<div id="pf1" class="pf w0 h0" data-page-no="1"><div class="pc pc1 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://files.passeidireto.com/2702439d-216d-41ef-b39f-73961d006a41/bg1.png"><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 ls0 ws1">progetto </div><div class="t m0 x2 h3 y2 ff1 fs1 fc1 sc0 ls0 ws1">didattica in rete</div><div class="t m1 x3 h4 y3 ff1 fs2 fc2 sc0 ls0">progetto</div><div class="t m1 x4 h5 y4 ff1 fs3 fc2 sc0 ls0 ws1">didattica in rete </div><div class="t m0 x5 h6 y5 ff2 fs4 fc3 sc0 ls0 ws1">P<span class="_0 blank"></span>olitecnico di <span class="_0 blank"></span>T<span class="_1 blank"></span>orino, <span class="ff3 ws0">febbraio</span> 2007</div><div class="t m0 x6 h6 y6 ff2 fs4 fc3 sc0 ls0 ws1">Dipartimento di Meccanica</div><div class="t m0 x7 h7 y7 ff4 fs5 fc3 sc0 ls0 ws1">Eser<span class="_0 blank"></span>citazioni di </div><div class="t m0 x8 h7 y8 ff4 fs5 fc3 sc0 ls0 ws1">elementi costruttivi delle macchine</div><div class="t m0 x1 h8 y9 ff2 fs6 fc3 sc0 ls0 ws1">Massimiliano A<span class="_0 blank"></span>valle</div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,-17.958000,-18.156300]}'></div></div> <div id="pf2" class="pf w2 h9" data-page-no="2"><div class="pc pc2 w2 h9"></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf3" class="pf w2 h9" data-page-no="3"><div class="pc pc3 w2 h9"><div class="t m0 x9 ha ya ff5 fs7 fc4 sc0 ls1 ws2">ESER<span class="_0 blank"></span>CIT<span class="_1 blank"></span>AZIONI DI ELEMENTI COSTR<span class="_0 blank"></span>UTTIVI</div><div class="t m0 xa ha yb ff5 fs7 fc4 sc0 ls1 ws2">DELLE MA<span class="_0 blank"></span>CCHINE</div><div class="t m0 xb hb yc ff6 fs8 fc4 sc0 ls2 ws3">Massi<span class="_0 blank"></span>milian<span class="_2 blank"> </span>o Avalle</div><div class="t m0 xc hc yd ff7 fs9 fc4 sc0 ls3 ws1">Dipartimento di Meccanica</div><div class="t m0 xd hc ye ff7 fs9 fc4 sc0 ls3 ws1"> </div><div class="t m0 xe hc yf ff7 fs9 fc4 sc0 ls4 ws4">II facoltà di Ingegneria - sede di Vercelli</div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf4" class="pf w2 h9" data-page-no="4"><div class="pc pc4 w2 h9"><div class="t m0 xf hd y10 ff8 fsa fc4 sc0 ls0 ws1">Massimiliano A<span class="_0 blank"></span>valle</div><div class="t m0 xf hd y11 ff8 fsa fc4 sc0 ls0 ws1">Esercitazioni di elementi costruttivi delle macchine</div><div class="t m0 xf hd y12 ff8 fsa fc4 sc0 ls5 ws5">È vietata la riproduzione, anche parziale, con qualsiasi mezzo e\ufb00<span class="_3 blank"></span><span class="ls0 ws6"> ettuato, <span class="_4 blank"></span>compresa </span></div><div class="t m0 xf hd y13 ff8 fsa fc4 sc0 ls0 ws1">la fotocopia, anche ad uso interno o didattico, non autorizzata.</div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf5" class="pf w2 h9" data-page-no="5"><div class="pc pc5 w2 h9"><div class="t m0 x10 he y14 ff8 fsb fc4 sc0 ls0">3</div><div class="t m0 x11 ha ya ff9 fs7 fc4 sc0 ls1">INDICE</div><div class="t m0 x12 hf y15 ff8 fsc fc4 sc0 ls0 ws1">\ue003. <span class="_5 blank"> </span>\uf763\uf761\uf76c\uf763\uf76f\uf76c\uf76f \uf764\uf769 \uf761\uf773\uf773\uf769 \uf765\uf764 \uf761\uf76c\uf762\uf765\uf772\uf769 <span class="_1 blank"></span>....................................................<span class="_6 blank"> </span>\ue00e</div><div class="t m0 x12 hf y16 ff8 fsc fc4 sc0 ls0 ws1">\ue00f. <span class="_5 blank"> </span>\uf773\uf774<span class="_0 blank"></span>\uf761<span class="_4 blank"></span>\uf774\uf76f \uf764\uf769 \uf774\uf765\uf76e\uf773\uf769\uf76f\uf76e\uf765 <span class="_4 blank"></span>..............................................................