E 0 0

376

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the other is by notching the under side of the tie-beam, andcutting the wall-plate in a reverse form to fit it. Thismethod is far preferable to the other, as it is not liable to bedrawn, which the other is very subject to when the timbershrinks.

Ridge-tree, a piece of timber fixed in the vertex of a roof,where the common rafters meet on each side of it ; the upperedge of it is higher than the rafters, for the purpose of fixingthe lead which goes over it to cover the ends of the slates inthe upper course.

Straps, thin pieces of iron running across the junction oftwo or more parts of a truss, or frame of carpentry, branch-ing out from the intersection in the direction of the severalpieces, for the purpose of securing them to each other. Theyought always to be double, viz., one strap on each side ; andtheir ends strongly bolted to each of the pieces.

The uses of these various parts will be illustrated in whatfollows ; and here, it may be proper to observe, that thoughevery one of the parts above defined may be found in thesame roof, it is not necessary that a complete roof shouldhave them all; the introduction of many of them depends onthe distance of the walls, the contour of the roof, the par-titions below, the quantity of head-room wanted in theI garrets, &c.

Other names of timbers will be fully illustrated in the• descriptions of other roofs, in due order of succession.

Before we proceed to the construction of roofing, it will benecessary to show upon what principles a body, or piece oftimber, may be supported in various positions.

Proposition I. Plate I. {of Mechanical Carpentry)Figure 1 .—If a heavy body, A b c d, be suspended by anytwo inclined strings, d e and C r, in a vertical plane, a rightline drawn through the intersection, perpendicular to the hori-zon, will pass through the centre of gravity of the body.

It is shown by the writers on mechanics, that if any threeforces act upon a point, or a body, their directions will tendto the same point, or be parallel to each other. It is wellknown that every body acts with its full force in one pointonly, viz., its centre of gravity, and in a direction perpen-dicular to the horizon: therefore, if a body be sustained atE and f, it will revolve round these points, till the line, gh,passing through the intersection, h, of the two strings, d eand c f, and the centre of gravity, g, becomes perpendicularto the horizon.

Corollary 1.—Hence, if any body be supported by twostrings, it may also he supported by two planes perpendicularto those strings, provided the two points of the body sup-ported be in the direction of the strings ; for every bodyacting upon a plane acts in a line perpendicular to thatplane.

Corollary 2. —Hence, also, a body may be supported bytwo props in any two directions that it may be supported bystrings, provided the surface of the body, at the points ofcontact, or the ends of the props, be planes, at right anglesto the strings.

Corollary 3.—Hence, all the properties that have beendemonstrated of three forces acting upon a body supposed tobe void of weight, will equally flow from a heavy body sup-ported by two strings, by substituting the weight of thebody for the middle force ; and hence, if the direction of anyforce supporting a heavy body be given, the other may easilybe found.

Proposition II.— Given the position in which a bodyshould be placed, and the position of a plane supporting thebody at one end; to find the position of another plune to sup-port it at another given point, and to find the pressure on theplanes, the weight of the body being given.

v ert‘ cfll

Through the centre of gravity of the body draw a ^ on th eline, and through any point on which the body resgiven plane, draw a line perpendicular to that pl a ” e ’ t0the vertical line : from the intersection draw a n ^ (]ra' v 'other point, which is to be supported; from that P®^ e jiret'a plane at right angles to this line, which will betion of the plane required. , aI1 ee o1>

To find the intensity of the forces, take any * ain; fr< ,a)the vertical line, to represent the weight of the o -pan *''the intersection; on that line, as a diagonal, COIB P‘ e jj n esp er ”lelogram, whose sides are in the directions of .i e of ^pendicular to the supporting planes ; and the s repr eSS "parallelogram, perpendicular to either plane, wnthe force on that plane. . p on * 1

Example 1. Figure 2. —Let the body, A bC d, _ , ff e, ® Cltop of the wall, it c at c, so as to touch the lowei o jj r e«'

nf flip hrtfltr fit t)int nnirit n « it ia tvintiirP.ri tO

end at *

a” 1

the

of the body, at that point, c ; it is required totion of a plane that will support the lower e 1 — pPto find the pressure of the body on the wall aplane. j ra w 4 ‘

Through the centre of gravity, G, of the body, . n f B>- - ‘ " R > J dire®'

□ -- o — j * /

vertical lineGF; draw cf perpendicular to ° ’ t ],e

and draw b i perpendicular to f b ; and b i will

in* 1

Jse

tion of the plane required. On the vertical lm ’ j e teFit to represent the weight of the body, and co^ f 0 ice° aparallelogram, iini; then f n will represent ^ r( , c 8 ct> I1, j.the wall-head, in the direction f c ; and F L the^ F>perpendicular to the plane, or in the direction ^ j. e qui re ’the vertical and horizontal thrusts on the wall at tpe" ^draw np perpendicular to fg, meeting it ,n f j,

force, f n, will be resolved into two forces, E p a t jjat w* 1 * ^will represent the horizontal part of the force, vl j^ r j Z on \ a 9pushes the wall in a direction parallel t0 . t ", e wn war<J 3f p the other part, which tends to press it do ^

direction perpendicular to the horizon. . - A j? C P>

Example 2. Figure 3.—Let the sloping bo J^ nC [des ' v ..supported by a wall at its lower end, D, whic ^ g raV1 .fd,the surface of the body, and let G be the cen t ipP er e „Jeit is required to cut a notch out of the body, a is 10

c, so that it may rest upon the top of a wa1, ^ a jj 9 . p ,to fit the notch, and to find the pressure on 1 * pe,rp e ”Draw the vertical line, ge ; from d draw t0 it; 1 9 (,lar to D c ; join e c, and make C f at right an^.^ be at j 0 .

the notch hcf being cut, the body, Ab C D, ^ par® j

To find the pressure on the walls, comple ^ j t9 g (

gram, eiki, having a given angle, D E C, , i;

on the given line, e G ; then, if K E represent p

the dire ctI °" c £,

the body, i e will represent the pressure m direct'® „ii

upon the wall at d, and ld the pressure mThe horizontal and perpendicular pressure reg0 iv*-omay be found, as in the preceding example, L . c j, is p e

t* . i n _ i _ _ • , _ ___ /MlA Ot » V

upo n , 6 pi'<r e acl ‘oW ,0t L,e»'

tli®

of the forces, i e and l e, into two ; ° ne - . , 0 j t . tdicular to the horizon, and the other para e aB1 ple, t 1 ^jtyScholium . —It must be observed in tins ce ntr e & jnotch, which is cut out at C, will remove ' a j ter th e

+ 1 onrl /»nTlSeQUeB l V..

slopf

nearer to the lower end, d, and consequeni‘/^ ree) t ve r/c f ; but as this can only be in a very sma ° p0 t(librium will hardly be affected by it, wnesmall. , ,, ror „ersof*

Example^. Figure 4 .—Let one of t e wb'® 1 0 t ° a ,body, a b c d, rest upon the top of a wa a n otcb, a ,v»*

level : it is required to find the position re9 t up 0 .

of the upper end, c, so that the body ® ^

ide to fit the notch. 0 t C H 111 ,, tl' e

Let the small part, f c n, be so cut 1 por ted gpdic 11to the horizon, then the body will be s ^ 0l( perPwalls at c and d. For, draw D b 6 K ’