Ft. Ins.
/left
537
516 495
7 6 {
Iright .. ..
483
508 530
deft
930
937 895
15 <
Iright
883
920 891
22 6 / 6
1,187
1,225 1,205
Bright ..
1,183
1,180 1,166
30 <
1,337
1,375 1,880
'right
1,337
1,341 1,350
37 6 \
1,387
1,415 1,416
(right . .
1,387
1,415 1,415
Chief maxima (10 inches from centre) .
1416-5
drawn through the extremities of some and slightly above those of the others,
will give the maxima at every point.
As the funicular polygon and wheel diagram are applicable to a great variety
of spans, the constructions above indicated should be made on tracing-paper, or,
better still, "butter" paper, as this paper is sufficiently transparent for the
purpose, and presents a better surface for drawing upon than ordinary
tracing paper.
MOVING LOADS ON RAILWAY UNDERBRIDGES.
71
CHAPTER VII.
DIAGRAMS OF MAXIMUM SHEAR.
THE shear at any point of a beam acted upon by a uniform moving load is a
maximum when the load extends over the segment of the beam between the
point and the farther abutment ; and further, this maximum shear is
Fig. 76.
proportional to the ordinate, at the point, of a parabola B C D, or A C f I, Fig. 76,
If, for any given type-train and span, the parabola corresponding to the " equiva-
lent uniform live load derived from the maximum shear " be drawn, the shear as
given by the ordinate to this parabola for any point along the span will be found
72 MOVING LOADS ON RAILWAY UNDERBRIDGES.
to be less, and sometimes very considerably less, than the maximum shear
produced at the point by the actual wheel-loads.
For example, in Fig. 76, the maximum positive shears produced by an engine
of the type shown in Fig. 78 as it moves across the span from right to left are
given by the ordinates to the full line B E G F, the maximum negative shears
by the ordinates to the line A E' G' I, and the shears due to the " equivalent
uniform load " by the ordinates to the parabolas BCD and A C' I. In this case,
therefore, we see that everywhere, except at the abutments, the actual maximum
shears are greater than the shears as given by the parabola ; the excess, C' G',
at the centre amounting to about 21 per cent.
Similarly, for the type-train shown in Figs. 78 and 79, and a span of 75 feet,
the diagram of maximum shears, Fig. 80, shows that the maximum shears due
to the actual wheel-loads are again everywhere greater than the shears
corresponding to the equivalent uniform load. In this case, at the centre of the
Fig. 77,
A
odd'
~) O O Qg O O
\
SB
A
span, the actual maximum, when the axle-loads are taken, is equal to 24 2 tons,
while the shear due to the equivalent uniform load is only 21-7 tons, or
10 3 per cent. less. It is obvious, therefore, that while tables of " equivalent
uniform live loads " derived from the maximum shears may be of great value in
making the preliminary calculations, it will often be necessary to draw out the
actual diagrams of maximum shears, or at least determine the maximum shear
at the centre and at one or two other points along the span, before preparing the
working drawings.
The work of constructing diagrams of maximum shear may be considerably
facilitated by noting the following observations: In Fig. 76 the maximum
positive shear for any point under B E occurs when the first wheel a is over the
point, and for the remainder of the span it occurs when the first heavy wheel c
is over the point. Again, assuming the engine as running backwards from left
to right, the maximum negative shear for the part above A E' occurs at any
MOV IN a LOADS ON RAILWAY VNDERBRIDGES. 73
point when the first wheel, i, is over the point, and for the remainder of the span
it occurs when the wheel e is over the point. This would seem to indicate
that the maximum shear at any point will occur when either a or c is over
the point if the engine be running forwards, or under i or e if running
backwards. This point may be investigated analytically as follows :
To determine which of the wheel-loads at a given point D will give the
maximum shear at the point, assume the train-load, in the first instance, to have
moved so far along the span from the farther abutment B (Fig. 77) as to bring
the front wheel W L over the point ; and let K be the total load on the beam,
and x the distance of its centre of gravity from B. The supporting force at A
98
and shear just in front of W l will then be = R.-j- (1) ; and so long, therefore, as
no other load comes on or goes off* the beam, the shear will vary as x, or
according to the ordinates of a straight line.
