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NCEES PE Agricultural and Biological Engineering Strength of Materials: practice questions

NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)

Original practice questions written to the published syllabus · not past papers
10worked here, free
18written to this exam in this subject
12syllabus units

About NCEES PE Agricultural and Biological Engineering

NCEES PE Agricultural and Biological Engineering is a licensing and registration exam run by NCEES (ASABE is the content partner), for candidates in USA. Computer-based, 85 questions, 9.5-hour appointment (8.5 hours of exam). Once a year (next: 28 Oct 2026). Closed book with the electronic NCEES handbook and supplied searchable ASABE standards; SI and US customary units.

Source: ncees.org.

Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.

Strength of Materials in NCEES PE Agricultural and Biological Engineering

Not a stand-alone section; the methods are used inside Facilities & Structures (beams, columns, trusses and bins) and Machinery Systems (shafts, pins and frames) questions, with ASABE standards supplied.

How members carry load: stress, strain, bending, torsion, deflection and buckling, used in machine and structure design.

Our 18 NCEES PE Agricultural and Biological Engineering questions in this subject: 0 easy, 12 medium, 6 hard, across 12 units.

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

These 10 were written to NCEES PE Agricultural and Biological Engineering and are not shown on any other exam's page.

Question 1 · medium

A farm-machinery test engineer compares a mild-steel and a cast-iron specimen after tension tests. Which observation correctly identifies the mild-steel specimen?

  1. It shows a flat granular fracture with almost no necking at all
  2. It shows a cup-and-cone fracture after visible necking
  3. It breaks without any yield plateau on its load–extension record
  4. It has less elongation at fracture than the cast-iron specimen
Answer and working

B. It shows a cup-and-cone fracture after visible necking

Ductile mild steel necks and breaks with a cup-and-cone surface after large elongation. A flat, granular break with little elongation and no yield plateau is typical of brittle grey cast iron.

Question 2 · medium

One arm of a sprayer boom is modelled as a 5 m cantilever carrying a uniformly distributed load of 100 N/m plus a 300 N end nozzle assembly. The bending moment at the root is:

  1. 2,750 N·m
  2. 1,812 N·m
  3. 4,000 N·m
  4. 688 N·m
Answer and working

A. 2,750 N·m

M = WL + wL²/2 = 300 × 5 + 100 × 25/2 = 1500 + 1250 = 2750 N·m. Using wL²/8 applies the simply supported result to a cantilever. 1,812 N·m uses wL²/8 for the boom load, 4,000 N·m puts the whole 800 N at the tip, and 688 N·m uses the simply supported forms WL/4 + wL²/8.

Question 3 · hard

A tubular PTO drive shaft with outer diameter 50 mm and inner diameter 40 mm transmits 1 kN·m. The maximum shear stress is closest to:

  1. 40.7 MPa
  2. 69.0 MPa
  3. 83.5 MPa
  4. 55.2 MPa
Answer and working

B. 69.0 MPa

τ_max = 16TD/[π(D⁴ − d⁴)] = 16 × 10⁶ × 50/[π(50⁴ − 40⁴)] = 69.0 MPa at the outer surface. Ignoring the bore gives 40.7 MPa; the stress at the inner surface is 55.2 MPa. Using cubes instead of fourth powers (16T/[π(D³ − d³)]) gives 83.5 MPa.

Question 4 · medium

A failed steel PTO shaft is sent for inspection. It broke under torsional overload across a flat surface perpendicular to its axis, after visible twisting. This indicates:

  1. A brittle tensile failure on the plane of maximum principal stress
  2. A buckling failure caused by axial thrust from the implement
  3. A fatigue failure started by a keyway, since all torsion failures are 45°
  4. A ductile shear failure on the plane of maximum shear stress
Answer and working

D. A ductile shear failure on the plane of maximum shear stress

In torsion the maximum shear acts on transverse (and longitudinal) planes, and ductile steel fails in shear there, with prior twisting. Brittle materials such as cast iron or chalk fail in tension on a 45° helix. A fatigue fracture would show beach marks, not gross twisting.

Question 5 · medium

A rectangular steel tool bar deflects too much under draft load, making tines work at uneven depth. With the same width and material, doubling the section depth reduces the deflection to:

  1. One-quarter of the original
  2. One-half of the original
  3. One-sixteenth of the original
  4. One-eighth of the original
Answer and working

D. One-eighth of the original

Deflection ∝ 1/I and I = bd³/12, so doubling d multiplies I by 8 and cuts deflection to 1/8. Depth is the cheapest way to stiffen a bar; bending stress falls to 1/4 because Z ∝ d².

