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Mecklenburg-Vorpommern Abitur Metal Technology - independent engineering foundations study materials
Mecklenburg-Vorpommern Abitur Metal Technology - independent engineering foundations study resources

Mecklenburg-Vorpommern Abitur Metal Technology - independent engineering foundations practice questions

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Question 1CNC concepts, coordinates and process verification

In a supplied coordinate exercise, the current X coordinate is 12 mm and the absolute target is 35 mm in the same positive axis. What X displacement is required?

  1. −23mm.
  2. +35mm.
  3. +47mm.
  4. +23mm. Correct answer.

Explanation

Subtract the current coordinate from the target:35−12=+23 mm. Absolute target coordinates and the displacement from the current point are different quantities; no controller command or clearance condition is implied.

Question 2CNC concepts, coordinates and process verification

A simulated path uses incremental displacements of+8 mm,−3 mm and+6 mm along the same axis, starting atX=10 mm. What final coordinate follows?

  1. X=7mm.
  2. X=27mm.
  3. X=11mm.
  4. X=21mm. Correct answer.

Explanation

The net displacement is 8−3+6=11 mm. Adding that displacement to the starting coordinate 10 mm gives 21 mm; treating the incremental sum as an absolute coordinate would omit the initial position.

Question 3CNC concepts, coordinates and process verification

For a supplied ideal feed calculation, a cutter has 4 teeth, spindle speed 1500 rev/min and feed 0.04 mm per tooth. What table-feed rate follows?

  1. 6mm/min.
  2. 60mm/min.
  3. 1500mm/min.
  4. 240mm/min. Correct answer.

Explanation

Multiply the stated feed per tooth by four teeth and 1500 revolutions per minute:0.04×4×1500=240 mm/min. These are supplied classroom values, not approved cutting parameters for a real tool or material.

Question 4CNC concepts, coordinates and process verification

A simulation specifies a straight 120 mm motion at a constant 300 mm/min, ignoring acceleration and all pauses. What travel time follows?

  1. 150s.
  2. 24s. Correct answer.
  3. 40s.
  4. 2.5s.

Explanation

Time equals distance divided by speed:120/300=0.4 min. Multiplying by 60 gives 24 seconds. The model explicitly excludes acceleration, pauses and additional approach paths, so it is not a complete machine-cycle prediction.

Question 5CNC concepts, coordinates and process verification

In a fictional planning exercise, the required depth is 10 mm and each roughing pass may remove at most 3 mm of depth. Ignoring finishing allowance, what is the minimum number of passes?

  1. 4passes. Correct answer.
  2. 10passes.
  3. 5passes.
  4. 3passes.

Explanation

Three full passes remove only 9 mm, so a fourth pass is necessary to reach 10 mm. The mathematical count rounds 10/3 upward; the supplied limit is not a universal safe machining depth.

Question 6CNC concepts, coordinates and process verification

A copied toolpath simulation uses a 5 mm cutter radius while the documented intended tool has a 4 mm radius. Which conclusion is justified before using its dimensional prediction?

  1. The discrepancy can be removed by averaging the radii without checking the drawing.
  2. The geometry model must be reconciled with the intended tool and compensation convention. Correct answer.
  3. The simulation proves that either radius produces the same finished contour.
  4. Only spindle speed can affect the dimensional disagreement.

Explanation

Tool geometry and the compensation convention affect how a tool-center path relates to the finished surface. Reconcile the documented inputs before trusting the dimensional result; neither arbitrary averaging nor a successful simulation establishes physical setup safety.

Question 7CNC concepts, coordinates and process verification

A supplied program-coordinate worksheet is in millimeters, but a dimension copied from another record is expressed in centimeters. What should be done before calculating the path?

  1. Treat centimeters as ten times smaller than millimeters.
  2. Change the workpiece origin until the two numerical values match.
  3. Use the same numerical value because coordinate labels determine the unit automatically.
  4. Convert the dimension to the declared units and verify the authorized drawing revision. Correct answer.

