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Hamburg Abitur Vocational Technology - independent engineering foundations study materials
Hamburg Abitur Vocational Technology - independent engineering foundations study resources

Hamburg Abitur Vocational Technology - independent engineering foundations practice questions

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Question 1Aerodynamic forces and elementary stability models

Use the supplied model L=0.5ρv²SC_L, withρ=1.2 kg/m³,v=20 m/s,S=2 m²andC_L=0.5. What lift magnitude follows?

  1. 120N.
  2. 24N.
  3. 480N.
  4. 240N. Correct answer.

Explanation

Substituting the supplied values gives 0.5×1.2×400×2×0.5=240 N. This is a conditional aerodynamic model result, not an approved aircraft load or an operating limit.

Question 2Aerodynamic forces and elementary stability models

In a supplied aerodynamic model, density, reference area and lift coefficient remain fixed while airspeed rises from 10 m/s to 15 m/s. By what factor does lift change?

  1. 3.0.
  2. 0.667.
  3. 2.25. Correct answer.
  4. 1.5.

Explanation

With density, area and coefficient fixed, modeled lift is proportional to speed squared. The ratio is(15/10)²=2.25; actual coefficient changes or other conditions would require a different calculation.

Question 3Aerodynamic forces and elementary stability models

A supplied comparison gives lift 600 N and drag 40 N for one condition. What is the lift-to-drag ratio?

  1. 24.
  2. 0.067.
  3. 15. Correct answer.
  4. 560.

Explanation

Divide like force units:600/40=15. This dimensionless ratio describes the stated aerodynamic condition; it does not by itself establish aircraft range, structural safety or a complete flight-performance prediction.

Question 4Aerodynamic forces and elementary stability models

After a small modeled pitch disturbance, an initial aerodynamic moment acts back toward the original attitude. Which property does that observation directly describe?

  1. Absence of any aerodynamic drag.
  2. Guaranteed decay of every later oscillation.
  3. A static restoring tendency. Correct answer.
  4. An unchanged center of gravity for every possible loading.

Explanation

The direction of the initial response describes static stability. Whether the subsequent motion decays, persists or grows requires dynamic information, so an initial restoring moment alone does not guarantee damping.

Question 5Aerodynamic forces and elementary stability models

Why is equal transit time of upper and lower air parcels not a valid general explanation of airfoil lift?

  1. Only gravity determines the aerodynamic force on a wing.
  2. Pressure forces cannot contribute to aerodynamic lift.
  3. The separated parcels are not required to reunite at the trailing edge at the same time. Correct answer.
  4. A wing must always have zero angle of attack to generate lift.

Explanation

The equal-transit premise imposes an unsupported timing requirement on air parcels. Pressure distributions and airflow momentum describe aerodynamic interaction; rejecting that premise does not reject pressure forces or the role of geometry and angle of attack.

Question 6Aerodynamic forces and elementary stability models

Two wind-tunnel reports calculate a lift coefficient from the same measured lift, density and speed, but one uses twice the reference area. What coefficient does that report obtain?

  1. Twice the coefficient of the other report.
  2. Four times the coefficient of the other report.
  3. The same coefficient regardless of area convention.
  4. Half the coefficient of the other report. Correct answer.

Explanation

RearrangingL=0.5ρv²SC_LgivesC_L=L/(0.5ρv²S). Doubling the reference area with the other values fixed halves the reported coefficient. Comparisons must therefore use consistent reference definitions.

Question 7Statics, strength and mechanical transmission

Uniform line load q = 2.0 kN/m on Einfeldträger L = 4.0 m. What is M_max?

  1. 2 kN·m
  2. 8 kN·m
  3. 4 kN·m Correct answer.
  4. 16 kN·m

Explanation

M_max = q L² / 8 = 2 × 16 / 8 = 4 kN·m.

