Triarticulada DWG Section for AutoCAD

Triarticulada DWG Section for AutoCAD

Truss calculations; plants; sections and construction details of truss calculation .

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

felipe garcia zubieta, developed exercise of design of the truss: the type of floor that I am going to use is a cooperating sheet of the Eurocol commercial house the thickness chosen is of with a height forging of so we obtain by means of the data of the own weight of which it will be of we calculate the loads: permanent: own weight of the structure: own weight of the forged: weight covered: variables: use: snow els: for a qelu of a length of truss of the distance between straps in our case match with the points where the weight of the in the scheme of the truss coincides with the points where they fall the different we will place the trusses each having therefore porticos in the ship of m., enunciado says that we will use as profiles belts as maximum in this case we will use the ipe because it has more we will also use a steel whose resistance is to find the length of the dimensions of one of them as a biapoyada beam. as we do not know the light of the but we have the inertia we can solve it of the soguiente form: we calculate qelu but without taking into account the own weight of the structure since we are calculating the belts. qelu: we calculate one of the belts that more load of forged therefore is of which receives each side mf fyd we can size a length of belt of but we will use so that it squares better in the given dimensions. we could place the trusses in the direction to save us but due to the entrance of trucks located in the face noert we will have to place them in direction since the distance between the pillars that fasten these trunks can not be greater of the entrance needs a free passage of m. now we check whether it meets this allowable rigidity. we know that the permissible shaft is that the permissible shaft of the tire is for that we will use the lesser loads. ix ipe therefore if it meets we solve the reactions: kn shell ra rb kn kn, the cuts: the cuts: ra pa, rapa pb, rapa, Dimension: lower string: as it is pulled it is not necessary to see if it complies. mmax, upper cord: as the upper cord is compressed we will have to check if it complies. mmax if it meets diagonal amounts: vmax compression if it complies, pb kn, trusses, trusses, trusses, making a cut in the middle doing symmetry ha ha kn hb kn ha hb, trusses, trusses, trusses, rapa, calculations of the stiffness of the bars of the stiffness of the bars for greater ease in the calculation we

Raw text data extracted from CAD file:

Felipe Garcia Zubieta Np:51539 24 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 1,5 3 2,33 3 3 3 3 3 3 3 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 2,33 3 1,5 3 2,33 2,33 4,71 8 400mm 17mm 29,9mm 215,1mm 190,4mm 135mm 10,2mm 142,5mm 9,2mm 110mm 220mm 9,2mm 220mm 110mm 140mm 73mm 6,9mm 4,7mm 36,5mm 165mm 330mm 11,5mm 220mm 110mm 9,2mm \OM EJERCICIO DESARROLLADO DE DISEÑO DE LA CERCHA: El tipo de forjado que voy a utilizar es una chapa colaborante de la casa comercial Eurocol 60, y el espesor elegido es de e=1mm y con una altura o canto de forjado de 12cm. Asi obtenemos mediante las tablas, el dato del peso propio del forjado, que va a ser de 2,21Kn/m2 Calculamos las cargas: - Permanentes: Peso Propio de la estructura: 0,25 KN/m2 Peso Propio del Forjado: 2,21 KN/m2 Peso Cubierta: 1 KN/m2 - Variables: Uso: 2 KN/m2 Nieve (Leon): 1,2 KN/m2 Q els: (0,25 + 2,21 + 1) + 2 + 1,2 x 0,7 = 6,3 KN/m2 Q elu:(0,25 + 2,21 + 1) x 1,35 + 2 x 1,5 + 1,2 x 0,7 x 1,5 = 8,931 KN/m2 Para una Qelu de 8,931 KN/m2 y una longitud de cercha de 24 m la distancia entre correas en nuestro caso coincidra con los puntos donde recae el peso de la cubierta( en el esquema de la cercha coincide con los puntos donde caen las distintas P) Vamos a colocar las cerchas cada 2,333 m teniendo por tanto 13 porticos en la nave de 28 m. Pa= 8,931 x 1,5/cos15º x 2,333 = 32,356 KN \A1;El enunciado dice que vamos a utilizar como correas perfiles IPE, como maximo de 140. En este caso utilizaremos el IPE 140 debido a que tiene mayor inercia, y también utilizaremos un acero S-275 cuya resistencia es de 27,5KN/cm2. Para hallar la longitud de las correas, dimensionamos una de ellas como una viga biapoyada. Como desconocemos la luz de la correa, pero contamos con la inercia de esta, podemos resolverla de la soguiente forma: IPE140------------> W = 77,3 cm3 Calculamos Qelu pero sin tener en cuenta el peso propio de la estructura ya que estamos calculando las correas. Qelu: (2,21 + 1)x 1,35 + 2 x1,5 +1,2 x0,7 x1,5 = 8,59 KN/m2 8,59 KN/m2 x3,105= 26,68 KN/m2 calculamos una de las correas que más carga de forjado recibe, por lo tanto es de las que recibe 1,5/cos15ºm a cada lado = TOTAL= 1,5/cos15º x2 = 3.105 Mf = Q L2/8 = 26,68 xL2)/8 = fyd \A0;\U+2265 M / W ; M= WxFyd= 77,3 cm3 x 25 KN/cm2= 1932,5= 19,325 KN/m 19,325= (26,68 xL2)/ 8; L=\A0;\U+221A 19,325 x8/ 26,68 = 2.41 m Podemos dimensionar una longitud de correa de 2.41 m pero utilizaremos 2,333m para que cuadre mejor en las cotas dadas. Podríamos colocar las cerchas en dirección Norte-Sur para ahorrarnos alguna, pero debido a la entrada de camiones situada en la cara Noerte tendremos que colocarlas en dirección Este-Oeste ya que la distancia entre los pilares que sujeten dichas cerchas no puede ser mayor de 2.41 m y