<span class="_7 blank"> </span>\ue012\ue00e</div><div class="t m0 x12 hf y17 ff8 fsc fc4 sc0 ls0 ws7">\ue012. \uf763\uf772\uf769\uf774\uf765\uf772\uf769 <span class="_8 blank"></span>\uf764\uf769 <span class="_8 blank"></span>\uf772\uf76f<span class="_0 blank"></span>\uf774<span class="_2 blank"> </span>\uf774\uf775\uf772\uf761 <span class="_9 blank"></span>............................................................<span class="_7 blank"> </span>\ue014\ue00e</div><div class="t m0 x12 hf y18 ff8 fsc fc4 sc0 ls0 ws7">\ue015. \uf766<span class="_0 blank"></span>\uf761<span class="_0 blank"></span>\uf774\uf769\uf763\uf761 <span class="_8 blank"></span>\uf76d\uf775\uf76c<span class="_4 blank"></span>\uf774\uf769\uf761\uf773\uf773\uf769\uf761\uf76c\uf765 <span class="_a blank"></span>............................................................<span class="_7 blank"> </span>\ue018\ue019</div><div class="t m0 x12 hf y19 ff8 fsc fc4 sc0 ls0 ws7">\ue00e. \uf765\uf766\uf766\uf765\uf774\uf774\uf76f <span class="_8 blank"></span>\uf764\uf769 <span class="_8 blank"></span>\uf769\uf76e\uf774<span class="_0 blank"></span>\uf761<span class="_0 blank"></span>\uf767\uf76c\uf769\uf76f <span class="_b blank"></span>.........................................................<span class="_7 blank"> </span>\ue003\ue01b\ue019</div><div class="t m0 x12 hf y1a ff8 fsc fc4 sc0 ls0 ws7">\ue014. \uf773\uf76f\uf76c\uf769\uf764\uf769 <span class="_8 blank"></span>\uf761\uf773\uf773\uf769\uf761\uf76c\uf773\uf769\uf76d\uf76d\uf765\uf774\uf772\uf769\uf763\uf769 <span class="_c blank"></span>...................................................<span class="_7 blank"> </span>\ue003\ue00f\ue01c</div><div class="t m0 x12 hf y1b ff8 fsc fc4 sc0 ls0 ws7">\ue01c. \uf766\uf76f\uf772\uf77a\uf761\uf76d\uf765\uf76e\uf774\uf769 <span class="_c blank"></span>......................................................................<span class="_7 blank"> </span>\ue003\ue015\ue01c</div><div class="t m0 x12 hf y1c ff8 fsc fc4 sc0 ls0 ws7">\ue018. \uf763\uf76f\uf76c\uf76c\uf765\uf767\uf761\uf76d\uf765\uf76e\uf774\uf769 <span class="_8 blank"></span>\uf766\uf769\uf76c\uf765\uf774\uf774<span class="_0 blank"></span>\uf761<span class="_0 blank"></span>\uf774\uf769 <span class="_d blank"></span>...................................................<span class="_7 blank"> </span>\ue003\ue014\ue019</div><div class="t m0 x12 hf y1d ff8 fsc fc4 sc0 ls0 ws1">\ue019. <span class="_5 blank"> </span>\uf761\uf763<span class="_0 blank"></span>\uf763\uf76f\uf770\uf770\uf769\uf761\uf76d\uf765\uf76e\uf774\uf769 \uf764\uf769 \uf766\uf76f\uf772\uf76d\uf761 <span class="_0 blank"></span>.................................................<span class="_7 blank"> </span>\ue003\ue018\ue019</div><a class="l" data-dest-detail='[7,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:106.154000px;bottom:446.469000px;width:319.999000px;height:13.846000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail='[37,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:106.154000px;bottom:428.777000px;width:316.922000px;height:11.539000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail='[67,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:104.615000px;bottom:408.777000px;width:321.538000px;height:14.615000