Question 8Statics, strength and mechanical transmission

Self-weight of a beam is modelled as q = 0.5 kN/m on L = 6 m Einfeldträger. What midspan moment does self-weight alone produce?

  1. 0.75 kN·m
  2. 2.25 kN·m Correct answer.
  3. 1.5 kN·m
  4. 9.0 kN·m

Explanation

M = q L² / 8 = 0.5 × 36 / 8 = 2.25 kN·m.

Question 9Statics, strength and mechanical transmission

Combine loads on an Einfeldträger: center live load F = 6 kN and uniform dead load q = 1 kN/m on L = 4 m. What is total M_max at midspan (superposition)?

  1. 4 kN·m only
  2. M_F + M_q = 6 + 2 = 8 kN·m Correct answer.
  3. 12 kN·m
  4. 24 kN·m

Explanation

M_F = F L/4 = 6; M_q = q L²/8 = 1×16/8 = 2; sum = 8 kN·m at midspan for these symmetric cases.

Question 10Statics, strength and mechanical transmission

The moment of a force about a point is calculated as:

  1. Force × the perpendicular distance from the point to the line of action of the force Correct answer.
  2. Force × the total length of the member
  3. Force ÷ the distance to the point
  4. Force × the distance measured along the line of action of the force

Explanation

M = F · a, where a is the perpendicular (shortest) distance from the reference point to the line of action. This perpendicular distance is called the lever arm (Hebelarm), and the unit is Nm or kNm.

Question 11Statics, strength and mechanical transmission

How many independent equilibrium conditions are available for a rigid body in a plane (2D) force system?

  1. Two: ΣFx = 0 and ΣFy = 0
  2. Three: ΣFx = 0, ΣFy = 0 and ΣM = 0 Correct answer.
  3. Four, one for each quadrant
  4. Six: three force and three moment equations

Explanation

A rigid body in a plane has three degrees of freedom (two translations and one rotation), so equilibrium requires exactly three equations: ΣFx = 0, ΣFy = 0 and ΣM = 0 about any chosen point.

Question 12Statics, strength and mechanical transmission

A movable (roller) support - verschiebliches Auflager - on a beam provides how many unknown reaction components?

  1. One, acting perpendicular to the sliding surface Correct answer.
  2. Two, one horizontal and one vertical
  3. Three, including a restraining moment
  4. None; it carries no load

Explanation

A roller support blocks only the displacement perpendicular to its sliding surface, so it contributes exactly one unknown reaction force in that direction. It allows both movement along the surface and rotation.

Question 13Statics, strength and mechanical transmission

A simply supported beam has one pinned support and one roller support. How many reaction components does it have, and is it statically determinate?

  1. Two components; statically indeterminate
  2. Three components; statically determinate Correct answer.
  3. Four components; statically indeterminate to the first degree
  4. Three components; kinematically unstable

Explanation

Pinned support = 2 components, roller support = 1 component, total 3 unknowns. In a plane there are exactly 3 equilibrium equations, so the reactions can be found from statics alone: the system is statically determinate (statisch bestimmt).

Question 14Statics, strength and mechanical transmission

Two forces act at the same point: F1 = 300 N horizontally and F2 = 400 N vertically. What is the magnitude and direction of the resultant?

  1. 500 N at 53.1° to the horizontal Correct answer.
  2. 700 N at 45°
  3. 500 N at 36.9° to the horizontal
  4. 100 N at 53.1° to the horizontal

Explanation

For perpendicular forces R = √(300² + 400²) = √250 000 = 500 N. The angle to the horizontal is α = arctan(F2/F1) = arctan(400/300) = 53.13°.

Question 15Statics, strength and mechanical transmission

A simply supported beam AB of span 6.0 m carries a single point load F = 24 kN at 2.0 m from support A. What are the support reactions A and B?

