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Calibre RAG MCP Server

by ispyridis
chunk_163.json1.55 kB
{ "id": "chunk_163", "text": "Fig.20-Ferraillage en paniers \nCadres vert lcaux \nFig.21-Ferraillage variante \"poutre\" \nEXEMPLE: Poteau earn~ centre avec moment \nDonnees: \nPoteau: 0,50 m x 0,50 m, charge ELU P = 3 MN, moment Mx = 0,4 MNm, \ndiametre des pieux D \n1 \n= D\n2 \n= D = \n0,60 m, /,\n28 \n= 25 MPa, f e = 500 MPa \nResultats: \nEntre axe des pieux : e = 1 ,5 (D \n1 \n+ D\n2\n) = 1 ,80 m \nEfforts dans les \npieux: FP\n1 \n= P 12-Mxl e = 1,5-0,4 I 1,8 = 1,278 MN \net \nFP\n2 \n= p 12 + M>:f e = 1,5 + 0,4 I 1,8 = 1,722 MN _ \nDimensions: \nA= 2 (D\n1 \n+ D\n2\n) + 0,30 = 2,70 met B = Max[Dt; D2] + 0,30-0,90 m \na\n1 \n= 0,5 a Fp\n2\n1 (FP\n1 \n+ Fp\n2\n) \n= 0,25 x 1,722 I 3 = 0 ,1435 m \na2 \n= 0,5 a FP\n1 \n1 (FP\n1 \n+ Fp\n2\n) = 0,25 X 1,278 I 3 = 0,1065 m \nd1 = d\n2 \n= e I 2 = 0,90 m \n421 \n\nA-\n1 \n= d\n1\n-\na\n1 \n= 0,90-0,1435 = 0,7565 met \n~ = ~-~ = 0,90-0,1065 = 0,7935 m \nBras de levier \nz = 1,3 Max [A.\n1 \n; Ail = 1,032 m \ntg \ne\n1 \n= z I A-\n1 \n= 1,035 I 0,7565 = 1,368 et tg e\n2 \n= z I A.2 = 1,035/0,7935 == 1,304 \nTraction dans Je tirant : \nT = Max[FP\n1 \ncotg e\n1 \n; \nFP\n2 \ncotg e\n2\n] \n=Max [1,278 I 1,368; 1,722/1,304] = 1,32 MN", "metadata": { "book_id": 34848, "title": "BASIC-STRUCTURAL-Henry-Thonier-Tome-2", "authors": "Unknown", "project": "basic_structural", "content_source": "ocr", "content_length": 756676, "chunk_index": 163, "line_start": 5320, "line_end": 5409, "has_formulas": false, "has_tables": false } }

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