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail='[91,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:106.154000px;bottom:390.316000px;width:316.922000px;height:14.615000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail='[111,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:105.384000px;bottom:371.854000px;width:319.230000px;height:12.308000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail='[129,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:103.846000px;bottom:353.393000px;width:319.999000px;height:15.384000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail='[149,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:103.846000px;bottom:334.931000px;width:321.538000px;height:13.077000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail='[171,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:103.846000px;bottom:317.239000px;width:320.768000px;height:13.077000px;background-color:rgba(255,255,255,0.000001);"></div></a><a class="l" data-dest-detail='[191,"Fit"]'><div class="d m2" style="border-style:none;position:absolute;left:104.615000px;bottom:298.008000px;width:319.230000px;height:14.616000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf6" class="pf w2 h9" data-page-no="6"><div class="pc pc6 w2 h9"></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf7" class="pf w2 h9" data-page-no="7"><div class="pc pc7 w2 h9"><img fetchpriority="low" loading="lazy" class="bi x13 y1e w3 h10" alt="" src="https://files.passeidireto.com/2702439d-216d-41ef-b39f-73961d006a41/bg7.png"><div class="t m0 x10 he y14 ff8 fsb fc4 sc0 ls0">5</div><div class="t m0 x14 ha y1f ff9 fs7 fc4 sc0 ls1 ws2">1.<span class="_0 blank"></span> CALCOL<span class="_0 blank"></span>O DI ASSI ED<span class="_0 blank"></span> ALBERI</div><div class="t m0 x13 h11 y20 ffa fsd fc4 sc0 ls0 wsa">Determinar<span class="_0 blank"></span>e le reazioni vin<span class="_0 blank"></span>colari e le caratteristi<span class="_0 blank"></span>che di sollecitazion<span class="_0 blank"></span>e dell\u2019albero su </div><div class="t m0 x13 h11 y21 ffa fsd fc4 sc0 ls0 wsb">due supporti m<span class="_0 blank"></span>ostrato in \ufb01<span class="_e blank"></span> <span class="_2 blank"> </span>gura e calcolar<span class="_0 blank"></span>e la tension<span class="_0 blank"></span>e equivalente secon<span class="_0 blank"></span>do V<span class="_0 blank"></span>on Mises </div><div class="t m0 x13 h12 y22 ffa fsd fc4 sc0 ls0 wsc">nella sezi<span class="_0 blank"></span>one di calettamen<span class="_0 blank"></span>to della ru<span class="_0 blank"></span>ota dentata (<span class="ffb wsd">d =</span> 30 mm). Scegli<span class="_0 blank"></span>ere un ac-</div><div class="t m0 x13 h11 y23 ffa fsd fc4 sc0 ls0 wse">ciai<span class="_0 blank"></span>o da boni\ufb01<span class="_e blank"></span> <span class="_1 blank"></span>ca in grado di gar<span class="_0 blank"></span>antire la r<span class="_0 blank"></span>esistenza dell\u2019alber<span class="_0 blank"></span>o con coef\ufb01<span class="_f blank"></span><span class="wsf"> cien<span class="_0 blank"></span>te <span class="_7 blank"> </span>di </span></div><div class="t m0 x13 h11 y24 ffa fsd fc4 sc0 ls0 ws1">sicur<span class="_0 blank"></span>ezza 2.5. Potenza tr<span class="_0 blank"></span>asmessa: 50 kW a 1500 giri/min.