  1. A = 8 kN, B = 16 kN
  2. A = 16 kN, B = 8 kN Correct answer.
  3. A = 12 kN, B = 12 kN
  4. A = 18 kN, B = 6 kN

Explanation

Taking moments about B: A · 6.0 = 24 · 4.0 → A = 16 kN. Vertical equilibrium: B = 24 − 16 = 8 kN. The reaction is always larger at the support NEARER the load, because that support has the shorter lever arm to resist with.

Question 16Statics, strength and mechanical transmission

A simply supported beam of span L = 8.0 m carries a uniformly distributed load q = 5 kN/m over its whole length. What is the maximum bending moment?

  1. 20 kNm
  2. 40 kNm Correct answer.
  3. 80 kNm
  4. 160 kNm

Explanation

For a full-span UDL on a simply supported beam, Mmax = q·L²/8 = 5 × 8²/8 = 5 × 64/8 = 40 kNm, occurring at midspan where the shear force passes through zero. The reactions are q·L/2 = 20 kN each.

Question 17Statics, strength and mechanical transmission

For a simply supported beam of span 6.0 m with a single point load of 24 kN at 2.0 m from A (reactions A = 16 kN, B = 8 kN), what is the maximum bending moment and where does it occur?

  1. 32 kNm at the load point, 2.0 m from A Correct answer.
  2. 48 kNm at midspan
  3. 32 kNm at midspan
  4. 16 kNm at the load point

Explanation

For a single point load the shear force is constant on each side and changes sign exactly at the load, so the bending moment peaks there. M = A · a = 16 kN × 2.0 m = 32 kNm. Checking from the other side gives the same value: 8 kN × 4.0 m = 32 kNm.

Question 18Statics, strength and mechanical transmission

In a beam, at which location does the bending moment reach a local maximum or minimum?

  1. Where the shear force is greatest
  2. Where the shear force passes through zero or changes sign Correct answer.
  3. Always at midspan
  4. Always directly at the supports

Explanation

The shear force is the derivative of the bending moment: dM/dx = V. Extreme values of M therefore occur where V = 0. In a shear-force diagram every crossing of the zero line marks a peak in the moment diagram.

Question 19Statics, strength and mechanical transmission

A cantilever of length 3.0 m carries a uniformly distributed load q = 4 kN/m over its full length plus a point load F = 10 kN at the free end. What is the bending moment at the fixed support?

  1. 30 kNm
  2. 48 kNm Correct answer.
  3. 39 kNm
  4. 66 kNm

Explanation

The UDL has resultant q·L = 12 kN acting at the centroid, 1.5 m from the support, giving q·L²/2 = 4 × 9/2 = 18 kNm. The end load gives F·L = 10 × 3.0 = 30 kNm. Superposition: M = 18 + 30 = 48 kNm.

Question 20Statics, strength and mechanical transmission

A simply supported beam AB of span 10.0 m carries a UDL q = 6 kN/m over the whole span plus a point load F = 20 kN at 2.5 m from A. What is the reaction at A?

  1. 40 kN
  2. 45 kN Correct answer.
  3. 35 kN
  4. 50 kN

Explanation

By superposition. UDL: total 6 × 10 = 60 kN, symmetric, so 30 kN at each support. Point load: A receives 20 × 7.5/10 = 15 kN (the lever arm to the FAR support divided by the span). Total A = 30 + 15 = 45 kN. Check: B = 30 + 5 = 35 kN, and 45 + 35 = 80 kN = 60 + 20.

Question 21Statics, strength and mechanical transmission

A simply supported beam of span 6.0 m carries a UDL of 8 kN/m over the left half only (from 0 to 3.0 m). Determine the reaction at A and the position and value of the maximum bending moment.