la entrada necesita un paso libre de 8.5 m. Ahora comprobamos si cumple esta rigidez admisible. Sabemos que la flecha admisible es de 5ql4/384EI y que la flecha admisible de la cubierta es de L/300. Para eso utilizaremos las cargas minoradas. Qels= 2,21 + 1 + 2 + (1,2 x0,7)= 6,05KN/m2 x(3/cos15º)m= 18,785KN/m Ix - IPE 140= 541cm4 ƒ =5qL4/384EI = 5 x18,785 xL(2333,3mm)4/ 384 x210000 x5410000= 6,38mm ƒ 2333,3/300= 7,77mm 0,00777m \U+2265 0,00638 por lo tanto si CUMPLE \L RESOLVEMOS LAS REACCIONES: \l\U+2211fv=0 Ra+Rb-2Pa\C250;-\C86;7Pb\C30; \C250;Ra+Rb= \C1;2(32,356) \C250;+ \C86;7(64,713)\C250; Ra+Rb= 64,713 + 453 Ra+Rb= 517,713-------> RA=517,713 - 258,856 = 258,856 KN =RA \fCentury Gothic|b0|i0|c0|p34;\L \l\U+2211\fMS Shell Dlg|b0|i0|c0|p0;M=0 Rbx24 \C1;-32,356x24 \C86;-64,731x(21+18+15+12+9+6+3)\C250;=0 RA = RB = 258,856 KN RB= (776,544 + 5437,404) /24 RB= 258,856 KN L \A1;\LREALIZAMOS LOS CORTES:LOS CORTES: \A1;\L \A1;\L \A1;\L \U+2211Fv=0 Q + RA - Pa = 0 Q= -258,856 +32,356= -226,5 \U+2211M3=0 \A1; M+PaX -258,856X +329,312 x =0 \A1; M= \A1;226,5X--------------------------> X= 0 ; M= 0 \A1; \A1;X= 3 ; M= 679,5\A1;\L RA \U+2211Fv=0 Q + RA - Pa - Pb =0 Q = -258,856 +32,356 +64,713 = -161,78 \U+2211M5 =0 \A1; M = -32,356X -64,713(X-3) +258,856X \A1; M = -32,356X -64,713X +194,139 +258,856X = \A1;161,787X + 194,139 ----> X =3 ; M = 679,5 \A1; \A1;X =6 ; M = 1164,861\A1;\L RA X Q M X RA Q M x \U+2211Fv=0 Q + RA - Pa - 2Pb =0 Q= -97,067 \U+2211M=0 \A1; M = -32,356X +258,856X -64,713(X-3) \A1;-64,713(X-6) \A1; M= 226.5X -64,713X +194,139 -64,713X \A1;+388,278 \A1; M= 97,074X +582,417 \A1;--------------------------> X =6 ; M= 1164,861 \A1; \A1;X =9 ; M= 1456,083\A1;\L RA x DIMENSIONADO: Cordon INFERIOR: como esta traccionado no hace falta ver si cumple. N7-9 o N11-13= Mmax A\U+2265 N/Fyd A\U+2265 1071,887/25= 42,87cm2---> IPE270:(i=11,2cm4, A=45,90cm2) Cordon SUPERIOR: como el cordon superior esta comprimido tendremos que comprobar si cumple. N17-19= Mmax A\U+2265 N/Fyd A\U+2265 1442,347/25= 57,7cm2---> IPE330:(i=13,7cm4, A=62,6cm2) \U+03BB=lk/i= 310,5cmx1/13,7=22,66 \U+03BBe= 86,81 \U+03BBt=22,66/86,81=0,26---->curva C---> 0,96=X 25\U+2265 1442,347/(0,96x62,6)= 24 SI CUMPLE \LDimensionado Diagonales y Montantes: \lN1-4= -618,484 = Vmax (-) Compresion A\U+2265 N/Fyd A\U+2265 618,484/25= 24,739cm2---> IPE220:(i=9,11cm4, A=33,4cm2) \U+03BB=lk/i= 378,26cmx1/9,11=41,521 \U+03BBe= 86,81 \U+03BBt=41,521/86,81=0,47---->curva C---> 0,83=X 25\U+2265 618,484/(0,83x33,4)= 22,31 SI CUMPLE Pb = 8,931 x 3/cos15º x 2,333 = 64,713 KN Pa 5 CERCHAS (porticos) 5 5 5 5 5 5 5 CERCHAS (porticos) Pb Pa Pb Pb Pb Pb Pb Pb Pa Pb Pa Pb Pb Pb Pb Pb Pb RA RB 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Pa