthesis in the second member of that equation represents the
load between the panel _p_ and the left end of the span, while the
second parenthesis represents the load in panel _p_ itself.
_l_
_W₁ + W₂ + ... + Wₙ_ = - ----(_W₁ + W₂_ + etc.)
_i_
_l_(_m + i_)
+ (_W₃ + W₄_ + etc.) --------------- (33)
_pi_.
It will be noticed in equation (33) that the quantity _m_ shows in
what panel the inclined web member whose greatest stress is desired
is located, and it is important to observe that panel carefully. If,
for instance, the vertical member _KD_ were in question, the point _O_
would be located at the intersection of the panel _NK_ and the lower
chord of the bridge. In other words, the point _O_ must be at the
intersection of the two chord members belonging to the same panel in
which the web member is located.
=98. Application of Criterions for both Chord and Web Stresses.=—The
criterion, equation (33), belongs to web members only. If it is desired
to find the position of moving load which will give the greatest chord
stresses in any panel, equation (27), already established for beams,
is to be used precisely as it stands, the quantity _l′_ representing
the distance from one end of the span to the panel-point about which
moments are taken.
If the desired positions of the moving load for greatest stresses have
been found by equations (27) and (33), those stresses themselves are
readily found by taking moments about panel-points for chord members
and about the intersection-points _O_, Fig. 24, for web members.
These operations are simple in character and are performed with
great facility. Tabulations and diagrams are made for given systems
of loading by which these computations are much shortened and which
enable the numerical work of any special case to be performed quickly
and with little liability to error. These tabulations and diagrams and
other shortening processes may be found set forth in detail in many
publications and works on bridge structures. They constitute a part of
the office outfit of civil engineers engaged in structural work.
The criterion, equation (27), for the greatest bending moments in a
bridge is applicable to any truss whatever, whether the chords are
parallel or inclined, but it is not so with equation (33). If the
chords of the trusses are parallel, the quantity _i_ in equation (33)
becomes infinitely great, and the equation takes the following form:
_l_
_W₁ + W₂ + ... + Wₙ_ = ---- (_W₃ + W₄_ + etc.) (34)
_p_
Ordinarily the span _l_ divided by the panel length _p_ is equal to the
number of panels in the span. Hence equation (34) shows, in the case
of parallel or horizontal chords, that when the moving load is placed
for the greatest web stress in any panel, the total load on the bridge
is equal to the load in that panel multiplied by the total number of
panels.
=99. Influence Lines.=—A graphical method, known as that of “influence
lines,” is used for determining the greatest shears and bending moments
caused by a train of concentrated weights passing along a beam or
bridge-truss. Obviously it must express in essence that which has
already been shown by the formulæ which determine positions of moving
loads for the greatest shears and bending moments. In reality it is the
application of graphical methods which have become so popular to the
determination of the greatest stresses in beams and bridges.
=100. Influence Lines for Moments both for Beams and Trusses.=—It is
convenient to construct these influence lines for an arbitrary load
which may be considered a unit load; the effect of any other load will
then be in proportion to its magnitude. The results determined from
influence lines drawn for a load which may be considered a unit can,
therefore, be made available for other loads by multiplying the former
by the ratio between any desired load and that for which the influence
lines are found.
[Illustration: FIG. 25.—Bending Moment in a Simple Beam.]
_AB_ in Fig. 25 represents a beam simply supported at each end, so
that any load _g_ resting upon it will be divided between the points
of support, according to the law of the lever. Let it be desired to
determine the bending moment at the section _X_ produced by the load
_g_ in all of its positions as it passes across the span from _A_ to
_B_. Two expressions for the bending moment must be written, one for
the load _g_ at any point in _AX_, and the other for the load at any
point in _BX_. The expression for the first bending moment is
_z_
_M = g_ ----(_l-x_), (_a_)
_l_
and that for the latter
_l-z_
_M′ = g_ ---- _x_. (_b_)
_l_
As shown in the figure, _z_ and _x_, the latter locating the section at
which the bending moments are to be found, are measured to the right
from _A_. Equation (_a_) shows that if the quantity _g_(l-_x_) be laid
off, by any convenient scale, as _BK_ at right angles to _AB_, _XC_
will represent the moment _M_ by the same scale when _x = z_ or when
_z_ has any value between 0 and _x_. Similarly will _AD_ be laid off
at right angles to _AB_ by the same scale as before, to represent _gx_.
