Bélidor, Bernard Forest de, La science des ingenieurs dans la conduite des travaux de fortification et d' architecture civile

Table of Notes

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            <pb o="27" file="0311" n="322" rhead="LIVRE IV. DES EDIFICES MILITAIRES."/>
            <p>
              <s xml:id="echoid-s6962" xml:space="preserve">Prenant pour modele la Solive de 3 pieds de longueur fur 6
                <lb/>
              pouces en quarré, qui porte un poids de 64500 liv. </s>
              <s xml:id="echoid-s6963" xml:space="preserve">l’on aura
                <lb/>
              a
                <emph style="sub">3</emph>
              = 216, 6 = 3, m = 64500. </s>
              <s xml:id="echoid-s6964" xml:space="preserve">Si preſentement la poutre, dont il
                <lb/>
              eſt queſtion, a 24 pieds de longueur, que le côté ſur lequel elle doit
                <lb/>
              être poſée ſoit de 12 pouces, & </s>
              <s xml:id="echoid-s6965" xml:space="preserve">que le poids qu’elle doit porter pour
                <lb/>
              n’être pas en danger de ſe rompre, ſoit de 70000. </s>
              <s xml:id="echoid-s6966" xml:space="preserve">il faut doubler ce
                <lb/>
              poids, pour les raiſons que j’ai dit ci-devant, & </s>
              <s xml:id="echoid-s6967" xml:space="preserve">alors il ſera conſideré
                <lb/>
              comme étant de 140000. </s>
              <s xml:id="echoid-s6968" xml:space="preserve">Ainſi nous aurons donc d = 24, c = 12,
                <lb/>
              & </s>
              <s xml:id="echoid-s6969" xml:space="preserve">n = 140000; </s>
              <s xml:id="echoid-s6970" xml:space="preserve">c’eſt pourquoi il n’eſt plus queſtion que de ſuivre ce
                <lb/>
              qu’enſeigne la formule, c’eſt-à-dire multiplier les valeurs de d & </s>
              <s xml:id="echoid-s6971" xml:space="preserve">de
                <lb/>
              n l’une par l’autre, & </s>
              <s xml:id="echoid-s6972" xml:space="preserve">le produit 3360000. </s>
              <s xml:id="echoid-s6973" xml:space="preserve">par la valeur de a
                <emph style="sub">3</emph>
              , c’eſt-
                <lb/>
              à-dire par 216 pour avoir 725760000. </s>
              <s xml:id="echoid-s6974" xml:space="preserve">= dna
                <emph style="sub">3</emph>
              qu’il faut diviſer par
                <lb/>
              le produit des trois nombres qui expriment la valeur de b, c, m, le-
                <lb/>
              quel donnera 2322000. </s>
              <s xml:id="echoid-s6975" xml:space="preserve">= b, c, m; </s>
              <s xml:id="echoid-s6976" xml:space="preserve">& </s>
              <s xml:id="echoid-s6977" xml:space="preserve">le quotient ſera 312, dont il
                <lb/>
              faut extraire la racine quarrée qu’on trouvera de 17 pouces, 7 lignes,
                <lb/>
              11 points, pour la hauteur verticale de la poutre.</s>
              <s xml:id="echoid-s6978" xml:space="preserve"/>
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              <s xml:id="echoid-s6979" xml:space="preserve">Si la hauteur verticale étoit donnée, & </s>
              <s xml:id="echoid-s6980" xml:space="preserve">qu’on voulût trouver l’é-
                <lb/>
              paiſſeur horiſontale, nommant cette épaiſſeur x; </s>
              <s xml:id="echoid-s6981" xml:space="preserve">& </s>
              <s xml:id="echoid-s6982" xml:space="preserve">l’autre c; </s>
              <s xml:id="echoid-s6983" xml:space="preserve">& </s>
              <s xml:id="echoid-s6984" xml:space="preserve">tout
                <lb/>
              le reſte avec les mêmes lettres, alors la formule ſe changeroit en
                <lb/>
              celle-ci {dna
                <emph style="super">3</emph>
              /bmcc} = x.</s>
              <s xml:id="echoid-s6985" xml:space="preserve"/>
            </p>
            <p>
              <s xml:id="echoid-s6986" xml:space="preserve">Enfin, ſi les deux dimenſions de l’équariſſage étoient données, & </s>
              <s xml:id="echoid-s6987" xml:space="preserve">
                <lb/>
              qu’on voulût ſçavoir quelle doit être la longueur d’une poutre pour
                <lb/>
              caſſer ſous l’éfort du poids n; </s>
              <s xml:id="echoid-s6988" xml:space="preserve">nommant c, la hauteur verticale, f l’é-
                <lb/>
              paiſſeur horiſontale; </s>
              <s xml:id="echoid-s6989" xml:space="preserve">nous ſervant toûjours du même modele, nous
                <lb/>
              aurons encore m, n, : </s>
              <s xml:id="echoid-s6990" xml:space="preserve">: </s>
              <s xml:id="echoid-s6991" xml:space="preserve">{a
                <emph style="sub">3</emph>
              /b}, {ccf/x} d’où l’on tire cette formule, après a-
                <lb/>
              voir dégagé l’inconnuë {bccfm/na
                <emph style="super">3</emph>
              } = x.</s>
              <s xml:id="echoid-s6992" xml:space="preserve"/>
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            <p>
              <s xml:id="echoid-s6993" xml:space="preserve">Comme de toutes les ſituations qu’on peut donner à une piéce de
                <lb/>
              bois par rapport à ſa longueur, il n’y en a point où elle ait moins de
                <lb/>
              force, que quand elle eſt poſée horiſontalement, il eſt à propos d’e-
                <lb/>
              xaminer ce qui arrive quand elle eſt poſée obliquement.</s>
              <s xml:id="echoid-s6994" xml:space="preserve"/>
            </p>
            <p>
              <s xml:id="echoid-s6995" xml:space="preserve">Si l’on conſidere la poutre AB poſée ſur deux apuis, dont l’un eſt
                <lb/>
                <note position="right" xlink:label="note-0311-01" xlink:href="note-0311-01a" xml:space="preserve">
                  <emph style="sc">Fig</emph>
                . 8.</note>
              beaucoup plus élevé que l’autre, il eſt conſtant que le poids D qui
                <lb/>
              ſeroit ſuſpendu dans le milieu de ſa longueur, n’agiſſant point ſelon
                <lb/>
              une direction perpendiculaire au bras de levier, fera d’autant moins
                <lb/>
              d’effort pour rompre cette poutre, que l’angle CFG formé par l’o-
                <lb/>
              bliquité de la poutre, & </s>
              <s xml:id="echoid-s6996" xml:space="preserve">la ligne horiſontale FG aprochera davantage
                <lb/>
              de valoir un droit, juſques-là que ſi la poutre étoit perpendiculai-
                <lb/>
              re à l’horiſon, c’eſt-à-dire que l’angle CFG fût effectivement droit, </s>
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