</div><div class="t m0 x15 h13 y25 ffc fsd fc3 sc0 ls6">250</div><div class="t m0 xe h13 y26 ffc fsd fc3 sc0 ls7">50</div><div class="t m0 x16 h13 y25 ffc fsd fc3 sc0 ls8 ws8">15 0</div><div class="t m0 x17 h14 y27 ffb fsa fc3 sc0 ls0">d</div><div class="t m0 x13 h15 y28 ff9 fsa fc4 sc0 ls0 ws9">Soluzione</div><div class="t m0 x13 hd y29 ff8 fsa fc4 sc0 ls9 ws10">Lo schema corrisponde al caso più semplice di un albero con ruota dentata e può </div><div class="t m0 x13 hd y2a ff8 fsa fc4 sc0 ls0 ws11">rappresentare l<span class="_0 blank"></span>\u2019albero di ingresso o di uscita di un riduttore di velocità ad ingra-</div><div class="t m0 x13 hd y2b ff8 fsa fc4 sc0 ls0 ws1">naggi con ruote dentate a denti diritti.</div><div class="t m0 x13 hd y2c ff8 fsa fc4 sc0 ls0 ws12">In questo caso non sono fornite indicazioni sul motore o utilizzator<span class="_0 blank"></span>e collegato </div><div class="t m0 x13 hd y2d ff8 fsa fc4 sc0 ls0 ws13">all\u2019estremità destra (in \ufb01<span class="_3 blank"></span> gura) dell<span class="_0 blank"></span>\u2019albero. Si può quindi ritener<span class="_0 blank"></span>e che il motore o </div><div class="t m0 x13 hd y2e ff8 fsa fc4 sc0 ls0 ws14">l\u2019utilizzatore non applichino alcun carico trasv<span class="_0 blank"></span>ersale ma si riducano a opporre una </div><div class="t m0 x13 hd y2f ff8 fsa fc4 sc0 ls0 ws15">coppia torcente. Altrimenti, nel caso in cui all\u2019estr<span class="_0 blank"></span>emità dell\u2019albero sia presente </div><div class="t m0 x13 hd y30 ff8 fsa fc4 sc0 ls0 ws16">un<span class="_4 blank"></span>\u2019altra r<span class="_2 blank"> </span>uota dentata, una puleggia, una cinghia o una catena, si dovrà tener<span class="_0 blank"></span>e </div><div class="t m0 x13 hd y31 ff8 fsa fc4 sc0 ls0 ws1">conto del carico trasversale applicato e del momento \ufb02<span class="_3 blank"></span><span class="ws17"> ettente conseguente.</span></div><div class="t m0 x18 h16 y32 ffd fsa fc5 sc0 ls0 ws1">Es. 1.1<span class="_10 blank"></span>Es. 1.1</div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf8" class="pf w2 h9" data-page-no="8"><div class="pc pc8 w2 h9"><img fetchpriority="low" loading="lazy" class="bi x19 y33 w4 h17" alt="" src="https://files.passeidireto.com/2702439d-216d-41ef-b39f-73961d006a41/bg8.png"><div class="t m0 xf he y14 ff8 fsb fc4 sc0 ls0">6</div><div class="t m0 xf h18 y34 ff8 fsd fc4 sc0 ls0 ws1">\uf763\uf761\uf76c\uf763\uf76f\uf76c\uf76f \uf764\uf769 \uf761\uf773\uf773\uf769 \uf765\uf764 \uf761\uf76c\uf762\uf765\uf772\uf769</div><div class="t m0 xf hd y35 ff8 fsa fc4 sc0 ls0 ws1b">Si può in ogni caso risolver<span class="_0 blank"></span>e l\u2019albero come in questo esercizio e, mediante il prin-</div><div class="t m0 xf hd y36 ff8 fsa fc4 sc0 ls0 ws1c">cipio della sovrapposizione degli e\ufb00<span class="_3 blank"></span> etti, risolvere successivamente il problema te-</div><div class="t m0 xf hd y37 ff8 fsa fc4 sc0 ls0 ws1d">nendo conto solamente del carico esterno sommando i diagrammi ottenuti dai </div><div class="t m0 xf hd y38 ff8 fsa fc4 sc0 ls0 ws1">due casi di carico.</div><div class="t m0 xf hd y39 ff8 fsa fc4 sc0 ls0 ws1e">Innanzitutto occorre tracciar<span class="_0 blank"></span>e uno schema del componente e determinare le forze </div><div class="t m0 xf hd y3a ff8 fsa fc4 sc0 ls0">agenti.</div><div class="t m0 x1a h14 y3b ffb fsa fc3 sc0 lsa">y<span class="lsd v1">ab</span></div><div class="t m0 x1b h19 y3c ffb fsa fc3 sc0 lsb">r<span class="ls0 v2">z</span></div><div class="t m0 x1c h14 y3d ffe fsa fc3 sc0 ls0 ws18">B<span class="_11 blank"></span>A</div><div class="t m0 xf h18 y3e ffd fsd fc4 sc0 ls0 ws1">Fig. 1.1 <span class="ff8">\u2013 Schema geometrico dell\u2019albero<span class="_0 blank"></span>.