Then when _x = z_ the expression for _M′_ will have the same value _XC_
as before. Hence if the lines _AC_ and _CB_ be drawn as parts of _AK_
and _DB_, any vertical intercept between _AB_ and _ACB_ will represent
the bending at _X_ produced by the load _g_ when placed at the point
from which the intercept is drawn. The lines _AC_ and _CB_ are the
influence lines for the bending moments produced by the load _g_ in its
passage across the span _AB_. It is to be observed that the influence
lines are continuous only when the positions of the moving load are
consecutive. In case those positions are not consecutive the influence
lines are polygonal in form.
If there are a number of loads _g_ resting on the span at the same
time, the total bending moments produced at _X_ will be found by taking
the sum of all the vertical intercepts between _AB_ and _ACB_, drawn at
the various points where those loads rest. The influence lines drawn
for a single load, therefore, may be at once used for any number of
loads.
The load _g_ is considered as a unit load. If the vertical intercepts
representing the bending moments by the scale used are themselves
represented by _y_, and if _W_ represent any load whatever, the general
expression for the bending moment at _X_, produced by any system of
loads, will be
_l_
---- ∑_Wy._ (_c_)
_g_
If this expression be written as a series, the general value of the
bending moment will be the following:
_l_
_M_ = --- (_W₁y₁ + W₂y₂ + W₃y₃_ + etc.). (_d_)
_g_
The effect of a moving train upon the bending moment at any given
section is thus easily made apparent by means of influence lines. It is
obvious that there will be as many influence lines to be drawn as there
are sections to be considered. In the case of a truss-bridge there will
be such a section at every panel-point.
A slight modification of the preceding results is to be made when the
loads are applied to the beam or truss at panel-points only.
In Fig. 25 let 1, 2, 3, 4, 5, 6, and 7 be panel-points at which loads
are applied, and let the load _g_ be located at the distance _z′_ to
the right of panel-point 5, also let the panel length be _p_. The
reactions at 5 and 6 will then be
_p-z′_ _z′_
_R₅ = g_ -------- and _R₆ = g_ ------.
_p_ _p_
The reactions at _A_ will then be
_l-z_
_R = g_ ------.
_l_
Hence the moment at any section _X_ in the panel in question will be
[ _l-x z′_]
_M = Rx-R₅_{_z_′-(_z-x_)} = _g_[ ---- _z_-(_z-z′ + p-x_)---]. (_e_)
[ _l p_ ]
Remembering that _z-z′_ is a constant quantity, it is at once clear
that the preceding expression is the equation of a straight line,
with _M_ and _z_ or _z′_ the variables. If _z′ = 0_, equation (_e_)
becomes identical with equation (_a_), while if _z′ = p_, it becomes
identical with equation (_b_). Hence the influence line for the panel
in which the load is placed, as 5-6, is the straight line _KL_. It is
manifest that when the load _g_ is in any other panel than that in
which the section _X_ is located, the effect of the two reactions at
the extremities of that panel will be precisely the same at the section
as the weight itself acting along its own line of action. Hence the
two portions _AK_ and _BL_ of the influence line are to be constructed
as if the load were applied directly to the beam or truss, and in the
manner already shown. The complete influence line will then be _AKLB_,
and it shows that the existence of the panel slightly reduces the
bending at any section within its limits. The panel 5-6, as treated,
is that of a beam in which the bending moment will, in general, vary
from point to point. If _AB_ were a truss, however, _X_ would always be
taken at a panel-point, and no intercept between panel-points, as 5 and
6, would be considered.
=101. Influence Lines for Shears both for Beams and Trusses.=—The
influence lines for shears in a simple beam, supported at each end,
can be drawn in the manner shown in Fig. 25_a_. In that figure _AB_
represents a non-continuous beam with span _l_ supported from _A_. The
reaction at _A_ will be
_l-z_
_R_ = ----- _g_.
_l_
[Illustration: FIG. 25_a_.—Shear in a Simple Beam.]
