Construction Load – Schemes DWG Detail for AutoCAD

Construction load – Schemes – Details

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

loading scheme, pp bo do pe hº aº total wall, Wall, rb kg, loading scheme, Wall, pp bo do pe hº aº muro, mba, mbc, Hyperrestatic resolution by cross method, calculation of the perfect embedding pairs, independent section, mba, mba, mba, mba tm, mba bb mm mba tm, mba, mba total tm tm tm tm, section, mbc, mba tm, calculation of angular stiffness cte, discarding the collaboration of the plates as the same bo do cte, section, mba ej, mbc ej, section, calculation of distribution coefficients, rb tm rb, kba, balance imbalance in the knots, deseq tm tm tm eq tm tm eq tm tm, calculation of reactions by breakdown of bars, section, goes tm goes, section, rm tmtm, mmax tm, xo tm, mo izq mmax mo izq tm, mo ‘der mmax mo’ der tm change of sign tm, calculation of wide beams of the beams, bm bm lo cmx cm cm, bm reduced point load bm cm cm, bm bm lo cmx cm cm, Verification of the height of the reinforcement, beam, cm cm cm bm tm, kh efect kx, cm cm cm cm collaboration case ks, as ks tm, It is adopted with, hangers, beam, cm cm cm bm tm, kh efect kx, cm cm cm cm collaboration case ks, as ks tm, It is adopted with, hangers, verification support, cm cm cm bm tm, kh efect ks, as ks tm, of additional levanto is adopted additional, additional, support for, of cutting we begin to verify by the highest value of cut support, The maximum thickness of the cut is allowed for a reduced cutting armor. It should be checked that the toba to stirrups according to the area of ​​the stirrups is used to calculate the torsion absorbed by the stirrups as ts tst zone tst, sep. adopted, of the torsion absorber by calculation of total traction of td kg slab of the section necessary to double with support kg kg bo.z area cut in the zone the standard allows to cover the stirrups that it equals exceeds the to the stirrups adopted to support the It is not necessary to bend bars. left support, two cutting bo.z in zone the norm allows to reduce tobc to adopt stirrups for sep adopt stirrups kg as it is not necessary to bend bars support right zone of cutting bo.z enough to cover with stirrups is not nesario to bend bars

Raw text data extracted from CAD file:

ESQUEMA DE CARGA V 12 (12 / 45 ) q pp = bo x do x pe Hº Aº = 0.12 m x 0.45 m x 2400 kg/m² = 130 kg/m q muro = 0.10 m x (3.00 m - 0.45 m )x 1400 kg/m² = 357 kg/m q total V24 = 130 kg/m + 357 kg/m + 615 kg/m + 493 kg/m = 1595 kg/m 493 615 357 130 q V24 q muro q pp l = 3.93 m l = 3.93 m 1595 kg / m RB RB = \Lq x l\l = \L1595 kg / m x 3.93 m\l 2 2 RB = 3134 kg ESQUEMA DE CARGA V5 - V4 ( 12 / 35 ) 101 371 384 431 420 746 746 L5 L2 L6 L3 V24 = 3134 kg q muro q pp 3.00 m 4.46 m 2.10 m q pp = bo x do x pe Hº Aº = 0.12 m x 0.35 m x 2400 kg/m² = 101 kg/m q muro = 0.10 m x (3.00 m - 0.35 m )x 1400 kg/m² = 371 kg/m q1 = 101 kg/m + 371 kg/m + 384 kg/m + 420 kg/m = 1276 kg/m \U+2248 1.28 t/m q1 = 101 kg/m + 371 kg/m + 384 kg/m + 746 kg/m = 1602 kg/m \U+2248 1.60 t/m q1 = 101 kg/m + 371 kg/m + 431kg/m + 746 kg/m = 1649kg/m \U+2248 1.65 t/m q1 q2 q3 2.10 m 2.46 m 0.90m 3.13 t 1.28 tm 1.60 tm 1.65 tm A B C MBA MBC RESOLUCION DEL HIPERESTATICO POR METODO DE CROSS 1 - CALCULO DE LOS PARES DE EMPOTRAMIENTO PERFECTO TRAMO A B (tramos independientes ) MBA 1.28 tm MBA 0.32 tm 3.00 m a = 2.10 m 3.00 m b = 0.90 m MBA a = 2.10 m 3.00 m b = 0.90 m 3.13 t MBA = \L+ q x q x l\L² = + \L1.28 t/m x 3m t/m x 3m \L² = 1.44 tm 8 8 MBA = +(\L q x q x b\L²\L )(2 - B )² B =\L bb = \L0.40 mm = 0.30m 8 l 3.00m MBA = +(\L 0.32 x 0.32 x 0.40\L²\L )(2 - 0.30 )² = 0.09 tm 8 \W3.99960; MBA = +(\L p x a p x a )(1 - \U+03AC² ) \U+03AC =\L a a = \H1.00005x;\L2.10 MBA TOTAL \H1.5x; = 1.44 tm + 0.09 tm + 1.68 tm = 3.21 tm TRAMO B C MBC 1.65 tm 2.46 m MBA = -\L q x q x l\L² = + \L1.65 t/m x 2.46 m t/m x 2.46 m \L² = - 1.25 tm 8 8 2 - CALCULO DE LAS RIGIDECES ANGULARES (E CTE \H1.5x;) Descartando la colaboracion de las placas como V4 y V5 El mismo bo y do = J CTE TRAMO A B MBA = -\L 3 EJ3 EJ = \L 3 3 = 1 l 3.00 m MBC = -\L 3 EJ3 EJ = \L 3 3 = 1.22 l 2.46 m TRAMO B C \U+03A3MB\H1.5x; = 1 + 1.22 = 2.22 3 - CALCULO DE LOS COEFICIENTES DE DISTRIBUCION \U+03A3MA\H1.5x; = 0 0 = 2.69 t x 1.05 m + 3.13 t x 2.10 m + 1.44 t x 2.55 m - RB x 3.00m + 2.328 tm RB = 5.13 t KBA \H1.5x;= M\H0.6667x;BA\H1.5x; = \L 1 \l= 0.45 \U+03A3M\H0.6667x;B\H1.5x; 2.22 \H0.9999x; K\H0.6667x;BC \H1.5x;= M\H0.6667x;BC\H1.5x; = \L1.22 \l= 0.55 \U+03A3M\H0.6667x;B\H1.5x; 2.22 \H0.9999x; \U+03A3\H1x; = 0.45 + 0.55 = 1 4 - DESEQUILIBRIO Y EQUILIBRIO EN LOS NUDOS Deseq = 3.21 tm - 1.25 tm = + 1.95 tm Eq = -1.96 tm x 0.45 m = - 0.882 tm Eq = -1.96 tm x 0.55 m = - 1.078 tm 0.45 A B C 0.55 +3.21 -1.25 -0.882 -1.078 +2.328tm -2.328tm 5 - CALCULO DE LAS REACCIONES POR DESGLOSE DE BARRAS 2.10 m 0.90m 3.13 t TRAMO A B 2.328 tm 1.28 tm 1.60 tm 2.69 t 1.44 t 3.00 m 5.13 t 2.13 t \U+03A3MB\H1.5x; = 0 0 = VA x 3.00 m - 2.69 t x 1.95 m - 3.13 t x 0.40 m - 1.44 t x 0.45 m + 2.328 