TRUMAN P. YOUNG & ASSOCIATES : ENGINEERS : CINCINNATI, OHIO LHC Lid Calculations May 22, 2015 Job No. 2015009

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LHC LID DESIGN CRITERIA Loads Design Loads per ASTM C857-07 – Standard Practice for Minimum Structural Design Loading For Underground Utility Structures Design Live Load: AASHTO HS-20 Design Vehicle Wheel Load = 16,000 Lb Tire footprint = 10” x 20” Per ASTM C857 Use 30% Impact factor Wheel Load with Impact = 1.3 * 16,000 = 20,800 lb/wheel Materials Concrete Strength: f’c = 5,000 PSI Deformed wire mesh yield strength: Specify Fy = 75,000 PSI Design per AASHTO Standard Specification for Highway Bridges 17th Ed. (Referenced in C857-07)

TRUMAN P. YOUNG & ASSOCIATES : ENGINEERS : CINCINNATI, OHIO LHC Lid Calculations May 22, 2015 Job No. 2015009

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Determination of effective slab width to resist wheel loads Case A – Reinforcement perpendicular to wheel loads. The wheel is travelling along the top of the trench parallel to the trench length. The AASHTO LFD Specification uses the following formula: MLL = [(S+2)/32] x P where S = Span, P = Wheel load Case B – Reinforcement parallel to wheel loads. The wheel is crossing the trench. The AASHTO LFD specification uses the following formula to distribute wheel loads: E = (4 + 0.06 x S) where E = Distribution width The AASHTO LFD specification requires an edge beam be provided to resist a live load moment equal to 0.10 x P x S which is approximately equal to 40% of the simple span wheel load moment. Assume that edge beam consists of 12” at each end of lid length. Design this beam for 40% of wheel load. The AASHTO specification does not require a shear check when slab is designed for flexure per the above criteria.

TRUMAN P. YOUNG & ASSOCIATES : ENGINEERS : CINCINNATI, OHIO LHC Lid Calculations May 22, 2015 Job No. 2015009

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LID LOADING DIAGRAMS

TRUMAN P. YOUNG & ASSOCIATES : ENGINEERS : CINCINNATI, OHIO LHC Lid Calculations May 22, 2015 Job No. 2015009

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LHC96xx-xx LID DESIGN Nominal Trench Width W = 96” Center to Center span = 96” + 6” = 102” = 8.5’ Lid Length = 60” Min. Lid thickness = 8.0” d = 8.0”-1” (Cover) -0.25” (1/2 bar diameter) = 6.75” Case A Loading Dead Load = (150) (8.0/12) = 100 lb./ft = .00833 k/in Dead load moment = (.00833) (102)2/8 = 10.8 in-k /ft Live load moment = (8.5+2)/32 x 16,000 x 1.3 x 12/1000 = 81.9 in-k /ft Mu = (1.3) (10.8) + (2.17) (81.9) = 192 in-k /ft = 16.0 ft-k/ft Case B Loading Dead load moment = 10.8 in-k /ft Live load moment = (10.4) (102/2) – (2.08) (5)2/2 = 504 in-k /wheel E = 4 + .06 x 8.5 = 4.51 ft 504/4.51 ft. = 111 in-k /ft Mu = (1.3) (10.8) + (2.17) (111) = 255 in-k /ft = 21.25 ft-k/ft Kn = (21.25) (12,000) / (0.9 x 12 x (6.75)2 ) = 518 Ρn = .0073 As req. = (.0073) (12) (6.75) = .591 sq-in. /ft D20 @ 4” As = 0.60 sq-in. /ft Edge Beam MLL = 0.40 x 504 = 202 in-k Mu = (1.3) (10.8) + (2.17) (202) = 452 in-k = 37.7 ft-k Kn = (37.7) (12,000) / (0.9 x 12 x (6.75)2 ) = 919 Ρn = .0140 As req. = (.0140) (12) (6.75) = 1.13 sq-in. D20 @ 2” As = 1.20 sq-in. Longitudinal distribution reinforcement Use AASHTO requirement of 67% of required main reinforcement. (0.67) (0.59) = 0.40 sq-in. /ft Use D10 @ 3” As = 0.40 sq-in. /ft

