FE Other Disciplines (NCEES Fundamentals of Engineering) — All Questions
33 questions
An axial tensile load of 10,000 N acts on a bar with cross-sectional area 0.002 m^2. What is the normal stress?
- a.20 MPa
- b.5 MPa✓
- c.0.5 MPa
- d.50 MPa
Normal stress sigma = P / A = 10,000 N / 0.002 m^2 = 5,000,000 Pa = 5 MPa. Stress is force per unit area and does not depend on the bar's length.
A simply supported beam of span 4 m carries a single concentrated load of 8 kN at midspan. What is the maximum bending moment?
- a.32 kN·m
- b.16 kN·m
- c.4 kN·m
- d.8 kN·m✓
For a central point load on a simply supported beam, the maximum moment occurs at midspan and equals P·L / 4 = (8 kN)(4 m) / 4 = 8 kN·m. The moment diagram is triangular, peaking under the load.
A steel bar (E = 200 GPa) 2 m long with cross-sectional area 1x10^-4 m^2 carries an axial load of 10 kN. What is its elongation?
- a.1 mm✓
- b.2 mm
- c.0.5 mm
- d.0.1 mm
Axial elongation delta = P·L / (A·E) = (10,000 x 2) / (1x10^-4 x 200x10^9) = 20,000 / 2x10^7 = 0.001 m = 1 mm. Stiffer materials (higher E) and larger areas reduce elongation.
A 2 m long bar stretches 2 mm under load. What is the axial strain?
- a.1
- b.0.001✓
- c.0.01
- d.0.0005
Strain = deformation/original length = 0.002 m/2 m = 0.001. Strain is dimensionless, expressing the fractional change in length.
A material with Young's modulus 200 GPa experiences an axial strain of 0.001 within its elastic range. What is the resulting stress?
- a.2 MPa
- b.20 MPa
- c.200 MPa✓
- d.200 GPa
Hooke's law: stress = E·strain = 200x10^9 x 0.001 = 200x10^6 Pa = 200 MPa. Stress is proportional to strain only within the linear elastic region.
A shear force of 5000 N acts over a cross-sectional area of 0.001 m^2. What is the average shear stress?
- a.50 MPa
- b.5 MPa✓
- c.0.5 MPa
- d.2.5 MPa
Average shear stress tau = V/A = 5000/0.001 = 5,000,000 Pa = 5 MPa. Shear stress acts parallel to the cross-section, unlike normal stress which acts perpendicular to it.
A component has a yield strength of 250 MPa and is designed to a working stress of 100 MPa. What is the factor of safety?
- a.0.4
- b.2.5✓
- c.150
- d.2.0
Factor of safety = failure (yield) stress/allowable working stress = 250/100 = 2.5. A factor above 1 provides margin against uncertainties in loads, materials, and analysis.
A solid circular shaft 0.05 m in diameter transmits a torque of 1000 N·m. What is the maximum shear stress at its surface?
- a.40.7 MPa✓
- b.81.5 MPa
- c.20.4 MPa
- d.4.07 MPa
Torsion formula tau = T·r/J, with J = pi·d^4/32 = pi(0.05)^4/32 = 6.14x10^-7 m^4 and r = 0.025 m. tau = 1000 x 0.025/6.14x10^-7 = 40.7 MPa. Shear stress is maximum at the outer radius.
What is the polar moment of inertia of a solid circular shaft of diameter d?
- a.pi·d^4/64
- b.pi·d^4/32✓
- c.pi·d^4/12
- d.pi·d^3/16
For a solid circular cross-section, J = pi·d^4/32. (The area moment of inertia about a diameter is pi·d^4/64, exactly half of J, since J = Ix + Iy for the two perpendicular axes.)
A rectangular beam (width 0.05 m, height 0.1 m) carries a bending moment of 10 kN·m. What is the maximum bending stress?
- a.120 MPa✓
- b.240 MPa
- c.60 MPa
- d.12 MPa
Bending stress sigma = M·c/I, with I = b·h^3/12 = 0.05(0.1)^3/12 = 4.17x10^-6 m^4 and c = 0.05 m. sigma = 10,000 x 0.05/4.17x10^-6 = 120 MPa. Maximum stress occurs at the extreme fiber farthest from the neutral axis.
What is the section modulus of a rectangular cross-section 0.05 m wide and 0.1 m tall about its horizontal centroidal axis?
