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UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza M ét odo delPór t i co Equi val ent e Losa si n Vi ga Di seño de un si st em a de pi so de t i po pl aca pl ana,m edi ant e elm ét odo de Di seño elpór t i co equi val ent e. Se pr oyect a un edi f i ci o de of i ci na ut i l i zando un si st em a de pi so t i po pl aca pl ana con l a di sposi ci ón de col um nas.No se per m i t en vi gas,ábacos nicol um nas con capi t el es.Las car gas l at er al es est ar án r esi st i das porm ur os de cor t ant e,l as col um nas ser án cuadr adas de 18" x18"y l a al t ur a de pi so es 10f t ,peso de l os t abi ques no est r uct ur al es 20l b/ f t 2,sobr ecar ga de ser vi ci o 45l b/ f t 2,usando unos m at er i al es con l as si gui ent es car act er i st i cas:f y=60000psi ,f ´c=4000psi par a l osas y f ' c=6000psipar a l as col um nas. ≔ f'cc 6000 ―― lb in2 ≔ f'cl 4000 ―― lb in2 ≔ SS 45 ―― lb ft2 ≔ C1 18 in ≔ C2 18 in ≔ hpiso 10 ft ≔ Wtabiques 20 ―― lb ft2 ≔ rec ― 3 4 in ≔ ϕ 0.5 in ≔ γc 150 ―― lb ft3 1. Di seño pr el i m i narpar a " h"de l a l osa. a)Cont r olde f l echa. Par a un si st em a de pl aca pl ana,l a m i ni m a al t ur a "h" con ar m adur a gr ado 60 es: ≔ L1 20 ft ≔ L2 16 ft ≔ h = ― L1 30 8 in I nt ent ar em os con un espesorde l osa en t odos l os panel es de 8i n ≔ d = - - h rec ϕ 6.75 in Peso pr opi o de l a l osa. ≔ Wp.p = ⋅ γc h 100 ―― lb ft2 Car ga per m anent e m ayor ada ≔ WD = ⋅ 1.2 ⎛ ⎝ + Wp.p Wtabiques ⎞ ⎠ 144 ―― lb ft2 Sobr ecar ga M ayor ada ≔ WL = ⋅ 1.6 SS 72 ―― lb ft2 ≔ Wu = + WD WL 216 ―― lb ft2
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza b)Cor t ant e en una di r ecci ón l osa- Col um na. Par a elcor t ant e en una di r ecci ón consi der aruna f r anj as de 12i n de ancho a un di st anci a "d" en l a di r ecci ón m as l ar ga. ≔ la = - - ― L1 2 ― C1 2 d 8.688 ft ≔ bw 1 ft ≔ ϕ 0.75 ≔ Vu = ⋅ ⋅ Wu la 1 ft 1876.5 lb ≔ Vc = ⋅ ⋅ ⋅ ⋅ 2 ‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 bw d 10245.78 lb ≔ ϕVc = ⋅ ϕ Vc 7684.335 lb > ϕVc Vu c)Cor t ant e en dos di r ecci ones Elcor t ant e en dos di r ecci ones va hacerelr esul t ado m enorde l as ecuaci ones en l a t abl a 22. 6. 5. 2.del ACI ≔ β = ― C1 C2 1 Rel aci ón ent r e l ado l ar go y l ado cor t o. ≔ αs 40 ≔ λs = ‾‾‾‾‾‾‾‾ 2 ―――― 2 + 1 ―― ⋅ d ― 1 in 10 1.093 = λs 1 ≔ λ 1 ≔ bo = + ⋅ 2 ⎛ ⎝ + C1 d⎞ ⎠ ⋅ 2 ⎛ ⎝ + C2 d⎞ ⎠ 99 in Per i m et r o de secci ón Cr í t i ca. ≔ ϕVc ⋅ ϕ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 4 λs λ ‾‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 bo d ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎛ ⎜ ⎝ + 2 ― 4 β ⎞ ⎟ ⎠ λs λ ‾‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 bo d ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎛ ⎜ ⎜ ⎝ + 2 ―― ⋅ αs d bo ⎞ ⎟ ⎟ ⎠ λs λ ‾‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 bo d ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ≔ ϕVc = min( (ϕVc) ) 126791.523 lb ≔ Ac = - ⋅ L1 L2 ⎛ ⎝⎛ ⎝ + C1 d⎞ ⎠ ⎛ ⎝ + C2 d⎞ ⎠⎞ ⎠ 315.746 ft2 ≔ Vu = ⋅ Wu Ac 68201.156 lb
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza ≥ ϕVc Vu = Resultado “ok” 2)El em ent o delpór t i co equi val ent e Det er m i nar l os f act or es de di st r i buci on de m om ent os y l os m om ent os en l os ext r em os em pot r ados par a l os el em ent os delpór t i co equi val ent e par a anal i zar el pór t i co equi val ent e par ci al ut i l i zar em os el pr ocedi m i ent o de di st r i buci ón de m om ent os. Los f act or es de r i gi dez,l os coef i ci ent es de cont i nui dad y l os coef i ci ent es par a l os m om ent os em pot ados en l os ext r em os ( FEM ) se det er m i nan usando l as t abl as descr i t as en l a pr esent aci ón. a)Ri gi dez f l exi onalde l as vi gas pl acas en am bos ext r em os. ≔ CN1 C1 ≔ CN2 C2 ≔ r1 = ―― CN1 L1 0.075 ≔ r2 = ―― CN2 L2 0.094 Li m i t e i nf er i or ; ≔ r11 0 ≔ Knf1 4 Li m i t e super i or ; ≔ r12 0.1 = r1 0.075 ≔ Knf2 4.18 ≔ KNF = - Knf1 ―――――――― ⋅ ⎛ ⎝ - r11 r1 ⎞ ⎠ ⎛ ⎝ - Knf1 Knf2 ⎞ ⎠ ⎛ ⎝ - r11 r12 ⎞ ⎠ 4.135 Secci ón t r ansver salde l a Vi ga pl aca; ≔ Is = ――― ⋅ L2 h3 12 8192 in4 ≔ Ecs = ⋅ ⋅ 57000 ‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 3604996.533 ―― lb in2
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza ≔ Ksb = ―――― ⋅ ⋅ KNF Ecs Is L1 508814017.272 ⋅ lb in Elcoef i ci ent e de Cont i nui dad ( Cc)Par a Vi ga Pl aca. Li m i t e i nf er i or ; ≔ r11 0 = r2 0.094 ≔ Cn1 0.5 Li m i t e super i or ; ≔ r12 0.1 ≔ Cn2 0.51 ≔ CN = - Cn1 ――――――― ⋅ ⎛ ⎝ - r11 r2 ⎞ ⎠ ⎛ ⎝ - Cn1 Cn2 ⎞ ⎠ ⎛ ⎝ - r11 r12 ⎞ ⎠ 0.509 Coef i ci ent e par a l os ext r em os em pot r ados ( FEM ) Li m i t e i nf er i or ; ≔ r11 0 = r1 0.075 ≔ mn1 0.0833 Li m i t e super i or ; ≔ r12 0.1 ≔ mn2 0.0847 ≔ mN = - mn1 ――――――― ⋅ ⎛ ⎝ - r11 r1 ⎞ ⎠ ⎛ ⎝ - mn1 mn2 ⎞ ⎠ ⎛ ⎝ - r11 r12 ⎞ ⎠ 0.08435 ≔ FEM = ⋅ ⋅ ⋅ mN Wu L2 L1 2 116605.44 ⋅ lb ft b)Ri gi dez f l exi onalde l as col um nas en am bos ext r em os,Kc. Ut i l i zando l as t abl as A7, ≔ ta 4 in ≔ tb 4 in ≔ Hc = - hpiso h 112 in Al t ur a de col um na. ≔ rc = ―― hpiso Hc 1.071
