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

Table of Notes

< >
[Note]
Page: 608
[Note]
Page: 608
[Note]
Page: 608
[Note]
Page: 608
[Note]
Page: 609
[Note]
Page: 609
[Note]
Page: 609
[Note]
Page: 609
[Note]
Page: 609
[Note]
Page: 609
[Note]
Page: 609
[Note]
Page: 609
[Note]
Page: 609
[Note]
Page: 610
[Note]
Page: 610
[Note]
Page: 610
[Note]
Page: 610
[Note]
Page: 610
[Note]
Page: 610
[Note]
Page: 610
[Note]
Page: 611
[Note]
Page: 611
[Note]
Page: 611
[Note]
Page: 611
[Note]
Page: 612
[Note]
Page: 612
[Note]
Page: 612
[Note]
Page: 612
[Note]
Page: 613
[Note]
Page: 613
< >
page |< < (9) of 695 > >|
    <echo version="1.0RC">
      <text xml:lang="fr" type="free">
        <div xml:id="echoid-div137" type="section" level="1" n="85">
          <div xml:id="echoid-div305" type="section" level="2" n="103">
            <p>
              <s xml:id="echoid-s6466" xml:space="preserve">
                <pb o="9" file="0293" n="304" rhead="LIVRE IV. DES EDIFICES MILITAIRES."/>
              dent au fibre HI, & </s>
              <s xml:id="echoid-s6467" xml:space="preserve">ainſi de tous les autres qui auront des bras de
                <lb/>
              leviers, plus ou moins grands, ſelon qu’ils ſeront éloignés du point
                <lb/>
              d’apui; </s>
              <s xml:id="echoid-s6468" xml:space="preserve">d’où il s’enſuit que les bras de leviers ſont en progreſſion
                <lb/>
              arithmetique, de même que les fibres qui leur répondent, & </s>
              <s xml:id="echoid-s6469" xml:space="preserve">que
                <lb/>
              les progreſſions de part & </s>
              <s xml:id="echoid-s6470" xml:space="preserve">d’autre vont ſe terminer à zero au point
                <lb/>
              A; </s>
              <s xml:id="echoid-s6471" xml:space="preserve">l’on peut donc dire à cauſe des triangles ſemblables, que le
                <lb/>
              produit du bras de levier AB, par le fibre BC, ſera celui du bras
                <lb/>
              de levier AH, par le fibre HI, comme le quarré de AB, eſt au
                <lb/>
              quarré de AH, & </s>
              <s xml:id="echoid-s6472" xml:space="preserve">que par conſequent l’effort de tous les fibres,
                <lb/>
              relativement à leurs bras de leviers, diminuent en venant vers le
                <lb/>
              point d’apui, dans la raiſon des quarrés, des termes d’une progreſ-
                <lb/>
              ſion arithmetique; </s>
              <s xml:id="echoid-s6473" xml:space="preserve">ainſi l’effort de tous les fibres étant répandu
                <lb/>
              dans le triangle ABC, ne ſera que le tiers de ce qu’il ſeroit, s’il
                <lb/>
              étoit réuni aux extrêmités B, & </s>
              <s xml:id="echoid-s6474" xml:space="preserve">C, des bras de leviers AB, & </s>
              <s xml:id="echoid-s6475" xml:space="preserve">AC;
                <lb/>
              </s>
              <s xml:id="echoid-s6476" xml:space="preserve">puiſque la ſomme de tous les quarrés de la progreſſion ne vaut
                <lb/>
                <note position="right" xlink:label="note-0293-01" xlink:href="note-0293-01a" xml:space="preserve">V. le C.
                  <lb/>
                Art. 366.</note>
              que le tiers du produit du plus grand quarré, par la grandeur qui
                <lb/>
              exprime la quantité des mêmes quarrés; </s>
              <s xml:id="echoid-s6477" xml:space="preserve">c’eſt pourquoi dans la ſui-
                <lb/>
              te on pourra ſans difficulté ſupoſer que la force de tous les fi-
                <lb/>
              bres eſt réunie à l’extrêmité du bras de levier, qui répond à la
                <lb/>
              puiſſance reſiſtante, quand au lieu d’admettre cette puiſſance telle
                <lb/>
              qu’elle eſt effectivement, on n’en ſupoſera que le tiers.</s>
              <s xml:id="echoid-s6478" xml:space="preserve"/>
            </p>
            <p>
              <s xml:id="echoid-s6479" xml:space="preserve">Preſentement, pour juger de la force du bois, commençons par
                <lb/>
                <note position="right" xlink:label="note-0293-02" xlink:href="note-0293-02a" xml:space="preserve">
                  <emph style="sc">Fig</emph>
                . 2.</note>
              examiner ce qui lui arrive quand il vient à ſe rompre; </s>
              <s xml:id="echoid-s6480" xml:space="preserve">ainſi ima-
                <lb/>
              ginons que l’on a poſé une poutre ou ſolive AC, ſur deux apuis,
                <lb/>
              il eſt conſtant que ſi on la charge dans ſon milieu d’un poids con-
                <lb/>
              ſiderable, la face ſuperieure ſortira de l’allignement horiſontal,
                <lb/>
              pour former un angle qui ſera d’abord un peu curviligne, & </s>
              <s xml:id="echoid-s6481" xml:space="preserve">qui
                <lb/>
              deviendra toûjours plus ſenſible, à meſure que le poids exercera
                <lb/>
              davantage ſa peſanteur, juſqu’à ce que les deux moitiés BA, & </s>
              <s xml:id="echoid-s6482" xml:space="preserve">
                <lb/>
              BC, ſe ſépareront dans le moment que la ſolive ſe rompra. </s>
              <s xml:id="echoid-s6483" xml:space="preserve">Or re-
                <lb/>
              marqués qu’au commencement les fibres qui ſont le long de la
                <lb/>
              ligne EF, dans la face ſuperieure, paroîtront ſe ſerrer, pendant
                <lb/>
              que ceux qui ſont opoſés dans la face inferieure s’alongeront & </s>
              <s xml:id="echoid-s6484" xml:space="preserve">
                <lb/>
              commenceront à ſe ſeparer: </s>
              <s xml:id="echoid-s6485" xml:space="preserve">ainſi, quand la force qui les uniſſoit
                <lb/>
              devient inferieure à la puiſſance qui agit, ils rompent tous preſque
                <lb/>
              dans le même inſtant, mais avant cela ils ſe ſont trouvés d’autant
                <lb/>
              plus tendus les uns que les autres, qu’ils étoient plus éloignés de
                <lb/>
              la ligne EF, que l’on peut regarder comme le point d’apui, com-
                <lb/>
              mun aux deux leviers recourbés HEA, & </s>
              <s xml:id="echoid-s6486" xml:space="preserve">GEC, car tout ce que
                <lb/>
              nous avons vû ci-devant ſe retrouve dans la deuxiéme figure; </s>
              <s xml:id="echoid-s6487" xml:space="preserve">la
                <lb/>
              difference eſt ſeulement, que la ſolive ayant une épaiſſeur déter- </s>
            </p>
          </div>
        </div>
      </text>
    </echo>
Impressum