SELCCIONE PERFIL MAS LIGERODATOS

Pu 1600 Lex 11Mntx 210 Ley 11Mnty 100 Lb 11fy 50

SOLUCION

❶

ASUMIMOS BB1 1.1 Mux 231B2 1 Muy 100

ASUMIMOS m&n tablastablas m 1.45 Pu(eqv) 2134.95

n 2 Le 11

❷DATOS

ᶲcPn 1940rx 6 Ag 38ry 3 Zx 120Lb 11 Cbx 1.25Lp 14.1 Cby 1Lr 66.7 M1x/M2x -0.48

M1x/M2x -0.5❸ ECUACION DE ITERACCION

ECUACION 1

❸ HALLAMOS MOMENTOS AMPLIFICADOS

PU/ᶲcPn

𝑃𝑢/(∅𝑐𝑃𝑛)≥0.2 →𝑃𝑢/(∅𝑐𝑃𝑛)+8/9(𝑀𝑢𝑥/(∅𝑏𝑀𝑛𝑥)+𝑀𝑢𝑦/(∅𝑏𝑀𝑛𝑦))≤1𝑃𝑢/(∅𝑐𝑃𝑛)≤0.2 →𝑃𝑢/(2∅𝑐𝑃𝑛)+(𝑀𝑢𝑥/(∅𝑏𝑀𝑛𝑥)+𝑀𝑢𝑦/(∅𝑏𝑀𝑛𝑦))≤1𝑀𝑢=𝛽1∗𝑀𝑛𝑡

𝛽2=𝐶𝑚/(1−Σ𝑃𝑢/Σ𝑃𝑒)

〖𝑃𝑢〗 _𝑒𝑞𝑣=𝑃𝑢+𝑀𝑢𝑥(𝑚)+𝑀𝑢𝑦(𝑢)

𝐶𝑚=0.6−0.4(𝑀1/𝑀2)

Cmx 0.792 Pex 22471.7 B1xCmy Pey 5617.93 B1y

𝛽1( 𝑝𝑜𝑟𝑡𝑖𝑐𝑜𝑠 𝑎𝑟𝑟𝑖𝑜𝑠𝑡𝑟𝑎𝑑𝑜𝑠)𝛽2( 𝑝𝑜𝑟𝑡𝑖𝑐𝑜𝑠 𝑛𝑜 )𝑎𝑟𝑟𝑖𝑜𝑠𝑡𝑟𝑎𝑑𝑜𝑠

𝛽1=𝐶𝑚/(1−𝑃𝑢/𝑃𝑒)

𝛽2=𝐶𝑚/(1−Σ𝑃𝑢/Σ𝑃𝑒)𝑃𝑒=(𝜋^2∗29000∗𝐴𝑔)/((〖𝐾𝑙 /𝑟)〗 ^2 )

𝐶𝑚=0.6−0.4(𝑀1/𝑀2)(−𝑀1/𝑀2) 𝐶.𝑠𝑖𝑚𝑝𝑙𝑒(+ 1𝑀 / 2)𝑀 𝐶. 𝑑𝑜𝑏𝑙𝑒

SECCIONW14*159

ECUACION 10.82

𝑃𝑒=(𝜋^2∗29000∗𝐴𝑔)/((〖𝐾𝑙 /𝑟)〗 ^2 )