  1. A = 18 kN; Mmax = 20.25 kNm at x = 2.25 m from A Correct answer.
  2. A = 12 kN; Mmax = 36 kNm at midspan
  3. A = 18 kN; Mmax = 18 kNm at x = 3.0 m, the end of the loaded length
  4. A = 6 kN; Mmax = 20.25 kNm at x = 2.25 m from A

Explanation

The load resultant is 8 × 3 = 24 kN acting at 1.5 m from A. ΣM about B: A × 6.0 = 24 × 4.5 → A = 18 kN, so B = 6 kN. In the loaded region V(x) = 18 − 8x, which is zero at x = 2.25 m. There M = 18 × 2.25 − 8 × 2.25²/2 = 40.5 − 20.25 = 20.25 kNm.

Question 22Statics, strength and mechanical transmission

Two parallel downward forces act on a rigid beam: F1 = 200 N at x = 0 and F2 = 300 N at x = 2.0 m. Where does their resultant act?

  1. 1.0 m from F1
  2. 1.2 m from F1 Correct answer.
  3. 0.8 m from F1
  4. 1.5 m from F1

Explanation

R = 200 + 300 = 500 N. By the moment theorem (Momentensatz), the moment of the resultant about a point equals the sum of the moments of the individual forces. About the position of F1: 500 · x = 200 · 0 + 300 · 2.0 → x = 600/500 = 1.2 m. The resultant always lies nearer the larger force.

Question 23Building systems, load paths, heat and moisture

An opaque wall has U = 0.28 W/(m²·K). If insulation improves U to 0.14 W/(m²·K) with the same area and ΔT, how does steady transmission heat loss through that wall change?

  1. It halves Correct answer.
  2. It doubles
  3. It stays the same
  4. It becomes zero

Explanation

Q̇ ∝ U for fixed A and ΔT, so cutting U from 0.28 to 0.14 halves transmission heat loss.

Question 24Building systems, load paths, heat and moisture

Estimate the daily space-heating energy to cover transmission losses only: H_T = 80 W/K, average ΔT = 15 K over 24 h. (Ignore ventilation and gains.)

  1. 1.2 kWh/day
  2. 28.8 kWh/day Correct answer.
  3. 120 kWh/day
  4. 288 kWh/day

Explanation

Q̇ = 80 × 15 = 1200 W = 1.2 kW; E = 1.2 kW × 24 h = 28.8 kWh/day.

Question 25Building systems, load paths, heat and moisture

A flat roof (Flachdach) of A = 60 m² has U = 0.20 W/(m²·K). Compared with an older roof at U = 0.50 W/(m²·K) for the same A and ΔT = 20 K, how much lower is the heat-loss rate with the improved roof?

  1. 120 W lower
  2. 600 W lower
  3. 900 W lower
  4. 360 W lower Correct answer.

Explanation

ΔQ̇ = (U_old − U_new)·A·ΔT = (0.50 − 0.20)×60×20 = 0.30×1200 = 360 W lower.

Question 26Building systems, load paths, heat and moisture

Convert 2.5 kW of continuous heat loss into joules over 1 hour.

  1. 2.5 J
  2. 9,000,000 J Correct answer.
  3. 9,000 J
  4. 90,000 J

Explanation

2.5 kW = 2500 W; energy in 1 h = 2500 J/s × 3600 s = 9,000,000 J (also 2.5 kWh).

Question 27Building systems, load paths, heat and moisture

A thermal bridge adds an extra linear transmittance Ψ = 0.10 W/(m·K) along a length ℓ = 12 m. At ΔT = 20 K, what additional heat-loss rate does the bridge cause?

  1. 2.4 W
  2. 240 W
  3. 24 W Correct answer.
  4. 0.24 W

Explanation

Q̇_tb = Ψ · ℓ · ΔT = 0.10 × 12 × 20 = 24 W.

Question 28Building systems, load paths, heat and moisture

Indoor air temperature is 21 °C and outdoor temperature is −3 °C. What ΔT should you use in Q̇ = U·A·ΔT?

  1. 18 K
  2. 21 K
  3. 3 K
  4. 24 K Correct answer.

