Mechanics of materials on the FE, stress by load type
This area takes the reactions you found in statics and asks what they do inside the part: 9–14 questions on stress, strain and deflection under axial, bending, torsion, shear and thermal loads. Bridge and building members are left to the civil exam's structural area; the general theory is here.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Mechanical
questions on the exam9–14
worked problems here16
of 14 knowledge areas8
Checked
What the area covers
The specification lists the load cases one by one. Grouped:
Shear and moment diagrams for beams with point, distributed and couple loads.
Stress transformation and Mohr's circle: principal stresses, maximum shear, rotated elements.
Stress and strain by load type: axial, bending, torsion, transverse shear and temperature change.
Combined loading: a shaft or bracket carrying two or more of the above at once.
Deformations: elongation, angle of twist, beam deflection and slope.
Columns: elastic buckling and the effect of end conditions.
Statically indeterminate systems: adding a compatibility equation when equilibrium alone runs out.
16 mechanics of materials problems
0 / 16 answered · 0 correct
Question 1 of 16 · Mechanics of Materials · 9–14 items
A simply supported beam of length L = 6 m carries a linearly varying distributed load w(x) = 12x kN/m, where x is the distance from the left support. At what location x does the maximum bending moment occur?
Given
Span L = 6 m (simply supported)
Distributed load w(x) = 12x kN/m
Find
Location of maximum moment x (m)
Solution
W = ½(72 kN/m)(6 m) = 216 kN at 4 m from A → RA = 216 × 2/6 = 72 kN
V(x) = 72 − 6x2 = 0 → x = √12 = 3.46 m
FE Reference Handbook, Mechanics of Materials: shear and moment relations (dM/dx = V).
This writes V = 72 − 12x2, forgetting that the load up to x is the area ½(12x)(x) = 6x2.
Midspan is the peak only for a symmetric load; the triangular load pushes the peak toward the heavy end.
RA = 72 kN; the shear V = 72 − 6x2 is zero at x = √12 = 3.46 m, where the moment peaks.
4 m is the centroid of the triangular load, not the point of zero shear.
Taking RA as half the total load (108 kN) ignores that the triangular load sits closer to support B.
Question 2 of 16 · Mechanics of Materials · 9–14 items
A beam carries the concentrated loads listed in the table below. What discontinuities appear in the internal shear (V) and bending moment (M) diagrams at x = 2 m and x = 5 m, respectively?
Position
Applied load
x = 2 m
15 kN point load
x = 5 m
40 kN·m concentrated moment
Solution
dV/dx = −w and dM/dx = V: a point force P → ΔV = P; a concentrated couple M0 → ΔM = M0.
A point force does not step the moment, and a pure couple has no net force, so V is continuous at x = 5 m.
Concentrated actions cause steps, not slope changes; a kink in V would come from a change in distributed load.
This swaps the effects: forces step the shear, couples step the moment.
A point load does change V abruptly; a kink in M at x = 2 m is only its side effect.
A concentrated force makes the shear diagram step by 15 kN, and a concentrated couple makes the moment diagram step by 40 kN·m.
Question 3 of 16 · Mechanics of Materials · 9–14 items
A solid circular shaft 2 m long is fixed at both ends and subjected to a torque of 1000 N·m at its center. If the shaft diameter is 50 mm and the modulus of rigidity is 80 GPa, what is the angle of twist at the center?
FE Reference Handbook, Mechanics of Materials: torsion.
Halving the correct twist again double-counts the symmetry; each 1 m half carries 500 N·m.
By symmetry each half carries 500 N·m over 1 m: θ = TL/(GJ) = 500 × 1/(80 × 109 × 6.14 × 10−7) = 0.0102 rad.
This puts the full 1000 N·m on a 1 m half; the two fixed ends share the torque equally.
This is TL/(GJ) with the full torque and the full 2 m length, as if one end were free.