Pb Pa Pb Pb Pb Pb Pb Pb RA RB 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 5 5 5 5 5 5 5 5 1 2 Pa 1 2 3 4 1 2 3 4 5 6 5 CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 Pa Pa Pb Pb Pb \U+2211Fh = 0 Realizando un corte por la mitad y haciendo simetria \U+2211M10 =0 32,356 x12 \C250;+\C1; \C86;64,731 x(9+6+3) \C250;+Ha x4,7154 -Ra x12 = 0 Ha = (-388,272 -1165,158 +3106,272)/4,7154 Ha = 329,312 KN Hb = -329,312 KN Ha + Hb = 0 CERCHAS (porticos) 5 5 5 5 5 5 5 CERCHAS (porticos) 5 CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) CERCHAS (porticos) 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 Ha Hb 5 Ha Hb Q M Pa Pb Pb Pb Q M \U+2211Fv=0 Q + RA - Pa - 3Pb =0 Q= -32,354 \U+2211M=0 \A1; M = -32,356X +258,856X -64,713(X-3) \A1;-64,713(X-6) -64,713(X-9) \A1; M= 226.5X -64,713X +194,139 -64,713X \A1;+388,278 -64,713X +582,417 \A1; M= 32,361X +1164,834 \A1;--------------------------> X =9 ; M= 1456,083 \A1; \A1;X =12 ; M= 1553,166 \A1;\L -226,5 -161,78 -97,067 -32,354 226,5 161,78 97,067 32,354 \A1;Mmax=1553,166 \A1;1456,083 \A1;1164,861 \A1; 679,5 \OQ N2-4 N1-4 N1-3 \U+2211M4 =0 \A1; 32,356 x3 -258,856 x3 +N1-3 x 1,4489= 0 \A1; N1-3= +469 Tracci\A1;ó\A1;n \A1;\L \U+2211M3 =0 \A1; -N1-4 x1,1897 -N2-4 x1,4489 +32,365 x3 \A1;-258,856 x3 = 0 \A1; N2-4= -976,8 Compresi\A1;ó\A1;n \A1;\L \U+2211Fy =0 \A1; 258,856 sen75 -32,365 sen15 +N1-4 sen23 = \A1;0 \A1; N1-4= -250,035 +8,376 /sen23 \A1; N1-4= -618,477 Compresi\A1;ó\A1;n \A1; \A1;\L y x CALCULO DE LOS ESFURZOS DE LAS BARRASDE LOS ESFURZOS DE LAS BARRAS \L Para mayor facilidad en el cálculo abatimos la cercha.\A1; \A1;\L RA Pa 1 2 3 4 N4-6 N3-6 N3-5 5 6 RA Pa 1 2 3 4 Pb \U+2211Fy =0 \A1; 258,856 sen75 -32,356 sen15 -64,713 sen15 \A1;+N3-6 sen23 = 0 \A1; N3-6= -250,035 +25,123 /sen23 \A1; N3-6= -575,618 Compresi\A1;ó\A1;n \A1; \A1;\L \U+2211M6 =0 \A1; 32,356 x6 -258,856 x6 +64,713 x3 +N3-5 x \A1;1,4489= 0 \A1; N3-5= +803,925 Tracci\A1;ó\A1;n \A1;\L \U+2211M5 =0 \A1; -N3-6 x1,1897 -N4-6 x1,4489 + 32,365 x6 \A1;-258,856 x6 +64,713 x3 = 0 \A1; N4-6= -1276,568 Compresi\A1;ó\A1;n \A1;\L Pa Pb Pb RA 1 2 3 4 6 N6-8 N5-8 N5-7 7 8 5 \U+2211Fy =0 \A1; 258,856 sen75 -32,356 sen15 -64,713 sen15 \A1;-64,713 sen15 +N5-8 sen23 = 0 \A1; N5-8= -250,035 +41,871 /sen23 \A1; N5-8= -532,752 Compresi\A1;ó\A1;n \A1; \A1;\L \U+2211M8 =0 \A1; 32,365 x9 -258,856 x9 +64,713 x6 +64,713 x3 \A1;+N5-7 x 1,4489= 0 \A1; N5-7= +1004,9 Tracci\A1;ó\A1;n \A1;\L \U+2211M7 =0 \A1; -N5-8 x1,1897 -N6-8 x1,4489 + 32,365 x9 -258,856 x9 \A1;+64,713 x6 +64,713 x3 = 0 \A1; N6-8= -1442,347 Compresi\A1;ó\A1;n \A1;\L 5 RA Pa Pb Pb Pb 1 2 3 4 6 7 8 5 N8-10 N7-10 N7-9 9 10 \U+2211Fy =0 \A1; 258,856 sen75 -32,356 sen15 -64,713 sen15 \A1;-64,713 sen15 -64,713 