Let _X_ be the section at which the shear for various positions of _g_
is to be found. When _g_ is placed at any point between _A_ and _X_ the
shear _S_ at the latter point will be
_z_
_S_ = _R_ = _g_ = -_g_ ----; (_f_)
_l_
but when the load is placed between _B_ and _X_ the shear becomes
_z_
_S_′ = _R_ = _g-g_ ----. (_h_)
_l_
Obviously these two values of the shear are equations of two parallel
straight lines, that represented by equation (_f_) passing through
_A_, and that represented by equation (_h_) passing through _B_, the
constant vertical distance between them being _g_. Hence let _BF_ be
laid off negatively downward and _AG_ positively upward, each being
equal to _g_ by any convenient scale. The ordinates drawn from the
various positions 1, 2, 3 ... 6 of _g_ on _AB_ to _AD_ and _BC_ will
be the shears at _X_ produced by the load _g_ at any point of the
span, and determined by equations (_f_) and (_h_). The influence line,
therefore, for the section _X_ will be the broken line _ADCB_. When
_g_ is at _X_ the sign of the shear changes, since the latter passes
through a zero value.
If a train of weights _W₁_, _W₂_, _W₃_, etc., passes across the span,
the total shear at _X_ will be found by taking the sum of the vertical
intercepts between _AB_ and _ADCB_, drawn at the positions occupied by
the various single weights of the train. If those single weights are
expressed in terms of the unit load _g_, the shear _S_ will have the
value
1
_S_ = ---- ∑ _Wy_;
_g_
_y_ being the general value of the intercept between _AB_ and the
influence line. The latter shows that the greatest negative shear at
_X_ will exist when the greatest possible amount of loading is placed
on _AX_ only, while the greatest positive shear at the same section
will exist when _BX_ only is loaded. If _BX_ is the smaller segment of
span, the latter shear is called the “counter-shear,” and the former
the “main shear.”
If the loads are applied at panel-points of the span only, the
treatment is the same in general character as that employed for bending
moments. In Fig. 25_a_ let 4 and 5 be the panel-points between which
the load _g_ is found, and let the panel length be _p_. Also, let _z′_
be the distance of the weight _g_ from panel-point 4. The reactions at
_A_ and 4 will then be
_l - z p - z′_
_R_ = ------_g_ and _R₄_ = ----------- _g_.
_l p_
The shear at the section _X_ for any position of the weight _g_ will
then be
(_z′ z_)
_S_ = _R - R₄_ = _g_(--- - ---). (_k_)
(_p l_)
As this is the equation of a straight line, with _S_ and _z_ or _z′_
for the coordinates, the influence line for the panel in which the
section _X_ is located will be the straight line represented by _KL_ in
Fig. 25_a_.
If _z′_ is placed equal to 0 and _p_ successively, then will equation
(_k_) become identical with equations (_f_) and (_h_) in succession.
The shears at points 4 and 5 will therefore take the same values as if
the loads were applied directly to the beam. For the reasons stated in
connection with the consideration of bending moments, loads in other
panels than that containing the section for which the influence line
is drawn will have the same effect on that section as if they were
applied directly to the beam or truss. Hence _AKLB_ is the complete
influence line for this case.
It is evident that there must be as many influence lines drawn as there
are sections to be discussed. Also, if _g_ is taken as some convenient
unit, i.e., 1000 or 10,000 pounds, it is clear that the labors of
computation will be much reduced.
=102. Application of Influence-line Method to Trusses.=—In considering
both the bending moments and shears when the loads are applied at
panel-points, it has been assumed, as would be the case in an ordinary
beam, that the bending moments as well as the shears may vary in the
panel; but this latter condition does not hold in a bridge-truss.
Neither bending moment nor shear varies in any one panel. Yet the
influence lines for moments and shears are to be drawn precisely as
shown in Figs. 25 and 25_a_. The section _X_ will always be found at
a panel-point, and no intercept drawn within the limits of the panel
adjacent to that section carrying the load _g_ is to be used. This
method will be illustrated by the aid of Fig. 25_b_.
The employment of influence lines may be illustrated by determining
the moment and shear in a single section of the truss shown in Fig.
24, which is reproduced in Fig. 25_c_, when carrying the moving load
exhibited in Fig. 25_b_, although its use may be much extended beyond
this simple procedure.