Question 4 of 16 · Mechanics of Materials · 9–14 items
A channel-section beam has a web 8 mm thick. At a certain section, the first moment of area is Q = 30,000 mm3, and the moment of inertia of the entire cross section is I = 2.5 × 106 mm4. If the beam carries a vertical shear force of 40 kN, what is the shear stress in the web?
Multiplying VQ/(It) by 1.5 applies the rectangular-section peak factor on top of the exact formula: 60 × 1.5 = 90 MPa.
Using half the web thickness (4 mm) doubles the stress; the cut passes through the full 8 mm web.
VQ/I = 480 N/mm is the shear flow; dividing by the web thickness t is missing.
Question 5 of 16 · Mechanics of Materials · 9–14 items
An unsymmetrical T-beam has the section properties shown below. The beam carries a positive internal bending moment of 25 kN·m. What is the maximum compressive bending stress in the cross section?
FE Reference Handbook, Mechanics of Materials: beam stresses (flexure formula).
This uses the whole 150 mm depth as c; c is measured from the neutral axis.
This is the stress at the bottom fiber, 105 mm from the centroid, which is in tension under positive moment.
This uses c = 75 mm, half the depth; for a T-section the neutral axis is not at mid-height.
Positive moment compresses the top fiber, 150 − 105 = 45 mm above the centroid: σ = 25 × 106 × 45/12.5 × 106 = 90 MPa.
Measuring c from mid-height (105 − 75 = 30 mm) mixes the geometric middle with the neutral axis, which passes through the centroid.
Question 6 of 16 · Mechanics of Materials · 9–14 items
A simply supported beam carries a uniformly distributed load of 10 kN/m across its 6-meter span. What is the maximum bending moment in kN·m?
Given
Uniform load w = 10 kN/m
Span L = 6 m
Find
Maximum bending moment Mmax (kN·m)
Solution
Mmax = wL2/8 = (10 kN/m)(6 m)2/8 = 45 kN·m
FE Reference Handbook, Mechanics of Materials: simply supported beam slopes and deflections.
wL2/2 = 180 is the fixed-end moment of a cantilever, not the midspan moment of a simply supported span.
This is PL/4 with the whole 60 kN load concentrated at midspan; spreading it uniformly halves the moment.
60 kN is the total load wL; a moment needs a lever arm.
Mmax = wL2/8 = 10 × 36/8 = 45 kN·m at midspan.
Question 7 of 16 · Mechanics of Materials · 9–14 items
A member is subjected to biaxial stress with σx = 60 MPa (tension) and σy = 20 MPa (compression). If there is no shear stress, what is the maximum shear stress in the material?
Given
Normal stress σx = 60 MPa (tension)
Normal stress σy = 20 MPa (compression)
Shear stress τxy = 0
Find
Maximum in-plane shear stress τmax (MPa)
Solution
τmax = (σ1 − σ2)/2 = (60 − (−20))/2 = 40 MPa
FE Reference Handbook, Mechanics of Materials: Mohr's circle.
(60 − 20)/2 drops the sign of the compressive stress; −20 MPa widens the circle.
σ1/2 would hold only if the other principal stress were zero; here it is −20 MPa.
With σ1 = 60 MPa and σ2 = −20 MPa, τmax = (σ1 − σ2)/2 = 40 MPa, the radius of Mohr's circle.
80 MPa is σ1 − σ2, the diameter of Mohr's circle; the maximum shear is the radius.
Question 8 of 16 · Mechanics of Materials · 9–14 items
A loaded bracket experiences an in-plane stress state where σx = 40 ksi, σy = 10 ksi, and τxy = -20 ksi. What is the orientation of the principal plane nearest to the positive x-axis?
2θp = −53.1° → θp = −26.6° (the other plane: 63.4°)
FE Reference Handbook, Mechanics of Materials: principal stresses.
This adds 180° to 2θp without halving; physical principal planes are 90° apart, not 180°.
63.4° is also a principal plane, but it is the one farther from the x-axis.
