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Analysis: $\\$ Data: Dimensions, $f_c,k, f_y , Ast$ $\\$ Find $M_u$ [ i.e. M.R Moment of resistance]

Mumbai University > CIVIL > Sem 7 > Limit State Method

Marks: 10 M

1 Answer
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Steps:

  1. To find, the depth of Actual N.A.

    $C_u=T_u \\ 0.36f_ckbX_u=0.87{\times}f_y{\times}Ast \\ \therefore X_u=\frac{0.87f_yAst}{0.36f_ckb}$

    $X_{umax}$ = $0.53 d⟹For \ F_{e250} \\ 0.48 d⟹For \ F_{e415} \\ 0.46d⟹For \ F_{e500}$

  • When $X_u \lt X_{umax}⟹$ Under reinforced

    use

    $M_u=T_u{\times}L_a \\ M_u=0.87f_yAst{\times}(d-0.42X_u) \\ OR \\ M_u=C_u{\times}L_a \\ M_u=0.36{\times}f_ck{\times}b{\times}X_u{\times}(d- X_u)$

    • When $X_u \gt X_{umax}⟹$ over reinforced

      It is not permitted as per IS:456

      Restrict X $X_u=X_{umax}$

      use

      $M_{umax}=C_u{\times}L_c \\ M_{umax}=0.36{\times}f_ck{\times}b{\times}X_{umax}{\times}(d-0.42X_{umax})$

      Important note

    Steal grade $X_{umax}$ $M_{umax}$
    1. $F_{e250}$ $0.53 d$ $0.149f_ckbd^2$
    2. $F_{e415}$ $0.48 d$ $0.138 f_ckbd^2$
    3. $F_{e500}$ $0.46 d$ $0.133f_ckbd^2$

    1. Step

      Data:- $B.M., f_ck , f_y$

      Find: Dimension, Ast=?

      Steps:-

      1) $M_u/M_{umax}=1.5{\times}B.M \\ Assume \ \ b = 230 \ \ mm \ \ to \ \ 300 mm \\ M_{umax}=.............fckbd^2 \ \ (depend \ \ on \ \ steel \ \ grade) \\ d=?$

      2) $C_u=T_u \\ \ \ \ \ 0.36f_ckbX_u=0.87{\times}f_y{\times}Ast \\ \ \ \ \ X_u=?$

      3) $Ast=\Bigg(\frac{0.5f_ckbd}{f_y}\Bigg){\times}\Bigg(\sqrt{1-\frac{4.6M_u}{f_ckbd^2}}\Bigg) \\ Ast=?$

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