sen15 +N7-10 sen23 = 0 \A1; N7-10= -250,035 +58,62 /sen23 \A1; N7-10= -489,889 Compresi\A1;ó\A1;n \A1; \A1;\L \U+2211M10 =0 \A1; 32,365 x12 -258,856 x12 +64,713 x9 +64,713 x6 \A1;+64,713 x3 +N7-9 x 1,4489= 0 \A1; N7-9= +1071,887 Tracci\A1;ó\A1;n \A1;\L \U+2211M9 =0 \A1; -N7-10 x1,018 -N8-10 x1,4489 + 32,365 x11,567 -258,856 x11,567 \A1;+64,713 x8,567 +64,713 x5,567 +64,713 x2,567 = 0 \A1; N8-10= -1406,416 Compresi\A1;ó\A1;n \A1;\L y x y x y x \A1; -976,8 \A1; +469 \A1; +803,925 \A1; +1004,9 \A1; +1071,887 \A1; -1406,416 \A1; -1442,347 \A1; -1276,568 \A1; -618,484 \A1; -575,618 \A1;-532,752 \A1; -489,889 \A1; -976,8 \A1; +469 \A1; -618,484 \A1; +803,925 \A1; -1276,568 \A1; -575,618 \A1; +1004,9 \A1; -1442,347 \A1;-532,752 \A1; +1071,887 \A1; -1406,416 \A1; -489,889 RA 1 3 4 Pb N1-2 N1-2= +64,713-258,856= -194,143\A1; Compresi\A1;ó\A1;n\A1;\L \A1; -471,758 \A1; -471,758 N4-6 5 6 RA Pa 1 2 4 Pb N1-3 N3-4 \U+2211M5 =0 \A1; -258,856 x6 +32,365 x6 +64,713 x3 -N4-6 \A1;x1,4489 +N3-4 x3 = 0 \A1; N3-4 = +258,856 x6 -32,365 x6 -64,713 x3 \A1;+N4-6 x1,4489 /3 \A1; N3-4= -228,27 Compresi\A1;ó\A1;n \A1;\L 3 5 6 RA Pa 1 2 4 Pb \U+2211M7 =0 \A1; -258,856 x9 +32,365 x9 +64,713 x6 +64,713 x3 \A1;-N6-8 x1,4489 +N5-6 x3 = 0 \A1; N5-6 = +258,856 x9 -32,365 x9 -64,713 x6 \A1;-64,713 x3 +N6-8 x1,4489 /3 \A1; N5-6= -211,271 Compresi\A1;ó\A1;n \A1;\L 3 N6-8 N3-5 N5-6 7 8 Pb 6 RA Pa 1 2 4 Pb 3 Pb N8-10 N5-7 N7-8 10 Pb 5 7 8 9 \U+2211M9 =0 \A1; -258,856 x11,567 +32,365 x11,567 +64,713 x8,567 +64,713 x5,567 +64,713 \A1;x2,567 -N8-10 x1,4489 +N7-8 x2,567 = 0 \A1; N7-8 = +258,856 x11,567 -32,365 x11,567 -64,713 x8,567 -64,713 x5,567 \A1;-64,713 x2,567 +N8-10 x1,4489 /2,567 \A1; N7-8= -166,235 Compresi\A1;ó\A1;n \A1;\L \A1; -228,27 \A1; -211,271 \A1; -166,235 N1-4 N1-3 y x RA Pa 1 2 3 N1-2 \U+2211M3 =0 \A1; -258,856 x3 +32,365x3 -N1-4 x1,1897 -N1-2 x3 \A1;= 0 \A1; N1-2 = -258,856 x3 +32,365x3 -N1-4 x1,1897 \A1;/3 \A1; N3-4= -471,758 Compresi\A1;ó\A1;n \A1;\L \A1; -166,235 \A1; -211,271 \A1; -228,27 6 RA Pa 1 2 4 Pb 3 Pb N8-10 10 Pb 5 7 8 9 N7-10 N9-10 \U+2211Fy =0 \A1; 258,856 sen75 -32,356 sen15 -64,713 sen15 -64,713 sen15 \A1;-64,713 sen15 +N7-10 sen23 +N9-10 sen60 = 0 \A1; N9-10= +442,05 Tracci\A1;ó\A1;n \A1; \A1;\L \A1; +442,05 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 Pa Pb Pb Pb Pb Pb Pb Pb Pa

Language	Spanish
Drawing Type	Section
Category	Construction Details & Systems
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File Type	dwg
Materials	Steel
Measurement Units
Footprint Area
Building Features
Tags	autocad, calculation, calculations, construction, details, DWG, plants, section, sections, stahlrahmen, stahlträger, steel, steel beam, steel frame, structure en acier, structures, truss