Week 7 RC PDF

# Advantages and Drawbacks of RC - Steel - Hot rolled - Cold worked - Concrete Mix ## Stress distribution in a beam | Ecu | 0.003 | |---|---| | | | |...

# Advantages and Drawbacks of RC - Steel - Hot rolled - Cold worked - Concrete Mix ## Stress distribution in a beam | Ecu | 0.003 | |---|---| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | - **TS500** - $φ_{min} = 7-12mm$ - **S** - $S_{min} = 0.8 \frac{f_{yd}}{f_{cd}}$ - $S_{max} = 0.85p \frac{f_{yd}}{f_{cd}}$ - $≤ 0.02$ - **As** - $p = \frac{A_s}{b.w.d}$ - **Tension failure** ( $p < p_b$ ) - $ε_s ≥ ε_{syd}$ --> $σ_s = f_{yd}$ (Akma perilmesine ulaşmış.) - $A_{sfyd} = 0.85 f_{cd} a.b.w$ - $a = \frac{A_{sfyd}}{0.85f_{cd}b.w}$ - $M_r = p_b.b.w.d^2.f_{yd}(1-0.599. \frac{p.f_{yd}}{f_{cd}})$ - **Balanced failure** ( $p = p_b$ ) - $ε_s= ε_{sgd} = \frac{f_{yd}}{ε_s}$ - $p_b = 0.85 k_1 \frac{f_{cd}.600}{f_{yd}.600+f_{yd}}$ - $M_b = p_b.b.w.d^2.f_{yd}(1-0.599. \frac{p.f_{yd}}{f_{cd}})$ - **Compression Failure** ( $p > p_b$ ) - $ε_{cu} = 0.003$ iken $τ_s < f_{yd}$ durumunda. - $A_{st} = 0.85 f_{cd} b.w.a$ - $σ_s = 600 k_1 \frac{d-a}{a}$ < $f_{yd}$ (N/mm²) - ( $0.85 f_{cd} (b.w) a² + (600g) a - 600 A_{sk}.d = 0$ --> a bulunun. - $M_r = 0.85 f_{cd} b.w.a (d-0.5a)$ # Example: Calculate $M_r$ and reinforcement quantity for rectangular-section with dimensions of 25x55cm and the reinforcement ratio $p = 0.01$. - **$S420$ is used.** - **Gizelge 7.1** - **a) C16** $M_r$ = ? ($k_1$ = 0.85) - $f_{yd} = 420$ = 365.22 N/mm² - $k_1 = 1.15$ - $d = 550 - 40 = 510$ - $p_b = 0.85 f_{cd} \frac{k_1. 600}{f_{yd}.600+f_{yd}} = 0.85 \frac{16/1.5.0.85.600}{365.22.600+365.22} = 0.013/1$ - $p < p_b$ Tension failure - $M_r = p_b.b.w.d^2.f_{yd}(1-0.599. \frac{p.f_{yd}}{f_{cd}})$ = $0.01.250.510.365.22 (1-0.59.0,01.365.22) = 189,5 . 10^6 Nmm$ = $189,5 kNm //$ - **b) C20** $M_r$ = ? ($k_1$ = 0.85) - $p_b = 0.85. \frac{20/1.5.0.85.600}{365.22. 600+365.22} = 0.01611$ - $p < p_b$ Tension failure - $M_r = 0.01.250.510.365.22 (1-0.59.0,01.365.22) = 199,1. 10^6 Nmm/ = 199,1 kNm //$ - **c) C25** $M_r$ = ?( $k_1$ = 0.85) - $p_b = 0.85 \frac{25/1.5..0.85..600}{365.22.600+365.22} = 0.02$ - $p ≤ p_b$ Tension failure - $M_r = 0.01.250.510.365.22 (1-0.59.0,01.365.22) = 206,8. 10^6 Nmm/ = 206,8 kNm //$ - **C20** - $f_{cd} = 1.25$ (2.25'lik artış) - **C16** - $f_{cd} = 1.25$ (Increment in the strength of concrete), $M_r = 189,5 kNm$, $a = 1.05$ (25) - **C25** - $f_{cd} = 1.56$ (1.56), $a = 1.09$ (29) - **C16** - $M_r = 199.1 kNm$, $a = 1.09$ (29) - $M_r = 189.5 kNm$, $a = 1.05$ (25) - The strength of concrete does not much effect on the moment of resistance capacity but steel does. # Example: - C25/30 - B420C - Stirup - $φ 10/90$ - **a) Examine** the RC-section in terms of conformity to TS500 - **b) Define** the type of failure? - $f_{ck} = 25 N/mm^2$, $f_{ctk} = 1.8 N/mm^2$ - $f_{yk} = 420 N/mm^2$ - $γ_{mc} = 1.15$ - $γ_{ms} = 1.15$ - $f_{cd} = 16.7 N/mm^2$ - $f_{yd} = 365.22 N/mm^2$ - $p_b = 0.85. \frac{25/1.15.600}{ 420/1.15.600+365.22} = 0.0205$ - **Tension reinf:** 2012 + 2422 = $A_s = 986 mm²$ - **Comp. remf:** 4612 = $A_s = 452 mm²$ - **Stirrup:** $A_{sw} = 2x (40^2) = 157.1 mm^2$ - **$p = \frac{A_s}{b.w.d} = \frac{ 986}{250.470}=0.0084$** (Tension reinf ratio) - **$p' = \frac{A_s}{b.w.d} = \frac{452}{250.470} = 0.0038$** (Comp. remf ratio) - **$p_w = \frac{A_s}{b.w.s} = \frac{158}{250.470} = 0.0070$** - **Check:** - $p - p' = 0.0046 < 0,85p_b \checkmark$ - $p_tr = 0.0084 < 0.02 \checkmark$ - $p ≥ 0.8 \frac{f_{ctd}}{f_{ck}f_{yd}} \checkmark$ - $p_w ≥ 0.3 \frac{f_{ctd}}{f_{ynd}} = 0.3. \frac{1.2}{365,22} = 0,001 \checkmark$ - **Tensile Failure** - $\checkmark$ - $ \checkmark$