Galicia PAU Technology and Engineering II common mistakes
Galicia PAU Technology and Engineering II · Understand the traps and how to avoid them.
When calculating a tensile stress, substituting the load in kgf or the diameter in mm directly into σ=F/A and reporting the result in MPa without converting to newtons and square metres, so the answer is off by a factor of about 10 (or 9.81).
Galician textbooks and workshop data sheets often give loads in kp/ kgf and sections in mm², and candidates carry the number straight into the formula, forgetting that 1 MPa = 1 N/mm² only when the force is in newtons.
What to do instead
Before substituting, write a one-line unit statement: F in N (multiply kgf by 9.81), A in mm², then σ comes out directly in N/mm² = MPa. Practise the identity 1 N/mm² = 1 MPa until it is automatic, and always state units next to every intermediate result.
Confusing the elastic section modulus W with the second moment of area I when applying the bending formula, or using the full height h instead of h/2 as the distance to the extreme fibre.
Both symbols appear together in the same bending problems and both depend on the same cross-section; candidates memorise σ = M·y/I but then plug in I where W = I/y is required, or forget that y is measured from the neutral axis to the outermost fibre.
What to do instead
Always write the pair explicitly: I for the section (mm⁴), y_max = h/2 from the neutral axis, W = I/y_max (mm³), σ_max = M/W. Compute I and W for the rectangle, circle and hollow circle once by hand and keep that sheet as a reference during revision.
Resolving a statics problem without first drawing the free-body diagram, so reaction components, unknown forces in bars and the direction of friction or cable tensions are guessed rather than derived.
Under exam time pressure candidates jump to equilibrium equations, and sign errors in the reactions then propagate silently through the whole problem; the diagram feels like wasted time.
What to do instead
Make the FBD a mandatory first step: isolate the body, draw every force with an arrow and a label, mark the chosen positive axes, then write ΣFx=0, ΣFy=0, ΣM=0. Check the result by taking moments about a second point; if the reactions do not satisfy it, the diagram or a sign is wrong.
Reporting a mechanism's degrees of freedom or a gear train's output speed without checking the Grübler/Kutzbach count or the sign/direction of each stage, giving a plausible number with the wrong direction or a locked mechanism.
Candidates count links and joints from memory of a diagram and forget that a redundant constraint or a higher pair changes the count; in gear trains they multiply ratios but ignore that an internal mesh reverses rotation while an external one does not.
What to do instead
Use DOF = 3(n−1) − 2j₁ − j₂ with n counted as moving links only, and verify the result against the physical intuition (a four-bar has 1 DOF). For trains, draw an arrow on each shaft as you go, multiply the tooth ratios stage by stage, and state the final direction explicitly in the answer.
Treating a real voltage or current source as ideal when computing a Thévenin equivalent or a voltage divider, ignoring internal resistance, and then being surprised that the measured terminal voltage differs from the calculated one.
Classroom exercises usually present ideal sources, so candidates apply the ideal divider formula to a circuit that includes a source resistance or a load, and the systematic error is invisible unless the numbers are checked against the circuit.
What to do instead
Before solving, mark on the schematic which elements are inside the source and which are the load. If a source resistance is given, redraw it as an explicit series resistor, then apply the divider or Thévenin method to the resulting network. Sanity-check with the open-circuit and short-circuit limits.
In three-phase or transformer problems, mixing line and phase quantities, or applying the √3 factor to the wrong magnitude, so power comes out three times too large or too small.
Star and delta connection formulas look symmetric and candidates memorise only one version; the distinction between line voltage and phase voltage, and between active, reactive and apparent power, blurs under pressure.
What to do instead
Write the connection at the top of the page (star: U_L = √3·U_f, I_L = I_f; delta: U_L = U_f, I_L = √3·I_f) and only then substitute. Compute P = √3·U_L·I_L·cos φ for a balanced load and cross-check with 3·U_f·I_f·cos φ; the two must agree.
Reading a ladder diagram or truth table and writing the Boolean expression without applying De Morgan or without distinguishing normally-open from normally-closed contacts, so the logic implemented is the complement of the intended one.
Ladder logic is drawn for relay practice where a closed contact conducts, while Boolean algebra treats a variable as true when energised; the two conventions conflict and candidates translate mechanically.
What to do instead
Annotate every contact with the variable and its state (X, X̄) before writing the expression, then simplify with Boolean algebra and De Morgan. Verify by simulating the truth table row by row, and check the fail-safe direction: a stop button must de-energise the circuit.
Designing a pneumatic circuit that cannot move because both sides of a double-acting cylinder are pressurised simultaneously, or omitting the exhaust/flow-control valve so the actuator strokes uncontrollably.
Candidates focus on getting the cylinder to extend and forget the return path, the 5/2 versus 3/2 valve distinction, and that speed control in pneumatics is done by meter-out throttling on the exhaust, not by restricting the supply.
What to do instead
Build the circuit in a fixed order: supply → filter-regulator → directional valve → actuator, and trace the exhaust path for each valve position. Use a 5/2 valve for double-acting cylinders, place flow-control valves on the exhaust lines (meter-out) for smooth motion, and verify each position by asking where the air goes.
Selecting a material by a single headline property (highest strength, lowest density) instead of matching the full requirement set, and ignoring manufacturability, corrosion or cost constraints that the problem statement implies.
Ashby-style charts and property tables invite one-number comparisons, and candidates treat the material choice as a maximisation problem rather than a constraint-satisfaction problem with a cost objective.
What to do instead
List the constraints first (must withstand load X without yielding, must resist the environment, must be weldable or machinable), screen materials against them, then rank the survivors by the relevant index (σ/ρ, stiffness, cost per kg). Justify the final choice in one sentence naming the deciding constraint.
In energy and efficiency questions, computing efficiency as output/input with mismatched units (kWh in, kJ out) or forgetting that the overall efficiency of a series chain is the product of stage efficiencies, not the sum or the average.
Energy problems mix thermal, mechanical and electrical units, and candidates add efficiencies because they think of them as percentages of the same whole; the compounding is counter-intuitive.
What to do instead
Convert everything to watts or joules before dividing, and write η_total = η₁·η₂·…·ηₙ explicitly for a chain. Check that the result is below the smallest individual efficiency and below 100 %; if not, the units or the multiplication is wrong.