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Question 1

Statics

For the simply supported beam shown, a 12 kN point load acts 3 m from the left support on a 6 m span. What is the vertical reaction at the right support, B?

A.3.0 kN
B.6.0 kN
C.9.0 kN
D.12.0 kN

Show solution

Answer · B

Take moments about the left support so the left reaction drops out. The right reaction satisfies R_B(6 m) = (12 kN)(3 m). Therefore R_B = 36/6 = 6.0 kN. The left reaction is also 6.0 kN because the point load is centered, but the requested reaction at B is 6.0 kN.

Question 2

Fluid Mechanics

The Reynolds number for flow in a pipe is 1,800. The flow regime is best described as:

A.Turbulent flow
B.Laminar flow
C.Transitional flow
D.Supercritical flow

Show solution

Answer · B

Step 1: Recall Reynolds number thresholds for pipe flow

Re < 2,100 -> Laminar flow

2,100 < Re < 4,000 -> Transitional flow

Re > 4,000 -> Turbulent flow

Step 2: Classify

Re = 1,800 < 2,100 -> Laminar flow (Answer B)

Note: "Supercritical flow" is an open-channel concept (Froude number), not a pipe flow regime.

Question 3

Geotechnical Engineering

A soil sample has a liquid limit of 45 and a plasticity index of 22. According to the Unified Soil Classification System (USCS), this soil is most likely classified as:

A.CL - Lean clay
B.CH - Fat clay
C.ML - Silt
D.SM - Silty sand

Show solution

Answer · A

Step 1: Check Liquid Limit

LL = 45 < 50 -> Soil is "low plasticity" (L designation).

Step 2: Plot on the Casagrande plasticity chart

The A-line equation is: PI = 0.73 x (LL - 20) = 0.73 x 25 = 18.25

Given PI = 22 > 18.25 -> the soil plots above the A-line -> Clay (C).

Step 3: Classify

Low-plasticity clay = CL - Lean Clay

Reference: USCS Classification per ASTM D2487.

Question 4

Transportation Engineering

The minimum stopping sight distance depends primarily on:

A.Lane width and shoulder width
B.Design speed and driver reaction time
C.Pavement type and drainage
D.Traffic volume and truck percentage

Show solution

Answer · B

Step 1: Recall the SSD formula

SSD = Reaction Distance + Braking Distance

SSD = V x t + V^2 / (2 x a)

Step 2: Identify the variables

V = design speed, t = perception-reaction time, a = deceleration rate.

Step 3: Determine the controlling factors

Both terms depend on design speed. The reaction distance also depends on driver reaction time.

Reference: AASHTO Green Book, Stopping Sight Distance section.

Question 5

Statics

For the cantilever beam shown, an 8 kN point load acts 4 m from the fixed support. What is the magnitude of the fixed-end moment reaction?

A.8 kN·m
B.16 kN·m
C.32 kN·m
D.64 kN·m

Show solution

Answer · C

The fixed support must resist the moment caused by the point load about the wall. For a point load on a cantilever, the fixed-end moment magnitude is M = P a, where a is the load distance from the fixed support. Substituting P = 8 kN and a = 4 m gives M = 32 kN·m. The vertical reaction is 8 kN, but the moment reaction is 32 kN·m.

Question 6

Fluid MechanicsFill-in-Blank

A tank contains water (gamma = 9.81 kN/m^3) to a depth of 5 m. What is the hydrostatic pressure at the bottom of the tank?

Your answer:kPa

Show solution

Answer · 49.05 kPa

Acceptable range · 48.5–49.5 kPa

Step 1: Identify the formula

Hydrostatic pressure: P = gamma x h

Step 2: Substitute values

P = 9.81 kN/m^3 x 5 m = 49.05 kPa

Hydrostatic pressure increases linearly with depth in a fluid of constant density.

Reference: FE Reference Handbook - Fluid Mechanics, Hydrostatic Pressure.

Question 7

Geotechnical Engineering

A strip footing is 1.5 m wide on a cohesive soil with c = 80 kPa and phi = 0 degrees. Using Terzaghi's bearing capacity equation (Nc = 5.7, Nq = 1.0, Ngamma = 0), the ultimate bearing capacity is most nearly:

A.456 kPa
B.380 kPa
C.570 kPa
D.228 kPa

Show solution

Answer · A

Step 1: Recall Terzaghi's bearing capacity equation

q_ult = c x Nc + q x Nq + 0.5 x gamma x B x Ngamma

Step 2: Identify given values

c = 80 kPa, Nc = 5.7, surface footing surcharge q = 0, and Ngamma = 0.

Step 3: Substitute and compute

q_ult = (80)(5.7) = 456 kPa

Reference: FE Reference Handbook - Geotechnical, Shallow Foundation Bearing Capacity.

Question 8

Statics

For the simply supported beam shown, a 6 kN/m uniformly distributed load acts over the full 3 m span. What is the maximum bending moment in the beam?

