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

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        <div xml:id="echoid-div101" type="section" level="1" n="66">
          <pb o="52" file="0078" n="80" rhead="LA SCIENCE DES INGENIEURS,"/>
        </div>
        <div xml:id="echoid-div102" type="section" level="1" n="67">
          <head xml:id="echoid-head84" xml:space="preserve">PROPOSITION PREMIERE.
            <lb/>
            <emph style="sc">Proble’me</emph>
          .</head>
          <p style="it">
            <s xml:id="echoid-s1303" xml:space="preserve">40. </s>
            <s xml:id="echoid-s1304" xml:space="preserve">Ayant le Profil ABCD, d’un mur élevé à plomb des
              <lb/>
            deux côtés & </s>
            <s xml:id="echoid-s1305" xml:space="preserve">ſoutenu par des contreforts repréſentés par le rec-
              <lb/>
            tangle AEFC, on demande ſi une puiſſance Q, agiſſoit de A,
              <lb/>
            en B, pour renverſer ce mur du côté du parement, ou une autre
              <lb/>
            P, de A, en E, pour le renverſer du côté des contreforts, quel
              <lb/>
            eſt le raport de la réſiſtance du mur dans ces deux cas, ou ce qui
              <lb/>
            eſt la même choſe, le raport de la puiſſance Q, à la puiſſance P,
              <lb/>
            ſupoſant qu’elles agiſſent chacune en particulier.</s>
            <s xml:id="echoid-s1306" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1307" xml:space="preserve">Conſiderés la Figure 5. </s>
            <s xml:id="echoid-s1308" xml:space="preserve">qui repréſente le Plan de la Maçonne-
              <lb/>
              <note position="left" xlink:label="note-0078-01" xlink:href="note-0078-01a" xml:space="preserve">
                <emph style="sc">Fig</emph>
              . 4.
                <lb/>
              & 5.</note>
            rie du Profil qui eſt au-deſſus, dont les contreforts ſont rectan-
              <lb/>
            gles & </s>
            <s xml:id="echoid-s1309" xml:space="preserve">égaux dans ce Plan, l’on ſupoſe que l’épaiſſeur LI, des
              <lb/>
            contreforts eſt égale à l’épaiſſeur C D, de la muraille; </s>
            <s xml:id="echoid-s1310" xml:space="preserve">que leur lon-
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            gueur FC, eſt double de leur épaiſſeur, & </s>
            <s xml:id="echoid-s1311" xml:space="preserve">que leur diſtance CL,
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            ou IK, eſt double de la longueur FC, ainſi nommant l’épaiſſeur
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            CD, ou LI, a; </s>
            <s xml:id="echoid-s1312" xml:space="preserve">FC, ſera 2a, & </s>
            <s xml:id="echoid-s1313" xml:space="preserve">CL, ou IK, ſera 4a; </s>
            <s xml:id="echoid-s1314" xml:space="preserve">quant à la
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            hauteur AC, de la muraille & </s>
            <s xml:id="echoid-s1315" xml:space="preserve">des contreforts, nous la nomme-
              <lb/>
            rons b, cela poſé, ab, ſera la valeur du rectangle AD, ramaſſé
              <lb/>
            dans le poids N, qui eſt ſuſpendu dans le milieu de la ligne CD,
              <lb/>
            & </s>
            <s xml:id="echoid-s1316" xml:space="preserve">2ab, ſera la valeur du rectangle EC: </s>
            <s xml:id="echoid-s1317" xml:space="preserve">or comme cette muraille
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            n’a point de longueur déterminée, nous n’y aurons point égard;
              <lb/>
            </s>
            <s xml:id="echoid-s1318" xml:space="preserve">cependant les contreforts étant à une certaine diſtance, & </s>
            <s xml:id="echoid-s1319" xml:space="preserve">ne for-
              <lb/>
            mant point de maſſif continu, comme la muraille fait dans ſa lon-
              <lb/>
            gueur, on ne peut pas dire que 2ab, expriment la valeur des contre-
              <lb/>
            forts, puiſque pour cela il faudroit qu’il n’y eût point d’intervalle
              <lb/>
            entr’eux; </s>
            <s xml:id="echoid-s1320" xml:space="preserve">il faut donc réduire la valeur des contreforts, de façon
              <lb/>
            qu’on puiſſe la conſiderer comme ſi elle régnoit ſur toute la lon-
              <lb/>
            gueur du mur: </s>
            <s xml:id="echoid-s1321" xml:space="preserve">pour cela l’on n’a qu’à diviſer 2ab, par 5, & </s>
            <s xml:id="echoid-s1322" xml:space="preserve">l’on aura
              <lb/>
            {2ab/5} égal à l’expreſſion du poids M, qu’on doit regarder comme
              <lb/>
            équivalant à tous les contreforts réünis enſemble dans un des points
              <lb/>
            de la ligne G M, tirée du centre de gravité.</s>
            <s xml:id="echoid-s1323" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1324" xml:space="preserve">Preſentement, il faut réünir le poids M, au poids N, enſorte
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            qu’il péſe autant en H, qu’il péſe en G, par raport au point d’apui </s>
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