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza Li m i t e super i orde ; ―― hpiso Hc ≔ r1 1.10 Li m i t e i nf er i orde ; ―― hpiso Hc ≔ r2 1.05 ≔ Kab1 5.09 ≔ Kab2 4.52 ≔ KAB = - Kab2 ――――――― ⋅ ⎛ ⎝ - r2 rc ⎞ ⎠ ⎛ ⎝ - Kab2 Kab1 ⎞ ⎠ ⎛ ⎝ - r2 r1 ⎞ ⎠ 4.764 ≔ Ec = ⋅ ⋅ 57000 ‾‾‾‾‾‾‾ 2 ⋅ f'cc ―― in2 lb ―― lb in2 4415201.015 ―― lb in2 ≔ Ic = ――― ⋅ C2 C1 3 12 8748 in4 ≔ Kc = ⋅ KAB ―― ⋅ Ec Ic hpiso 1533471847.842 ⋅ lb in c)Ri gi dez t or si onalde l os el em ent os t or si onal es Kt ≔ x 8 in ≔ y 18 in ≔ C = ⋅ ⎛ ⎜ ⎝ - 1 ⋅ 0.63 ― x y ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ⋅ x3 ― y 3 ⎞ ⎟ ⎠ 2211.84 in4 ≔ Kt = ――――― ⋅ ⋅ 9 Ecs C ⎛ ⎜ ⎜ ⎝ ⋅ L2 ⎛ ⎜ ⎝ - 1 ― C2 L2 ⎞ ⎟ ⎠ 3 ⎞ ⎟ ⎟ ⎠ 502175801.184 ⋅ lb in
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza d)Ri gi dez de l a col um na equi val ent e ≔ Kec = ―――― ⋅ ⋅ ⋅ 2 Kc 2 Kt + ⋅ 2 Kc ⋅ 2 Kt 756587176.717 ⋅ lb in Donde cor r esponde a dos el em ent os t or si onal es,uno a cada l ado de l a col um na,y Kt Kc cor r esponde a l a col um na super i or e i nf er i or ,en l a uni on de un pi so i nt er m edi o y col um nas i nt er i r or . e)Fact or es de di st r i buci ón Uni on ext er i or ; ≔ FDAB = ――― Ksb + Ksb Kec 0.402 ≔ FDDC FDAB Uni on I nt er i or ; ≔ FDBA = ――――― Ksb + + Ksb Ksb Kec 0.287 ≔ FDBC FDBA ≔ FDCB FDBA ≔ FDDC FDBA Col um na ext er i or ; ≔ FDCE = ――― ―― Kec 2 + Ksb Kec 0.299 Col um na I nt er i or ; ≔ FDCI = ――――― ―― Kec 2 + + Ksb Ksb Kec 0.213 3.Anal i si s de pór t i co par ci aldelpór t i co equi val ent e Det er m i nar l os m om ent os negat i vos y posi t i vos par a l a vi ga pl aca usando el m ét odo de di st r i buci ón de m om ent o.Debi do a que l a sobr ecar ga de ser vi ci o no es m ayorque t r es cuar t a par t es de l a car ga per m anent e de ser vi ci o,Se asum e que l os m om ent os de di seño ocur r en en t odas l as secci ones cr í t i cas consi der ado en t odas l os cl ar os l a t ot al i dad de l as car gas m ayor adas.
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza ≔ r = ――――― SS + Wp.p Wtabiques 0.375 < r ― 3 4 OK Usarelpr i m ercaso de car ga a)Car ga M ayor adas y m om ent os en l os ext r em o H Car ga M ayor ada. = Wu 216 ―― lb ft2 H M om ent os en l os ext r em os em pot r ados par a l a vi ga pl aca. = FEM 116605.44 ⋅ lb ft b)Di st r i buci on de m om ent os. H Coef i ci ent e de cont i nui dad par a l a col um na: = r1 1.1 ≔ CAB1 0.54 = rc 1.071 = r2 1.05 ≔ CAB2 0.57 ≔ CAB = - CAB1 ―――――――― ⋅ ⎛ ⎝ - r1 rc ⎞ ⎠ ⎛ ⎝ - CAB1 CAB2 ⎞ ⎠ ⎛ ⎝ - r1 r2 ⎞ ⎠ 0.55714 Not a: Los m om ent os ant i hor ar i os que act uan en l os ext r em os delel em ent o se consi der an posi t i vos.Los m om ent os delt r am o posi t i vo se det er m i nan apar t i rde l a si gui et e expr esi ón. L= I zqui er da y R= Der echa. Si endo M o elm om ent o a l a m i t ad de l a l uz par a una vi ga si m pl e. Not a;Sil os m om ent os en l os ext r em os no son i gual es,elm axi m o m om ent o delt r am o no se pr oduce en l a m i t ad de l a l uz per o su val ores si m i l aralcor r espondi ent e a l a de l a l uz. M om ent o posi t i vo en elt r am o A- B ( Ext er i or ) ≔ Mo = ―――― ⋅ ⋅ Wu L2 L1 2 8 172800 ⋅ lb ft
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza 4.M om ent os de Di seño. Se gr af i can l os m om ent os posi t i vos y negat i vos m ayor ados par a elsi st em a de l osa en l a di r ecci on de anal i si s.Los m om ent os de di seño negat i vo se t om an en l a car a de l os apoyos r ect i l i neos per o a una di st anci a no m ayorque 0. 175L1,apar t i rde l os cent r os de l os apoyos. H Car a de l os apoyos; ≔ r1 = ― C2 2 0.75 ft ≔ r2 = ⋅ 0.175 L2 2.8 ft se puede obser var que l a m i t ad de l a car a de l a col um na es m enor que l a di st anci a pr opuest a porelACI318 14 ≔ w = ⋅ Wu L2 3456 ― lb ft M om ent o Posi t i vo en elt r am o A- B,( Vano Ext er no) ; ≔ Mmitad = - Mo ⎛ ⎜ ⎝ ――――――――― + ⋅ 72260 lb ft ⋅ 133000 lb ft 2 ⎞ ⎟ ⎠ 70170 ⋅ lb ft
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza M om ent o Posi t i vo en elt r am o B- C,( Vano I nt er no) ; ≔ Mmitad = - Mo ⎛ ⎜ ⎝ ――――――――― + ⋅ 120700 lb ft ⋅ 120700 lb ft 2 ⎞ ⎟ ⎠ 52100 ⋅ lb ft Par a det er m i nar l os m om ent os en l as car as de l as col um nas sol o l e r est am os el ar ea pr opor ci onaldelcor t ant e ( M ét odo de l a sum a de ár ea)
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza 5.M om ent o Tot alport r am o. En l os si st em as de l osas com pr endi do dent r o de l as l i m i t aci ones delACI318 14,l os m om ent os r esul t ant e se pueden r educi r en una pr opor ci on t alque no sea necesar i o que l a sum at or i a num er i ca de l os m om ent os posi t i vos y elpr om edi o delm om ent o negat i vo sea m ayorque: ≔ ln = - L1 C1 18.5 ft ≔ MO = ―――― ⋅ ⋅ Wu L2 ln 2 8 147852 ⋅ lb ft Tr am os ext r em os: ≔ M = + ⋅ 70170 lb ft ――――――――― + ⋅ 55530 lb ft ⋅ 105850 lb ft 2 150860 ⋅ lb ft Tr am os I nt er no: ≔ M = + ⋅ 52100 lb ft ――――――――― + ⋅ 95050 lb ft ⋅ 95050 lb ft 2 147150 ⋅ lb ft 6.Di st r i buci ón de l os m om ent os de di seño en l as f r anj as de l a vi ga pl aca. Los m om ent os m ayor ados negat i vos y posi t i vos en l as secci ones cr í t i cas se pueden di st r i bui ra l a f r anj a de col um na y l as dos sem i f r anj as i nt er m edi as de l a vi ga pl aa de acuer do con l as pr opor ci ones especi f i cadas en l os ar t i cul os delm ét odo di r ect o delACI318 2014. = MFCNETE 55530 ⋅ lb ft = MFCPCTE 42084 ⋅ lb ft = MFCNITE 79387.5 ⋅ lb ft = MFINETE 0 ⋅ lb ft = MFIPCTE 28056 ⋅ lb ft = MFINITE 26462.5 ⋅ lb ft = MFCPTI 31278 ⋅ lb ft = MFCNTI 71287.5 ⋅ lb ft = MFIPTI 20852 ⋅ lb ft = MFINTI 23762.5 ⋅ lb ft Est os por cent aj e de coef i ci ent es de di st r i buci ón de m om ent o son de Pl acas Pl anas.