Further, with wheel W l over the point D, let the loads (of which R is the
resultant) on the beam be W lt W 2 , W 3 . . . W n-1 , and let the first load off the
beam be W n and z its distance to the right of B ; then, if z be less than a lt .
the shear at D, as the train moves towards the left until W 2 comes over
the point, will increase to
Of the two wheels W 1 and W 2 therefore, W 2 at D will give the greater or less
shear at the point according as the expression
I' + W,,^ - * - - ( 3 )
L iiL !
is positive or negative.
Again, with wheel W 2 over D, let R' be the total load on the beam, and
x' the distance of its centre of gravity from B ; and let W n+1 be the first load
off the beam, and z f its distance to the right of B. Then, if z' be less than ( 2 ),
the distance between the wheels W 2 and W 3 , of these two wheels at D, W 3 will
give the greater or less shear according as
is positive or negative.
74
MOVING LOADS ON RAILWAY UNDERBRJDOES.
Now for the train of six- wheels coupled engines shown in Figs. 78 and 79,
moving from right to left, W l = 9 tons = W 2 , W 3 = 18 J tons, % = 6 '5 feet,
and a 2 = 7*25 feet. Hence, for this case, expressions (3) and (4) cannot be
respectively positive and negative, or, in other words, W 2 can never give a
Fig. TV?.
9 TOMS 9
I - '-"T - 75*0-..I 5 -' -~j '**** !*/*To
HI -i-4 h't t H-'-H i 'i
maximum shear at any point. In the same way it can' easily'be shown (see
page 49) that W 3 (or wheel c) at D will give a greater shear at that point than
will W 4 , and W 4 a greater shear than W 5 , and so on.
When the train shown in Fig. 78, therefore, is moving from right to left,
the maximum shear at any point along the span will occur when either wheel [a
MOVING LOADS ON EAILWAY UNDERBRIDGES. 75
or wheel c is over the point ; and in the same way, when the train is moving
from left to right, Fig. 79, it can easily be shown that maximum shear will
always occur under either wheel i or wheel e.
DIAGRAMS OF MAXIMUM SHEAR CONSTRUCTION.
On the diagram already drawn and used in determining the greatest
maximum bending moment, etc., add the load steps L M M' N P Q . . . ,
Pig. 78, and then, on a sheet of tracing paper placed over the diagram, draw, at
a distance apart equal to one-tenth of the span, a series of vertical lines, one
of which is made to pass through the centre of wheel c. These lines will cut
the funicular polygon in points 7', 6', 5' ... Now assume the girder placed
with its left terminal below wheel c; the loads on the girder will then be
c, d, e . . . a', b f , c', and d' t and the corresponding diagram of bending moments
will be C D E . . . C' D' 0'. Through point in the force polygon draw O o
parallel to the closing line C o' to cut the line of loads in o, and through o draw
the horizontal line o" o" under the girder.
The diagram of positive shears will then be o" N N' P Q . . . T T' ; o" N
representing the shears at the left terminal, and p P, q Q . . . the shears at the
distances of 1, 2, . . .of the span from that terminal. If the girder be
now advanced to the left, under the loads, through a distance of one-tenth span
the new diagram of bending moments will be V C D E . . . B' C' 1' ; and if
a line O 1 be drawn parallel to the closing line 1' 1' to cut the line of loads in 1,