This is 2θp with the sign of τxy dropped.
Dropping the sign of τxy gives +26.6°; negative shear turns the principal plane clockwise from x.
tan 2θp = 2τxy/(σx − σy) = −40/30, so 2θp = −53.1° and θp = −26.6°; the other principal plane is 90° away at 63.4°.
Question 9 of 16 · Mechanics of Materials · 9–14 items
A circular shaft of diameter d is subjected to a torque T. What is the maximum shear stress in the shaft?
Solution
τmax = Tc/J = T(d/2)/(πd4/32) = 16T/(πd3)
FE Reference Handbook, Mechanics of Materials: torsion.
Using c = d instead of d/2 doubles the stress.
τmax = Tc/J with c = d/2 and J = πd4/32 gives 16T/(πd3).
This drops the factor 16 that comes from c/J = (d/2)/(πd4/32).
This uses c = d/4 or doubles J; the result is half the true maximum.
Question 10 of 16 · Mechanics of Materials · 9–14 items
A bar of length L is fixed at both ends and experiences a uniform temperature increase ΔT. If the coefficient of thermal expansion is α and the modulus of elasticity is E, what is the thermal stress developed in the bar?
Solution
δT = αLΔT, ε = δT/L = αΔT
σ = Eε = EαΔT (compressive for heating)
FE Reference Handbook, Mechanics of Materials: thermal deformations.
Fully blocked thermal strain α·ΔT becomes mechanical strain, so σ = E·α·ΔT, independent of length.
Without α the result has units of modulus × temperature, not stress.
α·ΔT·L is the free elongation; dividing by L again gives strain, so L cancels from the stress.
α·ΔT is the thermal strain; multiplying by E turns it into stress.
Question 11 of 16 · Mechanics of Materials · 9–14 items
A thin-walled tube with radius r and wall thickness t is subjected to a torque T. What is the shear stress in the tube wall?
Solution
τ = T/(2Amt), Am = πr2 → τ = T/(2πr2t)
FE Reference Handbook, Mechanics of Materials: torsion (thin-walled tubes).
An extra factor of 2 in the denominator halves the stress; the formula is τ = T/(2Amt).
Shear flow q = T/(2Am) with Am = πr2, and τ = q/t = T/(2πr2t).
This divides by a length–thickness product only; torque needs a lever arm, so r must appear squared.
Dropping the 2 from q = T/(2Am) doubles the stress.
Question 12 of 16 · Mechanics of Materials · 9–14 items
A beam bends about both principal axes. Ix = 4 × 106 mm4 with extreme-fiber distance cx = 50 mm, and Iy = 2 × 106 mm4 with cy = 75 mm. If the beam carries bending moments of 10 kN·m about the x-axis and 5 kN·m about the y-axis, what is the maximum bending stress?
FE Reference Handbook, Mechanics of Materials: beam stresses (flexure formula).
This is the Mx term alone; bending about y also stresses the same corner.
This is the My term alone; both normal stresses act on the corner fiber.
Combining the two normal stresses as √(1252 + 187.52) treats them as perpendicular vectors; normal stresses on one fiber add algebraically.
At the corner where both bending stresses have the same sign, σ = 10 × 106(50)/(4 × 106) + 5 × 106(75)/(2 × 106) = 125 + 187.5 = 312.5 MPa.
Question 13 of 16 · Mechanics of Materials · 9–14 items
A simply supported beam of length 6 m is subjected to a uniform load of 5 kN/m. The beam has a modulus of elasticity E = 200 GPa and a moment of inertia I = 4000 cm4. Calculate the midspan deflection.
wL4/(384EI) is the midspan deflection of a fixed-fixed beam; simple supports deflect five times as much.
Question 14 of 16 · Mechanics of Materials · 9–14 items
A rectangular beam with width b and height h is subjected to a shear force V. According to the shear stress formula for rectangular beams, what is the maximum shear stress in the cross-section?
