TIC . T.J,C. TIC. T.1•C. 'T •1 TIC. 16-13 18-0 18 -0 12-17 I1-10 14- 10 14-18 See also:

• EXPRESS (through the French from the past participle of the Lat. exprimere, to press out, by transference used of representing objects in painting or sculpture, or of thoughts, &c. in words)

_7 _ 56_ 6=9t; " ~I~ I 10=6"_ _ _ 7=_6 _ 7_=6" _ _ 14=5i ~_ _IO--O_'- 7 4# _ _ 65_6 oer _ Buffers - _ T- ' T T T T TJ 16.64- 12.25 16.0 17-18 27.5 27.5 Passenger Engine, Cal. Ry. In See also:

• AUSTRIA
• AUSTRIA, LOWER (Ger. Niederosterreich or Osterreich unter der Enns, " Austria below the river Enns ")
• AUSTRIA, UPPER (Ger. Oberosterreich or Osterreich ob der Enns, " Austria above the river Enns ")

Also See also:

• LONG, GEORGE (1800-1879)
• LONG, JOHN DAVIS (1838– )

Again, rapidly changing forces, due to the moving parts of the engine which are unbalanced vertically, See also:

• ACT (Lat. actus, actum)

Dead load in per cent 10 20 30 40 50 70 90 of See also:

• TOTAL

E. xvi.) and by F. E. Turneaure (Trans. Am. Soc. C. E. xli.). The latter used a recording,deflectometer and two recording extensometers. The observations are difficult, and the inertia of the See also:

• INSTRUMENT (Lat. instrumentum, from insincere, to build up, furnish, arrange, prepare)

(3) This pereeutage decreases rapidly for longer spans, becoming about 25 % for 75-ft. spans. (4) The increase per cent of See also:

• BOOM

The weight of main girders increases with the span, and there is for any type of bridge a limiting span beyond which the dead load stresses exceed the assigned limit of working stress. Let W1 be the total live load, Wf the total flooring load on a bridge of span 1, both being considered for the See also:

• PRESENT

Let wi, w2 be the weights of main girders per ft. run for a live load p per ft. run and spans 11, l2. Then w2/wl =1 [12/li + (12/11)2] Now let wi', w2' be the girder weights per ft. run for spans ii, l2, and live loads p' per ft. run. Then w2'/w2 = G (1 +4p'/p) w2'/1 = Is [12/11+ (12/li)2] (I +4p'/p) • A partially rational approximate See also:

• FORMULA
• FORMULA (Lat. diminutive of forma, shape, pattern, &c., especially used of rules of judicial procedure)

See also:

• ANEMOMETER (from Gr. aveµos, wind, and . rpov, a measure)

For a plate girder bridge of less height than the train, the wind is to be taken to act on a surface equal to the projected See also:

• AREA

In an interesting address to. the See also:

• BRITISH

Ratio 1 -{-p Values of f, tons per sq. in. P 1+2a Iron. Mild Steel. All dead load 0 1.00 7.5 9.0 •25 0.83 6.2 7.5 '33 0.78 5.8 7'0 .50 0'75 5.6 6.8 .66 011 5.3 5.9 Live load = Dead load 1 . o0 0.66 4'9 5.9 2•0o 0.60 4'5 5'4 4 00 0 56 4'2 50 All live load o•5o 3.7 4'5 Bridge sections designed by this rule differ little from those designed by formulae based directly on See also:

• WOHLER, FRIEDRICH (1800-1882)

Launhardt, based on an empirical expression for WOhler's See also:

• LAW
• LAW (0. Eng. lagu, M. Eng. lawe; from an old Teutonic root lag, " lie," what lies fixed or evenly; cf. Lat. lex, Fr. loi)
• LAW, JOHN (1671—1729)
• LAW, WILLIAM (1686-1761)

Putting the values of F in (1) and solving for f,,,,, ., we get for the breaking stress of a bar subjected to repetition of varying stress, fmax• = u(1 + (t -u)¢/u) [Stresses of same sign.] fads.=u(1+(u-s)(/u) [Stresses of opposite sign.] The working stress in any case is fa... divided by a factor of safety. Let that factor be 3. Then Wohler's results for iron and Bauschinger's for steel give the following equations for tension or thrust:- Iron, working stress, f =4.4 ('I +}0) Steel, ,, =5.87 (1+10). In these equations 4' is to have its + or - value according to the case considered. For shearing stresses the working stress may have o•8 of its value for tension. The following table gives values of the working stress calculated by these equations:- Working Stress for Tension or Thrust by Launhardt and Weyrauch Formula. . ¢ Working Stress f, 1 +-- tons per sq. in. 2 Iron. Steel. All dead load . . . . 1 •o 1 •g 6.6o 8.8o 0.75 1.375 6.05 8.07 0'50 1.25 5.50 7.34 All live load .

