Mechanics of Materials
7–11 of 110 itemsNCEES draws between 7 and 11 of the 110 from this area on any legal paper. Its outline lists 4 sub-topics; the bank holds 37 verified templates against them, which expand to 1,147 questions.
Sub-topics, from the NCEES outline
- AShear and moment diagrams
- BStresses and strains (e.g., diagrams, axial, torsion, bending, shear, thermal)
- CDeformations (e.g., axial, torsion, bending, thermal)
- DCombined stresses, principal stresses, and Mohr's circle
5 worked problems
Real questions from the bank, at the depth the exam asks them. Pick an answer first: the reasoning opens when you do, along with why each wrong answer is tempting.
Question 1 of 5
A 6-ft member stretches 0.18 in under load. The normal strain is most nearly:
Member 6 ft long that stretches 0.18 inches under load; strain is that change divided by the original length in the same units. Show the answer and the reasoning
Why B
Given: delta = 0.18 in, L = 6 ft = 72 in. Relation: epsilon = delta/L, both in the same unit. Substitute: epsilon = 0.18/72 Result: epsilon = 0.00250.
Why A is tempting
Converted the length from feet to inches twice over.
Why C is tempting
Left the original length in feet while the elongation is in inches. Strain is dimensionless only when both are in the same unit.
Why D is tempting
Reported the elongation rather than the strain.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 2 of 5
A beam overhangs a support by 13 ft. The overhang carries a 10-kip load at its tip and 1 kip/ft of uniform load. The magnitude of the moment over the support is most nearly:
Beam overhanging its support by 13 ft, carrying a 10 kip load at the free end and a uniform load of 1 kip per foot on the overhang. Show the answer and the reasoning
Why D
Given: Overhang a = 13 ft, tip load P = 10 kips, w = 1 kip/ft. Relation: The overhang acts as a cantilever from the support: M = P a + w a^2/2 (hogging). Substitute: M = 10(13) + 1(13)^2/2 Result: M = 214.5 ft-kips.
Why A is tempting
Ignored the concentrated load at the tip.
Why B is tempting
Ignored the distributed load on the overhang.
Why C is tempting
Used wL^2/8, which belongs to a simply supported span rather than a cantilevered overhang.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 3 of 5
A plane stress element has σx = 25 ksi, σy = 20 ksi and τxy = 4 ksi. The maximum principal stress is most nearly:
Plane stress element with a normal stress of 25 ksi on the x faces, 20 ksi on the y faces, and a shear stress of 4 ksi. Show the answer and the reasoning
Why C
Given: sigma_x = 25 ksi, sigma_y = 20 ksi, tau_xy = 4 ksi. Relation: sigma_1 = C + R, C = (sigma_x + sigma_y)/2, R = sqrt(((sigma_x - sigma_y)/2)^2 + tau_xy^2). Substitute: C = 22.50 ksi, R = 4.717 ksi Result: sigma_1 = 27.22 ksi.
Why A is tempting
Gave the minimum principal stress. Both are valid roots, but the maximum takes the plus sign.
Why B is tempting
Added the shear stress directly instead of combining it with the normal stress difference.
Why D is tempting
Left out the factor of one half on the stress difference inside the radical.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 4 of 5
A solid circular shaft 1.5 in in diameter carries a torque of 10 in-kips. The maximum torsional shear stress is most nearly:
Solid circular shaft 1.5 in in diameter carrying a torque of 10 inch-kips; the shear stress grows linearly from zero at the axis to a maximum at the outer fibre. Show the answer and the reasoning
Why D
Given: T = 10 in-kips, D = 1.5 in. Relation: tau = Tr/J with r = D/2 and J = pi D^4/32, which reduces to 16T/(pi D^3). Substitute: tau = 16(10)/(pi x 1.5^3) Result: tau = 15.09 ksi.
Why A is tempting
Used the thin-walled tube formula T/(2 pi r^2 t) with t = r. A solid shaft is not a thin tube; its stress follows Tr/J.
Why B is tempting
Halved the coefficient.
Why C is tempting
Raised the diameter to the fourth power. That belongs to J itself, not to the stress.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 5 of 5
A short composite column carries 175 kips. It contains 2 in^2 of steel (E = 29,000 ksi) and 45 in^2 of concrete (E = 4,000 ksi), bonded so both shorten equally. The load carried by the steel is most nearly:
Short composite column carrying 175 kips through two materials bonded to shorten together, with areas of 2 and 45 square inches and moduli of 29000 and 4000 ksi. Show the answer and the reasoning
Why A
Given: P = 175 kips; steel 2 in^2 at 29,000 ksi, concrete 45 in^2 at 4,000 ksi. Relation: Equal strain, so each material takes load in proportion to AE: P_s = P(AE)_s/sum(AE). Substitute: P_s = 175(58,000)/(58,000 + 180,000) Result: P_s = 42.65 kips.
Why B is tempting
Divided by the other material stiffness rather than by the total.
Why C is tempting
Gave the share carried by the other material.
Why D is tempting
Shared by modulus alone, ignoring how much of each material is present.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Four mistakes that cost the question
Each of these lands on an answer that is offered, so it costs the question outright. The minutes are our estimate of the time each one burns on top, against an average of 2.9 minutes a question (320 minutes for 110).
Costs the question and about 4 min
Taking the maximum moment at midspan by habit
The largest positive moment sits where the shear diagram crosses zero, which is midspan only for symmetric loading, and on a beam with an overhang the largest magnitude may be the negative moment over the support. Sketch the shear diagram before reading off any moment.
Costs the question and about 2 min
Using the diameter where the formula wants the radius
Torsion and bending stresses use c, the distance to the extreme fibre: half the diameter or half the depth. The polar moment of a solid shaft is πd⁴/32 with the diameter or πr⁴/2 with the radius. Mixing the two forms puts the answer 2 or 16 times off.
Costs the question and about 3 min
Reading a principal stress where the question asks for maximum shear
On Mohr's circle the centre is (σx + σy)/2 and the radius is the square root of ((σx − σy)/2)² + τxy². The principal stresses are the centre plus and minus the radius; the maximum in-plane shear is the radius alone. The usual slips are dropping the /2 inside the root and reporting centre plus radius for shear. Angles on the circle are twice those on the element.
Costs the question and about 3 min
Sharing load between steel and concrete by area
In a short composite column the two materials strain together, so they share load in proportion to E times A, not A alone. Steel is roughly eight times stiffer than ordinary concrete, so splitting by area understates the steel's share several times over.
Questions about Mechanics of Materials
How is Mechanics of Materials different from Structural Engineering?
Mechanics of materials is stress, strain and deformation in one member: axial, torsion, bending, shear and combined stress. Structural engineering builds on it with whole structures, load combinations, indeterminacy and code design in steel and concrete. Shear and moment diagrams belong to both.
Do I need Mohr's circle, or are the formulas enough?
Either gives the same principal stresses and maximum shear. The formulas are quicker for a single number; the circle is quicker when the question asks for an orientation or for several quantities at once. Settle on one sign convention for shear before exam day and use it every time.
What unit slip costs the most in this area?
Feet against inches. Moments come in ft-kips while section properties come in in⁴, so a bending stress needs the moment multiplied by 12. Leave the 12 out and the stress is twelve times too small, an answer that is offered.
Which formulas should I be able to use without looking up?
σ = P/A, σ = Mc/I, τ = Tc/J, τ = VQ/(Ib) and δ = PL/(AE). They appear in nearly every question in the area, and looking each one up costs time you will want for the shear and moment diagrams.
How many study hours this area is worth · Using the handbook under the clock