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    Steel Design Tutoring

    Member design and connections under the AISC specification, taught one limit state at a time using your own homework and exams.

    Steel design is the first course where you choose a member instead of just analyzing one. The AISC specification is the rulebook, and the course teaches you to work through it. Every member has several limit states, and the capacity is the smallest one. A tension member can yield or rupture. A column can buckle in several ways. A beam can reach its plastic moment or fail earlier by lateral-torsional buckling. The skill is knowing which checks apply and where to find the equations quickly.

    Sessions follow your textbook, usually Segui, McCormac, or Salmon and Johnson, along with the AISC Steel Construction Manual tables you are allowed to use. We practice both LRFD and ASD, since many courses teach LRFD first and then show ASD as a variation. Sizing problems get a repeatable routine: assume, check every limit state, adjust, and verify. We also build a one-page map of the specification so you know where each check lives.

    What We Cover

    LRFD and ASD design philosophy and load combinations
    Tension members: gross and net area, shear lag, block shear
    Compression members: effective length, slenderness, flexural buckling
    Beams: compact sections, flexure, lateral-torsional buckling, and shear
    Beam-columns and the interaction equations
    Bolted connections: bearing, slip-critical, and bolt group behavior
    Welded connections: fillet welds, weld groups, and eccentric loads

    A typical steel design problem, worked

    A bolted tension member has a gross area Ag=4.5 in2A_g = 4.5\ \text{in}^{2} and a net area An=3.8 in2A_n = 3.8\ \text{in}^{2}. The shear lag factor is U=0.85U = 0.85. Steel is Fy=50 ksiF_y = 50\ \text{ksi} and Fu=65 ksiF_u = 65\ \text{ksi}. Find the LRFD design strength.

    Yielding on the gross section: ϕPn=0.90×Fy×Ag=0.90×50×4.5=202.5 kips\phi P_n = 0.90 \times F_y \times A_g = 0.90 \times 50 \times 4.5 = 202.5\ \text{kips} The resistance factor for yielding is 0.90.

    Rupture on the net section: first the effective net area, Ae=U×An=0.85×3.8=3.23 in2A_e = U \times A_n = 0.85 \times 3.8 = 3.23\ \text{in}^{2} Then ϕPn=0.75×Fu×Ae=0.75×65×3.23=157.5 kips\phi P_n = 0.75 \times F_u \times A_e = 0.75 \times 65 \times 3.23 = 157.5\ \text{kips} The resistance factor for rupture is 0.75.

    The design strength is the smaller value, so ϕPn=157.5 kips\phi P_n = 157.5\ \text{kips} and rupture governs. For ASD, the allowable strength is Pn/ΩP_n / \Omega, with Ω=2.00\Omega = 2.00 for rupture: 65×3.232.00=105.0 kips\dfrac{65 \times 3.23}{2.00} = 105.0\ \text{kips} A full solution would also check block shear at the connection.

    Where students usually get stuck

    • Stopping at the first limit state that passes. Every applicable limit state needs a check.
    • Effective length and the KK factor, especially for frames with sway
    • Choosing LbL_b and CbC_b in lateral-torsional buckling problems
    • Mixing LRFD factors with ASD safety factors in the same calculation
    • Getting lost in the specification. We practice finding the right chapter and table quickly.
    Reinforced concrete design tutoring

    Who this is for

    Civil engineering and architectural engineering students taking steel design, and structural engineering graduate students who need to refresh the basics. It also suits students preparing for the structural depth of the PE Civil exam. Bring your syllabus, your manual, and any past exams, and we work from there. Tell us which edition of the specification your course uses. Sessions are one-on-one, online nationwide, with in-person options around Richmond, Sugar Land, and Katy, TX.

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