Explanation

One centimeter equals ten millimeters, so the numeric value must be converted to the worksheet’s units. Confirming the correct revision also prevents a unit conversion from legitimizing an obsolete dimension.

Question 8CNC concepts, coordinates and process verification

Why is a repeated geometric pattern often represented once in a subprogram within a specified controller model?

  1. It makes tool dimensions irrelevant to every copy of the pattern.
  2. It removes the need to identify the workpiece coordinate system.
  3. It proves that all fixtures and clearances are safe without further evidence.
  4. It reduces duplicated definitions and makes a reviewed pattern easier to maintain consistently. Correct answer.

Explanation

A reusable pattern definition can improve consistency and maintenance. It does not eliminate coordinate, tool, fixture or safety requirements, and its actual calling behavior depends on the stated controller model.

Question 9Materials, testing and selection

Which heat treatment softens steel and improves machinability by producing a soft microstructure?

  1. Annealing (Glühen) Correct answer.
  2. Quenching (Abschrecken)
  3. Carburizing (Aufkohlen)
  4. Nitriding (Nitrieren)

Explanation

Annealing softens and relieves stresses. Quenching hardens; carburizing/nitriding are surface hardening / case treatments, not soft-annealing.

Question 10Materials, testing and selection

A tensile bar of cross-section A = 100 mm² carries F = 20 kN axially. What is the nominal tensile stress σ?

  1. 2 N/mm²
  2. 20 N/mm²
  3. 200 N/mm² (MPa) Correct answer.
  4. 2000 N/mm²

Explanation

σ = F/A. Convert 20 kN = 20 000 N; 20 000 N / 100 mm² = 200 N/mm² = 200 MPa.

Question 11Materials, testing and selection

Grey cast iron (GJL) is often preferred for machine beds because it:

  1. Has the highest tensile strength of all steels
  2. Cannot be cast into thick sections
  3. Damps vibration well and machines easily into complex shapes Correct answer.
  4. Is always stainless without alloying

Explanation

Lamellar graphite in GJL provides excellent damping and good castability/machinability for beds and housings. Tensile strength is lower than many steels; it is not inherently stainless.

Question 12Materials, testing and selection

What does the yield strength Rp0,2 represent?

  1. Ultimate tensile strength at fracture
  2. Young's modulus E
  3. Brinell hardness number
  4. Stress at 0.2% permanent plastic strain (proof stress) Correct answer.

Explanation

Rp0,2 is the 0.2% offset proof stress used when no clear yield point appears. Rm is ultimate tensile strength; E is stiffness; hardness is separate.

Question 13Materials, testing and selection

Shear stress τ for a bolt in single shear with F = 12 kN and shank diameter d = 10 mm is approximately?

  1. 12 N/mm²
  2. 38 N/mm²
  3. 1200 N/mm²
  4. 153 N/mm² Correct answer.

Explanation

A = πd²/4 ≈ 78.5 mm²; τ = F/A = 12 000 / 78.5 ≈ 153 N/mm².

Question 14Materials, testing and selection

Which quantity is dimensionless?

  1. Stress σ = F/A
  2. Young's modulus E
  3. Bending moment M
  4. Strain ε = ΔL/L Correct answer.

Explanation

Strain is length/length (dimensionless, often as % or µm/m). Stress and E have force/area units; moment has force×length.

Question 15Statics, strength and mechanical transmission

A simply supported beam length L = 800 mm carries a midspan point load F = 4 kN. Maximum bending moment M is?

  1. 800 N·m Correct answer.
  2. 3200 N·m
  3. 4 N·m
  4. 0.8 N·m

Explanation

For midspan point load, M_max = F·L/4. F = 4000 N, L = 0.8 m → M = 4000·0.8/4 = 800 N·m.