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza 7.M om ent os en l as col um nas. El m om ent o no bal anceado de l as vi gas pl aca en l os apoyos del pór t i co equi val ent e se di st r i buye a l as col um nas r eal es por enci m a y por debaj o de l a vi ga pl aca de f or m a pr opor ci onala l as r i gi deces r el at i vas de l as col um nas. 8. Ver i f i car l a r esi st enci a a l a f l exi ón de l a l osa y l a r esi st enci a alcor t e en una col um na ext er i or . a)Ar m adur a t ot alde f l exi ón r equer i da par a l a f r anj a de di seño. Det er m i narl a ar m adur a r equer i da par a elm om ent o de l a f r anj a de col um na,en un t r am o ext er i or ,en una col um na ext er i or . ≔ Mu MFCNETE ≔ AnchoFranjaColum = ― L2 2 96 in ≔ b AnchoFranjaColum ≔ ϕ 0.90 Suponi endo que l a secci ón es cont r ol ada porl a t r acci ón. ≔ Rn = ――― Mu ⋅ ⋅ ϕ b d2 169.273 ―― lb in2 ≔ fy 60000 ―― lb in2 ≔ ρ = ⋅ ⋅ 0.85 ―― f'cl fy - 1 ‾‾‾‾‾‾‾‾‾‾ 2 - 1 ――― ⋅ 2 Rn ⋅ 0.85 f'cl ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0.0029 [ [ ] ] Elár ea delacer o r equer i do es: ≔ Asreq = ⋅ ⋅ ρ b d 1.876 [ [ ] ] in2 Elár ea m í ni m a es según 8. 6. 1. 1 ACI318 19: ≔ Asmin = ⋅ ⋅ 0.0018 b h 1.382 in2 Elár ea r equer i da es m enorque elár ea m í ni m a,Porl o t ant o ut i l i zam os elAsm i n: ≔ As = max⎛ ⎝ , Asreq Asmin ⎞ ⎠ 1.876 in2 La separ aci ón m axi m a según elACI318 19 en l a secci ón 8. 7. 2. 1,es de 2h o 18" ≔ S ⋅ 2 h 18 in [ [ ] ] ≔ S = min( (S) ) 16 in Col ocam os bar r as #4 ≔ Av 0.2 in2 ≔ Nbarras = ―― As Av 9.38 ≔ Nbarras = Ceil⎛ ⎝ , Nbarras 1⎞ ⎠ 10
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza # de bar r as N°4 en base a l a separ aci ón asum i da; ≔ Sreq 10 in Por l o t ant o se adopt an 10 bar r as N°4 en l a f r anj a de col um na cer ca de l a col um na ext er i or . Ver i f i cam os sise t r at a de una secci ón cont r ol ada porl a t r acci ón ≔ a = ――――― ⋅ ⋅ Nbarras Av fy ⋅ ⋅ 0.85 f'cl b 0.368 in ≔ β1 0.85 par a concr et os de 4000psi Suponi endo que; ≔ dt d Est o es conser vador ≔ C = ― a β1 0.433 in ≔ εt = - ⋅ ―― 0.003 C ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ dt 0.003 0.0438 [ [ ] ] ≔ Ec 29000000 ―― lb in2 ≔ εy = ― fy Ec 0.002 ≔ εty = + εy 0.003 0.0051 ≔ Losa ‖ ‖ ‖ ‖ ‖ ‖ ‖ if else ≥ εt εty ‖ ‖ ← Losa “Dúctil” ‖ ‖ ← Losa “Frágil” Con l os r esul t ados ant er i or es se obser vo que l a secci ón est a cont r ol ada porl a t r acci ón. Ver i f i cam os elacer o de r ef uer zo necesar i o en l a l osa cer ca de una col um na ext er i or par a t r ansf er enci a de m om ent o ent r e l a l osa y l a col um na según ACI318 19,secci ón 8. 4. 2. 2. 1. Fr acci ón de m om ent o no bal anceado t r ansf er i do porf l exi ón; El m om ent o que se va a t r ansm i t i r por f l exi ón cer ca de una col um na ext er i or es el m om ent o negat i vo ext er i or que se det er m i no par a un vano ext er i or ut i l i zando l os coef i ci ent e
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza Fr acci ón de m om ent o Tr ansf er i do por f l exi ón ACI318 19,Secci ón 8. 4. 2. 2. 2 Apl i cando elCASO "C" ≔ b1 = + C1 ― d 2 21.375 in ≔ b2 = + C2 d 24.75 in ≔ γf = ――――― 1 + 1 ⋅ ⎛ ⎜ ⎝ ― 2 3 ⎞ ⎟ ⎠ ‾‾‾ 2 ― b1 b2 0.617 Elval orde se puede aum ent arsegún elACI318 19. γf ≔ Mγf = ⋅ γf MFCNETE 34287.367 ⋅ lb ft Not a:Suponi endo com por t am i ent o de secci ón cont r ol ada port r acci ón det er m i narelar ea de ar m adur a r equer i da par a ⋅ γf Mu Elancho de l osa ef ect i vo donde ser á t r asf er i do elesf uer zo porf l exi ón ACI318 19,Secci ón 8. 4. 2. 23 ≔ befec = + C2 ⋅ 2 ( ( ⋅ 1.5 h) ) 42 in Coef i ci ent e de f l exi ón nom i nal ; ≔ Rn = ―――― Mγf ⋅ ⋅ ϕ befec d2 238.9 ―― lb in2