Question 6 · hard

A 100 mm × 40 mm steel flat-bar strut (E = 200 GPa), 2.5 m long, is pinned at both ends so that it can buckle in any direction. The Euler critical load is closest to:

  1. 168.4 kN
  2. 1,053 kN
  3. 42.1 kN
  4. 674 kN
Answer and working

A. 168.4 kN

Buckling occurs about the weaker axis: I_min = 100 × 40³/12 = 533,333 mm⁴. P_cr = π² × 200,000 × 533,333/2500² = 168.4 kN. Using I about the strong axis (3,333,333 mm⁴) overestimates by more than six times. The strong-axis value is 1,053 kN, the fixed–free value is 42.1 kN, and both ends fixed (L_e = L/2) gives 674 kN.

Question 7 · medium

The legs of an elevated farm water tank are slender pinned–pinned steel columns governed by Euler buckling. Adding horizontal bracing that holds each leg against sideways movement at mid-height raises the buckling load to:

  1. About twice the unbraced value
  2. About 1.4 times the unbraced value
  3. About four times the unbraced value
  4. The same value, since the leg length is unchanged
Answer and working

C. About four times the unbraced value

Bracing at mid-height halves the effective length; P_cr ∝ 1/L_e² rises fourfold, provided the braces are stiff enough and the leg stays elastic. If the new critical stress exceeds the yield range, crushing will govern instead.

Question 8 · medium

A welded air receiver of 800 mm diameter is to work at 1.5 MPa with an allowable stress of 100 MPa. The longitudinal welded joint efficiency is 80 percent. The minimum wall thickness (ignoring corrosion allowance) is:

  1. 6.00 mm
  2. 7.50 mm
  3. 3.750 mm
  4. 4.80 mm
Answer and working

B. 7.50 mm

The longitudinal joint weakens the hoop direction: t = pD/(2ση_l) = 1.5 × 800/(2 × 100 × 0.8) = 7.50 mm. Ignoring the efficiency gives 6.00 mm; multiplying by η instead of dividing gives 4.80 mm, which is unsafe. 3.750 mm uses the longitudinal-stress form pD/(4ση).

Question 9 · hard

A closed thin steel cylinder, 1 m internal diameter, 3 m long and 10 mm thick (E = 200 GPa, ν = 0.3), is pressurised to 1.5 MPa. The increase in its internal volume is closest to:

  1. 1.94 litres
  2. 1.68 litres
  3. 0.93 litres
  4. 0.88 litres
Answer and working

B. 1.68 litres

ε_v = (pD/4tE)(5 − 4ν) = (1.5 × 1000/(4 × 10 × 200,000)) × 3.8 = 7.125e-04. V = π × 1²/4 × 3 = 2.356 m³. ΔV = 7.125e-04 × 2.356 m³ = 1.679e-03 m³ = 1.68 L. Using the sphere factor 3(1 − ν) instead of (5 − 4ν) underestimates the change. Using (5 − 2ν) gives 1.94 litres, and taking the hoop strain pD/(2tE) alone as the volumetric strain gives 0.88 litres.

Question 10 · hard

Shank bolts that carry impact loads on a plough frame are sometimes turned down to the thread-root diameter along the shank. The reason is that this:

  1. Makes the shank stronger in static tension than the threaded part
  2. Raises the bolt's stiffness so that it deflects less under impact
  3. Moves the maximum stress from the threads to the head of the bolt
  4. Spreads strain energy along the bolt, lowering peak stress in impact
Answer and working

D. Spreads strain energy along the bolt, lowering peak stress in impact

Under impact, stress depends on the energy that must be absorbed per unit volume. A uniform smaller section stresses the whole length evenly, so more volume absorbs energy and the peak stress at the thread root falls. It reduces, not raises, stiffness and static strength.

A study plan for Strength of Materials in NCEES PE Agricultural and Biological Engineering

This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.

  1. First, the heavy units (50% of the 18 questions): Torsion of circular shafts (3, 17%); Columns: Euler and Rankine formulas (2, 11%); Deflection of beams (2, 11%); Theories of failure (2, 11%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Thin cylinders and spheres (2, 11%); Bending and shear stresses in beams (1, 6%); Combined stresses and Mohr's circle (1, 6%); Elastic constants and their relations (1, 6%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Shear force and bending moment diagrams (1, 6%); Strain energy (1, 6%); Stress, strain and Hooke's law (1, 6%); Thermal stresses (1, 6%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Bending and shear stresses in beams (1 hard of 1), Strain energy (1 hard of 1), Thermal stresses (1 hard of 1).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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