Madrid PAU Tech & Engineering II common mistakes
Madrid PAU Tech & Engineering II · Understand the traps and how to avoid them.
When a beam or frame is drawn with a roller, pin and fixed support, the candidate sums vertical forces and moments but never writes the horizontal equilibrium equation, so the horizontal reaction at the pin (or the horizontal component transmitted by a hinge) is omitted from the final answer.
Textbook problems are often vertical-load-only, so candidates build a habit of treating ΣFx as trivially zero. When an inclined load, a cable, or a horizontal thrust from an arch appears, the missing equation silently drops a reaction and every downstream internal-force value is wrong.
What to do instead
Before solving, list every unknown reaction with its assumed direction and count them against the three equilibrium equations (or six for two bodies). Always write ΣFx = 0 explicitly, even when it looks trivial, and state the sign convention (right/up/anticlockwise positive) at the top of the solution so markers can follow it.
The candidate computes normal stress as σ = F/A using the full applied load on a bolted or riveted lap joint, or on a member with a hole, instead of using the net cross-section, and also omits the bearing and shear checks on the fastener.
The formula σ = F/A is over-learned on plain bars. Joints require three separate checks (net-section tension, shear of the fastener, bearing on the plate), and candidates collapse them into one, producing an unsafe or over-conservative answer.
What to do instead
For any connection, draw the net section and write A_net = (w − n·d)·t. Then check in order: tension on the net section, τ = F/(n·A_fastener) for shear, and σ_bearing = F/(n·d·t) for bearing. State which check governs and why.
In a gear train or belt-and-pulley problem, the candidate multiplies or divides speed ratios in the wrong order and forgets that torque scales inversely with speed, so the output torque is reported as smaller than the input when the output turns more slowly.
The ratio i = z_driven/z_driver is memorised without a physical anchor. Candidates also forget that an idler gear changes direction but not ratio, and that a worm drive gives a large reduction in a single step.
What to do instead
Anchor every answer in energy: P = T·ω is conserved (minus efficiency), so if ω falls, T rises. Write the chain as a product of stage ratios with each stage labelled driver/driven, cancel idlers explicitly, then apply T_out = T_in · i · η and check the direction of rotation stage by stage.
When calculating the equivalent resistance of a mixed series-parallel network, the candidate reduces the network in a single pass from left to right without redrawing, and misidentifies which resistors are actually in parallel because two components share a node only through the supply.
Parallel means 'same two nodes', but candidates judge by visual layout on the page. After each reduction the topology changes, and without redrawing the next step is applied to the wrong pair.
What to do instead
Redraw the circuit after every reduction, marking nodes with letters (A, B, C) and labelling each resistor by the pair of nodes it connects. Two resistors are in parallel only if they share both node letters. Reduce until one equivalent resistance remains, then work backwards to find branch currents and verify with Kirchhoff's current law at each node.
The candidate applies Ohm's law to a real source (battery with internal resistance, or a sensor output) as V = I·R using only the external resistor, ignoring the internal resistance and the resulting voltage drop under load.
Ideal-source problems dominate practice, and the internal resistance is often a small number that candidates dismiss as negligible. The error shows up as a terminal voltage that does not match the measured value in the statement.
What to do instead
Always write the terminal voltage as V_terminal = ε − I·r and solve the loop equation ε = I(R + r). Check the result: the terminal voltage must fall as current rises, and open-circuit voltage must equal ε. Report the load power separately from the total power dissipated in the source.
When simplifying a logic function with Karnaugh maps or Boolean algebra, the candidate groups cells diagonally, groups an odd number of cells, or forgets to wrap groups around the map edges, and then presents a simplified expression that is not equivalent to the truth table.
The rules of grouping (powers of two, adjacency including wrap-around, don't-cares) are procedural and easy to half-remember, and candidates stop as soon as the expression 'looks shorter' without verifying it.
What to do instead
Enforce a verification step: substitute every input combination from the truth table into the simplified expression and confirm it matches. When grouping, mark each group with its literal term and check that every 1 is covered by at least one group and no group contains a 0. Treat don't-care cells as 1 only when it enlarges a group.
For a mechanism at a given crank angle, the candidate computes the instantaneous velocity of a point using v = ω·r with r taken as the distance from the fixed pivot to the point, rather than the perpendicular distance from the instant centre of rotation, and then reports the wrong direction.
The circular-motion formula is applied to a linkage as if every point rotated about the fixed joint. In a four-bar linkage the coupler point rotates about its own instant centre, and the perpendicular-distance rule is the only correct route.
What to do instead
Locate the instant centre by intersecting the lines through the two end velocities (each perpendicular to its radius). Then v = ω·(distance from instant centre), with direction perpendicular to that line. For accelerations, use the relative-acceleration equation with a normal component v²/r and a tangential component, and draw the vector polygon.
The candidate describes a closed-loop control system but cannot distinguish the sensor, the actuator and the controller in a block diagram, and labels the setpoint as the measured variable, so the error signal is computed with the wrong sign or the feedback is drawn as feedforward.
Everyday language blurs 'control' and 'measure'. Candidates memorise the words 'sensor, actuator, controller' without mapping them onto the physical elements of the described process, and sign conventions in the summing junction are rarely practised.
What to do instead
For each system, write four named lines: setpoint (desired value), error = setpoint − measured, controller output, and measured variable from the sensor. Then name the physical device for each block (e.g. thermistor for measurement, relay or motor for actuation, PLC for control). Draw the feedback arrow from the measured variable back to the summing junction with an explicit minus sign.
In technical drawing, the candidate omits or misplaces dimensions (dimensioning to hidden lines, missing the overall dimension, duplicating a dimension already implied), and applies scale or tolerance notation incorrectly on the drawing.
Candidates focus on the geometry of the views and treat dimensioning as an afterthought, and the standard rules (no redundant dimensions, dimensions outside the outline, tolerance and fit notation) are learned passively rather than practised on complete drawings.
What to do instead
Finish every drawing with a dimensioning pass: one overall length, width and height; each feature dimensioned once only; no dimension on a hidden line; tolerances written as nominal with upper and lower deviation or an ISO fit code. State the scale in the title block and check that every dimension in the drawing can be measured from the views as drawn.
When asked to select a material for a component, the candidate lists properties in the abstract ('strong', 'light', 'cheap') without linking them to the service conditions, and ignores the environmental impact or recyclability of the choice.
Selection is treated as a vocabulary test rather than a design decision. Candidates do not rank requirements (strength, stiffness, density, corrosion, cost, recyclability) or compare candidate materials against them, so the answer cannot be justified.
What to do instead
Use a short decision matrix: list the 3–4 governing requirements from the service conditions, weight them, score two or three candidate materials, and state the winner with the reason. Include one environmental criterion (recyclability, energy of production, or end-of-life) and note any trade-off explicitly, e.g. aluminium is lighter but more energy-intensive to produce than steel.