Answer :
To design a rectangular beam subjected to specified loads using C25 concrete and steel with a yield strength of 300 MPa, we need to determine the limiting depth ([tex]d_{lim}[/tex]) and the required area of steel reinforcement ([tex]A_s[/tex]). Let's walk through this step-by-step.
Step 1: Calculate the Total Factored Load
1.1 Unfactored Loads:
- Dead Load (DL) = 15 kN/m
- Live Load (LL) = 12 kN/m
1.2 Factored Loads:
Using partial safety factors:
- [tex]\gamma_{DL} = 1.35[/tex]
- [tex]\gamma_{LL} = 1.5[/tex]
Calculate the factored load per meter ([tex]w_u[/tex]):
[tex]w_u = 1.35 \times DL + 1.5 \times LL \\
w_u = 1.35 \times 15 + 1.5 \times 12 \\
w_u = 20.25 + 18 = 38.25 \text{ kN/m}[/tex]
Step 2: Determine the Maximum Bending Moment
For a simply supported beam, the maximum bending moment ([tex]M_u[/tex]) is at the center:
[tex]M_u = \frac{w_u \times L^2}{8} \\
M_u = \frac{38.25 \times 4^2}{8} \\
M_u = 76.5 \text{ kNm}[/tex]
Step 3: Calculate Limiting Depth ([tex]d_{lim}[/tex])
The limiting depth factor depends on the steel grade and concrete. Assume a beam section with a breadth [tex]b[/tex] of 230 mm (a common width for beams):
Using empirical values for C25 and [tex]f_yk = 300 \text{ MPa}[/tex], let's determine [tex]d_{lim}[/tex]. First we need the balanced moment capacity equation:
[tex]M_{bal} = R_c \cdot b \cdot d_{lim}^2[/tex]
For C25 concrete and steel yield strength [tex]f_yk[/tex] of 300 MPa, a standard value for the limiting depth can be used in calculations. Consideration of standard proportions results in:
[tex]d_{lim} \approx \frac{d}{1.2} \\
\text{Appropriate numbers can be derived through typical design codes like ACI or Eurocode. Here, we estimate based on typical cases.}[/tex]
Step 4: Calculate Required Area of Steel ([tex]A_s[/tex])
Using the moment of resistance equation:
[tex]A_s = \frac{M_u}{0.87 \times f_yk \times (d - 0.42\cdot d_{lim})} \\
d - 0.42\cdot d_{lim} \approx 0.75 \cdot d \\[/tex]
Substitute known [tex]d[/tex] once [tex]d_{lim}[/tex] is fully established:
[tex]A_s = \frac{76.5 \times 10^6}{0.87 \times 300 \times 0.75 \cdot d}[/tex]
The exact [tex]A_s[/tex] will depend on final dimensions adopted from iterative design steps, manageable only via specified code calculations confirming realistic assumptions and standard code values.
Note
This approach assumes typical sections and code provisions. Design requires consideration of minute material and geometric non-idealities and full referencing of comprehensive tables from relevant structural codes.
Conclusion:
- [tex]d_{lim}[/tex] provides a starting depth constraint.
- Solving for accurate [tex]A_s[/tex] involves detailed follow-up numerical design.
- For detailed reinforcement design, substantial reliance is on advanced numerical or software-aided calculations tailored to full realities of design codes.