A.4.50 kN·m
B.6.75 kN·m
C.9.00 kN·m
D.18.00 kN·m

Show solution

Answer · B

For a simply supported beam with a uniform load over the full span, the maximum moment occurs at midspan and equals wL^2/8. Substituting w = 6 kN/m and L = 3 m gives M_max = 6(3^2)/8. This is 54/8 = 6.75 kN·m. The total load is 18 kN, but that is not the bending moment.

Question 9

Statics

For the symmetric triangular truss shown, a 10 kN vertical load acts at the apex. What is the vertical reaction at each support?

A.2.5 kN
B.5.0 kN
C.7.5 kN
D.10.0 kN

Show solution

Answer · B

The truss geometry and loading are symmetric about midspan. The only external vertical load is 10 kN at the apex, so the two support reactions share the load equally. Each support reaction is 10/2 = 5.0 kN upward. This support-reaction result is the starting point for any method-of-joints member-force calculation.

Question 10

Transportation Engineering

A freeway segment has a free-flow speed of 60 mph and a density of 35 pc/mi/ln. The level of service (LOS) is most likely:

A.LOS A
B.LOS C
C.LOS D
D.LOS E

Show solution

Answer · C

Step 1: Recall LOS criteria for basic freeway segments

LOS is determined by density for basic freeway segments.

Step 2: Classify

Density = 35 pc/mi/ln falls within the LOS D range.

Reference: HCM Chapter 12 - Basic Freeway Segments.

Question 11

Mechanics of MaterialsMultiple Select

Which of the following factors affect the maximum deflection of a simply supported beam under a uniform load? Select all that apply.

Select all that apply

A.Modulus of elasticity (E)
B.Moment of inertia (I)
C.Span length (L)
D.Poisson's ratio (nu)
E.Load intensity (w)

Show solution

Answers · A, B, C, E

Step 1: Recall the deflection formula

For a simply supported beam with UDL: delta_max = 5wL^4 / (384EI)

Step 2: Identify the variables in the formula

E, I, L, and w appear in the formula. Poisson's ratio does not.

Correct answers: A, B, C, E.

Question 12

Mechanics of Materials

For the 2D stress element shown, σx = 50 MPa, σy = 30 MPa, and τxy = 20 MPa are applied in the directions shown. What is the maximum in-plane shear stress?

A.20.0 MPa
B.22.4 MPa
C.40.0 MPa
D.44.7 MPa

Show solution

Answer · D

Read the signs from the arrows: σx is tension, so σx = +50 MPa tension; σy is compression, so σy = -30 MPa compression. The τxy arrows show one shear orientation, but the shear term is squared in the Mohr-circle radius. The maximum in-plane shear stress is τmax = sqrt(((σx - σy)/2)^2 + τxy^2). Here (50 - (-30))/2 = 40 MPa, so τmax = sqrt((40)^2 + 20^2) = sqrt(2000) = 44.7 MPa.

Question 13

Fluid Mechanics

A rectangular channel is 4 m wide and carries a discharge of 12 m^3/s. The critical depth is most nearly:

A.0.97 m
B.1.25 m
C.0.75 m
D.1.50 m

Show solution

Answer · A

Step 1: Calculate unit discharge

q = Q / b = 12 m^3/s / 4 m = 3 m^2/s

Step 2: Apply the rectangular-channel critical depth formula

y_c = (q^2 / g)^(1/3)

Step 3: Substitute and compute

y_c = (9 / 9.81)^(1/3) = 0.97 m

Reference: FE Reference Handbook - Fluid Mechanics, Open Channel Flow.

Question 14

Structural Engineering

A W10x49 steel column (I = 272 in^4, A = 14.4 in^2) is 20 ft long with both ends pinned. Using E = 29,000 ksi, the Euler critical buckling load is most nearly:

A.1,355 kips
B.678 kips
C.339 kips
D.170 kips

Show solution

Answer · A

Step 1: Identify the Euler buckling formula

P_cr = pi^2 EI / (KL)^2

Step 2: Determine effective length

Both ends pinned -> K = 1.0. L = 20 ft = 240 in.

Step 3: Substitute values

P_cr = pi^2(29,000)(272) / (240)^2 = about 1,355 kips

Reference: FE Reference Handbook - Mechanics of Materials, Column Buckling.

Question 15

Geotechnical EngineeringDrag & Drop

Match each soil test to the property it measures by dragging each test to the correct property.

Items to match

Standard Proctor Test
Atterberg Limits Test
Consolidation Test
Direct Shear Test

Target slots

Maximum dry density and optimum moisture content
Plasticity characteristics (LL, PL, PI)
Compressibility and settlement parameters (Cc, Cv)
Shear strength parameters (c, phi)

Show solution

Correct matching

Standard Proctor Test → Maximum dry density and optimum moisture content
Atterberg Limits Test → Plasticity characteristics (LL, PL, PI)
Consolidation Test → Compressibility and settlement parameters (Cc, Cv)
Direct Shear Test → Shear strength parameters (c, phi)

Standard Proctor Test -> maximum dry density and optimum moisture content.

Atterberg Limits Test -> liquid limit, plastic limit, and plasticity index.

Consolidation Test -> compressibility and settlement parameters.

Direct Shear Test -> cohesion and friction angle using the Mohr-Coulomb failure criterion.

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