. . . o•0 1.005 44'44o 6-60 5'87 -0.25 0.875 3.85 5.14 -0.50 0'75 3.3o 4.40 -0.75 0.625 2.75. 3.67 Equal stresses + and - . -1.00 0.500 2.2o 2.93 Practical experience taught engineers that though 5 tons per sq. in. for iron, or 62 tons per sq. in. for steel, was safe or more than safe for long bridges with large ratio of dead to live load, it was not safe for short ones in which the stresses are mainly due to live Load, the weight of the bridge being small. The experiments of A. Wohler, repeated by Johann Bauschinger, Sir B. Baker and others, show that the breaking stress of a bar is not a fixed quantity, but depends on the range of variation of stress to which it is subected, if that variation is repeated a very large number of times. Let K be the breaking strength of a bar per unit of section, when it is loaded once gradually to breaking. This may be termed the statical breaking strength. Let kma:. be the breaking strength of the same bar when subjected to stresses varying from k,, . to kmie. alternately and repeated an indefinitely great number of times; k,,,;a. is to be reckoned + if of the same kind as k,aas. and - if of the opposite kind (tension or thrust). The range of stress is there-fore k,,,as.-kale., if the stresses are both of the same kind, and kaas.+k„ie., if they are of opposite kinds. Let o = k,,,.

.+=the range of stress, where 0 is always positive. Then See also:

• MILLER, HUGH (1802–1856)
• MILLER, JOAQUIN (CINCINNATUS HEINE) (1841- )
• MILLER, JOE (JosEPH or JosIAS( (1684-1738)
• MILLER, SAMUEL FREEMAN (1816-1890)
• MILLER, WILLIAM (1782-1849)
• MILLER, WILLIAM (1795-1861)
• MILLER, WILLIAM (1796–1882)
• MILLER, WILLIAM HALLOWES (18o1–188o)

The real nature of the action is not well understood, but the word fatigue may be used, if it is not considered to imply more than that the breaking stress under repetition of loading diminishes as the range of variation increases. To compare this with the previous table, (p =(A+B)/A=i+p. Except when the limiting stresses are of opposite sign, the two tables agree very well. In bridge work this occurs only in some of the bracing bars. It is a See also:

• MATTER

English bridge-builders are somewhat hampered in adopting rational limits of working stress by the rules of the Board of Trade. ,Nor do they all accept the guidance of Wohler's law. The following are some examples of limits adopted. For the Dufferin bridge (steel) the working stress was taken at 6.5 tons per sq. in. in bottom booms and diagonals, 6•o tons in top booms, 5.o tons in verticals and long compression members. For the See also:

• STANLEY (FAMILY)
• STANLEY, ARTHUR PENRHYN (18x5–1881)
• STANLEY, EDWARD (1779-1849)
• STANLEY, SIR HENRY MORTON (184o-1904)
• STANLEY, SIR WILLIAM (1548-163o)
• STANLEY, THOMAS (1625-1678)

The loading forces to be considered are vertical, the See also:

• HORIZONTAL

Let pi, p2 . . be the distances of the loads from a b, and p the distance of R1 from a b; then the bending moment at a b is M = Rlp—2(W1p1+W2p2 ...) where-the summation extends to all the loads to the left of a b. If the loads on the right of the section are considered the expressions are similar and give the same results. If Al Ac are the cross sections of the tension and compression flanges or chords, and h the distance between their See also:

• MASS (O.E. maesse; Fr. messe; Ger. Messe; Ital. messa; from eccl. Lat. missa)
• MASS, IN

There are to the left of the section the See also:

• EXTERNAL

Ri. .T. Bending Moment. to the left and right of the load are RI and —R2, and the distribution of shearing force is given by two rectangles. Bending moment increases uniformly from either abutment to the load, at which the bending moment is M = Rem = Rt(l —m). The distribution of bending moment is given by the ordinates of a triangle. Next let the girder carry a uniform load w per ft. run (fig. 40). The total load n tRl ' / / / / % % % / % /O///% • I Shear . -Re N. Ri 1 Bending Moment. • IRz -r A = C 9 Old( c x.