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza Cuant í a de acer o r equer i do; ≔ ρ = ⋅ ⋅ 0.85 ―― f'cl fy - 1 ‾‾‾‾‾‾‾‾‾‾ 2 - 1 ――― ⋅ 2 Rn ⋅ 0.85 f'cl ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0.00413 [ [ ] ] Acer o r equer i do; ≔ Asreq = ⋅ ⋅ ρ befec d 1.172 [ [ ] ] in2 Acer o m í ni m o; ≔ Asmin = ⋅ ⋅ 0.0018 befec h 0.605 in2 Acer o a ut i l i zar ; ≔ As = max⎛ ⎝ , Asreq Asmin ⎞ ⎠ 1.172 in2 Num er o de bar r as N°4; ≔ Av 0.2 in2 ≔ Nbarras = ―― As Av 5.858 ≔ Nbarras = Ceil⎛ ⎝ , Nbarras 1⎞ ⎠ 6 Usar6 bar r as N° 4 en l a zona de t r ansf er enci a porf l exi ón. Ver i f i cam os que sise t r at a de una secci ón cont r ol ada porl a t r acci ón ≔ a = ――――― ⋅ ⋅ Nbarras Av fy ⋅ ⋅ 0.85 f'cl befec 0.504 in ≔ C = ― a β1 0.593 in ≔ d dt ≔ εt = - ⋅ ―― 0.003 C ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ dt 0.003 0.031 [ [ ] ] Est á def or m aci ón es m ayorque l a def or m aci ón l i m i t e delACI318 19 ( ) ,porl o + εt 0.003 t ant o l a secci ón est á cont r ol ada porl a t r acci ón. "USAR 6 BARRAS #4, EN EL ANCHO EFECTI VO DE TRANSM I CI ÓN DE ESFUERZO A FLEXI ÓN" Pr oveer6 bar r as #4,concent r ando sei s de l as di ez bar r as en l a zona de t r ansf er enci a de un t ot alde 10 bar r as #4 en l a f r anj a de col um na, dent r o delancho de l a l osa de 42i n en l a t r ansm i ci ón porf l exi ón,sobr e l a col um na ext er i or .
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza Det er m i narl a ar m adur a r equer i da par a l a f r anj a i nt er m edi a Com o se pudo obser varque en vano ext er i or ,elm om ent o negat i vo ext er i orl o asum e t odo l a f r anj a de col um na,ent onces no l e cor r esponde nada a l a f r anj a i nt er m edi a.Per o se debe de pr opor ci onaruna ar ea de acer o m í ni m a. ≔ bFI = - L2 b 96 in Est e es elancho de l a f r anj a l as dos f r anj as i nt er m edi as. ≔ Asmin = ⋅ ⋅ 0.0018 bFI h 1.382 in2 ≔ Nbarras = ――― Asmin Av 6.912 ≔ Nbarras = Ceil⎛ ⎝ , Nbarras 1⎞ ⎠ 7 ≔ Smax ⋅ 2 h 18 in [ [ ] ] ≔ Smax = min( (S) ) 16 in SE ADOPTAN BARRAS #4 @ 16i n EN LAS DOS FRANJAS I NTERM EDI AS Ver i f i car l a t ensi ón de cor t e com bi nada en l a car a i nt er na de l a secci on cr í t i ca de t r ansf er enci a. H Esf uer zo de cor t e m ayor ado en l a col um na ext er i or . ≔ Vu = ⋅ Wu - ⋅ L2 ⎛ ⎜ ⎝ + ― L1 2 ― C1 2 ⎞ ⎟ ⎠ ⋅ ⎛ ⎜ ⎝ + C1 ― d 2 ⎞ ⎟ ⎠ ⎛ ⎝ + C2 d⎞ ⎠ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 36358.453 [ [ ] ] lb La f r acci ón de m om ent o no bal anceado t r ansf er i do por excent r i ci dad delcor t e debe sersegún ACI318 19,8. 4. 4. 2. 2: ≔ γv = - 1 γf 0.383 Elm om ent o que va hacer t r ansf er i do por excent r i ci dad delcor t e no debe ser m enor que 0. 30M o,Par a est e ej er ci ci o se debe apl i cardi cha secci ón ya que en l a col um na ext er i or ,hay un m om ent o negat i vo de 0. 26M o,l o cuales m enordelval orant es especi f i cado porl o cualse t r abaj ar á con elque sea m ayor .según elACI318 14,en l a secci ón 8. 10. 7. 3. ≔ MMTV = ⋅ 0.3 Mo 51840 ⋅ lb ft De l a si gui ent e f i gur a,l as pr opi edades de l a secci ón cr í t i ca par a una col um na de bor de soi ci t ada a f l exi on per pendi cul aralbor de.
UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza CASO "C" ≔ Ac = ⋅ ⎛ ⎝ + ⋅ 2 b1 b2 ⎞ ⎠ d 455.625 in2 Par a f aci l i t ar un poco l os cal cul os en M at hCad,i gual am os k = J/ C,l o cualpar t e de l a t abl a ant er i oren elcaso "C" H H H H Ac = ár ea de l a secci ón t r ansver salde concr et o que r esi st e t r ansf er eenci a de cor t e, i gualalper i m et r o "bo"m ul t i pl i cado porl a al t ur a ef ect i va. J = Pr opi edad de l a secci ón cr í t i ca anál oga al m om ent o de i ner ci a pol ar de l os segm ent os que com ponene elAc. C y C'= Di st anci as desde elej e bar i cent r i co de l a secci ón cr í t i ca y per i m et r o de l a secci ón cr í t i ca en l a di r ecci on de anal i si s consi der ada. k = no sal e en ni ngun codi go sol o es una var i abl e par a f aci l i t ar elcal cul o en M at hcad k=J/ C. ≔ k = ―――――――――――― + ⋅ ⋅ ⋅ 2 b1 2 d ⎛ ⎝ + b1 ⋅ 2 b2 ⎞ ⎠ ⋅ d3 ⎛ ⎝ + ⋅ 2 b1 b2 ⎞ ⎠ ⋅ 6 b1 3570.511 in3 Elesf uer zo port r ansf er enci a se det er m i na de acuer do alACI318 19,R8. 4. 4. 2. 3 ≔ vu = + ―― Vu Ac ―――― ⋅ γv MMTV k 146.449 [ [ ] ] ―― lb in2 La t ensi on alcor t e adm i si bl e; ≔ ϕ 0.75 ≔ ϕVc = ⋅ ⋅ ⋅ ϕ 4 ‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 189.737 ―― lb in2 > ϕVc Vu ‥ Ok .

