Illustrations.

Description.

Compression in Struts.

Compression in Beam.

Tension in Rods.

Amount of Reactions.

Table XVIII Trussed Beams 100233

Trussed Beam with one centre load

= w

AB = B C

Compression in B D

= + w

Compression in A B

= +w/2. AB/BD

Compression in B C same as in AB

Tension in A D

= - w/2. AD/BD

Tension in C D same as in A D

p = w/2 q = w/2

Table XVIII Trussed Beams 100234

Trussed Beam with one load - w1

(not central,) at any point

A B <> BC

Compression in B D = +w1

Compression in A B

+ p. AB/BD

Compression in B C same as in A B, or

+q. BC/BD

Tension in A D = - p. AD/BD Tension in C D

= - q. CD/BD

p = w1. BC/AC q = w1. AB/AC

Table XVIII Trussed Beams 100235

Trussed Beam with uniform load

= u and one central strut.

A B = B C

Compression in B D = +5/8. u

Compression in A B = + 5/16. u. AB/BD

Compression in B C same as in A B

Tension in A D

= - 5/16.u. AD/BD Tension in C D same as in A D

p = u/2 q = u/2

Table XVIII Trussed Beams 100236

Trussed Beam with uniform load

= u and two struts, dividing beam into three equal parts.

AB=BC =CF

Compression in B D = +11/30. u

Compression in C E same as in B D

Compression in A B

=+11/30. u. AB/BD

Compression in B C and

Compression in C F same as in A B

Tension in A D

= -11/30. u. AD/BD

Tension in D E

= - compression in A B

Tension in F E same as in A D

p= u/2 q = u/2

Table XVIII Trussed Beams 100237

Trussed Beam with two pqual loads each = wl and two struts at equal distances from ends.

A B = C F

Compression in B D = + w1

Compression in C E same as in B D

Compression in A B = + w1. AB/BD

Compression in B C and

Compression in C F same as in A B

Tension in A D

= -w1.AD/BD

Tension in D E

= - compression in A B

Tension in F E same as in A D

p = w1 q=w1

Table XVIII Trussed Beams 100238

Trussed Beam with two unequal loads to, and w11 at any points.

Providing p smaller than w1 and q larger than w11

Compression in B D

= + p Compression in B E

+ (q-w11). BE/BD

Compression in C E

= +w11

Compression in A B = + P. AB/BD

Compression in B C same as in C F Compression in C F

= + q. CF/BD

Tension in A D

= -p. AD/BD

Tension in D E

= - compression in A B

Tension in F E

FE

=- q.FE/BD

p = (w1.BF+w11.CF)/AF q = (w1.AB + w11.AC)A F

Table XVIII Trussed Beams 100239

Trussed Beam with two unequal loads w1 and w11 at any points.

Providing p larger than w1 and q smaller than w11

Compression in B D

= +w1

Compression in C D = + (p-w1). CD/BD

Compression in C E

= + q

Compression in A B = + p. AB/BD

Compression in B C same as in A B Compression in C F

= + q.CF/BD

Tension in A D

= - p. AD/BD

Tension in D E = - compression in C F

Tension in F E = - q. FE/BD

p = (w1.BF+w11.CF)/AF q = (w1.AB+w11.AC)/AF

Where p - the amount of the left reaction, in pounds. q = the amount of right reaction, in pounds. " w, w1 w11 = concentrated loads, in pounds. " u = uniform load, in pounds, over whole beam.

" A B, B C, C F, B D, B E, CD, C E, A D, D E, F E = the length of longitudinal central axes of these pieces, and must all be expressed uniformly, that is, all expressed either in feet or inches.

The amounts of compression in either struts or beam-parts will he the total compression in each, expressed in pounds; to obtain the compression per square inch, divide the amount by the area of cross-Section of the strut or part.

The amounts of tension in rods will be the total tension in each part, expressed in pounds; to obtain the tension per square inch, divide the amount by the area of cross-section of rod.

