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A rectangular beam $230mm \times 550mm$ depth is subjected to a sagging bending moment of $40kNm,$ shear force of $30kN,$ and a twisting moment of $12kNm$ at a given section.

Design the reinforcement at he given section take load factor $1.5.$ Assume effective cover $50mm.$ Use $M20$ grade of concrete $F_{e415}$ grade of steel.

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Data $b=230 \\ D=550mm, \\ d_c=50mm \\ d=500mm \\ M=40kNm \\ V=30 kN \\ T=12 kNm \\ f_{ck}=20 N/mm^2\\ f_y=415 N/mm^2 \\ M_d=1.5×40=60kNm. \\ V_u=1.5×30=45kNm. \\ T_u=1.5×12=18kNm \\ Now, M_t=\dfrac {T_u×[1+D⁄b]}{1.7}= \dfrac{18×10^6×[1+550⁄230]}{1.7} \\ M_t=35.91kNm \\ M_u= M_d+M_t =60+35.91=95.91 kNm \\ Here\space M_d \gt M_t \\ \text{ Provide singly reinforced section } \\ A_{st}=(\dfrac{0.5 f_{ck} bd}{f_y} )×[1-\sqrt{1-\dfrac{4.6M_u}{f_{ck} bd^2 }}]\\ =(\dfrac{0.5× 20×230×500}{415})×[1-\sqrt{1-\dfrac{4.6×95.91×10^6}{20×230×500^2 }}]\\ Aster = 595.54 mm^2 \\ Provide \space 3-16 mm ∅=\gt Astp = 603mm^2.\\ Now,\space V_e=V_u+\dfrac{1.6T_u}b=45×10^3+\dfrac{1.6×18×10^6}{230} \\ V_e=170.21 kN\\ Z_{ue}=\dfrac{V_e}{bd}=\dfrac{170.21×10^3}{230×500}=1.48N / mm^2\lt2.8\space hence\space Safe \\ Pt = \dfrac{100×Astp}{b×d}=\dfrac{100×603}{230×500}=0.52\%\\ \text { By Interpolation } \\ 0.5 0.48 \\ 0.52 Z_{uc}=0.486 \\ 0.75 0.56 \\ V_{uc}=Z_{uc} bd=0.486×230×500=55.89kN\\ V_{u\space min}=0.4bd=0.4×230×500=46kN \\ V_e \gt V_{uc}+V_{u\space min} \\ \text { Hence design & provide shear reinforcement } \\ V_{us}=V_e-V_{uc}=170.21-55.89=114.32kN\\ \text { Assume, 10mm∅ 21 } ?? \\ a_{sv}=2×\dfrac\pi4×10^2=157mm^2 \\ Spacing \\ S_1=\dfrac {0.87f_y a_{sv} d}{V_{us}} =\dfrac {0.87×415×157×500}{114.32×10^3 }\\ S_1=248.04 225mm c/c \\ S_2=0.75d=0.75×500=375mm \\ S_3=300mm \\ Now \\ a_{sv}=\dfrac {T_u×S_4}{b_1 d_1×0.87f_y }+\dfrac {V_u×S_4}{2.5d_1×0.87f_y }$

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$$b_1=164mm$$

$157=\dfrac {18×10^6×S_4}{164×450×0.87×415}+\dfrac {45×10^3×S_4}{2.5×450×0.87×415} \\ S_4=199 mm c/c \\ Also \\ a_{sv} = (\dfrac{ Z_{ue} - Z_{uc}}{0.87 f_y }) b \times S_5 157=(\dfrac {1.48-0.486}{0.87×415})×230×S_5\\ S_5=247.9mm \\ Now \\ x_1=164+2 × 16⁄2+2×10/2=190mm \\ y_1=450+16/2+12/2+2×10/2=474mm \\ S_6=x_1=190mm \\ S_7=\dfrac{x_1+y_1}4=\dfrac{190+474}4 \\ S_7=166mm \approx 150mm c/c \\ \text {Provided 10mm∅ $2L_G$ stirrups @ 150mm c/c }$

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