</span></div><div class="t m0 xf hd y3f ff8 fsa fc4 sc0 ls0 ws1f">S<span class="_0 blank"></span>ull\u2019albero è presente una ruota dentata. Durante l<span class="_0 blank"></span>\u2019ingranamento le due r<span class="_2 blank"> </span>uote </div><div class="t m0 xf hd y40 ff8 fsa fc4 sc0 ls0 ws20">dell\u2019ingranaggio si scambiano una forza tangente al cerchio di base inclinata del-</div><div class="t m0 xf hd y41 ff8 fsa fc4 sc0 ls0 ws1">l\u2019angolo di pressione di funzionamento<span class="_0 blank"></span>.</div><div class="t m0 x1d h14 y42 ffb fsa fc3 sc0 ls0">F</div><div class="t m0 x1e h14 y43 ffb fsa fc3 sc0 ls0">F</div><div class="t m0 xf h18 y44 ffd fsd fc4 sc0 ls0 ws1">Fig. 1.2 <span class="ff8 ws19">\u2013</span><span class="lsc"> </span><span class="ff8">F<span class="_0 blank"></span>orze scambiate nell\u2019ingranaggio.</span></div><div class="t m0 xf hd y45 ff8 fsa fc4 sc0 ls0 ws21">N<span class="_0 blank"></span>on avendo informazioni più precise sull\u2019ingranaggio si deve supporr<span class="_0 blank"></span>e di trovarsi </div><div class="t m0 xf hd y46 ff8 fsa fc4 sc0 ls9 ws22">di fronte ad una coppia di ingranaggi senza spostamento del pro\ufb01<span class="_3 blank"></span><span class="ls0"> <span class="_12 blank"> </span>lo, per cui l\u2019an-</span></div><div class="t m0 xf hd y47 ff8 fsa fc4 sc0 ls0 ws23">golo di pressione di funzionamento sia uguale all\u2019angolo di pr<span class="_0 blank"></span>essione di taglio, </div><div class="t m0 xf hd y48 ff8 fsa fc4 sc0 ls0 ws24">che è normalmente pari a 20°. Dunque si deve studiar<span class="_0 blank"></span>e la r<span class="_2 blank"> </span>uota dentata solleci-</div><div class="t m0 xf hd y49 ff8 fsa fc4 sc0 ls0 ws25">tata, in un piano normale all\u2019asse dell<span class="_0 blank"></span>\u2019albero, da una forza tangente al cerchio </div><div class="t m0 xf hd y4a ff8 fsa fc4 sc0 ls0 ws1">di base ed inclinata di 20° gradi come mostrato in \ufb01<span class="_e blank"></span><span class="ws26"> gura </span></div><div class="t m0 x1d hd y4b ffa fsd fc4 sc0 ls0 ws1a">1.3<span class="ff8 fsa">.</span></div><a class="l" data-dest-detail='[9,"Fit"]'><div class="d m2" style="border-width:1.000000px;border-style:none;border-bottom-style:solid;border-color:rgb(0,0,255);position:absolute;left:271.500000px;bottom:53.148400px;width:41.667000px;height:11.000000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pf9" class="pf w2 h9" data-page-no="9"><div class="pc pc9 w2 h9"><img fetchpriority="low" loading="lazy" class="bi x1f y4c w5 h1a" alt="" src="https://files.passeidireto.com/2702439d-216d-41ef-b39f-73961d006a41/bg9.png"><div class="t m0 x10 he y14 ff8 fsb fc4 sc0 ls0">7</div><div class="t m0 x20 h18 y34 ff8 fsd fc4 sc0 ls0 ws1">\uf763\uf761\uf76c\uf763\uf76f\uf76c\uf76f \uf764\uf769 \uf761\uf773\uf773\uf769 \uf765\uf764 \uf761\uf76c\uf762\uf765\uf772\uf769</div><div class="t m0 x21 h14 y4d ffb fsa fc3 sc0 ls0">F</div><div class="t m0 x13 h18 y4e ffd fsd fc4 sc0 ls0 ws1">Fig. 1.3<span class="fff lsc"> </span><span class="ff8 ws19">\u2013</span><span class="fff lsc"> </span><span class="ff8">F<span class="_0 blank"></span>orza agente sul dente della r<span class="_2 blank"> </span>uota dentata.