0
Ultimate moment of resistance

Determine ultimate moment of resistance of doubly reinforced rectangular section of size 300 x 550 mm (effective) having tension reinforcement of 6 bars of 25 mm ${\phi}$ and compression reinforcement of 3 bars of 20 mm ${\phi}$. Use $M_{20}$ concrete, $F_{e415}$ steel and L.S.M.

Assume 50 mm effective cover to the compression reinforcement.

numericals • 2.2k  views
0
0
$\frac{dc}{d}$ 0.05 0.1 0.15 0.2
$f_{sc}$ 355.1 351.9 342.4 329.2

$b = 300 mm \\ D = 550 mm \\ d = 50 mm \\ To \ find \ M_u =?$

$Ast$ = $6-25mm \\ =6{\times}491=2946 \ mm^2$

$Asc=3-20mm{\phi}=3{\times}3.14=942mm^2 \\ f_ck=20N/mm^2 \\f_y=415N/mm^2 \\ \frac{d_c}{d}=\frac{50}{550}=0.09$

0.05 355.1
0.072 $F_{sc}$
0.1 351.9

$\therefore f_{sc}=353.69N/mm^2 \\ f_{cc}=0.446{\times}20=8.92N/mm^2$

To find Depth of actual N.A

$C_u=T_u \\ (C_{u1}+C_{u2})=(T_{u1}+T_{u2}) \\ (0.36f_ckbX_u)+(f_{sc}+f_{cc})Asc=0.87f_yAst \\ (0.36{\times}20{\times}300{\times}X_u)+(352.5-8.92){\times}942=(0.87{\times}415{\times}2946) \\ \therefore X_u=424.49 \ mm X_{u \max}=0.48d=0.48{\times}500=264 \ mm^2 \\ \therefore X_u \gt X_{u \max}$

Hence it is an over reinforced section.

Not permitted as per IS:456

Hence restrict $X_u = X_{u \max}$

$M_{u \max}=(C_u{\times}L_{a})+(C_{u2}{\times}L_{a2}) \\ X_{u \max}=[(0.36f_ckbX_{u \max}){\times}(d-0.42X_{u \max})]+\bigg[[(f_{sc}-f_{cc}){\times}Asc]{\times}(d-d_c)\bigg] \\ X_{u \max}=[(0.36{\times}20{\times}300{\times}264){\times}(550-0.42{\times}264)]+\bigg[[(352.5-8.92){\times}942]{\times}(550-50)\bigg] \\ \therefore X_{u \max}=412.24KNm$

0
Please log in to add an answer.

Continue reading

Find answer to specific questions by searching them here. It's the best way to discover useful content.

Find more