- a.8.33x10^-4 m^3
- b.4.17x10^-5 m^3
- c.8.33x10^-5 m^3✓
- d.1.67x10^-4 m^3
Section modulus S = I/c = b·h^2/6 = 0.05(0.1)^2/6 = 0.05 x 0.01/6 = 8.33x10^-5 m^3. Bending stress can be computed directly as sigma = M/S.
A bar with Poisson's ratio 0.3 is stretched to an axial strain of 0.001. What is the lateral (transverse) strain?
- a.-0.0001
- b.-0.001
- c.-0.0003✓
- d.-0.003
Lateral strain = -nu x axial strain = -0.3 x 0.001 = -0.0003. The negative sign shows the bar contracts laterally as it stretches axially, which is the Poisson effect.
A material has Young's modulus 200 GPa and Poisson's ratio 0.25. What is its shear modulus?
- a.100 GPa
- b.200 GPa
- c.80 GPa✓
- d.40 GPa
For an isotropic material, G = E/(2(1 + nu)) = 200/(2(1 + 0.25)) = 200/2.5 = 80 GPa. The three elastic constants E, G, and nu are interdependent for isotropic materials.
A fully restrained steel bar (E = 200 GPa, alpha = 12x10^-6 /degC) is heated by 50 degC. What thermal stress develops?
- a.240 MPa
- b.120 MPa✓
- c.60 MPa
- d.12 MPa
For full restraint, thermal stress sigma = E·alpha·(delta T) = 200x10^9 x 12x10^-6 x 50 = 120x10^6 Pa = 120 MPa (compressive). If the bar were free to expand, no stress would arise.
A pin-ended column (E = 200 GPa, I = 1x10^-6 m^4) is 2 m long. What is the Euler critical buckling load?
- a.1974 kN
- b.247 kN
- c.987 kN
- d.493 kN✓
Euler buckling load P_cr = pi^2·E·I/L^2 = pi^2 x 200x10^9 x 1x10^-6/2^2 = 9.87 x 200,000/4 = 493,000 N = 493 kN. Critical load drops with the square of the effective length.
A simply supported beam (E = 200 GPa, I = 1x10^-5 m^4) of span 4 m carries a 10 kN load at midspan. What is the maximum deflection?
- a.6.67 mm✓
- b.13.3 mm
- c.0.67 mm
- d.3.33 mm
Central-load deflection delta = P·L^3/(48·E·I) = 10,000 x 4^3/(48 x 200x10^9 x 1x10^-5) = 640,000/9.6x10^7 = 0.00667 m = 6.67 mm. Deflection grows with the cube of the span.
A cantilever beam (E = 200 GPa, I = 1x10^-5 m^4) of length 3 m carries a 5 kN load at its free end. What is the tip deflection?
- a.45 mm
- b.7.5 mm
- c.22.5 mm✓
- d.11.25 mm
Cantilever end deflection delta = P·L^3/(3·E·I) = 5000 x 3^3/(3 x 200x10^9 x 1x10^-5) = 135,000/6x10^6 = 0.0225 m = 22.5 mm. A cantilever deflects far more than a simply supported beam of the same span.
A cantilever beam 3 m long carries a 5 kN point load at its free end. What is the maximum bending moment?
- a.30 kN·m
- b.7.5 kN·m
- c.5 kN·m
- d.15 kN·m✓
For an end-loaded cantilever, the maximum moment occurs at the fixed support: M = P·L = 5 x 3 = 15 kN·m. The bending-moment diagram is triangular, largest at the wall.
A cantilever beam 3 m long carries a uniform load of 4 kN/m over its full length. What is the maximum bending moment?
- a.18 kN·m✓
- b.36 kN·m
- c.6 kN·m
- d.9 kN·m
For a uniformly loaded cantilever, maximum moment at the support = w·L^2/2 = 4 x 3^2/2 = 18 kN·m. The total load (12 kN) acts at the centroid 1.5 m from the wall, giving the same result.
A simply supported beam of span 4 m carries a uniform load of 6 kN/m. What is the maximum bending moment?
- a.48 kN·m
- b.6 kN·m
- c.12 kN·m✓
- d.24 kN·m
For a simply supported beam with a uniform load, maximum moment at midspan = w·L^2/8 = 6 x 4^2/8 = 12 kN·m. This peaks at the center, where shear is zero.
A simply supported beam of span 4 m carries a uniform load of 6 kN/m. What is the maximum shear force?
- a.6 kN
- b.12 kN✓
- c.24 kN
- d.30 kN
For a simply supported uniformly loaded beam, the maximum shear equals the support reaction = w·L/2 = 6 x 4/2 = 12 kN, occurring at the supports where shear is greatest.