tm VA = 2.13 t TRAMO B C 2.46 m 1.65 tm 2.328 tm 4.06 t 2.03 0.95 2.48 t 1.08 t 0.95 2.03 RM = M M = \L2.328 tmtm = 0.95 t \W4.00000; l 2.46 m 2.10 m 2.46 m 0.90m 3.13 t 1.28 tm 1.60 tm 1.65 tm 2.69 t 1.44 t 4.06 t 1 2 3 4 5 6 2.13 2.13 t 5.13 2.98 8.11 t 1.08 t 10.8 +2.13 +2.98 -0.56 -3.69 -5.13 -1.08 Xo X1 q M -2.328 M1 = 0 M6 = 0 M2 = M3 = +1.65 Mmax = 1.77 tm Mmax = 0.35 tm Xo = \L2.13 t \l= 1.66 tm 1.28 m M 2-3 = 2.13 t x 2.10 m - 2.69 t x 1.05 m M 2-3 = 1.65 tm 1.28 tm Xo 1.66m 2.13 t 2.13 X1 = \L1.08 t \l= 0.65 tm 1.65 m Mo izq = Mmax V4 = 2.13 t x 1.66 m - 2.13 t x 1.66m / 2 Mo izq = 1.77 tm o Mo' der = Mmax V5 = 1.08 t x 0.65 m - 1.08 t x 0.65 m / 2 Mo' der = - 0.35 tm cambio de signo + 0.35 tm 1.65 tm 0.65m 1.08 t 1.08 t o' CALCULO DE ANCHOS COLABORANTES DE LAS VIGAS bm V4 bm = lo lo = \L0.8 x 300 cmx 300 cm = 40 cm \W4.00000; 6 6 bm reducido x carga puntual = 0.6 x bm = 0.6 x 40 cm = 24 cm bm V5 bm = lo lo = \L0.8 x 246 cmx 246 cm = 65 cm \W4.00000; 3 6 VERIFICACION DE LA ALTURA Y DIMENCIONAMIENTO DE LA ARMADURA ( H21 - ADN420 ) VIGA 4 Kh = h (cm ) h (cm ) =\L 30 cm 30 cm = 11.04 \W4.00000; \U+221A M (tm )/ bm (m ) \U+221A 1.77 tm / 0.24 m Kh efect = 10.02 Kx = 0.209 X = Kx x h = 0.209 x 30 cm = 6.27 cm < \H1.412x; d L2 = 11 cm 1º caso de colaboracion Ks = 0.45 As = Ks x M (tm )x M (tm )=\L 0.45 m x 1.77 tm0.45 m x 1.77 tm = 2.66 cm² h(m ) 0.30 m Se adopta 3 d 12 con 3.39 cm² Perchas 2 d 6 3 d 12 VIGA 5 Kh = h (cm ) h (cm ) =\L 30 cm 30 cm = 40.88 \W4.00000; \U+221A M (tm )/ bm (m ) \U+221A 0.35 tm / 0.65 m Kh efect = 22.83 Kx = 0.089 X = Kx x h = 0.089 x 30 cm = 2.67 cm < \H1.412x; d L3 = 11 cm 1º caso de colaboracion Ks = 0.43 As = Ks x M (tm )x M (tm )=\L 0.43 m x 0.35 tm0.43 m x 0.35 tm = 0.50 cm² h(m ) 0.30 m Se adopta 2 d 12 con 2.26 cm² Perchas 2 d 6 2 d 12 VERIFICACION APOYO V4 - V5 Kh = h (cm ) h (cm ) =\L 30 cm 30 cm = 46.81 \W4.00000; \U+221A M (tm )/ bm (m ) \U+221A 2.328 tm / 0.12 m Kh efect = 6.42 Ks = 0.49 As = Ks x M (tm )x M (tm )=\L 0.49 m x 2.328 tm0.49 m x 2.328 tm = 3.80 cm² h(m ) 0.30 m De V4 levanto 1 d 12 = 1.13 cm² Adicional = 3.80 cm² - 1.13 cm² = 2.67 cm² Se adopta 3 d 12 = 3.39 cm² ( adicional ) 3.00 m 2.46 m 3 d 12 2 d 12 \W4.00000; Adicional 3 Ø 12 APOYO Adicional3 d 12 1 d 12 \LVERIFICACION M DE CORTE V4-V5 COMENZAMOS A VERIFICAR POR EL MAYOR VALOR DE CORTE \l(V4 APOYO DERECHO) Qmax=5.13t =5130KG Q0 BA = Qmax - Q/m X r Rinterior=c/2 + h/2 Rinterior=25cm/2+30cm/2=27.5cm=0.275m qo BA=5130KG-1600KG/m X 0.275m=4690KG \LCALCULO