Project Name: Proposed Five Storey Annex Design by: ELL
Owner: BFAR Mark: 2DB-2
Location: BFAR Complex, Vasra, Diliman, Quezon City Date: 05/07/2025
Check by: Erwin Lipardo
DESIGN OF CONCRETE BEAM Structural Excell Program
Material Properties: Reinforcement Details Left Mid Right
Concrete f'c = 30 MPa Main bar Ø = 16 mm
Reinforcing fy = 414 MPa Shear bar Ø = 10 mm
Stirrup fy = 276 MPa Torsion bar Ø = 12 mm
Section Properties Span type = One end continuous
Length = 6.00 m Width = 0.30 m Height = 0.50 m
Stress Result (Any FEA or Frame Analysis)
Bending Moment Left Section Mid Section Right Section
Top Location 161.91 kN-m 50.17 kN-m 116.29 kN-m
Bottom Location 10.78 kN-m 83.65 kN-m 51.04 kN-m
Top Rebar = 6 pcs Top Rebar = 2 pcs Top Rebar = 6 pcs
Shear Stress 134.88 kN 0.00 kN 135.29 kN
Ratio = 0.91 Ratio = 0.74 Ratio = 0.65
Torsion Moment 4.76 kN-m 0.00 kN-m 4.76 kN-m
Bottom Rebar = 3 pcs Bottom Rebar = 3 pcs Bottom Rebar = 3 pcs
Service Moment 107.27 kN-m 62.22 kN-m 86.44 kN-m
Ratio = 0.1 Ratio = 0.86 Ratio = 0.49
Check Dimensional Limits, ACI 318M-14 Special moment frames ACI 18.6.3
Minimum beam depth ( ACI Table 9.3.1.1 ) a.) 1. Least 2 pcs at both top and bot ACI 18.6.3.1
a.) One end continuous, Depth = L/18.5 = 0.32 Depth use is ok 2. Min bar, [(0.25*√fc)*bd/fy = 3 pcs, 1.4*bd/fy = 3pcs] ACI 9.6.1.2
Special moment frames ( ACI Sec 18.6.2.1) 3. Max bar, 0.025*bd = 17 pcs
a.) Minimum Clear span, 4 × Depth = 1.76m , Depth use is ok b.) Positive moment at joint face shall be at least one half the negative moment ... ACI 18.6.3.2
b.) Minimum Width, Min(0.20m, 0.30×H) = 0.20m , Width use is ok Remark
SUMMARY CALCULATIONS
Design for Flexural Reinforcement
Top Moment Left Mid Right Bot Moment Left Mid Right
Mu (kN.m) = 161.91 50.17 116.29 Mu (kN.m) = 10.78 83.65 51.04
β1 value = 0.84 0.84 0.84 β1 value = 0.84 0.84 0.84 Table 22.2.2.4.3
Remarks = Doubly! Doubly! Doubly! Remarks = Doubly! Doubly! Doubly!
ρmax = 0.01582 0.00478 0.01582 ρmax = 0.00744 0.00687 0.00744 Section 9.3.3.1
ρmin = 0.01294 0.00338 0.00929 ρmin = 0.00338 0.00535 0.00338 Section 9.6.1.2
ρused = 0.01003 0.00303 0.01003 ρused = 0.00455 0.00455 0.00455 Section 21.2.2
Reduction Fac = 0.9 0.9 0.9 Reduction Fac = 0.9 0.9 0.9 Table 9.2.1
Neutral Axis = 68.8 44 68.8 Neutral Axis = 60.3 47.4 60.3 Section 22.2.2.4.2
a= 57.5 36.77 57.5 a= 50.39 39.61 50.39 Section 22.2.2.4.1
Tensile Strain = 0.0163 0.0271 0.0163 Tensile Strain = 0.019 0.025 0.019 Section 21.2.2.1
ΦMn (kN.m) = 178.19 67.58 178.19 ΦMn (kN.m) = 103.43 96.75 103.43
Design for Stirrup Reinforcement Design for Combined Shear and Torsion Reinforcement
Determine shear strength provided by the concrete Compute the Cracking Torque
φVc= 94.28kN, Vs= 371.56kN, Vmax= 465.84kN 22.5.5.1 & 2 , Eq 9.5.3.1 Tu= min of (Tcr, Tu) = min of (19.26 kNm, 4.76 kNm) = 4.76 kNm Sec 22.7.5.1a
Shear Force Distance @50mm from support = 131.91 kN Compute the Thereshold Torsion
● Since Vu = 131.91 kN < Vmax = 465.84 kN, Shear Ratio = 0.283 Ts = Tcr/4 = φ × 0.083√fc (Acp²/Pcp) =4.82 kNm Sec 22.7.4.1a
The Beam Section is Adequate Since Tu = 4.76 kNm < Ts = 4.82 kNm, Torsion Effect can be neglected.
Compute the Minimum Shear Effective Area Sec 22.5.8
Av,used/S = Max {Av,req/S, Av,min/S (a) , Av,min/S (b) } = 0.40 mm²/m
Compute for the Stirrup Spacing
Spacing used = Min ((Av,prov / Av,min/S), MaxSpacing) = 100.00 mm
Shear Force Distance @2×Height from support = 87.94 kN
● Since Vu = 87.94 kN < Vmax = 465.84 kN, Shear Ratio = 0.19
The Beam Section is Adequate
Compute the Minimum Shear Effective Area, Vu < 0.5φVc Sec 22.5.8
Av,used/S = Max {Av,min/S (a) , Av,min/S (b) } = 0.38 mm²/m
Compute for the Stirrup Spacing
Spacing used = Min ((Av,prov / Av,min/S), MaxSpacing) = 150.00mm
Checking for Crack Width and Longterm Deflection
Exposure Condition = Dry Air or with Protective Membrane/ Coating
Check Crack Width Left Mid Right
Mcr = 42.45 kNm 42.45 kNm 42.45 kNm ACI 318M-19 Eq. 24.2.3.5b
Remark = Beam is Cracking Beam is Cracking Beam is Cracking
Crack Width, Remark = 0.261mm, Crack Width is ok! 0.284mm, Crack Width is ok! 0.210mm, Crack Width is ok! ACI 224R Table 4.1
Diameter Stress = Since fs = 232.16 MPa < 280.00 MPa, ok!Since fs = 252.35 MPa < 280.00 MPa, ok!Since fs = 187.08 MPa < 280.00 MPa, ok! ACI 224R Table 4.2
Spacing Stress = Since S = 50.00mm < 200.00mm, ok! Since S = 100.00mm < 200.00mm, ok! Since S = 50.00mm < 250.00mm, ok! ACI 224R Table 4.3
Check Longterm Deflection
Eff Moment of Inertia = 1,113.13 x10⁶ mm⁴ 780.67 x10⁶ mm⁴ 1,142.85 x10⁶ mm⁴ ACI 318M-19 Table 24.2.3.5
λ = ξ / ( 1 + 50ρ' ) = 1.67 1.76 1.67 ACI 318M-19 Eq. 24.2.4.1.1
25.00mm 25.00mm 25.00mm ACI 318M-19 Table 24.2.2.0
Δall = L / 240 =
ACI 318M-19 Eq. 9.5.2.5
ΔL=Δi×λ×lg/le = 0.00mm, Deflection is Safe 0.00mm, Deflection is Safe 0.00mm, Deflection is Safe