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Losa_Sin_VIga_2D_PE (1).pdf

  • 1. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza M ét odo delPór t i co Equi val ent e Losa si n Vi ga Di seño de un si st em a de pi so de t i po pl aca pl ana,m edi ant e elm ét odo de Di seño elpór t i co equi val ent e. Se pr oyect a un edi f i ci o de of i ci na ut i l i zando un si st em a de pi so t i po pl aca pl ana con l a di sposi ci ón de col um nas.No se per m i t en vi gas,ábacos nicol um nas con capi t el es.Las car gas l at er al es est ar án r esi st i das porm ur os de cor t ant e,l as col um nas ser án cuadr adas de 18" x18"y l a al t ur a de pi so es 10f t ,peso de l os t abi ques no est r uct ur al es 20l b/ f t 2,sobr ecar ga de ser vi ci o 45l b/ f t 2,usando unos m at er i al es con l as si gui ent es car act er i st i cas:f y=60000psi ,f ´c=4000psi par a l osas y f ' c=6000psipar a l as col um nas. ≔ f'cc 6000 ―― lb in2 ≔ f'cl 4000 ―― lb in2 ≔ SS 45 ―― lb ft2 ≔ C1 18 in ≔ C2 18 in ≔ hpiso 10 ft ≔ Wtabiques 20 ―― lb ft2 ≔ rec ― 3 4 in ≔ ϕ 0.5 in ≔ γc 150 ―― lb ft3 1. Di seño pr el i m i narpar a " h"de l a l osa. a)Cont r olde f l echa. Par a un si st em a de pl aca pl ana,l a m i ni m a al t ur a "h" con ar m adur a gr ado 60 es: ≔ L1 20 ft ≔ L2 16 ft ≔ h = ― L1 30 8 in I nt ent ar em os con un espesorde l osa en t odos l os panel es de 8i n ≔ d = - - h rec ϕ 6.75 in Peso pr opi o de l a l osa. ≔ Wp.p = ⋅ γc h 100 ―― lb ft2 Car ga per m anent e m ayor ada ≔ WD = ⋅ 1.2 ⎛ ⎝ + Wp.p Wtabiques ⎞ ⎠ 144 ―― lb ft2 Sobr ecar ga M ayor ada ≔ WL = ⋅ 1.6 SS 72 ―― lb ft2 ≔ Wu = + WD WL 216 ―― lb ft2
  • 2. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza b)Cor t ant e en una di r ecci ón l osa- Col um na. Par a elcor t ant e en una di r ecci ón consi der aruna f r anj as de 12i n de ancho a un di st anci a "d" en l a di r ecci ón m as l ar ga. ≔ la = - - ― L1 2 ― C1 2 d 8.688 ft ≔ bw 1 ft ≔ ϕ 0.75 ≔ Vu = ⋅ ⋅ Wu la 1 ft 1876.5 lb ≔ Vc = ⋅ ⋅ ⋅ ⋅ 2 ‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 bw d 10245.78 lb ≔ ϕVc = ⋅ ϕ Vc 7684.335 lb > ϕVc Vu c)Cor t ant e en dos di r ecci ones Elcor t ant e en dos di r ecci ones va hacerelr esul t ado m enorde l as ecuaci ones en l a t abl a 22. 6. 5. 2.del ACI ≔ β = ― C1 C2 1 Rel aci ón ent r e l ado l ar go y l ado cor t o. ≔ αs 40 ≔ λs = ‾‾‾‾‾‾‾‾ 2 ―――― 2 + 1 ―― ⋅ d ― 1 in 10 1.093 = λs 1 ≔ λ 1 ≔ bo = + ⋅ 2 ⎛ ⎝ + C1 d⎞ ⎠ ⋅ 2 ⎛ ⎝ + C2 d⎞ ⎠ 99 in Per i m et r o de secci ón Cr í t i ca. ≔ ϕVc ⋅ ϕ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 4 λs λ ‾‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 bo d ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎛ ⎜ ⎝ + 2 ― 4 β ⎞ ⎟ ⎠ λs λ ‾‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 bo d ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎛ ⎜ ⎜ ⎝ + 2 ―― ⋅ αs d bo ⎞ ⎟ ⎟ ⎠ λs λ ‾‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 bo d ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ≔ ϕVc = min( (ϕVc) ) 126791.523 lb ≔ Ac = - ⋅ L1 L2 ⎛ ⎝⎛ ⎝ + C1 d⎞ ⎠ ⎛ ⎝ + C2 d⎞ ⎠⎞ ⎠ 315.746 ft2 ≔ Vu = ⋅ Wu Ac 68201.156 lb
  • 3. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza ≥ ϕVc Vu = Resultado “ok” 2)El em ent o delpór t i co equi val ent e Det er m i nar l os f act or es de di st r i buci on de m om ent os y l os m om ent os en l os ext r em os em pot r ados par a l os el em ent os delpór t i co equi val ent e par a anal i zar el pór t i co equi val ent e par ci al ut i l i zar em os el pr ocedi m i ent o de di st r i buci ón de m om ent os. Los f act or es de r i gi dez,l os coef i ci ent es de cont i nui dad y l os coef i ci ent es par a l os m om ent os em pot ados en l os ext r em os ( FEM ) se det er m i nan usando l as t abl as descr i t as en l a pr esent aci ón. a)Ri gi dez f l exi onalde l as vi gas pl acas en am bos ext r em os. ≔ CN1 C1 ≔ CN2 C2 ≔ r1 = ―― CN1 L1 0.075 ≔ r2 = ―― CN2 L2 0.094 Li m i t e i nf er i or ; ≔ r11 0 ≔ Knf1 4 Li m i t e super i or ; ≔ r12 0.1 = r1 0.075 ≔ Knf2 4.18 ≔ KNF = - Knf1 ―――――――― ⋅ ⎛ ⎝ - r11 r1 ⎞ ⎠ ⎛ ⎝ - Knf1 Knf2 ⎞ ⎠ ⎛ ⎝ - r11 r12 ⎞ ⎠ 4.135 Secci ón t r ansver salde l a Vi ga pl aca; ≔ Is = ――― ⋅ L2 h3 12 8192 in4 ≔ Ecs = ⋅ ⋅ 57000 ‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 3604996.533 ―― lb in2
  • 4. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza ≔ Ksb = ―――― ⋅ ⋅ KNF Ecs Is L1 508814017.272 ⋅ lb in Elcoef i ci ent e de Cont i nui dad ( Cc)Par a Vi ga Pl aca. Li m i t e i nf er i or ; ≔ r11 0 = r2 0.094 ≔ Cn1 0.5 Li m i t e super i or ; ≔ r12 0.1 ≔ Cn2 0.51 ≔ CN = - Cn1 ――――――― ⋅ ⎛ ⎝ - r11 r2 ⎞ ⎠ ⎛ ⎝ - Cn1 Cn2 ⎞ ⎠ ⎛ ⎝ - r11 r12 ⎞ ⎠ 0.509 Coef i ci ent e par a l os ext r em os em pot r ados ( FEM ) Li m i t e i nf er i or ; ≔ r11 0 = r1 0.075 ≔ mn1 0.0833 Li m i t e super i or ; ≔ r12 0.1 ≔ mn2 0.0847 ≔ mN = - mn1 ――――――― ⋅ ⎛ ⎝ - r11 r1 ⎞ ⎠ ⎛ ⎝ - mn1 mn2 ⎞ ⎠ ⎛ ⎝ - r11 r12 ⎞ ⎠ 0.08435 ≔ FEM = ⋅ ⋅ ⋅ mN Wu L2 L1 2 116605.44 ⋅ lb ft b)Ri gi dez f l exi onalde l as col um nas en am bos ext r em os,Kc. Ut i l i zando l as t abl as A7, ≔ ta 4 in ≔ tb 4 in ≔ Hc = - hpiso h 112 in Al t ur a de col um na. ≔ rc = ―― hpiso Hc 1.071