Solution
τmax = VQ/(Ib) with Q = bh2/8, I = bh3/12 → τmax = 3V/(2bh)
FE Reference Handbook, Mechanics of Materials: transverse shear stress.
For a rectangle the shear stress is parabolic with its peak at the neutral axis, 1.5 times the average V/(bh).
A factor of 6 belongs to the section modulus bh2/6 in bending, not to shear stress.
V/(bh) is the average shear stress; the distribution peaks at the neutral axis.
This is a third of the true peak; the factor for a rectangle is 3/2, not 1/2.
Question 15 of 16 · Mechanics of Materials · 9–14 items
A slender steel column with pinned-pinned boundary conditions has a length of 3 m, a moment of inertia I = 8 × 10−6 m4, and a modulus of elasticity E = 200 GPa. Using Euler's formula, what is the critical buckling load?
FE Reference Handbook, Mechanics of Materials: columns (Euler's formula).
This uses K = 0.5 (fixed-fixed); pinned ends allow rotation, so the effective length is the full 3 m.
Dividing by L instead of L2 gives π2EI/L = 5.26 × 106 (in N·m, not N); the effective length is squared.
This uses K = 0.7 (fixed-pinned), which does not match pinned ends.
Pcr = π2EI/(KL)2 with K = 1: π2 × 200 × 109 × 8 × 10−6/9 = 1.75 × 106 N.
Question 16 of 16 · Mechanics of Materials · 9–14 items
A simply supported beam of length L carries a concentrated load P at the center. What is the maximum deflection if EI is constant?
Solution
δmax = PL3/(48EI) at midspan
FE Reference Handbook, Mechanics of Materials: simply supported beam slopes and deflections.
The midspan deflection of a simply supported beam under a central point load is PL3/(48EI).
5/384 is the coefficient for a uniform load (with wL4); a central point load gives PL3/(48EI).
PL3/(3EI) is the tip deflection of a cantilever with an end load.
L2 gives a slope-type quantity, not a deflection; PL2/(16EI) is the end slope of this beam.
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Common traps
Most wrong answers in this area use a correct formula on the wrong property or the wrong point of the section. Checked against the Mechanics of Materials chapter of the FE Reference Handbook and Gere and Goodno's Mechanics of Materials:
J for torsion, I for bending. The polar moment of a circular shaft is twice the area moment about a diameter; using one for the other is a factor-of-two error that lands on a listed option.
Top fiber or bottom fiber. Bending stress changes sign across the neutral axis. In an unsymmetrical section the two extreme fibers sit at different distances, so the larger stress is not automatically where the moment is drawn.
In-plane versus absolute maximum shear. Mohr's circle for the plane gives the in-plane value. When both in-plane principal values are tensile, or both compressive, the absolute maximum shear brings in the third principal stress, which is zero.
Angle doubling. Rotations on Mohr's circle are twice the rotation of the element. Report the element angle.
Thermal strain needs a restraint to become stress. A free bar just grows. Ask first whether anything stops it.
Effective length. Buckling depends on the end conditions through the effective length factor, not on the actual length alone.
Superposition. The Handbook carries standard beam cases. A beam with two loads is usually two table cases added, as long as the material stays elastic.
Engineering stress is the default
Unless a question says otherwise, stress is load over the original area and strain is change in length over the original length. True stress and true strain belong to large-deformation questions in material properties.
What mechanics of materials feeds
Statics supplies the reactions and the centroid and area-moment calculations. Mechanical design and analysis applies the same stresses to shafts, springs, bolts and pressure vessels, then adds failure theories and fatigue. The stress-strain diagram and material constants come from material properties and processing.
Bridge and building members are covered in the civil discipline's structural engineering area.
Yes. Standard beam cases are tabulated in the on-screen Handbook. You still need to pick the right support case and add cases together when a beam carries more than one load.
How many mechanics of materials questions are on the FE Mechanical exam?
9–14 questions, the same range as statics. With dynamics, these three areas form the largest block of the outline.
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