A tee beam section having effective depth of $450$ mm flange width of $1200$ mm widthof $300$ mm, slab thickness of $120$ mm. The tension reinforcement consist of $6$ bars of $25$ mm dia.

Determine ultimate moment of resistance of the beam section use $M20$ concrete , $F_{e415}$ steel and LSM

1 Answer

Data:- $b_f =1200 mm \\ D_f = 120 mm \\ b_w =300 mm \\ d = 450 mm \\ Ast = 6-25 mm \\ M20, F_{e415} \\ Assume X_u \lt D_f \text{[N.A lies in the flange]} \\ C_u =T_u \\ 0.36f_ckb_fX_u=0.87f_yAst \\ 0.36\times 20\times1200\times X_u =0.87 \times 415\times2945.24 \\ X_u=123.1 mm \\ X_u \gt D_f .... \text{ Assumption incorrect }\\ Case 2A : \dfrac 37 X_u \geq D_f \to \text{ N.A lies in web} \\ C_{u1}+C_{u2}=T_{u1}+T_{u2} \\ (0.36f_ckb_w X_u)+[0.446f_ck \times(b_f-b_w)\times D_f]=[0.87f_yAst]\\ 0.36\times20\times300\times X_u +[0.446\times20\times(1200-300)\times120]=[0.87\times415\times2945.24]\\ X_u=279.38 mm \space and \space X_{u\space max}=0.48d=216 mm\\ \dfrac 37 \times X_u=\dfrac 37 \times 46.43=19.89 \lt D_f\\ \text { Hence Assumption is incorrect.} \\ X_{u} \gt X_{u\space max} \to \text { Under-reinforced section } \\ \dfrac 37\times X_{u\space max}=86.4 mm \lt D_f\\ Follow Case 2B :-\\ y_f=(0.15X_{u\space max})+(0.65D_f)\\ y_f=(0.15\times X_u)+ (0.65\times120)\\ y_f=0.15 X_u +78\\ Now, \\ (C_{u1}+C_{u2})=(T_{u1}+T_{u2})\\ (0.36f_ckb_wX_u)+[0.446f_ck \times (b_f-b_w) \times y_f]=[0.87f_y Ast]\\ 0.36\times20\times300\times X_u +[0.446\times20\times(1200-300)\times0.15 \times X_u \times78]=[0.87\times415\times6\times491]\\ X_u=130.03 mm \\ \dfrac 37\times X_u=55.72mm \lt D_f \\ \text { Assumption is correct}\\ Now, X_{u\space max}=0.48d=0.48\times450=216 mm \\ X_u \lt X_{u\space max}.... \text {Hence under reinforced section }\\ M_u=(C_{u1}\times L_{a1})+ (C_{u2}\times L_{a2})\\ M_u=[(0.36f_ckb_wX_{u\space max})\times(d-0.42X_{u\space max})] + [0.446f_ck \times (b_f-b_w)\times y_f \times (d-\dfrac {y_f}2]\\ M_{u\space max}=[(0.36\times20\times300\times130.03)\times(450-0.42\times130.03)] + [0.446\times20\times(1200-300)\times(0.15\times130.03 +78)\times(450-\dfrac {0.15\times130.03+78}2]\\ M_{u\space max}=425.13 KNm$

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