Explanation

ΔT = 21 − (−3) = 24 K (temperature differences in °C and K have the same magnitude).

Question 29Building systems, load paths, heat and moisture

If H_T must not exceed 40 W/K and one wall already contributes U·A = 28 W/K, what is the maximum U for a new 20 m² element still within the budget (ignore bridges)?

  1. 0.40 W/(m²·K)
  2. 0.60 W/(m²·K) Correct answer.
  3. 1.40 W/(m²·K)
  4. 2.00 W/(m²·K)

Explanation

Remaining conductance = 40 − 28 = 12 W/K; U_max = 12 / 20 = 0.60 W/(m²·K).

Question 30Building systems, load paths, heat and moisture

Over a heating season of 200 days with average transmission loss 1.5 kW, approximately how much energy is lost to transmission?

  1. 300 kWh
  2. 1,500 kWh
  3. 7,200 kWh Correct answer.
  4. 30,000 kWh

Explanation

E ≈ 1.5 kW × (200 × 24) h = 1.5 × 4800 = 7,200 kWh.

Question 31Building systems, load paths, heat and moisture

A room’s opaque envelope sums to Σ(U A) = 35 W/K plus thermal-bridge equivalent 5 W/K. At ΔT = 22 K, transmission heat-loss rate is:

  1. 0.77 kW
  2. 1.10 kW
  3. 0.88 kW Correct answer.
  4. 8.80 kW

Explanation

H_T = 35 + 5 = 40 W/K; Q̇ = 40 × 22 = 880 W = 0.88 kW.

Question 32Measurement, feedback, PLC and automation

A sensor outputs 0–10 V for 0–100 °C linearly. What temperature corresponds to 6.5 V?

  1. 6.5 °C
  2. 65 °C Correct answer.
  3. 35 °C
  4. 100 °C

Explanation

Span: 10 V → 100 °C ⇒ 10 °C/V; 6.5 × 10 = 65 °C.

Question 33Measurement, feedback, PLC and automation

A current-loop transmitter is 4–20 mA for 0–50 bar. What pressure corresponds to 12 mA?

  1. 12 bar
  2. 30 bar
  3. 25 bar Correct answer.
  4. 50 bar

Explanation

Live zero: 4 mA = 0, 20 mA = 50 bar. Span 16 mA → 50 bar. At 12 mA, fraction = (12−4)/16 = 0.5 → 25 bar.

Question 34Measurement, feedback, PLC and automation

A motor nameplate is 400 V, 2.5 A, power factor 0.85, efficiency 0.90 (three-phase). Approximate electrical input power P_el = √3 · U · I · cosφ.

  1. 0.85 kW
  2. 2.50 kW
  3. 1.47 kW Correct answer.
  4. 0.90 kW

Explanation

P_el = 1.732 × 400 × 2.5 × 0.85 ≈ 1.732 × 850 ≈ 1472 W ≈ 1.47 kW.

Question 35Measurement, feedback, PLC and automation

For which input combinations does a two-input XOR (exclusive OR) gate produce a 1 at its output?

  1. Only when both inputs are 1
  2. Only when the two inputs are different Correct answer.
  3. Only when both inputs are 0
  4. For every combination except 0/0

Explanation

XOR outputs 1 exactly when the inputs are unequal: 0/1 and 1/0 give 1, while 0/0 and 1/1 give 0. It is therefore also called an inequality or antivalence gate and is the core of binary adders and parity checks.

Question 36Measurement, feedback, PLC and automation

What distinguishes a closed-loop control (Regelung) from an open-loop control (Steuerung)?

  1. A Regelung continuously measures the actual value and feeds it back for comparison with the setpoint, forming a closed loop Correct answer.
  2. A Regelung always responds faster than a Steuerung
  3. A Steuerung uses electricity while a Regelung uses pneumatics
  4. A Regelung works without any sensors

Explanation

The defining criterion is the structure of the action path. A Steuerung has an open action chain (offener Wirkungsablauf) - the effect of the output is never checked. A Regelung closes the loop (geschlossener Wirkungskreis): the actual value is measured, compared with the setpoint, and the deviation drives the correction, so disturbances are automatically counteracted.

