Calculate DWG Block for AutoCAD

Calculate DWG Block for AutoCAD

CALCULATE OF STEEL AND WOODEN STRUCTURE

Drawing labels, details, and other text information extracted from the CAD file (Translated from Spanish):

pvc, c.i.s., tile sheet, tendal, main beam, wood column, metal column, metal mezzanine, coastal, fb fv of timber timber of mts. mts. mts. between costanerasentre costaneras normal separation between using mts. spaces between costaneras costaneras loads loads lamina tile cm cv total load of de of de tendalesde tendales mt., of first check fb fb ok at ct pp second check cut fv fv vu ok lb third check per cm lw pul cm ok the three checks that were made the side of us reflects that is suitable for what we consider. of charges pcost lb ptotal pcost no. cost lb pcost ptotal venian osea now they are first checkup fb fb ok at ct pp, second check cut fv fv vu ok lb third check per cm cm lw ok check the three checks that we made the trend of reflects us that is suitable for what we consider. of charges pcost lb ptotal pcost no. cost lb pcost ptotal venian osea that are now first checkup fb fb ok at ct pp second checkup cut fv fv vu ok lb third check per per cm cm lw ok check the three checks you made the beam reflects us that it is suitable for what we consider it., critical column calculation of critical column wood column fc ratio less than, wood pv pto pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvt pvp lb pv ptotal last then at feet cu lb at cu ppc lb ft aseccion ok fc aseccion ok, steel reinforced steel, table ksi fb fb it is important to mention that the data do not match because there is one with the tables but for educational purposes the procedure is shown. critical metal column calculation of critical metal column column of loads of loads of loads pcol lb ptotal pcol no. col lb pcol ptotal cutotal at pounds that gives us for the metal column is one of however by the concept that the columns below must be more robust than the top we chose one: since the columns of wood of the second level have one

Raw text data extracted from CAD file:

PVC Ø 2" C.I.S. 4,5 6 8 A B C 1 2 2,8 3,25 TEJA + LAMINA TENDAL VIGA PRINCIPAL COLUMNA MADERA COLUMNA METAL ENTREPISO METAL COSTANERAS \W5.77095; fb = 1200 lb/pul\C5;² \W5.86042; fv = 120 lb/pul\C5;² \W7.12494; \W6.98601; \LCUBIERTA DE MADERADE MADERA \L-PENDIENTE DE 30%DE 30% 4.50 mts. x 0.30 = 1.35 mts. d = \U+221A(4.50²)+(1.35²) = 4.69 mts. Ó 15.38' \U+2261 16' \L-SEPARACI\L\C5;Ó\LN ENTRE COSTANERASENTRE COSTANERAS \L SEPARACIÓN NORMAL ENTRE 0.30 Y 0.80 \U+2192 USAR 0.60 mts. O 1.968' 16 / 1.968 = 8.13 \U+2261 8 ESPACIOS 16 / 8 = 2' \LSEPARACI\L\C3;Ó\LN ENTRE COSTANERAS 2'ENTRE COSTANERAS 2' \L \L-INTEGRACI\L\C5;Ó\LN DE CARGASDE CARGAS \L \W9.01636; LAMINA 3 lb/pie² \W8.96013; TEJA \L18 lb/pie\L² \W8.94990; CM 21 lb/pie² \W9.33232; CV \L12 lb/pie\L² \W5.99935; 33 lb/pie\C3;² \C3;\U+2192 CARGA TOTAL (CT) \L-PROPUESTA DE DE DE DE SEPARACI\L\C5;Ó\LN DE TENDALESDE TENDALES 1 mt. \C3;Ó 3.28' \LCOSTANERA DE 2"x2"DE 2"x2" \L PRIMER CHEQUEO FLEXI\C1;ÓN fb = 1200 lb/pulg\C1;² fb = M c / I = (1082.51)(1)/1.33 = 813.91 lb/pulg\C3;² \C3;\U+2192 OK M = w L\C7;² / 8 = (5.59)(39.36\C7;²) / 8 = 1082.51 lb/pul w = AT x CT + PP = (2)(33) + 1.10 = 67.10 lb/pie \C7;Ó 5.59 lb/pul PP= Asecci\C7;ón x \U+2113 = (0.16)(0.16)(43)= 1.10 lb/pie I = 1/12 b h³ = 1/12 (2)(2³) = 1.33 pul^ 4 SEGUNDO CHEQUEO CORTE fv = 120 lb/pulg\C1;² fv = 1.5 Vu / A = (1.5)(110.04)/4 = 41.26 lb/pulg\C3;² \C3;\U+2192 OK Vu= w L / 2 = (67.1)(3.28)/2 = 110.04 lb A = b h = (2)(2) = 4 pul\C7;² TERCER CHEQUEO DEFLEXI\C1;ÓN \C7;\U+0394 PER = L / 240 = 39.36/240 = 0.164 pul \C7;Ó 0.416 cm \C7;\U+0394 ACT= \L5 w Lw L\L^ 4 = \L(5)(5.59)(39.36^ 4) = \L67081408.25 = 0.08 pul \C3;Ó 0.20 cm \C3;\U+2192 OK 384 E I (384)(1.6x10^ 6)(1.33) 817152000 LOS TRES CHEQUEOS QUE SE LE REALIZO A LA COSTANERA DE 2"x2" NOS REFLEJA QUE ES ADECUADA PARA LO QUE LA CONSIDERAMOS. \LTENDAL DE 4"X8"DE 4"X8" \L INTEGRACI\C7;ÓN DE CARGAS Pcost = Asecci\C7;ón x \U+2113 x L = (0.16)(0.16)(43)(46) = 50.63 lb Ptotal = Pcost x No. cost = (50.63)(9) = 455.67 lb A = b x h = (46)(16) = 736 pies\C7;² Pcost = Ptotal / A = 0.62 lb/pie\C3;² \C3;Ó 1 lb/pie\C3;² VENIAN 33 lb/pie\C3;² OSEA QUE AHORA SON 34 lb/pie\C3;² PRIMER CHEQUEO FLEXI\C1;ÓN fb = 1200 lb/pulg\C1;² fb = M c / I = (46402.56)(4)/170.66 = 1087.60 lb/pulg\C3;² \C3;\U+2192 OK M = w L\C7;² / 8 = (10.07)(192\C7;²) / 8 = 46402.56 lb/pul w = AT x CT + PP = (3.28)(34) + 9.36 = 120.88 lb/pie \C7;Ó 10.07 lb/pul PP= Asecci\C7;ón x \U+2113 = (.33)(0.66)(43)= 9.36 lb/pie I = 1/12 b h³ = 1/12 (4)(8³) = 170.66 pul^ 4 SEGUNDO CHEQUEO CORTE fv = 120 lb/pulg\C1;² fv = 1.5 Vu / A = (1.5)(967.04)/32 = 45.33 lb/pulg\C3;² \C3;\U+2192 OK Vu= w L / 2 = (120.88)(16)/2 = 967.04 lb A = b h = (4)(8) = 32 pul\C7;² TERCER CHEQUEO DEFLEXI\C1;ÓN \C7;\U+0394 PER = L / 240 = 192/240 = 0.8 pul \C7;Ó 2.03 cm \C7;\U+0394 ACT= \L5 w Lw L\L^ 4 = \L(5)(10.07)(192^ 4) = \L6.84 x 10^ 10x 10^ 10 = 0.65 pul \C3;Ó 1.65 cm \C3;\U+2192 OK 384 E I (384)(1.6x10^ 6)(170.66) 1.04 x 10^ 11 LOS TRES CHEQUEOS QUE SE LE REALIZO A LA TENDAL DE 4"x8" NOS REFLEJA QUE ES ADECUADA PARA LO QUE LA CONSIDERAMOS. \LVIGA DE 8"X16"DE 8"X16" \L INTEGRACI\C7;ÓN DE CARGAS Pcost = Asecci\C7;ón x \U+2113 x L = (0.33)(0.66)(43)(46) = 149.84 lb Ptotal = Pcost x No. cost = (149.84)(15) = 2246.6 lb A = b x h = (46)(16) = 736 pies\C7;² Pcost = Ptotal / A = (2246.6)(736) = 3.05 lb/pie\C3;² \C3;Ó 3 lb/pie\C3;² VENIAN 34 lb/pie\C3;² OSEA QUE AHORA SON 37 lb/pie\C3;² PRIMER CHEQUEO