On a stress-strain diagram, the yield point marks:
- a.The onset of permanent (plastic) deformation✓
- b.The maximum stress
- c.The fracture point
- d.Zero stress
The yield point is where the material begins to deform permanently; beyond it, unloading no longer returns the specimen to its original length. Below yield, deformation is elastic and fully recoverable.
The ultimate tensile strength of a material is:
- a.The stress at zero strain
- b.The stress at first yield
- c.The maximum stress the material can withstand✓
- d.Half the yield stress
Ultimate strength is the highest stress on the stress-strain curve, reached before necking and fracture. It exceeds the yield stress, which marks the onset of permanent deformation.
Compared with a brittle material, a ductile material:
- a.Has no yield point
- b.Undergoes large plastic deformation before fracture✓
- c.Fails suddenly with little deformation
- d.Cannot be stretched at all
A ductile material (such as mild steel) undergoes substantial plastic deformation before fracture, giving warning of failure. A brittle material (such as cast iron or glass) fractures suddenly with little plastic strain.
The modulus of resilience of a material is defined as:
- a.The ultimate stress
- b.The slope of the curve
- c.The area under the plastic region of the stress-strain curve
- d.The area under the elastic region of the stress-strain curve✓
Modulus of resilience is the strain energy per unit volume absorbed up to the elastic limit, equal to the area under the elastic portion of the stress-strain curve. Toughness, by contrast, is the total area up to fracture.
A point is under uniaxial stress of 100 MPa (the other principal stress is zero). What is the maximum in-plane shear stress?
- a.25 MPa
- b.50 MPa✓
- c.100 MPa
- d.200 MPa
Maximum shear stress = (sigma1 - sigma2)/2 = (100 - 0)/2 = 50 MPa, occurring on planes at 45 degrees to the loading axis. This is why ductile materials in tension often shear along 45-degree planes.
A tension member has an allowable stress of 100 MPa and a cross-sectional area of 0.002 m^2. What is the maximum allowable axial load?
- a.50 kN
- b.400 kN
- c.200 kN✓
- d.100 kN
Allowable load P = sigma_allow x A = 100x10^6 x 0.002 = 200,000 N = 200 kN. This rearranges the axial stress relation sigma = P/A to solve for capacity.
An axially loaded bar carries a 10 kN load and elongates 1 mm elastically. How much strain energy is stored?
- a.10 J
- b.1 J
- c.5 J✓
- d.2.5 J
For linear-elastic behavior, strain energy U = (1/2)·P·delta = 0.5 x 10,000 x 0.001 = 5 J. It equals the triangular area under the load-deformation curve.
For a bar in pure uniaxial tension, the maximum shear stress occurs on planes oriented at what angle to the axis?
- a.90 degrees
- b.45 degrees✓
- c.30 degrees
- d.0 degrees
In uniaxial tension the maximum shear stress acts on planes at 45 degrees to the loading axis, where it equals half the axial stress. The maximum normal stress, in contrast, acts on the plane perpendicular to the axis (0 degrees).
Near a hole or sharp notch in a loaded member, the local stress is:
- a.Reduced to zero
- b.Unchanged from the average
- c.Amplified above the average (stress concentration)✓
- d.Reduced below the average
Geometric discontinuities like holes and notches concentrate stress, producing local peaks above the nominal (average) stress. The stress concentration factor quantifies this amplification and is critical to fatigue design.
On a stress-strain diagram, the modulus of elasticity (Young's modulus) is represented by:
- a.The strain at fracture
- b.The ultimate stress
- c.The area under the whole curve
- d.The slope of the initial (linear elastic) region✓
Young's modulus equals the slope of the straight-line elastic portion of the stress-strain curve, E = stress/strain. A steeper slope indicates a stiffer material.
For a circular shaft of radius r, length L, twisted through angle theta, the maximum shear strain at the surface is:
- a.T·L/(G·J)
- b.P/A
- c.r·theta/L✓
- d.M·c/I
The surface shear strain in torsion is gamma = r·theta/L, where theta is the total angle of twist. (T·L/(G·J) gives the angle of twist itself, M·c/I is bending stress, and P/A is axial stress.)
The bulk modulus of a material relates:
- a.Bending moment to curvature
- b.Pressure to volumetric strain✓
- c.Torque to angle of twist
- d.Shear stress to shear strain
Bulk modulus K = -pressure/volumetric strain measures a material's resistance to uniform (hydrostatic) compression. Shear modulus relates shear stress to shear strain, a distinct elastic property.