Question 37Measurement, feedback, PLC and automation

Applying De Morgan's theorem, which expression is equivalent to NOT (A AND B)?

  1. (NOT A) AND (NOT B)
  2. (NOT A) OR (NOT B) Correct answer.
  3. A OR B
  4. NOT (A OR B)

Explanation

De Morgan's rule states that negating a conjunction turns it into a disjunction of the negated terms: NOT(A AND B) = (NOT A) OR (NOT B). This is why a NAND function can be realised by an OR gate with inverted inputs, a substitution used constantly when converting between circuit forms in a Funktionsplan.

Question 38Measurement, feedback, PLC and automation

In a basic RS flip-flop built from two cross-coupled NOR gates, which input combination is not permitted?

  1. S = 0, R = 0
  2. S = 1, R = 0
  3. S = 0, R = 1
  4. S = 1, R = 1 Correct answer.

Explanation

With both inputs active the flip-flop is told to set and reset at the same time. Both outputs are forced to 0, so Q and NOT-Q are no longer complementary. Worse, when the inputs are released together, the state that results depends on tiny differences in gate delays - an unpredictable race condition.

Question 39Measurement, feedback, PLC and automation

How does a PLC (speicherprogrammierbare Steuerung, SPS) execute its program?

  1. Once only, at power-up
  2. Cyclically: read the inputs into the process image, execute the program, then write the output image to the outputs Correct answer.
  3. In parallel, with all instructions executed simultaneously
  4. Only when an operator presses a start button

Explanation

The PLC runs a fixed cycle: it samples all inputs into the process input image, works through the program using those frozen values, and finally transfers the process output image to the physical outputs. The cycle time, typically a few milliseconds, sets the worst-case reaction time of the whole controller.

Question 40Measurement, feedback, PLC and automation

In an automation system, which component converts a physical quantity such as pressure or temperature into an electrical signal?

  1. The actuator (Aktor)
  2. The sensor Correct answer.
  3. The CPU of the PLC
  4. The bus coupler

Explanation

The sensor sits at the interface between the process and the control system, converting a non-electrical quantity into an electrical signal the controller can read. Networked sensors that deliver their data directly onto a fieldbus are a key building block of the Internet of Things in production.

Question 41Measurement, feedback, PLC and automation

Simplify the Boolean function Y = A·B + A·(NOT B).

  1. Y = A Correct answer.
  2. Y = B
  3. Y = A + B
  4. Y = 0

Explanation

Factor out A: Y = A · (B + NOT B). Since B + NOT B = 1 always, Y = A · 1 = A. The output depends solely on A, so B is a redundant input and two gates can be deleted from the Funktionsplan.

Question 42Energy, thermal machines and environmental assessment

A flat-plate collector aperture A = 6.0 m² operates at mean irradiance G = 700 W/m² with collector efficiency η = 0.55. What is the useful thermal power output?

  1. 2.31 kW Correct answer.
  2. 1.54 kW
  3. 4.20 kW
  4. 0.55 kW

Explanation

P = η · G · A = 0.55 × 700 × 6.0 = 2,310 W ≈ 2.31 kW.

Question 43Energy, thermal machines and environmental assessment

A hot-water store holds 300 L ≈ 300 kg of water. Specific heat c ≈ 4180 J/(kg·K). Temperature rises by 30 K. Approximately how much thermal energy is stored?

  1. 3.76 kWh
  2. 37.6 kWh
  3. 10.45 kWh Correct answer.
  4. 125.4 kWh

Explanation

Q = m c ΔT = 300 × 4180 × 30 = 3.762×10⁷ J; ÷ 3.6×10⁶ J/kWh ≈ 10.45 kWh.