The moving load shown in Fig. 25_b_ is that of a railroad train
consisting of a uniform train-load of 4000 pounds per linear foot
drawn by two locomotives with the wheel concentrations shown; it is
a train-load frequently used in the design of the heaviest class of
railroad structures. If the criterion of equation (27) be applied to
this moving load, passing along the truss shown in Fig. 25_c_, from
left to right, it will be found that the greatest bending moment is
produced at the section _Q_ when the second driving-axle of the second
locomotive is placed at the truss section in question, as shown in Fig.
25_c_.
The unit load to be used in connection with the influence lines will be
taken at 10,000 pounds. Remembering that the panel lengths are each 30
feet, it will be seen that the panel-point _Q_ is 150 feet from _A_.
Hence the product _gx_ will be 1,500,000 foot-pounds. Similarly the
product _g_(l - _x_) will be 900,000 foot-pounds. Laying off the first
of these quantities, as _AD_, at a scale of 1,000,000 foot-pounds per
linear inch, and the second quantity, as _BK_, by the same scale, the
influence line _ACB_ can at once be completed. Vertical lines are next
to be drawn through the positions of the various weights, including
one through the centre of the uniform train-load 110 feet in length
resting on the truss. The vertical line through the centre of the
uniform train-load is shown at _O_. By carefully scaling the vertical
intercepts between _AB_ and _ACB_, and remembering that each of the
loads on the truss must be divided by 10,000, the following tabulated
statement will be obtained, the sum of the intercepts for each set
of equal weights being added into one item, and all the items of
intercepts being multiplied by 1,000,000:
.195 × 110 × .4 × 1,000,000 = 8,580,000 foot-pounds.
1.78 × 2.6 × ” = 4,628,000 ” ”
2.14 × 4 × ” = 8,560,000 ” ”
.485 × 2 × ” = 970,000 ” ”
1.525 × 2.6 × ” = 3,965,000 ” ”
.9 × 4 × ” = 3,600,000 ” ”
.12 × 2 × ” = 240,000 ” ”
----------
2⟌30,543,000 ” ”
----------
Moment for one truss = 15,271,500 ” ”
The lever-arm of _ef_, i.e., the normal distance from _Q_ to _ef_, is
39.7 feet. Hence the stress in _ef_ is
15,271,500
----------- = 384,700 pounds.
39.7
All the chord stresses can obviously be found in the same manner.
In order to place the same moving load so as to produce the greatest
shear at the same section _Q_, the criterion of equation (33) must be
employed. The dimensions of the truss shown in connection with Fig. 29
give the following data to be used in that equation: _i_ = 210 feet,
_m_ = 60 feet, and _p_ = 30 feet.
_l_(_m + i_) _l_
Hence ------------ = 10²/₇, --- = 1¹/₇.
_pi i_
[Illustration: FIG. 25_b_.]
[Illustration: FIG. 25_c_.]
[Illustration: FIG. 25_d_.]
Introducing these quantities into equation (33), and remembering
that the train moves on to the bridge from _A_, it would be found
that the second axle of the first locomotive must be placed at the
section _Q_, as shown in Fig. 25_d_, which exhibits the lower-chord
panel-points numbered from 1 to 7. The conventional unit load _g_
will be taken in this case at 20,000 pounds. It is represented as
_AG_ and _BF_ (Fig. 25_d_), laid off at a scale of 10,000 pounds per
inch. _K_ is immediately under panel-point 5 and _L_ is immediately
above panel-point 6, hence the broken line _AKLB_ is the influence
line desired. The vertical lines are then drawn from each train
concentration in its proper position, all as shown, including the
vertical line through the centre of the 54 feet of uniform train-load
on the left. The summation of all the vertical intercepts between _AB_
and the influence line _AKL_, having regard to the scale and to the
ratio between the various loads and the unit load _g_, will give the
following tabular statement:
.22 × 54 × .2 × 10,000 = 23,760 pounds.
2.2 × 1.3 × ” = 28,600 ”
3.02 × 2 × ” = 60,400 ”
.9 × 1 × ” = 9,000 ”
4.06 × 1.3 × ” = 53,780 ”
4.53 × 2 × ” = 90,060 ”
.5 × 1 × ” = 5,000 ”
--------
2⟌ 270,600 ”
--------
Shear for one truss = 135,300 ”
These simple operations illustrate the main principles of the method of
influence lines from which numerous and useful extensions may be made.