TRUMAN P. YOUNG & ASSOCIATES : ENGINEERS : CINCINNATI, OHIO LHC Lid Calculations May 22, 2015 Job No. 2015009

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TRENWA LHC48xx-xx LID DESIGN Nominal Trench Width W = 48” Center to Center span = 48” + 6” = 54”=4.5’ Lid Length = 60” Min. Lid thickness = 6.5” d = 6.5”-1” (Cover) -0.25” (1/2 bar diameter) = 5.25” Case A Loading Dead Load = (150) (6.5/12) = 81.3 lb./ft Dead load moment = (.081/12) (54)2/8 = 2.47 in-k /ft Live load moment = (4.5+2)/32 x 16,000 x 1.3 x 12/1000 = 50.7 in-k /ft Mu = (1.3) (2.47) + (2.17) (50.7) = 113 in-k /ft = 9.42 ft-k /ft Case B Loading Dead load moment = 2.4 in-k /ft Live load moment = (10.4) (54/2) – (2.08) (5)2/2 = 255 in-k /wheel E = 4 + .06 x 4.5 = 4.27 ft 255/4.27 ft. = 59.7 in-k /ft Mu = (1.3) (2.47) + (2.17) (59.7) = 133 in-k /ft = 11.1 ft-k /ft Kn = (11.1) (12,000) / (0.9 x 12 x (5.25)2 ) = 448 Ρn = .0064 As req. = (.0064) (12) (5.25) = 0.40 sq-in. /ft D12 @ 3” As = 0.48 sq-in /ft Edge beam MLL = 0.40 x 255 = 102 in-k Mu = (1.3) (2.47) + (2.17) (102) = 225 in-k = 18.8 ft-k /ft Kn = (18.8) (12,000) / (0.9 x 12 x (5.25)2 ) = 758 Ρn = .0111 As req. = (.0111) (12) (5.25) = 0.70 sq-in. /ft D12 @ 2” As = 0.72 sq-in /ft Longitudinal distribution reinforcement Use AASHTO requirement of 67% of required main reinforcement. (0.67) (0.40) = 0.27 sq-in. /ft D8 @ 3”

Note: Calculations for LHC48, 36, and 24 are for standard 6.5” thick lids. Lids are 8” thick for this project for dimensional purposes.

TRUMAN P. YOUNG & ASSOCIATES : ENGINEERS : CINCINNATI, OHIO LHC Lid Calculations May 22, 2015 Job No. 2015009

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TRENWA LHC36xx-xx LID DESIGN Nominal Trench Width W = 36” Center to Center span = 36” + 6” = 42”=3.5’ Lid Length = 60” Min. Lid thickness = 6.5” d = 6.5”-1” (Cover) -0.25” (1/2 bar diameter) = 5.25” Case A Loading Dead Load = (150) (6.5/12) = 81.3 lb./ft Dead load moment = (.081/12) (42)2/8 = 1.49 in-k /ft Live load moment = (3.5+2)/32 x 16,000 x 1.3 x 12/1000 = 42.9 in-k /ft Mu = (1.3) (1.49) + (2.17) (42.9) = 95.0 in-k /ft = 7.92 ft-k /ft Case B Loading Dead load moment = 1.49 in-k /ft Live load moment = (10.4) (42/2) – (2.08) (5)2/2 = 193 in-k /wheel E = 4 + .06 x 3.5 = 4.21 ft 193/4.21 ft. = 45.8 in-k /ft Mu = (1.3) (1.49) + (2.17) (45.8) = 101 in-k /ft = 8.42 ft-k /ft Kn = (8.42) (12,000) / (0.9 x 12 x (5.25)2 ) = 339 Ρn = .0047 As req. = (.0047) (12) (5.25) = 0.30 sq-in. /ft D10 @ 4” As = 0.30 sq-in /ft Edge Beam MLL = 0.40 x 193 = 77.2 in-k /ft Mu = (1.3) (1.49) + (2.17) (77.2) = 169 in-k /ft = 14.1 ft-k /ft Kn = (14.1) (12,000) / (0.9 x 12 x (5.25)2 ) = 568 Ρn = .0082 As req. = (.0082) (12) (5.25) = 0.52 sq-in. /ft D10 @ 2” As = 0.60 sq-in /ft Longitudinal distribution reinforcement Use AASHTO requirement of 67% of required main reinforcement. (0.67) (0.30) = 0.20 sq-in. /ft D8 @ 4” As = 0.24 sq-in. /ft

