chunk_412.json•1.57 kB
{
"id": "chunk_412",
"text": "= 1,35 x 7,62 = \n10,28 kNim\n2 \nLemomentsurappui 1 vautalors: M\n1 \n=-(1,906x 14,04 + 0,976 x 10,28) =-36,79 kNrnfm \nsoit M\n11 \n= 14,04 x (5,4-x) I 2-36,79 xI 5,4 = 31,09 x-7,02 x2 \ndont Ia derivee vaut: 31,09-14,04 x, qui s'annule pour x = 2,215 m \net le moment maximum \nMtlmax = 34,43 kNrnfm \nEn \ntravee 2, le moment maximum vaut : \nMamax = P2 L\n2 \nI 8 + (MI + M0 I 2 \navec M\n1 \n= M\n2 \n=-(1,906 X 10,28 + 0,976 X 14,04) =-33,30 kNmfm \nd'ou Mamax= M\n0 \n+ M\n1 \n= 17,88 kNrnfm \nLes abscisses x\n1 \net x\n2 \ndes points de moments nuls sont donnes par (Fig. 73) : \n\"'-• \n\"'•l-----\n----\n---\n----~ \n1 .. j \nFig. 73 -Abscissas des points de moments nuls \n-abscisse relative du point de moment maximum: :i\n0 \n= 0,5 + (M\n2\n-\nM\n1\n) \nI (8 M\n0\n) \n-Mmax = (1-X\n0\n) \n(4 M\n0\nx\n0 \n+ Mt) +x\n0 \nM2 \n-Xt = Xo-jMmaxl ( 4Mo) \n-x2 = Xo + JMmaxl ( 4Mo) \n556 \nAppui 0 Travee 1 \ncas 1 \np {kNim2) \n14,04 \nM (kNm/m) \n-10,24 \nXo \n0.4262 \nMmax {kNmlm) \n26,94 \nx1 \n0 ,0634 \n(m) (0,34) \nx2 \n0,7890 \n(m) \n(4,26) \nCas 2 \np (kN/m\n2\n) \n14,04 \nM (kNm/m) \n0,00 \nXo \n0.4101 \nMmax (kNm/m) \n34,43 \nx1 \n0,0000 \n(m) \n(0,00) \nx2 \n0,8203",
"metadata": {
"book_id": 34848,
"title": "BASIC-STRUCTURAL-Henry-Thonier-Tome-2",
"authors": "Unknown",
"project": "basic_structural",
"content_source": "ocr",
"content_length": 756676,
"chunk_index": 412,
"line_start": 14342,
"line_end": 14438,
"has_formulas": false,
"has_tables": false
}
}