  • 5. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza Li m i t e super i orde ; ―― hpiso Hc ≔ r1 1.10 Li m i t e i nf er i orde ; ―― hpiso Hc ≔ r2 1.05 ≔ Kab1 5.09 ≔ Kab2 4.52 ≔ KAB = - Kab2 ――――――― ⋅ ⎛ ⎝ - r2 rc ⎞ ⎠ ⎛ ⎝ - Kab2 Kab1 ⎞ ⎠ ⎛ ⎝ - r2 r1 ⎞ ⎠ 4.764 ≔ Ec = ⋅ ⋅ 57000 ‾‾‾‾‾‾‾ 2 ⋅ f'cc ―― in2 lb ―― lb in2 4415201.015 ―― lb in2 ≔ Ic = ――― ⋅ C2 C1 3 12 8748 in4 ≔ Kc = ⋅ KAB ―― ⋅ Ec Ic hpiso 1533471847.842 ⋅ lb in c)Ri gi dez t or si onalde l os el em ent os t or si onal es Kt ≔ x 8 in ≔ y 18 in ≔ C = ⋅ ⎛ ⎜ ⎝ - 1 ⋅ 0.63 ― x y ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ⋅ x3 ― y 3 ⎞ ⎟ ⎠ 2211.84 in4 ≔ Kt = ――――― ⋅ ⋅ 9 Ecs C ⎛ ⎜ ⎜ ⎝ ⋅ L2 ⎛ ⎜ ⎝ - 1 ― C2 L2 ⎞ ⎟ ⎠ 3 ⎞ ⎟ ⎟ ⎠ 502175801.184 ⋅ lb in
  • 6. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza d)Ri gi dez de l a col um na equi val ent e ≔ Kec = ―――― ⋅ ⋅ ⋅ 2 Kc 2 Kt + ⋅ 2 Kc ⋅ 2 Kt 756587176.717 ⋅ lb in Donde cor r esponde a dos el em ent os t or si onal es,uno a cada l ado de l a col um na,y Kt Kc cor r esponde a l a col um na super i or e i nf er i or ,en l a uni on de un pi so i nt er m edi o y col um nas i nt er i r or . e)Fact or es de di st r i buci ón Uni on ext er i or ; ≔ FDAB = ――― Ksb + Ksb Kec 0.402 ≔ FDDC FDAB Uni on I nt er i or ; ≔ FDBA = ――――― Ksb + + Ksb Ksb Kec 0.287 ≔ FDBC FDBA ≔ FDCB FDBA ≔ FDDC FDBA Col um na ext er i or ; ≔ FDCE = ――― ―― Kec 2 + Ksb Kec 0.299 Col um na I nt er i or ; ≔ FDCI = ――――― ―― Kec 2 + + Ksb Ksb Kec 0.213 3.Anal i si s de pór t i co par ci aldelpór t i co equi val ent e Det er m i nar l os m om ent os negat i vos y posi t i vos par a l a vi ga pl aca usando el m ét odo de di st r i buci ón de m om ent o.Debi do a que l a sobr ecar ga de ser vi ci o no es m ayorque t r es cuar t a par t es de l a car ga per m anent e de ser vi ci o,Se asum e que l os m om ent os de di seño ocur r en en t odas l as secci ones cr í t i cas consi der ado en t odas l os cl ar os l a t ot al i dad de l as car gas m ayor adas.
  • 7. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza ≔ r = ――――― SS + Wp.p Wtabiques 0.375 < r ― 3 4 OK Usarelpr i m ercaso de car ga a)Car ga M ayor adas y m om ent os en l os ext r em o H Car ga M ayor ada. = Wu 216 ―― lb ft2 H M om ent os en l os ext r em os em pot r ados par a l a vi ga pl aca. = FEM 116605.44 ⋅ lb ft b)Di st r i buci on de m om ent os. H Coef i ci ent e de cont i nui dad par a l a col um na: = r1 1.1 ≔ CAB1 0.54 = rc 1.071 = r2 1.05 ≔ CAB2 0.57 ≔ CAB = - CAB1 ―――――――― ⋅ ⎛ ⎝ - r1 rc ⎞ ⎠ ⎛ ⎝ - CAB1 CAB2 ⎞ ⎠ ⎛ ⎝ - r1 r2 ⎞ ⎠ 0.55714 Not a: Los m om ent os ant i hor ar i os que act uan en l os ext r em os delel em ent o se consi der an posi t i vos.Los m om ent os delt r am o posi t i vo se det er m i nan apar t i rde l a si gui et e expr esi ón. L= I zqui er da y R= Der echa. Si endo M o elm om ent o a l a m i t ad de l a l uz par a una vi ga si m pl e. Not a;Sil os m om ent os en l os ext r em os no son i gual es,elm axi m o m om ent o delt r am o no se pr oduce en l a m i t ad de l a l uz per o su val ores si m i l aralcor r espondi ent e a l a de l a l uz. M om ent o posi t i vo en elt r am o A- B ( Ext er i or ) ≔ Mo = ―――― ⋅ ⋅ Wu L2 L1 2 8 172800 ⋅ lb ft
  • 8. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza 4.M om ent os de Di seño. Se gr af i can l os m om ent os posi t i vos y negat i vos m ayor ados par a elsi st em a de l osa en l a di r ecci on de anal i si s.Los m om ent os de di seño negat i vo se t om an en l a car a de l os apoyos r ect i l i neos per o a una di st anci a no m ayorque 0. 175L1,apar t i rde l os cent r os de l os apoyos. H Car a de l os apoyos; ≔ r1 = ― C2 2 0.75 ft ≔ r2 = ⋅ 0.175 L2 2.8 ft se puede obser var que l a m i t ad de l a car a de l a col um na es m enor que l a di st anci a pr opuest a porelACI318 14 ≔ w = ⋅ Wu L2 3456 ― lb ft M om ent o Posi t i vo en elt r am o A- B,( Vano Ext er no) ; ≔ Mmitad = - Mo ⎛ ⎜ ⎝ ――――――――― + ⋅ 72260 lb ft ⋅ 133000 lb ft 2 ⎞ ⎟ ⎠ 70170 ⋅ lb ft
  • 9. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza M om ent o Posi t i vo en elt r am o B- C,( Vano I nt er no) ; ≔ Mmitad = - Mo ⎛ ⎜ ⎝ ――――――――― + ⋅ 120700 lb ft ⋅ 120700 lb ft 2 ⎞ ⎟ ⎠ 52100 ⋅ lb ft Par a det er m i nar l os m om ent os en l as car as de l as col um nas sol o l e r est am os el ar ea pr opor ci onaldelcor t ant e ( M ét odo de l a sum a de ár ea)
  • 10. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza 5.M om ent o Tot alport r am o. En l os si st em as de l osas com pr endi do dent r o de l as l i m i t aci ones delACI318 14,l os m om ent os r esul t ant e se pueden r educi r en una pr opor ci on t alque no sea necesar i o que l a sum at or i a num er i ca de l os m om ent os posi t i vos y elpr om edi o delm om ent o negat i vo sea m ayorque: ≔ ln = - L1 C1 18.5 ft ≔ MO = ―――― ⋅ ⋅ Wu L2 ln 2 8 147852 ⋅ lb ft Tr am os ext r em os: ≔ M = + ⋅ 70170 lb ft ――――――――― + ⋅ 55530 lb ft ⋅ 105850 lb ft 2 150860 ⋅ lb ft Tr am os I nt er no: ≔ M = + ⋅ 52100 lb ft ――――――――― + ⋅ 95050 lb ft ⋅ 95050 lb ft 2 147150 ⋅ lb ft 6.Di st r i buci ón de l os m om ent os de di seño en l as f r anj as de l a vi ga pl aca. Los m om ent os m ayor ados negat i vos y posi t i vos en l as secci ones cr í t i cas se pueden di st r i bui ra l a f r anj a de col um na y l as dos sem i f r anj as i nt er m edi as de l a vi ga pl aa de acuer do con l as pr opor ci ones especi f i cadas en l os ar t i cul os delm ét odo di r ect o delACI318 2014. = MFCNETE 55530 ⋅ lb ft = MFCPCTE 42084 ⋅ lb ft = MFCNITE 79387.5 ⋅ lb ft = MFINETE 0 ⋅ lb ft = MFIPCTE 28056 ⋅ lb ft = MFINITE 26462.5 ⋅ lb ft = MFCPTI 31278 ⋅ lb ft = MFCNTI 71287.5 ⋅ lb ft = MFIPTI 20852 ⋅ lb ft = MFINTI 23762.5 ⋅ lb ft Est os por cent aj e de coef i ci ent es de di st r i buci ón de m om ent o son de Pl acas Pl anas.