TRUMAN P. YOUNG & ASSOCIATES : ENGINEERS : CINCINNATI, OHIO LHC Lid Calculations May 22, 2015 Job No. 2015009

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TRENWA LHC24xx-xx LID DESIGN Nominal Trench Width W = 24” Center to Center span = 24” + 6” = 30” =2.5’ Lid Length = 60” Min. Lid thickness = 6.5” d = 6.5”-1” (Cover) -0.25” (1/2 bar diameter) = 5.25” Case A Loading Dead Load = (150) (6.5/12) = 81.3 lb./ft Dead load moment = (.081/12) (30)2/8 = 0.76 in-k /ft Live load moment = (2.5+2)/32 x 16,000 x 1.3 x 12/1000 = 35.1 in-k /ft Mu = (1.3) (0.76) + (2.17) (35.1) = 77.2 in-k /ft = 6.43 ft-k /ft Kn = (6.43) (12,000) / (0.9 x 12 x (5.25)2 ) = 259 Ρn = .0038 As req. = (.0038) (12) (5.25) = 0.24 sq-in. /ft D10 @ 4” As = 0.30 sq-in /ft Case B Loading Dead load moment = 0.76 in-k /ft Live load moment = (10.4) (30/2) – (2.08) (5)2/2 = 130 in-k /wheel E = 4 + .06 x 2.5 = 4.15 ft 130/4.15 ft. = 31.3 in-k /ft Mu = (1.3) (0.76) + (2.17) (31.3) = 68.9 in-k /ft = 5.79 ft-k /ft Edge Beam MLL = 0.40 x 130 = 52 in-k /ft Mu = (1.3) (0.76) + (2.17) (52) = 114 in-k /ft = 9.5 ft-k /ft Kn = (9.50) (12,000) / (0.9 x 12 x (5.25)2 ) = 383 Ρn = .0054 As req. = (.0054) (12) (5.25) = 0.34 sq-in. /ft D10 @ 3” As = 0.40 sq-in /ft Longitudinal distribution reinforcement Use AASHTO requirement of 67% of required main reinforcement. (0.67) (0.24) = 0.16 sq-in. /ft Use D8 @ 6” As = .160 sq-in. /ft

TRUMAN P. YOUNG & ASSOCIATES : ENGINEERS : CINCINNATI, OHIO LHC Lid Calculations May 22, 2015 Job No. 2015009

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Check minimum flexural reinforcement Per AASHTO Mn = 1.2 Mcr Mcr = fr x S Fr = 7.5 √5,000 = 530 PSI S = 12 x (6.5)2/6 = 84.5 1.2 Mcr = 1.2 x 530 x 84.5 x 1/1000 = 53.7 in-k /ft = 4.48 ft-k /ft Kn = (4.48) (12,000) / (0.9 x 12 x (5.25)2 ) = 180 Ρn = .0022 As min. = (.0022) (12) (5.25) = 0.14 sq-in. /ft Per ACI 318 As,min = (3) √5000 (12) (5.75) / 75,0000 = 0.195 sq-in. /ft (200) (12) (5.75)/75,000 = 0.184 sq-in. /ft D12 @ 6” As = 0.24 sq-in /ft D10 @ 6” As = 0.20 sq-in /ft

George Schurr

INSTEEL PART NUMBER 533-206198

George Schurr

INSTEEL PART NUMBER 533-206200

George Schurr

INSTEEL PART NUMBER 533-206195