Nepal Engineering Council Registration Exam — Agricultural Engineering · Nepal · Nepal Engineering Council
Original practice questions written to the published syllabus · not past papersNepal Engineering Council Registration Exam — Agricultural Engineering is a licensing and registration exam run by Nepal Engineering Council, for candidates in Nepal.
Official site: nec.gov.np.
We have not been able to confirm these details against the exam body's own specification, so fees, dates and pass marks are left out on purpose. Check the current format, eligibility, fees and dates on the official site.
How members carry load: stress, strain, bending, torsion, deflection and buckling, used in machine and structure design.
Our 19 Nepal Engineering Council Registration Exam questions in this subject: 11 easy, 7 medium, 1 hard, across 12 units.
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 Nepal Engineering Council Registration Exam and are not shown on any other exam's page.
A 100 m long steel cable hangs freely down a deep bore well, carrying only its own weight. With density 7850 kg/m³ and E = 200 GPa, its elongation under self-weight is closest to:
B. 1.93 mm
Self-weight elongation δ = ρgL²/(2E) = 7850 × 9.81 × 100²/(2 × 200 × 10⁹) = 1.925e-03 m = 1.93 mm. Forgetting the ½ (treating the whole weight as an end load) gives 3.85 mm. 0.193 mm and 19.3 mm are decimal slips.
The number of independent elastic constants needed to describe a homogeneous, isotropic, linear-elastic material is:
B. Two
Of E, G, K and ν only two are independent; the others follow from E = 2G(1 + ν) = 3K(1 − 2ν). Twenty-one is the count for a fully anisotropic material.
A steel bolt passes through a brass tube and the nut is snug with no initial stress. The assembly is then heated uniformly. Brass has the larger coefficient of expansion. After heating:
A. The brass tube is in compression and the steel bolt in tension
Brass wants to grow more than steel; the steel holds it back, putting brass in compression, and the brass stretches the steel, putting it in tension. The forces are equal and opposite because there is no external load.
For a simply supported beam carrying a uniformly distributed load over its whole span, the shear force and bending moment diagrams are respectively:
B. A sloping straight line and a parabola
dV/dx = −w (constant), so V is linear; dM/dx = V (linear), so M is a second-degree curve, a parabola. A parabola and cubic go with a uniformly varying load; a constant V and linear M go with point loads only.
A clockwise concentrated couple applied at a point on a simply supported beam causes, at that point:
A. A sudden jump in the bending moment diagram only
A pure couple adds no transverse force, so V is unchanged across it, but M jumps by the value of the couple. A point load causes a jump in V and a kink in M.
In a steel I-section beam of a farm building, most of the vertical shear force is carried by the:
A. Web
Shear stress τ = VAȳ/(Ib) is small in the wide flanges and large in the narrow web, which carries most of the shear force; the flanges carry most of the bending moment.
A 50 mm × 100 mm timber joist is used on edge (100 mm vertical) instead of flat. Its bending strength (section modulus) is increased by a factor of:
B. 2
Z = bd²/6: on edge 50 × 100²/6 = 83,333 mm³; flat 100 × 50²/6 = 41,667 mm³; ratio 2 (= d/b). The stiffness (I) ratio would be 4. 4 is the stiffness (I) ratio, 8 is (d/b)³, and 1 assumes strength depends only on area, which is unchanged.
Two shafts of different diameters are joined end to end (in series) and transmit a torque. Which statement is correct?
B. Both carry the same torque and the total twist is the sum of their twists
In series the same torque passes through each shaft, and the angles of twist add. Equal twist with torques adding is the parallel (compound) case.
According to the first moment-area theorem, the change in slope between two points on the elastic curve equals the:
C. Area of the M/EI diagram between the two points
First theorem: Δθ = ∫M/(EI)dx, the area under M/EI. The second theorem uses the first moment of that area to give the tangential deviation. The SFD area gives the change in bending moment.
An element is in a state of pure shear τ. The principal stresses and the orientation of the principal planes are:
C. +τ and −τ, on planes at 45° to the shear planes
Mohr's circle for pure shear is centred at the origin with radius τ, so σ₁ = +τ and σ₂ = −τ at 45° to the shear planes (90° on the circle). This is why brittle materials twisted to failure break on a 45° helix.
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.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
Do these early in revision so they are fast on the day.
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