R2=w(c+x) Xc cx=w (c+x)2, 1•-•...1 Zc t --•----,I 4 4 i 1 which is also the shearing force at C 1 for that position of the load. As the load travels, the shear at the See also:

• HEAD (in 0. Eng. heafad; the word is common to Teutonic languages; cf. Dutch hoofd, Ger. Haupt, generally taken to be in origin connected with Lat. caput, Gr. KerbOvi7)
• HEAD, SIR EDMUND WALKER, BART
• HEAD, SIR FRANCIS BOND, BART

43 shows the maximum shear at vertical sections due to a dead and travelling load, the latter advancing (fig. 43, a) from the left and (fig. 43, b) from the right abutment. Comparing the figures it will be seen that over a distance x near the See also:

• MIDDLE

Then the reaction at B and shear at C is Rn/i. Next let the loads advance a distance a so that W2 comes to C. Then the shear at C is R(n+a)/l—W,, plus any reaction d at B, due to any additional load which has come on the girder during the See also:

• MOVEMENT

But the shear may be greater when W2 is at C. In that case the shear at C becomes R(n+a)/l+d—Wi, if a>p, and R(n+a)/l+d—W,a/p, if a<p. If we neglect d, then the shear increases by moving W2 to C, if Ra/l > WI in the first case, and if Rail> Wla/p in the second case. 26. Greatest Bending Moment due to travelling concentrated Loads.—For the greatest bending moment due to a travelling live load, let a load of w per ft. run advance from the left abutment (fig. 47), and let its centre be at x from the left abutment. The reaction at B is 2wx2/l and the bending moment at any section C, at m from the left abutment, is2wx2/(l—m)/l, which increases as x increases till the span is covered. Hence, for uniform travelling loads, the bending moments 4 >r i k 1 are greatest when the loading is complete. In that case the loads on either side of C are proportional to m and l—m. In the case of a See also:

• SERIES (a Latin word from serere, to join)

The bending moment at C is M =W,x(l-m)/l+W2m {I — (x+a)/l f . If the loads are moved a iistance Ax to the right, the bending moment becomes M +AM = Wi (x+Ax) (1—m) /1+W2m { I — (x+Ax+a)/l{ Am=W1z x(l—m)/l-W20xm/l, and this is positive or Wi(l—m) >W2m, or if W,/m>W2/(l—m). But these are the average loads per ft. run to the left and right of C. Hence, if the average load to the left of a section is greater than that to the right, the bending moment at the section will be increased by moving the loads to the right, and See also:

• VICE

The greatest bending moment will occur with one of the greatest loads at the section, and when this further condition is satisfied. Let fig. 49 represent a beam with the series of loads travelling from the right. Let a b be is wl; the reactions at abutments, R, = R2 = iwl. The distribution of shear on vertical sections is given by the ordinates of a sloping line. The greatest bending moment is at the centre and = Ms = ewl2. At any point x from the abutment, the bending moment is M = iwx(i—x), an See also:

• EQUATION (from Lat. aequatio, aequare, to equalize)

W, W2 A( ).0 CQi.QB ' i 1 yl..b--,i k m 1-m 4 1-a c the section considered, and let W. be the load at a b when the bending moment there is greatest, and W,, the last load to the right then on the bridge. Then the position of the loads must be that which satisfies the condition l greater than W1+W2+ ... WA-' x less than WI+W2+ ..• . l Wz Wi+W2+ Fig. 50 shows the See also:

• CURVE (Lat. curvus, bent)

Till WI has advanced a distance a only one load is on the girder, and the curve A"F gives bending moments due to WI only; as WI advances to a distance a+b, two loads are on the girder, and the curve FG gives moments due to W1 and W2. GB" is the curve of moments for all three loads WI+W2+W2. Fig. 51 shows maximum bending moment curves for an extreme case of a short bridge with very unequal loads. The three lightly dotted parabolas are the curves of maximum moment for each of the loads taken separately. The three heavily dotted curves are curves of maximum moment under each of the loads, for the three loads passing over the bridge, at the given distances, from left to right. As might be expected, the moments are greatest in this case at the sections under the 15-ton load. The heavy continuous line gives the last-mentioned curve for the See also:

• REVERSE

Now, except for very short bridges and very unequal loads, a parabola can be found which includes the curve of maximum moments, This parabola is the curve of maximum moments for a travelling load uniform per ft. run. Let we be the load per ft. run which would produce the maximum momentsrepresented by this parabola. Then we may be termed the uniform load per ft. equivalent to any assumed set of concentrated loads. Waddell has calculated tables of such equivalent uniform loads. But it is not difficult to find we, approximately enough for practical purposes, very simply. Experience shows that (a) a parabola having the same See also:

• ORDINATE

27. See also:

• INFLUENCE (Late Lat. influentia, from influere, to flow in)