and through this latter point a horizontal line V 1" be drawn under the girder,
we shall obtain the new diagram of positive shears 1" M M' N . . . S S', in
which the shears at the left terminal and at the distances from it of 0*1, 0*2,
0*3 span . . . are now represented by 1"I, % "N,pi P. ... Advancing the girder
in this way through successive intervals of one-tenth span we obtain a series
of diagrams of positive shear corresponding to the base lines 0" 0", V V,
2" 2" ... each of which gives a value for the shear at the left terminal and
at the distances of O'l, 0-2, 0'3, 0-4, and 0*5 span from it. These values
can now be scaled from the diagrams and tabulated, and the greatest of
the values given for each point along the span taken and used in plotting
76 MOVING LOADS ON RAILWAY UNDERBRIDGES.
the diagram of maximum positive shears. A similar method may then be
adopted as shown in Fig. 79, for constructing the diagram of maximum negative
shears. As the maximum positive shear at any point, however, occurs when
either wheel c or wheel a is over the point, and the maximum negative shear
when either wheel e or wheel i is over the point, the construction of the
diagram is facilitated considerably by adopting the following procedure.
Under wheel c the shears at the left terminal, and at the distances from it
of 0-1, 0-2, 0-3, 0-4, and 0-5 span, are given by 0" N, ^N, n a N, ?i 3 N, w 4 N
and % N respectively ; or, in the force diagram, by 0-& c, l-b c, 2-b c, 3-1 c,
4:-bc, and 5-bc. To determine the shears under wheel a we may proceed as
follows : For the position of the girder represented by the base line 6"6" the
shear at the centre of the span is m 5 I, and for the position 7"7" it is 5 H.
Now during this change in the girder's position, if the total load upon the
girder remained the same the shear at the centre would, according to equation
(1), page 73, diminish at a uniform rate, as represented by the line ??i 5 5 ; and,
therefore, when wheel a was over the centre the shear at that point would be 1 5 L.
Similarly, at a distance of one-tenth span from the centre, when wheel a was
over the point the shear would be / 4 L, and so on. If, however, during the
change of position of the girder a new load came on, or a load went off the
girder, the shears under load a would not be given exactly by l & L, / 4 L, etc. ;
but if the steps through which the girder is advanced were small say, one-
twentieth of the span the errors thus introduced would be negligible.
Proceeding in the same way to determine the maximum (negative) shears for
the half girder on the right, the girder is first placed with its right terminal
under wheel e, Fig. 79, and then advanced towards the right through successive
intervals of one- tenth span. In this way the shears under wheel e are first
obtained, and then those under wheel i, the shears under wheel e at the right
terminal and at intervals of one-tenth span from it being given by 0"Q',
#1 Q'> <?2 Q' - <ls Q'> and those under wheel i by u 5 U' at the centre, and by
^4 U', % U' ... at intervals of one-tenth span from it.
For the example worked out and illustrated in Figs. 78, 79 and 80, the
maximum positive shears at the left terminal and at the distances from it of
O'l, 0*2, 0*3, and O a 4 span, occur under wheel c, and are represented by
MOVING LOADS ON RAILWAY UNDEliBRIDGES.
77
o" N, % N, ?& 2 N, % N, and 7i 4 N respectively ; at the centre, however, the
positive shear is a maximum when under wheel a, and is represented by 1 5 L,
Fig. 78.
The maximum negative shears at the right terminal and at intervals of
Fig. 79.
A"^\
CO
;*'
"***
S.C-6
6<*S.6
igakat
~^G.G~
i-7 3
-7.G-+
-7.6 -
VT-*
-*Afi-5.eH.j
,
,
'
i
!; ;
jl
||
y
one-tenth the span from it are given by o" Q', ^ Q', & Q' (under e), u 3 U', w 4 U',
and w 6 U' (under 'i).
The maximum shears having been thus determined for a number of points
along the span, the diagram can be plotted as shown in Fig. 80.
78
MOVING LOADS ON RAILWAY UNDERBRIDGES.
Instead of advancing the girder through successive intervals of one-tenth
span, as in the example given (or of one-twentieth span, where great accuracy
is required), an approximation to the true form of the diagram of maximum
shears, and one which would be sufficiently accurate for the great majority of
\
Fig. <sv;.
\
\
\
cases met with in practice, could be obtained by determining the maximum
shears at the abutments, at a quarter span, and at the centre, and then erecting
ordinates representing the maximum shears at these points and joining their
extremities by straight lines.
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