👉Beam Design — Ultimate Strength Design (ACI 318) (Metric)
💫Design Inputs (Example):
⏺Span L = 4.00 m,Beam width b = 0.300 m, Overall depth D = 0.500 m, Effective depth d = 0.452 m,Slab thickness = 0.125 m, Tributary width = 2.00 m
⏺Unit weight RCC = 25 kN/m³,
⏺Live load on slab = 3.00 kN/m²,
⏺Finishes = 0.5 kN/m
⏺Concrete f'c = 30.0 MPa,
⏺Steel fy = 500.0 MPa
⏺Load factors: w_u = 1.2D + 1.6L (ACI)
⏺Strength reduction factors: φ_flexure = 0.9,
⏺φ_shear = 0.75
Step 1: Calculate Loads (per meter length of beam)
Self weight of beam = b × D × γ = 0.300 × 0.500 × 25 = 3.750 kN/m
Slab load on beam = slab_thickness × γ × Ls = 0.125 × 25 × 2.00 = 6.250 kN/m
Live load on beam = LL_slab × Ls = 3.00 × 2.00 = 6.000 kN/m
Dead load (DL) = Self weight + Slab load +
Finishes = 3.750 + 6.250 + 0.50 = 10.500 kN/m
Factored UDL w_u = 1.2×DL + 1.6×LL = 1.2×10.500 + 1.6×6.000 = 22.200 kN/m
Step 2: Compute Factored Moment and Shear (Simply Supported Beam, UDL)
Maximum factored moment Mu = w_u L² / 8 = 22.200 × 4.00² / 8 = 44.400 kN·m
Maximum factored shear Vu = w_u L / 2 = 22.200 × 4.00 / 2 = 44.400 kN
Step 3: Required Nominal Moment Capacity
Design moment Mu = 44.400 kN·m
Required nominal moment Mn = Mu / φ = 44.400 / 0.9 = 49.333 kN·m
Step 4: Compute Required Steel Area As (Singly Reinforced Rectangular Section)
Solved for As (closed form quadratic) → Required As = 221.8 mm² (≈ 0.000222 m²)
Recommended using 20 mm bars (area per bar ≈ 314.16 mm²) → Number of 20 mm bars = 1 nos
Step 5: Shear Check (Simplified ACI Approximation)
Approximate concrete shear capacity Vc = 126.261 kN
φ·Vc = 94.696 kN
Maximum factored shear Vu = 44.400 kN
Check: Vu <= φ·Vc → OK
Step 6: Reinforcement & Detailing Notes
1. Use selected tensile reinforcement As distributed in available bars, ensure clear cover andspacing.
2. Provide shear stirrups if Vu > φ·Vc; otherwise minimum shear reinforcement per ACI.
3. Check development length, anchorage, and serviceability (deflection, crack width).
4. Always run full ACI checks (interaction, torsion if present) and use appropriate safety factors.
📜Note: This worked example is illustrative and simplified. Use full ACI 318 procedures and software for finaldesign.
#BeamLoadCalculation
#USDMethod
#ACI318Design
#StructuralEngineering
#CivilEngineeringWorld
#EngineeringDesign
#StructuralDesignTips
#ConcreteBeamDesign
#UltimateStrengthDesign
#LoadCalculationStepByStep
#BeamAnalysis
#SiteEngineerGuide
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@etconp
💫Design Inputs (Example):
⏺Span L = 4.00 m,Beam width b = 0.300 m, Overall depth D = 0.500 m, Effective depth d = 0.452 m,Slab thickness = 0.125 m, Tributary width = 2.00 m
⏺Unit weight RCC = 25 kN/m³,
⏺Live load on slab = 3.00 kN/m²,
⏺Finishes = 0.5 kN/m
⏺Concrete f'c = 30.0 MPa,
⏺Steel fy = 500.0 MPa
⏺Load factors: w_u = 1.2D + 1.6L (ACI)
⏺Strength reduction factors: φ_flexure = 0.9,
⏺φ_shear = 0.75
Step 1: Calculate Loads (per meter length of beam)
Self weight of beam = b × D × γ = 0.300 × 0.500 × 25 = 3.750 kN/m
Slab load on beam = slab_thickness × γ × Ls = 0.125 × 25 × 2.00 = 6.250 kN/m
Live load on beam = LL_slab × Ls = 3.00 × 2.00 = 6.000 kN/m
Dead load (DL) = Self weight + Slab load +
Finishes = 3.750 + 6.250 + 0.50 = 10.500 kN/m
Factored UDL w_u = 1.2×DL + 1.6×LL = 1.2×10.500 + 1.6×6.000 = 22.200 kN/m
Step 2: Compute Factored Moment and Shear (Simply Supported Beam, UDL)
Maximum factored moment Mu = w_u L² / 8 = 22.200 × 4.00² / 8 = 44.400 kN·m
Maximum factored shear Vu = w_u L / 2 = 22.200 × 4.00 / 2 = 44.400 kN
Step 3: Required Nominal Moment Capacity
Design moment Mu = 44.400 kN·m
Required nominal moment Mn = Mu / φ = 44.400 / 0.9 = 49.333 kN·m
Step 4: Compute Required Steel Area As (Singly Reinforced Rectangular Section)
Solved for As (closed form quadratic) → Required As = 221.8 mm² (≈ 0.000222 m²)
Recommended using 20 mm bars (area per bar ≈ 314.16 mm²) → Number of 20 mm bars = 1 nos
Step 5: Shear Check (Simplified ACI Approximation)
Approximate concrete shear capacity Vc = 126.261 kN
φ·Vc = 94.696 kN
Maximum factored shear Vu = 44.400 kN
Check: Vu <= φ·Vc → OK
Step 6: Reinforcement & Detailing Notes
1. Use selected tensile reinforcement As distributed in available bars, ensure clear cover andspacing.
2. Provide shear stirrups if Vu > φ·Vc; otherwise minimum shear reinforcement per ACI.
3. Check development length, anchorage, and serviceability (deflection, crack width).
4. Always run full ACI checks (interaction, torsion if present) and use appropriate safety factors.
📜Note: This worked example is illustrative and simplified. Use full ACI 318 procedures and software for finaldesign.
#BeamLoadCalculation
#USDMethod
#ACI318Design
#StructuralEngineering
#CivilEngineeringWorld
#EngineeringDesign
#StructuralDesignTips
#ConcreteBeamDesign
#UltimateStrengthDesign
#LoadCalculationStepByStep
#BeamAnalysis
#SiteEngineerGuide
#ConstructionEngineering
@etconp
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