1. **Problem statement:** We need to compute the depth of the compressive stress block in a beam with a stepped cross-section. 2. **Given data:** - Concrete compressive strength, $f_c' = 20.7$ MPa - Steel yield strength, $f_y = 415$ MPa - Modulus of elasticity of steel, $E_s = 200000$ MPa - Distance from extreme compression fiber to neutral axis, $C_c = 50$ mm - Total beam height = 500 mm - Beam width varies by steps with bottom width 300 mm and three 50 mm-wide steps each 50 mm tall - 4 steel reinforcement bars with 20 mm diameter 3. **Approach:** - The depth of the compressive stress block, $a$, is found using equilibrium: concrete compression = steel tension. - Using the simplified rectangular stress block, the compressive force $C = 0.85 f_c' a b$ where $b$ is the effective width. - Tensile force in steel $T = A_s f_y$ where $A_s$ is area of steel. 4. **Calculations:** - Calculate steel area: $A_s = 4 \times \pi \times (20/2)^2 = 4 \times \pi \times 10^2 = 4 \times 314.16 = 1256.64$ mm$^2$ - Assume effective width $b = 300$ mm (bottom width for maximum compression) - Set $C = T$, so $$0.85 \times 20.7 \times a \times 300 = 1256.64 \times 415$$ - Simplify: $$0.85 \times 20.7 \times 300 \times a = 1256.64 \times 415$$ $$5278.5 \times a = 521,005.6$$ - Solve for $a$: $$a = \frac{521,005.6}{5278.5} \approx 98.7 \text{ mm}$$ 5. **Interpretation:** The depth of compressive stress block is approximately 98.7 mm, which fits within the beam height and is consistent with the given geometry. **Final answer:** The depth of the compressive stress block $a \approx 98.7$ mm.