</span></div><div class="t m0 x13 hd y4f ff8 fsa fc4 sc0 ls0 ws1">Detto <span class="_7 blank"> </span><span class="ffb ws18">z</span><span class="ws2c"> l\u2019asse dell<span class="_0 blank"></span>\u2019albero, <span class="ffb ws18">x</span> la direzione tangenziale e <span class="ffb lse">y</span> la direzione radiale ci </span></div><div class="t m0 x13 hd y50 ff8 fsa fc4 sc0 ls0 ws2d">si trov<span class="_0 blank"></span>a di fronte ad una forza <span class="ff10 ws27">F</span><span class="ws2e"> risultante scambiata applicata sul cerchio primi-</span></div><div class="t m0 x13 hd y51 ff8 fsa fc4 sc0 ls0 ws1">tivo come nella \ufb01<span class="_3 blank"></span><span class="ws17"> gura successiva. </span></div><div class="t m0 x22 h14 y52 ffb fsa fc3 sc0 ls0">F</div><div class="t m0 x23 h14 y53 ffb fsa fc3 sc0 lsf">F<span class="fse ls10 v3">t</span><span class="ls0 ws18 v4">F<span class="fse v3">r</span></span></div><div class="t m3 x24 h1b y54 ffb fsf fc3 sc0 ls0">y</div><div class="t m0 x25 h14 y55 ffb fsa fc3 sc0 ls0">x</div><div class="t m0 x2 h1c y56 ff11 fsa fc3 sc0 ls0 ws28">A\ue000<span class="ff12 ls9 ws2f">= 2<span class="_2 blank"> </span>0<span class="_2 blank"> </span>°</span></div><div class="t m0 x13 h18 y57 ffd fsd fc4 sc0 ls0 ws1">Fig. 1.4<span class="fff"> \u2013 <span class="ff8">F<span class="_0 blank"></span>orza agente sulla r<span class="_2 blank"> </span>uota dentata.</span></span></div><div class="t m0 x13 hd y58 ff8 fsa fc4 sc0 ls0 ws30">La forza risultante <span class="ff10 ws27">F</span> può essere scomposta, per la successiva analisi, nelle due <span class="_12 blank"> </span>com-</div><div class="t m0 x13 h1d y59 ff8 fsa fc4 sc0 ls0 ws31">ponenti, la forza tangenziale <span class="ffb ls11">F<span class="fs10 ls12 v5">t</span><span class="ls0 ws1 v0"> </span></span><span class="v0">e la forza radiale <span class="ffb ls13">F<span class="fs10 ls0 ws29 v5">r</span></span>. Le due componenti di forza </span></div><div class="t m0 x13 hd y5a ff8 fsa fc4 sc0 ls0 ws1">possono essere espresse in funzione della risultante <span class="ff10 ws27">F</span> come segue:</div><div class="t m0 x12 h14 y5b ff13 fsc fc4 sc0 ls0 ws2a">F<span class="ffb fs10 ls14 v5">t</span><span class="ffa ls15 ws1"> <span class="ffb fsa ls9 ws2f">= F</span><span class="ls16"> </span></span><span class="ff12 ws2b">cos<span class="ffa ls16 ws1"> </span><span class="ff14">\u03b1</span></span></div><div class="t m0 x12 h1e y5c ffb fsa fc4 sc0 ls0 ws18">F<span class="fs10 ws29 v5">r</span><span class="ls9 ws2f v0"> = F <span class="ff12 fsc ls0 ws1">sin <span class="ff14">\u03b1</span></span></span></div><div class="t m0 x13 h1f y5d ff8 fsa fc4 sc0 ls0 ws32">P<span class="_0 blank"></span>oiché però la forza tangenziale <span class="ffb ls17">F<span class="fs10 ls0 ws29 v5">t</span></span><span class="v0"> può essere calcolata dalla coppia agente (sic-</span></div><div class="t m0 x13 hd y5e ff8 fsa fc4 sc0 ls0 ws33">come il suo braccio è pari al raggio primitivo <span class="ffb ws34">r = d / </span><span class="ws1">2 <span class="ffb ls9 ws35">= m z</span><span class="ls18 ws36"> / 2 della ruota) </span></span></div><div class="t m0 x13 hd y5f ff8 fsa fc4 sc0 ls0 ws1">come <span class="ffb ws18">F<span class="fs10 ls14 v5">t</span><span class="ws37 v0"> = C / r<span class="ff8 ws38"> (avendo indicato con </span><span class="ws18">C<span class="ff15 ls19 ws1"> </span><span class="ff8 ws38">la coppia agente) si calcolerà