Mkc iC am of travelling loads at fixed distances, let PI, P2, P2, . . . be the loads which at the moment considered are at distances ml, m2, . from the left abutment. Set off • these distances along AB and let yI, y2, . . be the corresponding ordinates of the influence curve (y = Ff) on the verticals under the loads. Then the moment at C due to all the loads is M.=PIyr+Psy2+ .. . 0 W3 A B i I {h b •j• a• d ----- I • a,b • i X 1 ~ } t rfl 1 The position of the loads which gives the greatest moment at C may be settled by the criterion given above. For a uniform travel-See also:

• LING, PER HENRIK (1776-1839)

With unit load in the position shown, the load at D' is (p-n)/p and that at E' is n/p. The moment of the load at C is m(l-x)/l-n(p-n)/p. This is the equation to the dotted line RS (fig. 52). If the unit load is at F', the reaction at B' and the shear at C' is m/l, positive if the shearing stress resists a tendency of the part of the girder on the right to move upwards; set up Ff =m/l (fig. 54) H on the vertical under the load- Repeating the process for other positions, we get the influence line AGHB, for the shear at C due to unit load anywhere on the girder. GC =x/l and CH =-(l-x)/i. The lines AG, HB are parallel. If the load is in the bay D'E' and is carried by a rail girder which distributes it to cross girders at D'E', the part of the influence line under this bay is altered. Let n (Fig. 55) be the distance of the load from D', xi the distance of D' from the left abutment, and p the length of a bay. The loads at D', E, due to unit weight on the rail girder are (p-n)/p and nip.

The reaction at B' is I(p-n)x1+n(x1+P)}/pl. The shear at C is the reaction at B' less the load at E', that is, {p(xi+n)-ml} /pl, which is the equation to the line DH (fig. 54). Clearly, the distribution of the load by the rail girder considerably alters ID' the distribution of shear due to a load in the bay x, n. in which the section See also:

• CON

But if the load is distributed to the bracing inter-sections by rail and cross girders, then the shear at C' will be greatest when the load extends to N, and will have the values wXADN and -wXNEB. An interesting See also:

• PAPER
• PAPER (Fr. papier, from Lat. papyrus)

F. Swain (Trans. Am. Soc. C.E. xvii., 1887), and another by L. M. See also:

• HOSKINS, JOHN (d. 1664)

T. Eddy (Trans. Am. Soc. C.E. xxii., 1890). Let hk (fig. 56) represent in magnitude and position a load W, at x from the left abutment, on a girder AB of span 1. See also:

• LAY

Similarly, po is the reaction at B and shear at any point of CB. The shaded rectangles represent the distribution of shear due to the load at C, while no may be termed the datum line of shear. Let the load move to D, so that its distance from the left abutment is x+a. Draw a vertical at D, intersecting fh, kg, in s and q. Then qr/ro=hk/hg or ro =W(l-x-a)/l, which is the reaction at A and shear at any point of AD, for the new position of the load. Similarly, rs=W(x+a)!l is the shear on DB. The distribution of shear is given by the partially shaded rectangles. For the application of this method to a series of loads Prof. Eddy's paper must be referred to. 29. Economic Span.—In the case of a bridge of many spans, there is a length of span which makes the cost of the bridge least. The cost of abutments and bridge flooring is practically See also:

• INDEPENDENT

Let P be the cost of one See also:

• PIER (older forms per or Pere, from Med. Lat. pera; the word for piers out of water. For river piers, where a firm, watertight is of obscure origin, and the connexion with Fr. Pierre, Lat. stratum is found at a moderate depth below the river-bed, the pctr

...1 of a too-ft. span erected, and P the cost of one pier with its See also:

• PROTECTION

Braced Girders.—A See also:

• FRAME

Let a frame consist of the five members AB, BD, DC, CA, CB (fig. 57), jointed at the A B points A, B, C and D, and all capable of resisting tension and compression. This frame will be rigid, i.e. it cannot be distorted without causing an alteration in the length of one or more of the members; but if from a change of temperature or any other cause b one or all of the members change their member, but will merely cause a change in the form of the frame. Such a frame as this cannot be self-strained. A workman, for instance, cannot produce a stress on one member by making some other member of a wrong length. Any error of this kind will merely affect the form of the frame; if, however, another member he introduced between A and I), then if BC be shortened Al) wit be strained so as to extend it, and the four other members will be compressed; if CB is lengthened AD will thereby be compressed, and the four other members extended; if the workman does not snake CB and AD of exactly the right length they and all the members will be permanently strained. These stresses will be unknown 1 the compression members being generally of See also:

• TIMBER

The cantilever and suspended girder types are as economical and See also:

• FREE

61, b). These may be used in either the upright or the inverted position. A multiple truss consists of a number of simple trusses, e.g. Bollman truss. Some timber bridges consist of queen-post trusses in the upright position, as shown diagrammatically in fig. 62, where the circles indicate points at which the flooring girders ''''\-41'r" /0 !x 11 V' transmit load to the main girders. See also:

• COMPOUND
• COMPOUND (from Lat. componere, to combine or put together)

On the Pacific See also:

• COAST (from Lat. costa, a rib, side)

65 shows a See also:

• COMMON

1864) the following See also:

• DEFINITION (Lat. definitio, from de-finire, to set limits to, describe)

67 a, lettered in the usual manner, correspond to the forces indicated by arrows in fig. 66, and lettered according to Bow's method. When all the forces are vertical, as will be the case in girders, the polygon of external forces will be reduced to two straight lines, fig. 67 b, superimposed and divided so that the length AX represents the load AX, the length AB the load AB, the length YX the reaction YX, and so forth. The line XZ consists of a series of lengths, as XA, AB . . . DZ, representing the loads taken in their order.' In subsequent diagrams the two reaction lines will, for the See also:

• SAKE

67. figure of the three lines YX, XE, EY in the frame, and represents the three forces in equilibrium at the point YXE of the frame. The direction of YX, being a thrust upwards, shows the direction in which we must go round the triangle YXE to find the direction of the two other forces; doing this we find that the force XE must act down towards the point YXE, and the force EY away from the same point. Putting arrows on the frame diagram to indicate the direction of the forces, we see that the member EY must pull and therefore act as a tie, and that the member XE must push and act as a strut. Passing to the point XEFA we find two known forces, the load XA acting downwards, and a push from the strut XE, which, being in compression, must push at both ends, as indicated by the arrow, fig. 66. The directions and magnitudes of these two forces are already drawn (fig. 67 a) in a fitting position to represent part of the polygon of forces at XEFA; beginning with the upward thrust EX, continuing down XA, and drawing AF parallel to AF in the frame we complete the polygon by drawing EF parallel to EF in the frame. The point F is determined by the intersection of the two lines, one beginning at A, and the other at E. We then have the polygon of forces EXAF, the reciprocal figure of the lines See also:

• MEETING (from " to meet," to come together, assemble, 0. Eng. metals ; cf. Du. moeten, Swed. mota, Goth. gamotjan, &c., derivatives of the Teut. word for a meeting, seen in O. Eng. Wit, moot, an assembly of the people; cf. witanagemot)

67 d as the complete reciprocal figure of the frame and forces upon it, and we see that each line in the reciprocal figure See also:

• MEASURES

69 a. The method of sections already described is often more convenient than the method of reciprocal fgures, and the method of influence lines is also often the readiest way of dealing with braced girders. See also:

• WORM

Consider any other point F of the curve, fig. 70, at a distance xfrom the vertex, the horizontal component of the resultant (tangent to the curve) will be unaltered; the vertical component V will be simply the sum of the loads between 0 and F, or wx. In the triangle FDC, let FD be tangent to the curve, FC vertical, and DChorizontal ; these three sides will necessarily be proportional respectively to the resultant tension along the chain at F, the vertical force V passing through the point D, and the horizontal tension at 0; hence H : V=DC : FC=wx2/2y : wx=x/2 : y, hence DC is the half of OC, proving the curve to be a parabola. The value of R, the tension at any point at a distance x from the vertex, is obtained from the equation R2 = H2+V2 = w2x4/4Y2 + w2x2, or, 2. . R=wx,l (I+xe/4y2). Let i be the See also:

• ANGLE (from the Lat. angulus, a corner, a diminutive, of which the primitive form, angus, does not occur in Latin; cognate are the Lat. angere, to compress into a bend or to strangle, and the Gr. ayKOS, a bend; both connected with the Aryan root ank-, to

Then the deviation y = DE of the neutral axis of the bent beam at any point D from the axis OX is given by the relation d2y MI dx2 ` EI' where M is the bending moment and I the amount of inertia of the beam at D, and E is the coefficient of See also:

• ELASTICITY

Thus if the horizontal scale be one-tenth of an inch to the foot, a= 120, and a beam See also:

• LOO (formerly called " Lanterloo," Fr. lanturlu, the refrain of a popular 17th-century song)

See also:

• RANKINE, WILLIAM JOHN MACQUORN (182o-1872)