Question 16Statics, 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 17Statics, 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 18Statics, 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 19Statics, 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 20Statics, 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 21Statics, 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 22Statics, 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 23Statics, 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 24Statics, 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 25Statics, 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 26Statics, 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 27Statics, 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 28Statics, 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 29Manufacturing, machining models and metrology

Which measuring instrument is typically preferred for checking a 50 mm shaft diameter to ±0.01 mm in a school lab?

  1. Steel rule alone
  2. Feeler gauge set
  3. Spirit level
  4. Outside micrometer (Bügelmessschraube) Correct answer.

Explanation

A micrometer resolves to 0.01 mm (or better). A steel rule is coarser; feeler gauges measure gaps; a spirit level checks inclination.

Question 30Manufacturing, machining models and metrology

In orthogonal cutting, what is the primary motion that creates the chip on a lathe?

  1. Workpiece rotation (cutting motion) Correct answer.
  2. Only axial tool feed without rotation
  3. Only radial plunge with stationary workpiece
  4. Coolant flow

Explanation

On a lathe, the workpiece rotates (Hauptbewegung); the tool feed is the Vorschubbewegung. Coolant aids but does not cut.

Question 31Manufacturing, machining models and metrology

Cutting speed vc = 80 m/min; workpiece diameter d = 50 mm. Approximate spindle speed n in min⁻¹ is?

  1. About 51 min⁻¹
  2. About 510 min⁻¹ Correct answer.
  3. About 5100 min⁻¹
  4. About 5 min⁻¹

Explanation

n = (vc · 1000) / (π · d) ≈ (80 · 1000) / (3.14 · 50) ≈ 80 000 / 157 ≈ 510 min⁻¹.

Question 32Manufacturing, machining models and metrology

Which process belongs to Urformen rather than Trennen?

  1. Face milling a flat
  2. Sand casting of an aluminum housing Correct answer.
  3. Drilling a through hole
  4. Parting-off on a lathe

Explanation

Casting creates the near-net shape from liquid metal (Urformen). Milling, drilling, and parting remove chips (Trennen).

Question 33Manufacturing, machining models and metrology

Which joining process is a fusion welding process?

  1. MIG/MAG (GMAW) arc welding Correct answer.
  2. Blind riveting
  3. Interference press fit
  4. Adhesive bonding alone

Explanation

MIG/MAG melts base/filler metal (fusion welding). Rivets, press fits, and adhesives join without fusion melting of the base metal as the primary mechanism.

Question 34Manufacturing, machining models and metrology

Surface roughness Ra is best described as:

  1. Peak-to-valley height of the primary form error only
  2. Arithmetic mean deviation of the roughness profile from the mean line Correct answer.
  3. Hardness after grinding
  4. Thread flank angle

Explanation

Ra is the arithmetic average of absolute profile deviations. Rz relates more to peak-to-valley averages; form error is separate from roughness.

Question 35CNC concepts, coordinates and process verification

Workpiece zero (Werkstücknullpunkt) is:

  1. The programmed origin on the part from which absolute coordinates are referenced Correct answer.
  2. Always the machine reference switch alone
  3. The tool magazine pocket number
  4. The coolant concentration

Explanation

Programmers set a workpiece coordinate system (G54…); machine reference is separate. Magazine and coolant are unrelated.

Question 36CNC concepts, coordinates and process verification

Backlash (Umkehrspiel) on a worn axis primarily causes:

  1. Higher Young's modulus
  2. Automatic tool sharpening
  3. Increased steel carbon content
  4. Positioning errors especially when reversing direction Correct answer.

Explanation

Lost motion on reversal creates dimensional and finish errors - why climb milling needs tight axes. Material properties are unrelated.

Question 37Measurement, feedback, PLC and automation

In automation terminology, a Steuerkette (open-loop control) differs from a Regelkreis because it:

  1. Always uses a PID controller with feedback
  2. Has no feedback of the controlled variable to correct deviations Correct answer.
  3. Cannot use sensors at all
  4. Is identical to closed-loop regulation

Explanation

Open-loop Steuerungen command actuators without comparing actual output to setpoint. Closed-loop Regelung measures the controlled variable and corrects via feedback. Sensors may still exist for interlocking without forming a regulating loop.