DE ToBA \l ToBA=QoBA/bo z Z=0.85h=0.85cmX30cm=25.5cm ToBA=4690/12cm X 25.5=15.33 (ZONA 2 CORTE se permiteuna armadura reducida de corte .ToBA REDUCIDO=(ToBA)²/To2 limite dezona2 (15.33kg/cm²)²/18 KG/cm²=13.05 KG/cm² Debe verificarse que el ToBA 0.40 To 13.05 0.40 X 15.33=6.13 kg/cm\L²\l \LAdoptamos estribos segun zona de corte(zona 2) \lZ2sep.max 25cm Z2sep.max 25cm 0.6do = 0.6x35cm=\L21cm \l se adopta estribos 0.6/20cm \LSE CALCULA LA TORSION QUE ABSORBEN LOS ESTRIBOS \l(TST) TST=NºDE RAMAS X As X TS =2 X 0.22400KG/cm²=5.6KG/CM² TST Zona 2 TST 0.25 To 560 0.25X15.33=3.83KG/CM² 6mm=0.28cm² 9.33 12 cm B0 X SEP. ADOPTADA 12cm 20 12cm \LCALCULO DE LA TORSION A ABSORBER POR BARRAS\l \LDOBLADAS(TD) TdBA=\lToreb BA-TST)-TST 9.33 372 dBA=13.05 KG/CM²-560KG/CM²=7.45KG/CM² CALCULO DE TRACCION TOTAL T (VOLUMEN DE TENSIONES\L \lT=a x td x bo=\L40cm x 3.72 losa x 12cm\l=893 kg 2 \LCALCULO DE LA SECCION NESESARIA A DOBLAR \lAs= T = 893KG =0.26cm T5X 2 2400KG/CM² X 2 \LLEVANTO 1 0.26 CON 1.13cm² 0.26cm²\l \LVIGA 4 APOYO IZQUIERDO\l Qmax=2.13T - 21.30 KG R COL.EXT=C/3+h/2 =25KG/3+30/2=23CM \L \lQoA=QMAX-Q/M X R QoA=2130 KG-1280KG/M X 0.23M =1836 KG ToA=QoA =1830KG =6 KG/CM² 75KG/CM² Bo.Z 12CMX21.5CM ZONA 1DE CORTE En la zona 1 la norma permite cubrir los estribos que que iguala o supera los 0.4 x6kg to 0.4 x 6kg=2.4kg/cm² con los estribos adoptados para el apoyo del 6 12cm =9.33kg/cm>2.4kg/cm² no es nesesario doblar barras. *\LVIGA 5 APOYO IZQUIERDO \lQmax=2.98T=\L24.80KG \lQoBC=Qmax=QoBC=29.80KG-1650KG/m X 0.275m=2526KG ToBC=QoBC = 2526KG =8.25KG/cm \LZONA DOS CORTE \l Bo.Z 12cmX25.5cm En zona 2 la norma permite reducir To ToBC REDUCIDO=(ToBC)²=(8.25KG/CM²)²=3.79 KG/CM To2 18KG/CM² ToBC>04 X To 3.76>04X8.25=3.30KG/M² Adopto estribos para 28 sep max<25cm <06xdo=06x35cm=219 adopto estribos 6CM 20CM T est=2 x0.28x 2400kg =5.60 kg 12cm x 20cm como 5.60>3.79kg/cm² no es nesesario doblar barras *\Lviga 5 apoyo derecho \lQmax 108KG=1080KG R EXT=0.23M QoC=Q/MXR QoE=1080 KG-1650 KG/MX0.23M=700.5KG ToC=QoC =700.5KG =2.29KG/CM² ZONA 1 DEL CORTE Bo.Z 12CMX25.50 BASTA CUBRIR 0.4XTo=040X2.29KG/CM²=0.92KG/M² CON ESTRIBOS 6 20 (5.60KG/CM)²CUBRO 0.92KG/CM² NO ES NESARIO DOBLAR BARRAS

Language	Spanish
Drawing Type	Detail
Category	Construction Details & Systems
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File Type	dwg
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Tags	autocad, construction, DETAIL, details, DWG, erdbebensicher strukturen, load, schemes, seismic structures, strukturen