  • 11. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza 7.M om ent os en l as col um nas. El m om ent o no bal anceado de l as vi gas pl aca en l os apoyos del pór t i co equi val ent e se di st r i buye a l as col um nas r eal es por enci m a y por debaj o de l a vi ga pl aca de f or m a pr opor ci onala l as r i gi deces r el at i vas de l as col um nas. 8. Ver i f i car l a r esi st enci a a l a f l exi ón de l a l osa y l a r esi st enci a alcor t e en una col um na ext er i or . a)Ar m adur a t ot alde f l exi ón r equer i da par a l a f r anj a de di seño. Det er m i narl a ar m adur a r equer i da par a elm om ent o de l a f r anj a de col um na,en un t r am o ext er i or ,en una col um na ext er i or . ≔ Mu MFCNETE ≔ AnchoFranjaColum = ― L2 2 96 in ≔ b AnchoFranjaColum ≔ ϕ 0.90 Suponi endo que l a secci ón es cont r ol ada porl a t r acci ón. ≔ Rn = ――― Mu ⋅ ⋅ ϕ b d2 169.273 ―― lb in2 ≔ fy 60000 ―― lb in2 ≔ ρ = ⋅ ⋅ 0.85 ―― f'cl fy - 1 ‾‾‾‾‾‾‾‾‾‾ 2 - 1 ――― ⋅ 2 Rn ⋅ 0.85 f'cl ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0.0029 [ [ ] ] Elár ea delacer o r equer i do es: ≔ Asreq = ⋅ ⋅ ρ b d 1.876 [ [ ] ] in2 Elár ea m í ni m a es según 8. 6. 1. 1 ACI318 19: ≔ Asmin = ⋅ ⋅ 0.0018 b h 1.382 in2 Elár ea r equer i da es m enorque elár ea m í ni m a,Porl o t ant o ut i l i zam os elAsm i n: ≔ As = max⎛ ⎝ , Asreq Asmin ⎞ ⎠ 1.876 in2 La separ aci ón m axi m a según elACI318 19 en l a secci ón 8. 7. 2. 1,es de 2h o 18" ≔ S ⋅ 2 h 18 in [ [ ] ] ≔ S = min( (S) ) 16 in Col ocam os bar r as #4 ≔ Av 0.2 in2 ≔ Nbarras = ―― As Av 9.38 ≔ Nbarras = Ceil⎛ ⎝ , Nbarras 1⎞ ⎠ 10
  • 12. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza # de bar r as N°4 en base a l a separ aci ón asum i da; ≔ Sreq 10 in Por l o t ant o se adopt an 10 bar r as N°4 en l a f r anj a de col um na cer ca de l a col um na ext er i or . Ver i f i cam os sise t r at a de una secci ón cont r ol ada porl a t r acci ón ≔ a = ――――― ⋅ ⋅ Nbarras Av fy ⋅ ⋅ 0.85 f'cl b 0.368 in ≔ β1 0.85 par a concr et os de 4000psi Suponi endo que; ≔ dt d Est o es conser vador ≔ C = ― a β1 0.433 in ≔ εt = - ⋅ ―― 0.003 C ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ dt 0.003 0.0438 [ [ ] ] ≔ Ec 29000000 ―― lb in2 ≔ εy = ― fy Ec 0.002 ≔ εty = + εy 0.003 0.0051 ≔ Losa ‖ ‖ ‖ ‖ ‖ ‖ ‖ if else ≥ εt εty ‖ ‖ ← Losa “Dúctil” ‖ ‖ ← Losa “Frágil” Con l os r esul t ados ant er i or es se obser vo que l a secci ón est a cont r ol ada porl a t r acci ón. Ver i f i cam os elacer o de r ef uer zo necesar i o en l a l osa cer ca de una col um na ext er i or par a t r ansf er enci a de m om ent o ent r e l a l osa y l a col um na según ACI318 19,secci ón 8. 4. 2. 2. 1. Fr acci ón de m om ent o no bal anceado t r ansf er i do porf l exi ón; El m om ent o que se va a t r ansm i t i r por f l exi ón cer ca de una col um na ext er i or es el m om ent o negat i vo ext er i or que se det er m i no par a un vano ext er i or ut i l i zando l os coef i ci ent e
  • 13. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza Fr acci ón de m om ent o Tr ansf er i do por f l exi ón ACI318 19,Secci ón 8. 4. 2. 2. 2 Apl i cando elCASO "C" ≔ b1 = + C1 ― d 2 21.375 in ≔ b2 = + C2 d 24.75 in ≔ γf = ――――― 1 + 1 ⋅ ⎛ ⎜ ⎝ ― 2 3 ⎞ ⎟ ⎠ ‾‾‾ 2 ― b1 b2 0.617 Elval orde se puede aum ent arsegún elACI318 19. γf ≔ Mγf = ⋅ γf MFCNETE 34287.367 ⋅ lb ft Not a:Suponi endo com por t am i ent o de secci ón cont r ol ada port r acci ón det er m i narelar ea de ar m adur a r equer i da par a ⋅ γf Mu Elancho de l osa ef ect i vo donde ser á t r asf er i do elesf uer zo porf l exi ón ACI318 19,Secci ón 8. 4. 2. 23 ≔ befec = + C2 ⋅ 2 ( ( ⋅ 1.5 h) ) 42 in Coef i ci ent e de f l exi ón nom i nal ; ≔ Rn = ―――― Mγf ⋅ ⋅ ϕ befec d2 238.9 ―― lb in2
