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154

'MEC

the angle formed by one piece and the produced direction ofthe other, its own strain, whether compression or extension,is of the same kind with that of the most remote of the othertwo, and opposite to that of the nearest. Thus, in Figure19, where a b is drawing, the remote piece, A c, is also draw-ing, while a d is pushing or resisting compression.

In these calculations the resistance of joints has no share ;and they must not be supposed to exert any force whichtends to prevent the angles from changing. The joints aresupposed to be perfectly flexible, or like compass joints ; thepin of which only keeps the pieces together when one ormore of the pieces draws or pulls ; of which description thecarpenter must always suppose all joints to be, when he cal-culates the thrusts and draughts of the different pieces of hisframes. The strains on joints, and their power to produceor balance them, are of a different kind, and require a verydifferent examination.

Seeing that the angles which the pieces make with eachother are of such importance to the magnitude and the pro-portion of the excited strains, it is proper to find out someway of readily and compendiously conceiving and expressingj this analogy.

i First, in general, the strain on any piece is proportional tothe straining force.

Secondly, the strain on any piece, A c, is proportional tothe sine of the angle which the straining force makes withthe other piece directly, and to the sine of the angle whichthe pieces make with each other inversely.

Forit is plain thatthe three pressures, Ae, AF, andAG, whichare exerted at the point A, are in the proportion of the linesAE,

1 A f, and f e (because f e is equal to a g). But because the sidesof a triangle are proportional to the sines of the opposite angles,the strains are proportional to the sines of the angles ape,a e f, and f a e. But the sine of a f e is the same with thesine of the angle cad, which the two pieces, AC and ad,i make with each other ; and the sine of A e f is the same withj the sine of e ad, which the straining piece ba makes withj the piece ac. Therefore we have this analogy : sine cad:

i

. sine e A d

sine ead = ae:af and af = ae X-. Now,

sine cad

the sines of angles are most conveniently conceived as deci-mal fractions of the radius, which is considered as unity.Thus sine 30" is the same thing with 0.5, or-^; and so ofothers. Therefore, to have the strain on AC , arising fromany load, A e, acting in the direction ae, multiply ae bythe sine of ead, and divide the product by the sine ofCAD.

This rule shows how great the strains must be when theangle cad becomes very open, approaching to 180 degrees.But when the angle cad becomes very small, its sine (whichis our divisor) is also very small; and we should expect avery great quotient in this case also. But we must observe,that in this case the sine of E a d is also very small: and thisis our multiplier. In such a case the quotient cannot exceedunity.

But it is unnecessary to consider the calculation by thetables of sines more particularly. The angles are seldomknown otherwise than by drawing the figure of the frame ofcarpentry. In this case, we can always obtain the measuresof the strains from the same scale with equal accuracy, bydrawing the parallelogram afeg.

Hitherto we have considered the strains excited at a only,

; as they affect the pieces on which they are exerted. But thej pieces, in order to sustain, or be subject to, any strain, mustI be supported at their ends C and d ; and we may consideri them as mere intermediums, by which these strains are made

to act on the points of support : therefore A f and A A ^also measures of the forces which press or pull at 0 anThe supports to he found for these points may be infin 1various ; but we shall attend only to such as depend onframing itself.

Figure 21 is a structure which very frequently ocC fwhere one beam, b a, is strongly pressed ( to the endanother, A d, which latter is prevented from yielding)well because it lies on a third, as because its end)is hindered from sliding backwards. It is indifferent l rwhat this pressure arises : we have represented it as ovvto a weight hung at B, while B is withheld from yielding ^a rod or rope hooked to the wall. The beam a d m aysupposed at full liberty to exert all its pressure on D, as 1 ^were supported on rollers lodged in the beam hd; butloaded beam b a presses both on the beam a d and onWe wish only to know what strain is borne by A D. , j jf

All bodies act on each other in the direction perpending pto their touching surfaces; therefore the support given byis in a direction perpendicular to it. We may therefore jply its place at A by a beam, A c, perpendicular to H D ’firmly supported at c. In this case, therefore, we may 1 ^a e, as before, to represent the pressure exerted by j.loaded beam, and draw eg perpendicular to ad, and ^parallel to it, meeting the perpendicular A C in F. Th® n ( | ]eis the strain compressing a d, and A f the pressure onbeam h d. . ■ Ja

It may be thought, that since we assume as a p rinthat the mutual pressures of solid bodies are exerted P er pre s, ;dicular to their touching surfaces, this balance of P re ® s t j, e ir jin framings of timbers, depends on the directions of v jbutting joints ; but it does not, as will readily appear by ^sidering the present case. Let the joint or abutment o |two pieces ba, ad, be mitred in the usual manner,direction f a f. Therefore, if A e be drawn perpendicn • ^A.f, it will be the direction of the actual pressure exerte ^ jthe loaded beam, b a, on the beam a d. But the

reactio^jj

a d, in the opposite direction At, will not balance the P 1 e o9 ite>of B A ; because it is not in the direction precisely °l ) I’j )gl ,ii)b A will therefore slide along the joint, and press on t ^ e p I .a'' rhd. A E represents the load on the mitre joint a.e e perpendicular to a e, and e f parallel to it. The p ,e t )]flae will be balanced by the reactions e A and J A :pressure a e produces the pressures a e and a f; of wtlmust be resisted by the beam h d, and A e by the be# pe

”” " ’ ”" n 'nle, ' f

The pressure A f not being perpendicular to h d, can 1 'fully resisted by it ; because, by our assumed prin c1 " p lC e-

reacts only in a direction perpendicular toTherefore, draw f k, f i parallel to H D, and perpe:

[•inc'Pits ton dic u •

it !

for 00

it. The pressure A f will be resisted by H d with't* ^ #p A ; but there is required another force, i a, to p r ® v j 9 lie®beam b a from slipping outwards. This must be » u ot \i&by the reaction of the beam d a. In like manner t 1force, a e, cannot be fully resisted by the beam A d, ° r j ¥

by the prop d, acting by the intervention of the be» tb e

the action of that prop is exerted through the beam ^ (]i<3direction da. The beam ad, therefore, is presse ^beam hd, by the force Ae, as well as by A f. '1° ” n a iid(^this pressure on h d is, draw eg perpendicular to ll ?’ 0 ^parallel to it, cutting e g in r. The forces g A andresist, and balance A e. . j c ]i a (

Thus we see, that the two forces, a e and a/, ' v a tequivalent to ae, are equivalent also to A k, At, K °’ fBut because a f and e e are equal and parallel, and ® - % j 1are also parallel, as also er and fk, it is evidentequal to r e, or to o f, and i A is equal to re, 0 ^ >

Therefore the four forces Ag, A o, A k, a i, are e< l u