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="22" file="0306" n="317" rhead="LA SCIENCE DES INGENIEURS,"/>
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              <s xml:id="echoid-s6809" xml:space="preserve">Supofant une poutre AB, de 24. </s>
              <s xml:id="echoid-s6810" xml:space="preserve">pieds de longueur, & </s>
              <s xml:id="echoid-s6811" xml:space="preserve">de 10.
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              <s xml:id="echoid-s6812" xml:space="preserve">
                <note position="left" xlink:label="note-0306-01" xlink:href="note-0306-01a" xml:space="preserve">
                  <emph style="sc">Fig</emph>
                . 12.</note>
              pouces ſur 14. </s>
              <s xml:id="echoid-s6813" xml:space="preserve">d’équariſſage, poſée de cant, & </s>
              <s xml:id="echoid-s6814" xml:space="preserve">ſerrée par ſes deux
                <lb/>
              bouts, on demande quel poids elle peut porter aux deux tiers de
                <lb/>
              ſa longueur, avant l’inſtant de ſe rompre; </s>
              <s xml:id="echoid-s6815" xml:space="preserve">pour cela, il faut com-
                <lb/>
              mencer par chercher la peſanteur du poids E, qu’elle portera dans
                <lb/>
              ſon milieu, & </s>
              <s xml:id="echoid-s6816" xml:space="preserve">on trouvera qu’il eſt de 73500. </s>
              <s xml:id="echoid-s6817" xml:space="preserve">livres: </s>
              <s xml:id="echoid-s6818" xml:space="preserve">or ſi on ſe
                <lb/>
              rapelle que l’action de ce poids eſt partagée en trois, dont un tiers
                <lb/>
              agit à l’extrêmité A, un autre à l’extrêmité B, & </s>
              <s xml:id="echoid-s6819" xml:space="preserve">le troiſiéme dans
                <lb/>
              le milieu D, l’on verra qu’aſin que la poutre ſoit chargée aux deux
                <lb/>
              tiers C, comme elle le ſeroit dans le milieu, avec le poids de
                <lb/>
              73500. </s>
              <s xml:id="echoid-s6820" xml:space="preserve">il faut que chaque bout ſoit tiré de la même façon, c’eſt
                <lb/>
              pourquoi je multiplie 24500. </s>
              <s xml:id="echoid-s6821" xml:space="preserve">qui eſt le tiers du poids E, par 12.
                <lb/>
              </s>
              <s xml:id="echoid-s6822" xml:space="preserve">qui eſt la longueur du bras de levier AD, ou BD, qui répond aux
                <lb/>
              extrêmités, & </s>
              <s xml:id="echoid-s6823" xml:space="preserve">diviſe le produit par les deux-tiers de la longueur
                <lb/>
              de la poutre, qui expriment alors le bras de levier CB, qui répond
                <lb/>
              au bout B, & </s>
              <s xml:id="echoid-s6824" xml:space="preserve">le quotient 18375. </s>
              <s xml:id="echoid-s6825" xml:space="preserve">eſt la partie du poids qui doit agir
                <lb/>
              à l’extrêmité C de ce levier, pour faire le même effet que le tiers du
                <lb/>
              poids E, fait en D pour avoir la partie du poids qui doit tirer l’au-
                <lb/>
              tre bout A, de la même façon que l’eſt le précédent, je multiplie
                <lb/>
              encore 24500. </s>
              <s xml:id="echoid-s6826" xml:space="preserve">par 12. </s>
              <s xml:id="echoid-s6827" xml:space="preserve">& </s>
              <s xml:id="echoid-s6828" xml:space="preserve">diviſe le produit par l’autre tiers AC, de
                <lb/>
              la longueur de la poutre, c’eſt-à-dire, par 8. </s>
              <s xml:id="echoid-s6829" xml:space="preserve">pour avoir 36750. </s>
              <s xml:id="echoid-s6830" xml:space="preserve">
                <lb/>
              qui eſt ce que l’on demande; </s>
              <s xml:id="echoid-s6831" xml:space="preserve">enfin comme les deux bouts ne
                <lb/>
              pouvoient être rompus ci-devant que par l’action du tiers qui agit
                <lb/>
              dans le milieu, il faut donc fupoſer que la poutre eſt encore char-
                <lb/>
              gée au point C, du poids de 24500. </s>
              <s xml:id="echoid-s6832" xml:space="preserve">ainſi ajoûtant ce nombre avec
                <lb/>
              les deux précedens, c’eſt-à-dire, avec 18375. </s>
              <s xml:id="echoid-s6833" xml:space="preserve">& </s>
              <s xml:id="echoid-s6834" xml:space="preserve">36750. </s>
              <s xml:id="echoid-s6835" xml:space="preserve">l’on aura
                <lb/>
              79625. </s>
              <s xml:id="echoid-s6836" xml:space="preserve">pour la valeur du poids G, que la poutre peut porter à
                <lb/>
              l’endroit C pour être chargée de la même façon qu’elle le ſeroit ſi
                <lb/>
              elle avoit porté dans ſon milieu le poids E de 73500. </s>
              <s xml:id="echoid-s6837" xml:space="preserve">qui n’eſt
                <lb/>
              ici qu’imaginaire, puiſqu’il en faut faire abſtraction, & </s>
              <s xml:id="echoid-s6838" xml:space="preserve">ne conſide-
                <lb/>
              rer la poutre chargée que du ſeul poids G.</s>
              <s xml:id="echoid-s6839" xml:space="preserve"/>
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            <p>
              <s xml:id="echoid-s6840" xml:space="preserve">Si on vouloit charger une poutre de pluſieurs poids, poſés à
                <lb/>
              differens endroits de ſa longueur, & </s>
              <s xml:id="echoid-s6841" xml:space="preserve">qu’on deſirât ſavoir quel rap-
                <lb/>
              port il y a de cette charge avec celle que la poutre peut porter
                <lb/>
              avant l’inſtant de ſe rompre, il faudra commencer par chercher
                <lb/>
              quel eſt le poids que cette poutre peut porter dans le milieu, en-
                <lb/>
              ſuite ſupoſer qu’on a réuni tous les poids dont il eſt queſtion dans
                <lb/>
              le même milieu, alors on pourra comparer ce poids avec celui
                <lb/>
              que la poutre eſt capable de ſoûtenir, & </s>
              <s xml:id="echoid-s6842" xml:space="preserve">l’on verra s’il eſt plus
                <lb/>
              grand ou plus petit, pour juger du parti qu’il faudra prendre.</s>
              <s xml:id="echoid-s6843" xml:space="preserve"/>
            </p>
            <p>
              <s xml:id="echoid-s6844" xml:space="preserve">Comme il ne conviendroit pas de charger les poutres de tout </s>
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