This is civil engineering concrete design practice. Everything needed is provided. Please answer all parts to the best of your ability. Provide thorough answer and explanation. This is only one question with a max of 3 parts due to guidelines. Centroid of a composite areay=Σy,Α;ΣΑ;First moment of area for a crosssectionQ = YAMoment of inertia for arectangle, about neutral x-axis1 = 1/2 bh³f=Parallel axis theoremI = ΣĪ + Ad2Linear stress distribution in across sectionAll limits for design will be provided as part of the given information for each problem.MyIMaximum moment for a simplysupported beam under adistributed loadMmaxn =W1²8Modular ratioEsEcAt ultimate, assumedcompression force in a concretesectionC = 0.85fc AcFor a rectangular sectionC = 0.85f ab, where a =At ultimate, assumed tensionforce in a concrete sectionT = AsfyNominal moment for a concretesectionMn = T(d - y) = C(d - y)For a rectangular sectionMn = T (d - ²) = C (d - 2)For a T-beam analysisM₁ = C, (d-+G(d-)= Asw fy (d − 2) + Assfy (d - 1/4)Percentage of reinforcing steelAsbdРFrom Aspbd, designcapacity, фMn, isþMnpM₁ = opbd²fy (1-P)From the above equation, p is0.85/2 (1fyp == B1C1-Where Rn1=Muobd²For doubly reinforced beamsMn = Asify (d-²) +As2fy (d-d')2Rn0.85 fWhere Aszfy = A'sfs, and As=As1 + As2Tensile strainEt=0.003d-cCStrain in compression steel(doubly reinforced)c-d'€ = 0.003-Yield strain (steel)=Ey ELoad Factors (Ch5.3)U = 1.2D + 1.6LModulus of rupture for aconcrete cross section (Ch19.2.3)fr = 7.52√ √fMinimum shrinkage andtemperature reinforcementarea (one-way slabs)= 0.0018bhAs =Shear capacityφVn = φV€ + φV 4. The section below (at right) shows a floor system consisting of T-beams that connect one-way slabs. Each T-beamhas a span length, In, of 40 feet (i.e., in/out of the page). The T-beam cross-section and dimensions are shown inmore detail at left. Complete parts a-c below.2.5"a. Compute bf, the flange width, according to ACI318-19 requirements. Table 6.3.2.1 is shown below. Note: Swrepresents the clear distance between webs.b. The neutral axis for this section is in the web. Compute the total area of steel, As, needed to resist bending inthe beam.Compute the depth of the concrete compression area, a, for this section.C.16"6"As?22"Wu = 6.15 k/ft (including self weight);Mu 1230 kft8"Flange locationTable 6.3.2.1-Dimensional limits for effectiveoverhanging flange width for T-beamsEach side ofwebOne side of web5 feetf'c = 3000 psi, B₁ = 0.85fy = 60000 psiW₁Effective overhanging flange width, beyond faceof webLeast of:Least of:8hS₁/21₁/86hS₁/2In/1216"

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This is civil engineering concrete design practice. Everything needed is provided. Please answer all parts to the best of your ability. Provide thorough answer and explanation. This is only one question with a max of 3 parts due to guidelines. Centroid of a composite areay=Σy,Α;ΣΑ;First moment of area for a crosssectionQ = YAMoment of inertia for arectangle, about neutral x-axis1 = 1/2 bh³f=Parallel axis theoremI = ΣĪ + Ad2Linear stress distribution in across sectionAll limits for design will be provided as part of the given information for each problem.MyIMaximum moment for a simplysupported beam under adistributed loadMmaxn =W1²8Modular ratioEsEcAt ultimate, assumedcompression force in a concretesectionC = 0.85fc AcFor a rectangular sectionC = 0.85f ab, where a =At ultimate, assumed tensionforce in a concrete sectionT = AsfyNominal moment for a concretesectionMn = T(d - y) = C(d - y)For a rectangular sectionMn = T (d - ²) = C (d - 2)For a T-beam analysisM₁ = C, (d-+G(d-)= Asw fy (d − 2) + Assfy (d - 1/4)Percentage of reinforcing steelAsbdРFrom Aspbd, designcapacity, фMn, isþMnpM₁ = opbd²fy (1-P)From the above equation, p is0.85/2 (1fyp == B1C1-Where Rn1=Muobd²For doubly reinforced beamsMn = Asify (d-²) +As2fy (d-d')2Rn0.85 fWhere Aszfy = A'sfs, and As=As1 + As2Tensile strainEt=0.003d-cCStrain in compression steel(doubly reinforced)c-d'€ = 0.003-Yield strain (steel)=Ey ELoad Factors (Ch5.3)U = 1.2D + 1.6LModulus of rupture for aconcrete cross section (Ch19.2.3)fr = 7.52√ √fMinimum shrinkage andtemperature reinforcementarea (one-way slabs)= 0.0018bhAs =Shear capacityφVn = φV€ + φV 4. The section below (at right) shows a floor system consisting of T-beams that connect one-way slabs. Each T-beamhas a span length, In, of 40 feet (i.e., in/out of the page). The T-beam cross-section and dimensions are shown inmore detail at left. Complete parts a-c below.2.5&#34;a. Compute bf, the flange width, according to ACI318-19 requirements. Table 6.3.2.1 is shown below. Note: Swrepresents the clear distance between webs.b. The neutral axis for this section is in the web. Compute the total area of steel, As, needed to resist bending inthe beam.Compute the depth of the concrete compression area, a, for this section.C.16&#34;6&#34;As?22&#34;Wu = 6.15 k/ft (including self weight);Mu 1230 kft8&#34;Flange locationTable 6.3.2.1-Dimensional limits for effectiveoverhanging flange width for T-beamsEach side ofwebOne side of web5 feetf'c = 3000 psi, B₁ = 0.85fy = 60000 psiW₁Effective overhanging flange width, beyond faceof webLeast of:Least of:8hS₁/21₁/86hS₁/2In/1216&#34;

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