la com-</span></span></span></span></div><div class="t m0 x13 hd y60 ff8 fsa fc4 sc0 ls0 ws1">ponente di forza radiale a partire da quest\u2019ultima:</div><div class="t m0 x12 h20 y61 ffb fsa fc4 sc0 ls0 ws18">F<span class="fs10 ws29 v5">r</span><span class="ls9 ws2f v0"> = F</span><span class="fs10 ws1 v5"> t</span><span class="ws1 v0"> <span class="ff12 fsc ws2b">tan</span> <span class="ff14 fsc">\u03b1</span></span></div></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div> <div id="pfa" class="pf w2 h9" data-page-no="a"><div class="pc pca w2 h9"><img fetchpriority="low" loading="lazy" class="bi x26 y62 w6 h21" alt="" src="https://files.passeidireto.com/2702439d-216d-41ef-b39f-73961d006a41/bga.png"><div class="t m0 xf he y14 ff8 fsb fc4 sc0 ls0">8</div><div class="t m0 xf h18 y34 ff8 fsd fc4 sc0 ls0 ws1">\uf763\uf761\uf76c\uf763\uf76f\uf76c\uf76f \uf764\uf769 \uf761\uf773\uf773\uf769 \uf765\uf764 \uf761\uf76c\uf762\uf765\uf772\uf769</div><div class="t m0 x27 h22 y63 ffb fs11 fc3 sc0 ls0">y</div><div class="t m0 x28 h22 y64 ffb fs11 fc3 sc0 ls0">a</div><div class="t m0 x29 h22 y65 ffb fs11 fc3 sc0 ls0">a</div><div class="t m0 x2a h22 y64 ffb fs11 fc3 sc0 ls0">b</div><div class="t m0 x29 h22 y66 ffb fs11 fc3 sc0 ls0">b</div><div class="t m0 x2b h22 y67 ffb fs11 fc3 sc0 ls0 ws39">F<span class="fs12 v6">r</span></div><div class="t m0 x1c h22 y68 ffb fs11 fc3 sc0 ls1a">F<span class="fs12 ls0 v6">t</span></div><div class="t m0 x2c h22 y69 ffb fs11 fc3 sc0 ls1b">F<span class="fs12 ls0 v6">t</span></div><div class="t m0 x1d h22 y6a ffb fs11 fc3 sc0 ls1c">F<span class="fs12 ls1d v6">t</span><span class="ls0 ws39 v1">F<span class="fs12 v6">r</span></span></div><div class="t m0 x2d h22 y6b ffb fs11 fc3 sc0 ls0">F</div><div class="t m0 x2e h22 y6c ffb fs11 fc3 sc0 ls1e">M<span class="fs12 ls0 v6">t</span></div><div class="t m0 x2f h22 y6d ffb fs11 fc3 sc0 ls0">z</div><div class="t m0 x30 h22 y6e ffb fs11 fc3 sc0 ls0">r</div><div class="t m0 x31 h22 y6f ffb fs11 fc3 sc0 ls0">z</div><div class="t m0 x32 h22 y70 ffb fs11 fc3 sc0 ls0">y</div><div class="t m0 x33 h22 y71 ffb fs11 fc3 sc0 ls0">x</div><div class="t m0 x34 h22 y72 ffb fs11 fc3 sc0 ls0">x</div><div class="t m0 x35 h22 y73 ffe fs11 fc3 sc0 ls0">B</div><div class="t m0 x11 h22 y74 ffe fs11 fc3 sc0 ls0">B</div><div class="t m0 x36 h22 y73 ffe fs11 fc3 sc0 ls0">A</div><div class="t m0 x11 h22 y75 ffe fs11 fc3 sc0 ls0">A</div><div class="t m0 xf hd y35 ff8 fsa fc4 sc0 ls0 ws3c">La coppia agente è in questo caso fornita implicitamente in quanto è fornita </div><div class="t m0 xf hd y36 ff8 fsa fc4 sc0 ls0 ws1">la potenza <span class="ffb ws18">P</span> e la velocità di rotazione <span class="ffb lse">n</span>. La coppia risulta:</div><div class="t m0 x1f h14 y76 ffb fsa fc4 sc0 ls9 ws2f">C = P / <span class="ff14 fsc ls0 v0">\u03c9</span></div><div class="t m0 xf hd y77 ff8 fsa fc4 sc0 ls0 ws1">A<span class="_0 blank"></span>vendo indicato con <span class="ff14 fsc ws3a">\u03c9</span><span class="ff16"> <span class="ff12 ls9 ws2f">= 2</span><span class="ff14 ws28">\u03c0<span class="ffb ws18">n<span class="ff12 ws3b">/60</span></span></span></span> la velocità angolare.