Loads, determinate and indeterminate analysis, steel and reinforced-concrete design.
NCEES lists these Structural Engineering topics for the FE Civil: analysis of statically determinate beams, columns, trusses and frames; deflection of determinate beams, trusses and frames; column analysis: buckling and boundary conditions; structural determinacy and stability; elementary indeterminate structures; loads, load combinations and load paths; design of steel components; design of reinforced-concrete components.
Choose any section below to get a free FE Civil practice problem. Work it with the FE Reference Handbook open, tap your answer, then check the solution.
Structures are designed for the worst of several combinations of dead (D), live (L), roof live (Lr), snow (S) and wind (W) load.
A column carries a dead load D = 200 kN, a floor live load L = 150 kN and a roof snow load S = 80 kN. There is no wind. Using the ASCE 7 LRFD strength combinations, what is the governing factored axial load?
1.4D = 280 kN. 1.2D + 1.6L + 0.5S = 240 + 240 + 40 = 520 kN. 1.2D + 1.6S + 1.0L = 240 + 128 + 150 = 518 kN. The largest governs: 520 kN.
Common trap: (B) Leaves out the 0.5S companion load.
Load CombinationsLRFDASDThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Statics alone solves a determinate structure: ΣF = 0 and ΣM = 0 give the reactions, and sections give the internal forces.
An 8 m simply supported beam carries a uniform load of 10 kN/m over its full length plus a 40 kN point load 2 m from support A. What is the maximum bending moment?
RA = 10(8)/2 + 40(6)/8 = 70 kN. Shear just right of the point load = 70 − 10(2) − 40 = 10 kN, still positive, so V = 0 at x = 2 + 10/10 = 3.0 m. M = 70(3.0) − 10(3.0)²/2 − 40(1.0) = 125 kN·m.
Common trap: (B) Takes the moment under the point load; the shear is still positive there.
Method of JointsMethod of SectionsShear and MomentThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Deflection formulas come straight from the Handbook.
A simply supported steel floor beam (E = 200 GPa) spans 9 m and carries a service dead load of 10 kN/m and a service live load of 15 kN/m. The live-load deflection is limited to L/360. What is the minimum required moment of inertia?
Allowable δ = L/360 = 9,000/360 = 25.0 mm. With w = 15 kN/m = 15 N/mm: I = 5wL⁴/(384Eδ) = 5(15)(9,000)⁴ / [384(200,000)(25.0)] = 2.56 × 108 mm⁴.
Common trap: (B) Uses dead plus live load against the live-load limit.
Beam DeflectionIndeterminate BeamsMoment DistributionThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
LRFD steel design compares a factored demand to a design strength φRn.
A 16 × 200 mm steel plate (Fy = 345 MPa, Fu = 450 MPa) is a tension member connected by one row of 2 bolts across its width, in 24 mm diameter holes. Take U = 1.0. What is the LRFD design tensile strength?
Ag = 16(200) = 3,200 mm². Yielding: 0.90(345)(3,200) = 993.6 kN. An = 3,200 − 2(24)(16) = 2,432 mm² = Ae. Fracture: 0.75(450)(2,432) = 820.8 kN. The smaller governs: 820.8 kN.
Common trap: (B) Checks only gross-section yielding; fracture governs.
Steel DesignEuler BucklingPlastic MomentThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
For a singly reinforced rectangular beam, the steel yields and the concrete stress block has depth a = Asfy / (0.85 f′cb).
A rectangular beam has b = 300 mm, d = 500 mm, As = 2,500 mm², fy = 420 MPa and f′c = 28 MPa (β1 = 0.85). What is the net tensile strain εt in the steel at nominal strength?
a = 2,500(420) / [0.85(28)(300)] = 147.1 mm. c = a/β1 = 147.1/0.85 = 173.0 mm. εt = 0.003(d − c)/c = 0.003(500 − 173.0)/173.0 = 0.00567. This is above 0.005, so the section is tension-controlled and φ = 0.90.
Common trap: (B) Uses c = a, forgetting to divide by β1.
Reinforced ConcreteBeam DesignColumn DesignThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
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