A singly reinforced rectangular beam is 450 mm (wide) and 560 mm (effective depth) with tensile reinforcement of 4 mild steel bars of 20 mm diameter. The effective cover to tensile reinforcement is 40 mm. Establish whether the section is under-reinforced or over-reinforced and calculate ultimate moment of resistance of beam. Use M20 concrete.
Given Data:
- Beam width (b): 450 mm
- Effective depth (d): 560 mm
- Diameter of tensile reinforcement bars: 20 mm
- Number of tensile bars (n): 4 bars
- Effective cover (d’): 40 mm
- Grade of concrete: M20
- Yield strength of mild steel (fy): 250 N/mm² (for mild steel)
- Characteristic strength of concrete (fck): 20 N/mm²
Step 1: Calculate Area of Tensile Reinforcement (Ast)
Area of steel (Ast)=n×π4×dbar2text{Area of steel (Ast)} = n times frac{pi}{4} times d_{bar}^2
Ast=4×π4×(20)2text{Ast} = 4 times frac{pi}{4} times (20)^2
Ast=4×314.16=1256.64 mm2text{Ast} = 4 times 314.16 = 1256.64 , text{mm}^2
Step 2: Calculate Limiting Area of Steel (Ast,lim)
The limiting area of steel (Ast,lim) is calculated using the following formula:
Ast,lim=0.36×fck×b×xu,lim/fytext{Ast,lim} = 0.36 times fck times b times x_{u,lim} / f_y
Where:
- fckfck = 20 N/mm²
- fyfy = 250 N/mm² (for mild steel)
- bb = 450 mm
- xu,limx_{u,lim} is the limiting depth of the neutral axis for under-reinforced sections.
For Fe-250 (mild steel), the limiting neutral axis depth factor for M20 concrete is 0.53d:
xu,lim=0.53×560=296.8 mmx_{u,lim} = 0.53 times 560 = 296.8 , text{mm}
Now, calculate the limiting steel area:
Ast,lim=0.36×20×450×296.8250text{Ast,lim} = frac{0.36 times 20 times 450 times 296.8}{250}
Ast,lim=3830.8 mm2text{Ast,lim} = 3830.8 , text{mm}^2
Step 3: Check Whether the Section is Under-Reinforced or Over-Reinforced
If Ast<Ast,limAst < Ast,lim, the section is under-reinforced. Otherwise, it is over-reinforced.
In this case:
- Ast=1256.64 mm2Ast = 1256.64 , text{mm}^2
- Ast,lim=3830.8 mm2Ast,lim = 3830.8 , text{mm}^2
Since Ast<Ast,limAst < Ast,lim, the section is under-reinforced.
Step 4: Calculate Ultimate Moment of Resistance (Mu)
For under-reinforced sections, the moment of resistance (Mu) can be calculated as:
Mu=0.87×fy×Ast×(d−0.42×xu)text{Mu} = 0.87 times f_y times text{Ast} times left( d – 0.42 times x_u right)
Where xux_u is the actual depth of the neutral axis:
xu=0.87×fy×Ast0.36×fck×bx_u = frac{0.87 times f_y times text{Ast}}{0.36 times fck times b}
xu=0.87×250×1256.640.36×20×450x_u = frac{0.87 times 250 times 1256.64}{0.36 times 20 times 450}
xu=1058.044/3240x_u = 1058.044 / 3240
xu=97.46 mmx_u = 97.46 , text{mm}
Now, calculate the ultimate moment:
Mu=0.87×250×1256.64×(560−0.42×97.46)text{Mu} = 0.87 times 250 times 1256.64 times (560 – 0.42 times 97.46)
Mu=0.87×250×1256.64×(560−40.93)text{Mu} = 0.87 times 250 times 1256.64 times (560 – 40.93)
Mu=273318×519.07text{Mu} = 273318 times 519.07
Mu=141.89×106 Nmmtext{Mu} = 141.89 times 10^6 , text{Nmm}
Mu=141.89 kNmtext{Mu} = 141.89 , text{kNm}
Final Answer:
- The beam section is under-reinforced.
- The ultimate moment of resistance of the beam is 141.89 kNm.
Let me know if you need further clarification or additional calculations!