</div><div class="t m0 xf hd y78 ff8 fsa fc4 sc0 ls0 ws3d">Si sono dunque identi\ufb01<span class="_3 blank"></span><span class="ws3e"> <span class="_12 blank"> </span>cate due componenti di forza agenti sulla r<span class="_2 blank"> </span>uota dentata ed </span></div><div class="t m0 xf h23 y79 ff8 fsa fc4 sc0 ls0 ws3f">occorre studiare i lor<span class="_0 blank"></span>o e\ufb00<span class="_e blank"></span><span class="ls9"> etti sull\u2019albero. La forza radiale <span class="ffb ls0 ws18">F<span class="fs10 ws29 v5">r</span><span class="ws1 v0"> <span class="ff8 ws3f">agisce trasversalmen-</span></span></span></span></div><div class="t m0 xf h24 y7a ff8 fsa fc4 sc0 ls0 ws40">te all\u2019asse dell<span class="_0 blank"></span>\u2019albero, mentre la forza tangenziale <span class="ffb ws18">F<span class="fs10 ws29 v5">t</span><span class="ls1f ws1 v0"> </span></span><span class="v0">è eccentrica. Occorre farne </span></div><div class="t m0 xf hd y7b ff8 fsa fc4 sc0 ls0 ws1">una trasposizione sul <span class="_0 blank"></span>piano <span class="ffb ws18">xz</span><span class="ws41"> che contiene l\u2019asse dell\u2019albero<span class="_0 blank"></span>. Secondo il teorema </span></div><div class="t m0 xf hd y7c ff8 fsa fc4 sc0 ls0 ws42">di trasposizione, una generica forza <span class="ffb lse">f</span> è equivalente al sistema formato da una </div><div class="t m0 xf hd y7d ff8 fsa fc4 sc0 ls0 ws43">forza agente parallelamente alla forza <span class="ffb ws1">f <span class="_2 blank"> </span></span>più una coppia, aventi rispettivamente il </div><div class="t m0 xf hd y7e ff8 fsa fc4 sc0 ls0 ws1">valore della forza<span class="ffb"> f</span> e del prodotto <span class="ffb">f r</span> do<span class="_0 blank"></span>ve <span class="ffb ws18">r</span> è la traslazione applicata.</div><div class="t m0 xf h18 y7f ffd fsd fc4 sc0 ls0 ws1">Fig. 1.5<span class="ff8"> \u2013 Sistema di forze agenti sull\u2019alber<span class="_0 blank"></span>o.</span></div><div class="t m0 xf hd y80 ff8 fsa fc4 sc0 ls0 ws44">N<span class="_0 blank"></span>ella \ufb01<span class="_e blank"></span> <span class="_12 blank"> </span>gura successiva si applica questo principio per sostituire alla forza originale </div><div class="t m0 xf h25 y81 ffb fsa fc4 sc0 ls0 ws18">F<span class="fs10 ws29 v5">t</span><span class="ws1 v0"> <span class="_0 blank"></span><span class="ff8 ws45">(in grigio) il sistema di azioni equivalente (in nero) formato da una forza di <span class="_2 blank"> </span>va-</span></span></div><div class="t m0 xf hd y82 ff8 fsa fc4 sc0 ls0 ws1">lore <span class="_2 blank"> </span><span class="ffb ws18">F<span class="fs10 ws29 v5">t</span></span><span class="ws46"> traslata del braccio <span class="ffb ws18">r</span> pari al raggio primitivo della ruota, più una <span class="_0 blank"></span>coppia </span></div><div class="t m0 xf h14 y83 ffb fsa fc4 sc0 ls9 ws1">M </div><div class="t m0 x26 h26 y84 ffb fs10 fc4 sc0 ls0 ws29">t<span class="fsa ws1 v7"> <span class="ff8">che ha valore <span class="ffb ws18">F<span class="fse ls20 v3">t</span><span class="ls21 ws1"> r<span class="_13 blank"> </span></span></span> ed è quindi proprio pari al valore della coppia agente <span class="ffb ws18">C</span>.</span></span></div><a class="l" data-dest-detail='[11,"Fit"]'><div class="d m2" style="border-width:1.000000px;border-style:none;border-bottom-style:solid;border-color:rgb(0,0,255);position:absolute;left:84.115200px;bottom:92.124000px;width:67.461800px;height:9.000000px;background-color:rgba(255,255,255,0.000001);"></div></a></div><div class="pi" data-data='{"ctm":[1.000000,0.000000,0.000000,1.000000,0.000000,0.000000]}'></div></div>
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