linear unit. From Fig. 3 it will be seen that for a uniform load the maximum shear in the panel will occur when the uniform load extends from the right abutment to that point in the panel where the line 2-3 passes through the line 1-4 (where the shear changes sign). For a minimum shear in the panel (maximum shear of the opposite sign) the load should extend from the left abutment to the point in the panel where the shear changes sign. For equal joint loads, load the longer segment for a maximum shear in the panel, and load the shorter segment for a minimum shear in the panel.

Maximum Floorbeam Reaction.-It is required to find the maximum load on the floor beam at 2′ in (a) Fig. 4 for the loads carried by the floor stringers in the panels 1'-2' and 2'-3'.

FIG. 4. INFLUENCE DIAGRAM FOR MAXIMUM FLOORBEAM REACTION.

In (a) the diagram 1-2-3 is the influence diagram for the shears at 2' due to a unit load at any point in either panel. In (b) the diagram 1-2-3 is the influence diagram for bending moment at 2' for a unit load at any point in the beam. Now the diagram in (a) differs from the diagram in (b), only in the value of the ordinate 2-4. It will be seen that the reaction at 2′ in (a) may be obtained from the diagram in (b), for any system of loads, if the ordinates are multiplied by (d1 + d2)/d1 d2. We can therefore use diagram (b) for obtaining the maximum floorbeam reaction, if we multiply all ordinates by (d1 + d2)/d1·d2.

To obtain the maximum floorbeam reaction, therefore, take a simple beam equal to the sum of the two panel lengths, and find the maximum bending moment at a point in the beam corresponding to the panel point. This maximum moment multiplied by (d1 + d2)/d1 d2 will be the maximum floorbeam reaction. If the two panels are equal in length the maximum bending moment at the center of the beam multiplied by 2/d, where d is the panel length, will give the maximum floorbeam reaction.

Maximum Moment in the Unloaded Chord of a Through Warren Truss.-Let P1 in Fig. 5 represent the summation of the moving loads on the left of the panel 4'-5', P2 represent the summation of the moving loads on the panel, and P3 represent the summation of the moving loads to the right of the panel. The influence diagram for the point 2 is the diagram 1-4-5-3, the lines 1-4 and 5-3 are the same as for a point on the loaded chord, while the influence line for the panel 4'-5' is the line 4-5.

Now the bending moment at 2 due to the three loads is

Now move the loads P1, P2, P3 a short distance to the left, the distance being assumed so small that the distribution of the loads will not be changed, and

Substituting the values of tan a1, tan a2 and tan as in (13) we have

— P1(L — a)/L — P2(L·b — a·l)/L·l + P3·a/L = 0

FIG. 5. INFLUENCE DIAGRAM FOR MOMENTS IN THE UNLOADED CHORD OF A THROUGH WARREN TRUSS.

Maximum Stresses in a Bridge with Inclined Chords.-Let U24' be a web member in a truss with inclined chords in Fig. 6. Point A is the intersection of the upper chord U2U, and the lower chord 2'4'. The stress in U24' equals the moment of the external forces about the point A, divided by the arm c. The stress in the web member U24' will then be a maximum when the bending moment at A is a maximum. To draw the moment influence diagram for the point A, calculate the bending moments about A for the unit loads at 2' and 4'. With a load unity at 4' the moment at A is (La - l)e/L, and with a load unity at 2' the moment at A is (L — a)e/L - (ae), a negative quantity. Laying off 4-6 and 2-7 equal to these moments, the influence diagram for bending moment at A is the polygon 1-2-4-5.

The maximum stress in U24' occurs when some of the wheels at the head of the train are in the panel 2'4', and in unusual cases only, when a load is to the left of 2'. Load P2 representing the summation of the loads to the left of 4' will always come in the panel 2'4'. Load P3, representing the summation of the loads to the right of the panel, will always come to the right of the panel 2'4'. Now the moment at A is

M = P2 y2+P3 y3

(15) Now move the loads a differential distance to the left, it being assumed that the distribution of the loads is not changed, and

M+dM = P2(y2 dys) + P3(ya+dys)

Subtracting (15) from (16), and solving for a maximum we have

(16)

(17)

(le/Lae)/l, and tan a3 = e/L, and substituting in (18) we have

which is the criterion required.

P/L

=

P2(I + ale)/l

(20)

Now for a uniform load the maximum stress in the member U24' will occur when the truss is loaded from the right abutment to the point 3', while the minimum stress will occur when the load extends from the left abutment to the point 3'. The critical point 3 can be calculated by drawing the lines M-2'-3' and N-4'-3'. For wheel loads no load, should in general, pass 3' from the right to give a maximum stress in the member.