222 safe building.

In Table XVII, pages 218 and 219, are given the various formulae for reactions, greatest bending moments and deflections, for the most usual cases of continuous girders. The architect can. if he wishes, neglect to allow for the additional strength and stiffness of continuous girders, as both are on the safe side. But he must never overlook the fact that the central reactions are much greater, or in other words, that the end supports carry less, and the central supports carry more, than when the girders are cut.

Bending moments can he figured, at any desired point along a continuous girder, as usual, subtracting from the sum of the reactions on one side multiplied by their respective distances from the point, the sum of all the weights on the same side, multiplied by their respective distances from the point. Sometimes the result will be negative, which means a reversal of the usual stresses and strains. Otherwise the rules and formulae hold good, the same as for other girders or beams. Table XVII gives all necessary information at a glance.

Strength is frequently added to a girder or beam by trussing it, as shown in Table XVIII, pages 220 and 221. One or two struts are placed against the lower edge of a beam and a rod passed over them and secured to each end of the beam; by stretching this rod the beam becomes the compression chord of a truss and also a continuous girder running over one or two supports. There must therefore be enough material in the beam to stand the compression, and in addition to this enough to stand the transverse strains on the continuous girder. If the loads are concentrated immediately over the braces, there will be no transverse strain whatever, but the braces will be compressed the full amount of the respective loads on each. In the case of uniform loads, transverse strains cannot be avoided, of course, but where loads are concentrated the struts should always be placed immediately under them. Even where loads are placed very unevenly, it is better to have the panels of the truss irregular, thus avoiding cross or transverse strains. This same rule holds good in designing trusses of any kind.

Table XVIII shows very clearly the amount and kind of strains in each part of trussed beams. Where there are two struts and they are of any length care must be taken by diagonal braces or otherwise, to keep the lower ends of braces from tipping towards each other. Theoretically they cannot tip, but practically, sometimes, they do. Care must be taken that the beam is braced sideways, or else it must be figured for its safety against lateral flexure (Formula 5.) Then it must have material enough not to shear off at supports, nor to crush its under side where lying support. The ends of rods must have sufficient bearing not to crush the wood. Iron shoes are sometimes used, bu1 if very large are apt to rot the wood. In that case it is well to have a few small holes in the shoes, to allow ventilation to end of timber. If iron straps and bolts are used at the end, care must be taken that the Strap does not tear apart at bolt holes; that it does not crush itself against bolts; that it docs not shear off the holts, and that it does not crush in the end of timber. Care must also be taken to have enough bolts, so that they do not crush the wood before them, and to keep the bolts from shearing out, that is tearing out the wood before them. In all trusses and trussed works the joints must be carefully designed to cover all these points. Many architects give tremendous sizes for timbers and rods in trusses, thus adding unnecessary weight, but when it comes to the joint, they overlook it, and then are surprised when the truss gives out. The next time they add more timber and more iron, till they learn the lesson. It must he remembered that the strength of a truss is only equal to the strength of its weakest part, be that part a member or only a part of a joint. This subject will be fully dealt with in the chapter on

Trusses.

The deeper the truss is made, that is, the further we separate the top and bottom chords, the stronger will it he; besides additional depth adds very much to the stiffness of a truss.

All trussed beams, and all trusses should be "cambered up," that is, built up above their natural lines sufficiently to allow for settling back into their correct lines, when loaded. The amount of the camber should equal the calculated deflection, For all beams, girders, etc., of uniform cross-section throughout, the deflection can be calculated from Formulae (37) to (42) according to the manner of loading. For wrought-iron beams and plate-girders of uniform cross-section throughout, the deflection can he calculated from the same formulae; where, however, the load is uniform and it is desired to simplify the calculation, the deflection can be quite closely calculated from the following Formula: to. a

Importance of Joints.

Depth

Desirable.

Deflection of Girders and Beams.

Uniform Cross-section and Load.