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    Reinforced Concrete Design Tutoring

    Beams, slabs, columns, and footings under ACI 318, explained step by step and practiced on problems from your own course.

    Reinforced concrete design combines two materials with very different behavior. Concrete is strong in compression and weak in tension, and steel bars supply the tension. The course teaches you how to size the section, place the bars, and make sure the member fails gradually rather than suddenly. The ACI 318 code gives the rules, and strength design is built around factored loads, strength reduction factors, and strain compatibility. Early topics feel mechanical. Later topics, such as columns and footings, combine many checks.

    Sessions use your textbook (often Wight, Nilson, or McCormac) and the code edition your professor assigns. We work on the flexure routine until it is automatic, and then extend it to shear, development length, slabs, and columns. Problems are done the way you will be tested: setup first, numbers second, and a reasonableness check at the end. We also keep a short list of the code provisions that cost students the most points.

    What We Cover

    ACI 318 strength design: load factors, phi factors, and strain limits
    Flexure of singly and doubly reinforced rectangular beams
    T-beams and L-beams, effective flange width
    Shear design: concrete contribution, stirrups, and spacing limits
    Development length, hooks, and splices
    One-way slabs, minimum thickness, and temperature steel
    Columns and interaction diagrams, footings, and serviceability

    A typical reinforced concrete problem, worked

    A rectangular beam has b=12 inb = 12\ \text{in}, effective depth d=21.5 ind = 21.5\ \text{in}, and As=3.00 in2A_s = 3.00\ \text{in}^{2} (three #9 bars). Concrete is fc′=4 ksif'_c = 4\ \text{ksi} and steel is fy=60 ksif_y = 60\ \text{ksi}. Find the design moment strength.

    Start with the depth of the equivalent stress block: a=Asfy0.85fc′b=3.00×600.85×4×12=18040.8=4.41 ina = \dfrac{A_s f_y}{0.85 f'_c b} = \dfrac{3.00 \times 60}{0.85 \times 4 \times 12} = \dfrac{180}{40.8} = 4.41\ \text{in}

    The nominal moment is Mn=Asfy(d−a2)=180×(21.5−2.21)=3,473 kip-inM_n = A_s f_y \left(d - \tfrac{a}{2}\right) = 180 \times (21.5 - 2.21) = 3{,}473\ \text{kip-in}, or 289.4 kip-ft.

    Check that the section is tension-controlled. For fc′=4 ksif'_c = 4\ \text{ksi}, β1=0.85\beta_1 = 0.85, so c=a/β1=5.19 inc = a / \beta_1 = 5.19\ \text{in}. The steel strain is 0.003×d−cc=0.003×16.315.19=0.00940.003 \times \dfrac{d - c}{c} = \dfrac{0.003 \times 16.31}{5.19} = 0.0094 which is above 0.005. So ϕ=0.90\phi = 0.90 and the design strength is ϕMn=0.90×289.4=260.5 kip-ft\phi M_n = 0.90 \times 289.4 = 260.5\ \text{kip-ft}

    Where students usually get stuck

    • Skipping the strain check and using the wrong ϕ\phi factor
    • Mixing kip-in with kip-ft in the same line of work
    • Doubly reinforced beams, where it is easy to lose track of whether the compression steel has yielded
    • Development length and shear spacing, where many small code provisions interact
    • Reading the interaction diagram for columns without understanding what each point represents
    Structural analysis tutoring

    Who this is for

    Civil and architectural engineering students in a first or second concrete design course, and graduates reviewing for the structural PE exam. It also suits students who did fine in steel design but feel unsure about concrete, since the logic is different. Bring your code edition and your last exam, and we pick up from there. Sessions are one-on-one, online nationwide, with in-person options around Richmond, Sugar Land, and Katy, TX.

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    In-person sessions across Richmond, Sugar Land, and Katy, or online with a shared whiteboard — evenings and weekends.

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