FLEXI\C1;ÓN fb = 1200 lb/pulg\C1;² fb = M c / I = (338392.08)(8)/2730.66 = 991.38 lb/pulg\C3;² \C3;\U+2192 OK M = w L\C7;² / 8 = (27.81)(312\C7;²) / 8 = 338392.08 lb/pul w = AT x CT + PP = (8)(37) + 37.74 = 333.74 lb/pie \C7;Ó 27.81 lb/pul PP= Asecci\C7;ón x \U+2113 = (0.66)(1.33)(43)= 37.74 lb/pie I = 1/12 b h³ = 1/12 (8)(16³) = 2730.66 pul^ 4 SEGUNDO CHEQUEO CORTE fv = 120 lb/pulg\C1;² fv = 1.5 Vu / A = (1.5)(110.04)/4 = 50.84 lb/pulg\C3;² \C3;\U+2192 OK Vu= w L / 2 = (333.74)(26)/2 = 4338.62 lb A = b h = (8)(16) = 128 pul\C7;² TERCER CHEQUEO DEFLEXI\C1;ÓN \C7;\U+0394 PER = L / 240 = 312/240 = 1.30 pul \C7;Ó 3.30 cm \C7;\U+0394 ACT= \L5 w Lw L\L^ 4 = \L(5)(27.81)(312^ 4) = \L1.31x10\L^ 12 = 0.78 pul \C3;Ó 1.98 cm \C3;\U+2192 OK 384 E I (384)(1.6x10^ 6)(2730.66)1.67X10^ 12 LOS TRES CHEQUEOS QUE SE LE REALIZO A LA VIGA DE 8"x16" NOS REFLEJA QUE ES ADECUADA PARA LO QUE LA CONSIDERAMOS. \L CALCULO DE COLUMNA CRITICA DE MADERACALCULO DE COLUMNA CRITICA DE MADERA \L \L \W6.18121; fc = 120 lb/pul\C5;² \W0.41948; RELACION L/d = MENOR \W0.41948; QUE 50 4.87 6 8 A B C 1 2 2,435 7 \LCOLUMNA DE MADERA DE 9"x9"DE MADERA DE 9"x9" \L INTEGRACI\C7;ÓN DE CARGAS Pv = Asecci\C7;ón x \U+2113 x L = (0.66)(1.33)(43)(26.24) = 990.43 lb Ptotal = Pv x No. v = (990.43)(2) = 1980.87 lb A = b x h = (26.24)(16) = 419.84 pies\C7;² Pv = Ptotal / A = 4.71 lb/pie\C3;² \C3;Ó 5 lb/pie\C3;²(CARGA ULTIMA ENTONCES 42) \L AT = b h = (7)(2.435) = 17.045 mts\C7;² \C7;Ó 183.37 pies\C7;² L = Ltotal-Lviga 4.15-0.40 = 3.75mts \C7;Ó 12.3 pies CU = 42 lb/pie\C7;² PPC= Asecci\C7;ón x \C7;\U+2113 x L=(0.75)(0.75)(43)(12.28) = 297 lb P = AT x CU + PPC = (183.37)(42)+297 = 7998.54 lb d = 0.75 pies Aseccion = b h (9)(9) = 81 pul\C7;² R L / d = 12.3/0.75 = 16.40 \C3;\U+2192 OK fc = P / Aseccion = 7998.54/81 = 98.74 lb/pulg\C3;² \C3;\U+2192 OK \W0.76684; \H1.68410x;\L\C5;ENTREPISO DE LOSACERO CON REFUERZO DE ACERO SEG\C7;ÚN GR\C7;ÁFICA W10x15 \C3;\U+2192 W10x15 CARACTER\C7;ÍSTICAS SEG\C7;ÚN TABLA A = 4.41 pul\L² I = 68.9 pul^ 4d = 9.90" c = 4.95" fy= 36 ksi M = fb I / c = (23.76)(68.9)/(4.95) = 330.72 k-in \C3;Ó 27.56 k-pie fb = (0.66)(36) = 23.76 ES IMPORTANTE MENCIONAR QUE LOS DATOS NO COINCIDEN POR HABER UNA CONFUSI\C7;ÓN CON LAS TABLAS PERO PARA EFECTOS EDUCATIVOS SE MUESTRA EL PROCEDIMIENTO. \L \L CALCULO DE COLUMNA CRITICA DE METALCALCULO DE COLUMNA CRITICA DE METAL \L Lcol = Ltotal-Pv= 3.25 - 0.25= 3.00mts \C3;Ó 9.84 pul \L \L-INTEGRACI\L\C7;Ó\LN DE CARGASDE CARGAS \L INTEGRACI\C7;ÓN DE CARGAS Pcol = Asecci\C7;ón x \U+2113 x L = (0.75)(0.75)(43)(12.3) = 297.50 lb Ptotal = Pcol x No. col = (297.50)(6) = 1785.03 lb A = b x h = (46)(14.76) = 678.96 pies\C7;² Pcol = Ptotal / A = (1785.03)(678.69) = 2.62 lb/pie\C3;² \C3;Ó 3 lb/pie\C3;² P= CUtotal x AT = 119 x 188.27 = 22404.13 libras \C7;Ó 22.40 K CUtotal= CUc+CUe+Pcol= 37 + 79 + 3= 119 lb/pie\C7;² AT= b h = 2.25 x 7 = 17.5 mts\C7;² \C7;Ó 188.27 pie\C7;² LA SECCI\C3;ÓN QUE NOS DA PARA LA COLUMNA DE METAL ES UNA DE W4x30 SIN EMBARGO POR EL CONCEPTO DE QUE LAS COLUMNAS DE ABAJO DEBEN SER MAS ROBUSTAS QUE LAS DE ARRIBA OPTAMOS POR UNA: \W7.07470; W10x229 \W8.99977; YA QUE LAS COLUMNAS DE MADERA DEL SEGUNDO NIVEL TIENEN UNA SECCI\C3;ÓN DE 9"x9".

Language	Spanish
Drawing Type	Block
Category	Calculations
Additional Screenshots
File Type	dwg
Materials	Steel, Wood
Measurement Units
Footprint Area
Building Features	Car Parking Lot
Tags	autocad, block, cálculate, DWG, steel, structure, wooden