  • 14. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza Cuant í a de acer o r equer i do; ≔ ρ = ⋅ ⋅ 0.85 ―― f'cl fy - 1 ‾‾‾‾‾‾‾‾‾‾ 2 - 1 ――― ⋅ 2 Rn ⋅ 0.85 f'cl ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0.00413 [ [ ] ] Acer o r equer i do; ≔ Asreq = ⋅ ⋅ ρ befec d 1.172 [ [ ] ] in2 Acer o m í ni m o; ≔ Asmin = ⋅ ⋅ 0.0018 befec h 0.605 in2 Acer o a ut i l i zar ; ≔ As = max⎛ ⎝ , Asreq Asmin ⎞ ⎠ 1.172 in2 Num er o de bar r as N°4; ≔ Av 0.2 in2 ≔ Nbarras = ―― As Av 5.858 ≔ Nbarras = Ceil⎛ ⎝ , Nbarras 1⎞ ⎠ 6 Usar6 bar r as N° 4 en l a zona de t r ansf er enci a porf l exi ón. Ver i f i cam os que sise t r at a de una secci ón cont r ol ada porl a t r acci ón ≔ a = ――――― ⋅ ⋅ Nbarras Av fy ⋅ ⋅ 0.85 f'cl befec 0.504 in ≔ C = ― a β1 0.593 in ≔ d dt ≔ εt = - ⋅ ―― 0.003 C ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ dt 0.003 0.031 [ [ ] ] Est á def or m aci ón es m ayorque l a def or m aci ón l i m i t e delACI318 19 ( ) ,porl o + εt 0.003 t ant o l a secci ón est á cont r ol ada porl a t r acci ón. "USAR 6 BARRAS #4, EN EL ANCHO EFECTI VO DE TRANSM I CI ÓN DE ESFUERZO A FLEXI ÓN" Pr oveer6 bar r as #4,concent r ando sei s de l as di ez bar r as en l a zona de t r ansf er enci a de un t ot alde 10 bar r as #4 en l a f r anj a de col um na, dent r o delancho de l a l osa de 42i n en l a t r ansm i ci ón porf l exi ón,sobr e l a col um na ext er i or .
  • 15. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza Det er m i narl a ar m adur a r equer i da par a l a f r anj a i nt er m edi a Com o se pudo obser varque en vano ext er i or ,elm om ent o negat i vo ext er i orl o asum e t odo l a f r anj a de col um na,ent onces no l e cor r esponde nada a l a f r anj a i nt er m edi a.Per o se debe de pr opor ci onaruna ar ea de acer o m í ni m a. ≔ bFI = - L2 b 96 in Est e es elancho de l a f r anj a l as dos f r anj as i nt er m edi as. ≔ Asmin = ⋅ ⋅ 0.0018 bFI h 1.382 in2 ≔ Nbarras = ――― Asmin Av 6.912 ≔ Nbarras = Ceil⎛ ⎝ , Nbarras 1⎞ ⎠ 7 ≔ Smax ⋅ 2 h 18 in [ [ ] ] ≔ Smax = min( (S) ) 16 in SE ADOPTAN BARRAS #4 @ 16i n EN LAS DOS FRANJAS I NTERM EDI AS Ver i f i car l a t ensi ón de cor t e com bi nada en l a car a i nt er na de l a secci on cr í t i ca de t r ansf er enci a. H Esf uer zo de cor t e m ayor ado en l a col um na ext er i or . ≔ Vu = ⋅ Wu - ⋅ L2 ⎛ ⎜ ⎝ + ― L1 2 ― C1 2 ⎞ ⎟ ⎠ ⋅ ⎛ ⎜ ⎝ + C1 ― d 2 ⎞ ⎟ ⎠ ⎛ ⎝ + C2 d⎞ ⎠ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 36358.453 [ [ ] ] lb La f r acci ón de m om ent o no bal anceado t r ansf er i do por excent r i ci dad delcor t e debe sersegún ACI318 19,8. 4. 4. 2. 2: ≔ γv = - 1 γf 0.383 Elm om ent o que va hacer t r ansf er i do por excent r i ci dad delcor t e no debe ser m enor que 0. 30M o,Par a est e ej er ci ci o se debe apl i cardi cha secci ón ya que en l a col um na ext er i or ,hay un m om ent o negat i vo de 0. 26M o,l o cuales m enordelval orant es especi f i cado porl o cualse t r abaj ar á con elque sea m ayor .según elACI318 14,en l a secci ón 8. 10. 7. 3. ≔ MMTV = ⋅ 0.3 Mo 51840 ⋅ lb ft De l a si gui ent e f i gur a,l as pr opi edades de l a secci ón cr í t i ca par a una col um na de bor de soi ci t ada a f l exi on per pendi cul aralbor de.
  • 16. UNI VERSI DAD NACI ONAL DE I NGENI ERI A Facul t ad de Tecnol ogí a de l a Const r ucci ón Depar t am ent o de Est r uct ur a I ng.Juan Car l os M endoza CASO "C" ≔ Ac = ⋅ ⎛ ⎝ + ⋅ 2 b1 b2 ⎞ ⎠ d 455.625 in2 Par a f aci l i t ar un poco l os cal cul os en M at hCad,i gual am os k = J/ C,l o cualpar t e de l a t abl a ant er i oren elcaso "C" H H H H Ac = ár ea de l a secci ón t r ansver salde concr et o que r esi st e t r ansf er eenci a de cor t e, i gualalper i m et r o "bo"m ul t i pl i cado porl a al t ur a ef ect i va. J = Pr opi edad de l a secci ón cr í t i ca anál oga al m om ent o de i ner ci a pol ar de l os segm ent os que com ponene elAc. C y C'= Di st anci as desde elej e bar i cent r i co de l a secci ón cr í t i ca y per i m et r o de l a secci ón cr í t i ca en l a di r ecci on de anal i si s consi der ada. k = no sal e en ni ngun codi go sol o es una var i abl e par a f aci l i t ar elcal cul o en M at hcad k=J/ C. ≔ k = ―――――――――――― + ⋅ ⋅ ⋅ 2 b1 2 d ⎛ ⎝ + b1 ⋅ 2 b2 ⎞ ⎠ ⋅ d3 ⎛ ⎝ + ⋅ 2 b1 b2 ⎞ ⎠ ⋅ 6 b1 3570.511 in3 Elesf uer zo port r ansf er enci a se det er m i na de acuer do alACI318 19,R8. 4. 4. 2. 3 ≔ vu = + ―― Vu Ac ―――― ⋅ γv MMTV k 146.449 [ [ ] ] ―― lb in2 La t ensi on alcor t e adm i si bl e; ≔ ϕ 0.75 ≔ ϕVc = ⋅ ⋅ ⋅ ϕ 4 ‾‾‾‾‾‾‾ 2 ⋅ f'cl ―― in2 lb ―― lb in2 189.737 ―― lb in2 > ϕVc Vu ‥ Ok .