Question 44Energy, thermal machines and environmental assessment

An inverter nameplate is 5.0 kW AC. Array DC power at a moment is 5.8 kW. Assuming the inverter clips at rated AC output (ignore efficiency band), what AC power is delivered?

  1. 5.8 kW
  2. 10.8 kW
  3. 0.8 kW
  4. 5.0 kW Correct answer.

Explanation

Clipping limits output to the inverter rating: 5.0 kW AC (DC excess is not converted).

Question 45Energy, thermal machines and environmental assessment

Collector optical efficiency η₀ = 0.78 and linear loss coefficient a₁ = 3.5 W/(m²·K). At G = 800 W/m² and (T_m − T_a) = 40 K, approximate η ≈ η₀ − a₁(T_m − T_a)/G.

  1. 0.605 Correct answer.
  2. 0.78
  3. 0.430
  4. 0.200

Explanation

η ≈ 0.78 − 3.5×40/800 = 0.78 − 140/800 = 0.78 − 0.175 = 0.605.

Question 46Energy, thermal machines and environmental assessment

A heat pump delivers 12 kWh heat while consuming 3 kWh electricity. What is COP = Q_out / W_el?

  1. 0.25
  2. 3
  3. 15
  4. 4 Correct answer.

Explanation

COP = 12 / 3 = 4. The evaluated result is 4, following the arithmetic shown.

Question 47Energy, thermal machines and environmental assessment

PV: A = 20 m², G = 750 W/m², η_mod = 0.17, η_inv = 0.95, wiring/other PR_extra = 0.98. Approximate AC power?

  1. 2.55 kW
  2. 3.00 kW
  3. 15.0 kW
  4. 2.38 kW Correct answer.

Explanation

P_AC ≈ 0.17×750×20 × 0.95 × 0.98 = 2550 × 0.931 ≈ 2374 W ≈ 2.38 kW.

Question 48Aerodynamic forces and elementary stability models

A simplified lift estimate uses L = ½ ρ v² S CL. With ρ = 1.225 kg/m³, v = 40 m/s, S = 12 m² and CL = 0.80, L is closest to which value?

  1. About 9.4 kN Correct answer.
  2. About 4.7 kN
  3. About 18.8 kN
  4. About 94 kN

Explanation

Compute dynamic pressure ½ρv² = 0.5·1.225·40² = 0.5·1.225·1600 = 980 Pa. Then L = 980·12·0.80 = 9408 N ≈ 9.4 kN. Halving or doubling CL or S produces the nearby distractors.

Question 49Engineering models, drawings, evidence and project decisions

In the European first-angle projection method used in German technical drawings, where is the view obtained by looking from the LEFT placed relative to the front view?

  1. To the right of the front view Correct answer.
  2. To the left of the front view
  3. Above the front view
  4. Below the front view

Explanation

In first-angle projection (Projektionsmethode 1) the object lies between the observer and the projection plane, so each view is thrown onto the plane BEYOND the object and therefore appears on the side opposite the viewing direction. Looking from the left places that view on the right; looking from above places the plan below.

Question 50Engineering models, drawings, evidence and project decisions

A shaft is dimensioned 40 ± 0.05 mm. What are the limits of size and the tolerance?

  1. 40.05 / 39.95 mm; tolerance 0.10 mm Correct answer.
  2. 40.05 / 39.95 mm; tolerance 0.05 mm
  3. 40.5 / 39.5 mm; tolerance 1.0 mm
  4. 40.05 / 40.00 mm; tolerance 0.05 mm

Explanation

Upper limit of size = 40 + 0.05 = 40.05 mm, lower limit = 40 − 0.05 = 39.95 mm. The tolerance (Maßtoleranz) is the DIFFERENCE between the two limits: 40.05 − 39.95 = 0.10 mm, i.e. twice the individual deviation.