The core ideas are flexural design, shear design, reinforcement limits, and serviceability behavior. Flexural strength is usually estimated from an equivalent rectangular stress block, where concrete compression balances steel tension. Shear design checks whether concrete alone is enough or whether stirrups must carry the remaining shear.
Deflection, cracking, cover, spacing, and development length help ensure that a beam performs well after it is strong enough.
Key Facts
- For a singly reinforced rectangular beam, force equilibrium is T = As fy and C = 0.85 fc' b a, so a = As fy / (0.85 fc' b).
- Nominal flexural strength for a singly reinforced rectangular beam is Mn = As fy (d - a/2).
- Design flexural strength must satisfy phi Mn >= Mu, where Mu is the factored moment demand.
- For tension-controlled flexure in many ACI-style checks, phi is commonly 0.90 when tensile strain is at least 0.005.
- Reinforcement ratio is rho = As / (b d), and it must stay between minimum and maximum limits to avoid brittle or under-reinforced behavior.
- Nominal shear strength is Vn = Vc + Vs, and the design check is phi Vn >= Vu.
- For vertical stirrups, shear reinforcement strength is commonly Vs = Av fy d / s, where s is stirrup spacing.
- The approximate concrete stress block depth is a = beta1 c, where c is the neutral axis depth and beta1 depends on concrete strength.
Vocabulary
- Factored load effect
- A demand such as Mu or Vu found by applying load factors to service loads for strength design.
- Nominal strength
- The calculated member capacity before applying the strength reduction factor phi.
- Strength reduction factor
- The factor phi used to reduce nominal strength for uncertainty in materials, dimensions, and failure mode.
- Effective depth
- The distance d from the extreme compression face of the beam to the centroid of the tension reinforcement.
- Reinforcement ratio
- The ratio rho = As / (b d), which measures how much tensile steel is provided in a beam section.
- Tension-controlled section
- A flexural section in which steel yields with large tensile strain before concrete compression failure.
Common Mistakes to Avoid
- Using h instead of d in flexural formulas is wrong because the effective depth is measured to the steel centroid, not to the bottom of the beam.
- Forgetting the strength reduction factor phi is wrong because design strength is phi Mn or phi Vn, not the nominal strength alone.
- Assuming more steel is always safer is wrong because excessive reinforcement can make the beam compression-controlled and more brittle.
- Mixing service loads with factored strength checks is wrong because Mu and Vu must come from the correct factored load combinations.
- Ignoring stirrup spacing and detailing limits is wrong because shear reinforcement must be both strong enough and properly distributed along the beam.
Practice Questions
- 1 A rectangular beam has b = 300 mm, d = 500 mm, As = 1800 mm^2, fc' = 28 MPa, and fy = 420 MPa. Find a and Mn using a = As fy / (0.85 fc' b) and Mn = As fy (d - a/2).
- 2 For a beam with Mu = 220 kN m and calculated Mn = 275 kN m, check whether the beam is adequate if phi = 0.90.
- 3 A beam has Vu = 180 kN, Vc = 95 kN, phi = 0.75, Av = 200 mm^2, fy = 420 MPa, and d = 450 mm. Find the required stirrup spacing s using phi (Vc + Av fy d / s) >= Vu.
- 4 Explain why a reinforced concrete beam is usually designed so that the steel yields before the concrete crushes.