FE Civil practice problems with solutions

Answer: A) 28.7

Equation: Area = ∫ f(x) dx from x1 to x2 = F(x2) - F(x1)

1. F(x) = ∫(x² + 7x + 6) dx = (1/3)x³ + (7/2)x² + 6x
2. F(2) = 28.67; F(0) = 0
3. Area = F(2) - F(0) = 28.7

Why the wrong choices are wrong:

• B used one trapezoid (average of the end heights times the width), not the integral.
• C raised each power but did not divide by the new power.
• D differentiated instead of integrating.

Handbook: FE Reference Handbook 10.6, Mathematics, Integral Calculus (definite integral), p. 49.

Answer: B) 12.9 kips (C)

Equation: Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive

1. Reactions, sum of moments about A = 0: Cy = P·a/L = (18.5 kips)(15 ft)/(55 ft) = 5.045 kips
2. Sum Fy = 0: Ay = P - Cy = 13.45 kips; sum Fx = 0: Ax = 0
3. Length BC = √((L - a)² + h²) = √(40² + 17²) ft = 43.46 ft
4. Joint C, sum Fy = 0: Cy + F_BC(h/BC) = 0, so F_BC = -Cy·BC/h = -(5.045 kips)(43.46 ft)/(17 ft) = -12.90 kips
5. F_BC = -12.90 kips, so BC carries 12.9 kips in compression (C)

Why the wrong choices are wrong:

• A resolved BC with the horizontal run L - a instead of the rise h at joint C (cosine for sine).
• C set the member force equal to the reaction without resolving it along the member.
• D has the right size but the wrong sense; equilibrium of joint C puts BC in compression.

Handbook: FE Reference Handbook 10.6, Statics, Plane Truss: Method of Joints, p. 98.

Answer: D) 16.1 mm

Equation: δt = αL(T - To)

1. δt = αL(T - To) = (10 × 10⁻⁶ /°C)(43.5 m × 1,000)(37°C) = 16.1 mm

Why the wrong choices are wrong:

• A used half the temperature change.
• B doubled the free expansion as if both ends moved by δ.
• C converted the temperature change to °F while using α per °C.

Handbook: FE Reference Handbook 10.6, Mechanics of Materials, Thermal Deformations, p. 131.

Answer: D) 139 kN

Equation: F = PC·A = γ·hc·A (pressure at the centroid times the area)

1. Depth to the centroid: hc = d + H/2 = 0.7 + 1.85 = 2.550 m
2. F = γ hc A = (9.81 kN/m³)(2.550 m)(1.5 m × 3.7 m) = 139 kN

Why the wrong choices are wrong:

• A used H/2 as the depth, ignoring the water above the gate.
• B used the depth of the top edge instead of the centroid.
• C used half the gate area.

Handbook: FE Reference Handbook 10.6, Fluid Mechanics, Forces on Submerged Surfaces and the Center of Pressure, p. 184.

Answer: A) 65.0 kPa

Equation: σ' = σ - u; σ = Σ(γ·thickness) down to the point; u = γw × (depth below the water table)

1. Depth to the middle of the clay: z = H1 + H2/2 = 4.5 + 1.75 = 6.25 m
2. Total stress: σ = γm·dw + γsat,sand·(H1 - dw) + γsat,clay·(H2/2) = (17.4)(1.5) + (18.4)(3.0) + (17.3)(1.75) = 111.6 kPa
3. Pore-water pressure: u = γw(z - dw) = (9.81 kN/m³)(4.75 m) = 46.60 kPa
4. σ' = σ - u = 111.6 - 46.60 = 65.0 kPa

Why the wrong choices are wrong:

• B computed the stress at the bottom of the clay layer instead of its middle.
• C is the total stress; the pore-water pressure u was not subtracted.
• D computed the stress at the top of the clay layer instead of its middle.

Handbook: FE Reference Handbook 10.6, Civil Engineering, Geotechnical, Effective stress, p. 269.