--- title: "Stress-strain curves, design values of stresses" subject: "Design of RCC Structure" chapter: 3 key: "semester_7/design_of_rcc_structure:stress-strain-curves-design-values-of" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/stress-strain-curves-design-values-of site: BCT Engineering --- # Stress-strain curves, design values of stresses Design of RCC Structure · Chapter 3: Limit State Method of design - Asked in: 2082 Baisakh, 3 marks; 2078 Bhadra, 2+2 marks; 2067 Asar (old course), 5 marks; 2068 Baisakh (old course), 5 marks (asked 4 times) - Video: [Idealized Stress Strain Curve of Concrete | Why 0.67 times fck | IS 456-2000](https://www.youtube.com/watch?v=v21ED4zZn6w) — All About Structural Analysis and Design - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/stress-strain-curves-design-values-of - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-3 ## Question Draw idealized stress-strain curves for both steel and concrete (explain the actual and idealized stress-strain diagrams) and discuss the design values of stresses. ## Short answer Concrete idealized curve: parabola to strain 0.002 then horizontal to 0.0035, peak design stress 0.446fck. Steel (HYSD) idealized curve: linear to 0.8fy then curve to fy with 0.002 plastic strain, design stress 0.87fy. Design values: concrete 0.446fck, steel 0.87fy. ## Answer #### Concrete **Actual curve:** stress rises almost linearly for the first part, then curves, reaches a peak at strain about 0.002, then falls (descending branch) until crushing at strain about 0.0035. It has no clear yield point. **Idealised curve (IS 456, Fig. 21):** a parabola from zero to strain 0.002, then a horizontal line (constant stress) up to the ultimate strain 0.0035 in bending. The peak stress is $0.67f_{ck}$ (to allow for the difference between cube and in-situ strength). The ultimate strain is 0.0035 in flexure and 0.002 in pure axial compression. $$ \sigma_c=0.446f_{ck}\left[2\frac{\varepsilon}{0.002}-\left(\frac{\varepsilon}{0.002}\right)^2\right]\quad(\varepsilon\le0.002),\qquad \sigma_c=0.446f_{ck}\quad(0.002\le\varepsilon\le0.0035) $$ #### Steel **Actual curve:** - Mild steel (Fe250): linear to a definite yield point, a yield plateau, then strain hardening. - HYSD / TMT bars (Fe415, Fe500): no definite yield point; the yield stress is taken as the 0.2% proof stress. **Idealised curve (IS 456, Fig. 23):** Hooke's law up to $0.8f_y$ (linear, $E_s=2\times10^5$ N/mm$^2$), then non-linear up to $f_y$ with inelastic strain of 0.002. For design, stresses are divided by $\gamma_{ms}=1.15$, giving $0.87f_y$. Yield strain for design: $\varepsilon_y=\dfrac{0.87f_y}{E_s}+0.002$ (0.00380 for Fe415). ```text Concrete (IS 456 Fig. 21) Steel (Fe415, Fig. 23) stress stress 0.67fck | _______________ fy | ______ actual | / | 0.87fy| __/ ______ design (design |/ | |_/ / 0.446fck) | |/ / +------------------> +----------------> strain 0.002 0.0035 Es=2x10^5 0.87fy/Es+0.002 ``` #### Design values of stresses | Quantity | Characteristic | Partial factor | Design value | |---|---|---|---| | Concrete (stress block) | $0.67f_{ck}$ | $\gamma_{mc}=1.5$ | $0.446f_{ck}$ | | Steel | $f_y$ | $\gamma_{ms}=1.15$ | $0.87f_y$ | | Loads | $F_k$ | $\gamma_f=1.5$ | $1.5F_k$ | The 0.446 factor (= 0.67/1.5) is applied to the stress block. The equivalent rectangular block used in design has an average stress $0.36f_{ck}$ acting over depth $x_u$, with its centroid at $0.42x_u$ from the compression edge. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the idealised IS 456 stress-strain curve for concrete in bending? A. Straight line up to 0.0035 B. Parabola up to strain 0.002, then constant stress up to 0.0035 C. Parabola up to 0.0035 with no flat part D. Constant stress from zero strain to 0.002 Answer: B. The curve is parabolic to 0.002 and then horizontal to the ultimate flexural strain 0.0035. (Key point: Idealised concrete curve) 2. Where does the factor 0.446 fck in the concrete stress block come from? A. 0.67 fck divided by gamma_mc = 1.5 B. fck divided by 1.15 C. 0.36 fck times 1.239 D. 0.87 fck divided by 2 Answer: A. The 0.67 fck allows for cube versus in-situ strength, and dividing by gamma_mc = 1.5 gives 0.446 fck. (Key point: Design value of concrete stress) 3. How is yield stress defined for HYSD/TMT bars which have no definite yield point? A. Stress at 0.1% strain B. 0.2% proof stress C. Ultimate tensile stress D. Stress at 0.0035 strain Answer: B. HYSD bars show no clear yield plateau, so the 0.2% proof stress is used as fy. (Key point: HYSD steel yield) 4. How is the design stress of steel obtained? A. fy / 1.5 = 0.67 fy B. fy / 1.15 = 0.87 fy C. 0.8 fy D. fy x 1.15 Answer: B. Dividing characteristic yield stress by gamma_ms = 1.15 gives the design value 0.87 fy. 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