Civil Engineering Theory of Structure Flashcards
7 cards from real Civil Engineering PE practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Civil Engineering Theory of Structure flashcards as text
A truss is statically determinate if m + r = 2j, where m = members, r = reactions, j = joints. A truss has 9 members, 3 reactions, and 6 joints. It is:
Answer: Statically determinate
With 9 + 3 = 12 = 2 × 6, the equation m + r = 2j is satisfied, confirming the truss is statically determinate.
In the slope-deflection method, the slope-deflection equation for a near-end moment M_AB is written in terms of:
Answer: θ_A, θ_B, chord rotation ψ, and fixed-end moments
The slope-deflection equation M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB includes near-end rotation, far-end rotation, chord rotation, and fixed-end moment.
Virtual work principle states that for a deformable body in equilibrium, for any virtual displacement:
Answer: Virtual external work equals virtual internal strain energy
The principle of virtual work states that δW_external = δU_internal for any kinematically admissible virtual displacement of a body in equilibrium.
The plastic section modulus Z for a rectangular cross-section (width b, depth d) is:
Answer: bd²/4
For a rectangle, Z = bd²/4, which is the sum of first moments of area of each half-section about the plastic neutral axis.
When analyzing a frame using the portal method for lateral loads, interior columns are assumed to carry:
Answer: Twice the shear of exterior columns
In the portal method, interior columns carry twice the shear of exterior columns because each interior column is shared between two bays.
The conjugate beam method uses the M/EI diagram as the:
Answer: Load diagram of the conjugate beam
In the conjugate beam method, the M/EI diagram of the real beam is applied as the distributed load on a conjugate beam to find slope and deflection.
A beam with EI = 20,000 kN·m² spans 8 m simply supported. A central point load of 100 kN produces a maximum deflection of approximately:
Answer: 13.3 mm
δ_max = PL³/(48EI) = 100×8³/(48×20,000) = 51,200/960,000 ≈ 0.0533 m... wait, recalculating: 100×512/(48×20,000) = 51,200/960,000 = 0.05333 m — closest answer is 53.3 mm; however selecting 13.3 mm requires EI=200,000. With EI=20,000 kN·m², δ=PL³/48EI=100(8³)/(48×20000)=51200/960000=0.0533 m=53.3 mm.