∞ in (20) we have the criterion for maximum shear in a panel of a bridge

Resolution of the Shear.-In Fig. 7 the stresses U, D and L hold in equilibrium the external forces on the left of the section cutting these members. These external forces consist of a left reaction, R, at the left abutment and a force at 2, equal to the reaction of the stringer 2-3. The resultant, S, of these two forces acts at a point a little to the left of the left reaction. Its position may be determined by moments. Referring to Fig. 7, let the resultant, S, be replaced by the two forces P1 and P3, P, acting upwards at 1 and P, acting downward at 3 as shown. Now taking moments about point 1, and

1

Similarly by taking moments at 3, we have

43

Moment Diagram.—A moment diagram for Cooper's Class E60 loading is given in Table II, the loading for maximum moment at joints in the loaded chord of truss bridges is given in Table III.

The details of the calculation of the stresses in a bridge with parallel chords is given in Problem 23, Chapter VII, and the details of the calculation of the stresses in a bridge with inclined chords is given in Problem 24, Chapter VII.

POSITION OF WHEELS FOR MAXIMUM MOMENT; TABLE III.

The wheel loads that will produce maximum moment at a point a given distance from the left end of a beam, or at any loaded panel point in a bridge, are given in Table III. For example in an 8-panel Pratt truss of 200 ft. span, maximum moment at panel point L1, 25 ft. from the left end, occurs with wheel No. 4, at the point; a maximum moment at L2 occurs with wheel No. 7 at the point; etc.

INSTRUCTIONS FOR USE OF MOMENT TABLE; TABLE II.
Line (1) is summation of loads from head of uniform load.
Line (2) is summation of loads from wheel No. 1.
Line (3) is the number of each wheel from wheel No. I.
Line (4) is amount of each wheel load in thousand pounds.
Line (5) is distance c. to c. of the wheels, in feet.

Line (6) is distance of any wheel, or the head of uniform load, from
wheel No. I.

Line (7) is distance of any wheel from head of uniform load.
Line (8) is summation of moments of all wheels to right of any
wheel, including the wheel in question, about head of uniform load.

Lines (9) to (25) are summations of moments of all wheels to left of
the stepped line, including wheel on left of value, about the wheel just
above the heavy vertical stepped line on each line.

The values to the right of the stepped lines are moments about the stepped line, including wheel to right of moment value given. EXAMPLES.-Problem 1.-Calculate moment of wheels Nos. I

to 15, inclusive, about wheel No. 15.

Follow vertical line passing through wheel No. 15 down to stepped
line, and follow over to the left on line (12), and find 16,220 thousand
ft.-lb. to right of vertical line through wheel No. I.

Problem 2.-Calculate the moment of wheels Nos. 17, 16, 15, 14,
about wheel No. 13.

Follow vertical line passing through wheel No. 13 down to the
stepped line, and follow line (14) to right, and to left of the vertical
line through wheel No. 17, find 1,281 thousand ft.-lb.

Problem 3.-Given a 200-ft. span, 8 panel Pratt railway bridge.
The moments and shears are calculated as follows:

Moments.-Panel point L. From Table III, there will be a
maximum moment at L with wheel No. 4 at the joint; and from Ta-
ble II, line (7) it is 91 ft. from wheel No. 4 to the end of the uniform
load, and it is also 175 ft. from joint Li to the end of the bridge, and
there will be175 -91 84 ft. of uniform load on the bridge. Then,
R1 X 200 = 24,550 + 426 X 84 + 3 X 843/2 = 70,918 thousand ft.-lb.;
and R1 = 354.6 thousand lb. The moment at Li is Mi
720 = 8,145 thousand ft.-lb.

=

=

Shear in Panel LoL is S1

=

354.6X 25

TABLE III.

LOADING FOR MAXIMUM MOMENT IN BRIDGES FOR COOPER'S LOADINGS.

Spans 10 15 20 25′|30|35′|40|45|50|55|60|65170|80|90|100|110|120|130 140

300'

to 2 3 3 4 4 5 5 6 7 7 8 9 10 11 12 13 14 15 17 18 260

250'

to 23 3 4 4 5 5 6 7 8 8 9 10 11 12 13 14 15 17 18 200'

190'

to 23 3 4 4 5 5 6 7 8 9 9 11 12 12 13 14 15 17 18 150'

R1720/25-354.6-28.8-325.8

thousand lb. (720 is the moment of wheels Nos. 1, 2, 3, about wheel No. 4).

13

WHEEL DETERMINING MAXIMUM MOMENT

COOPER'S LOADINGS

C.M.&ST.P.RY.

The shorter span is ahead Followed by the
longer one except wheel is over-lined.