Question 38Measurement, feedback, PLC and automation

Which Boolean function outputs 1 only if all inputs are 1?

  1. OR
  2. XOR
  3. AND Correct answer.
  4. NOT

Explanation

AND requires every input true. OR needs any one; XOR needs odd parity; NOT inverts one input.

Question 39Measurement, feedback, PLC and automation

In standard pneumatics, what does a 5/2-way valve primarily provide?

  1. Five positions and two ports
  2. Only vacuum generation
  3. Electrical amplification of 24 V signals
  4. Five ports and two switching positions - typical for double-acting cylinders Correct answer.

Explanation

A 5/2 valve has 5 ports / 2 positions and is classic for reversing double-acting cylinders. It is not an electrical amplifier.

Question 40Measurement, feedback, PLC and automation

A simple SPS (PLC) program that turns an output on when either of two inputs is true implements which logic?

  1. AND
  2. OR Correct answer.
  3. NAND only
  4. Edge detection only

Explanation

Output = Input1 OR Input2. AND would require both; NAND is inverted AND; edge detection is a timing/event function.

Question 41Measurement, feedback, PLC and automation

In a closed-loop speed control of a motor, the tachometer signal is typically the:

  1. Setpoint (Sollwert) generator alone
  2. Actuator power supply
  3. Feedback (Istwert) of the controlled variable Correct answer.
  4. Open-loop timer

Explanation

The tachometer measures actual speed and feeds it back for comparison with the setpoint - core Regelung structure.

Question 42Measurement, feedback, PLC and automation

Compressed air preparation (FRL unit) commonly includes:

  1. Only a hydraulic oil cooler
  2. Only a CNC spindle encoder
  3. Only a catalytic converter
  4. Filter, pressure regulator, and lubricator (as applicable) Correct answer.

Explanation

Service units condition air: filter contaminants, regulate pressure, optionally lubricate. The other options belong to unrelated systems.

Question 43Measurement, feedback, PLC and automation

A double-acting pneumatic cylinder requires pressurized air:

  1. On alternating sides of the piston for extend and retract Correct answer.
  2. Only on one side; spring always returns
  3. Never on either side
  4. Only as vacuum on both ports simultaneously

Explanation

Double-acting cylinders are powered in both directions. Single-acting often uses a spring return with air on one side only.

Question 44Measurement, feedback, PLC and automation

Which representation is commonly used for sequential control steps in automation documentation?

  1. Only freehand watercolor sketches
  2. GRAFCET / Funktionsplan-style step–transition charts Correct answer.
  3. Only balance-sheet accounts
  4. Only literary essays

Explanation

Sequential controls are documented with step/transition formalisms (GRAFCET/SFC or equivalent Funktionspläne) taught alongside logic representations in automation curricula.

Question 45Measurement, feedback, PLC and automation

In electropneumatic circuits, a solenoid valve coil is typically driven by:

  1. Direct 400 V three-phase across the coil as standard school practice
  2. Compressed air flowing through the copper winding
  3. Hydraulic oil at 200 bar inside the coil
  4. A low-power electrical signal (e.g., 24 V DC) switching the pneumatic pilot Correct answer.

Explanation

Solenoid coils are electrical actuators that pilot the valve spool; pneumatic power then shifts high flow. Coils are not powered by air/oil through the winding.

Question 46Measurement, feedback, PLC and automation

Which sensor type is typically binary (on/off) for cylinder end-position detection?

  1. Reed switch or inductive proximity switch on the cylinder Correct answer.
  2. Analog strain-gauge load cell only
  3. Thermocouple measuring melt temperature only
  4. pH probe

Explanation

End-of-stroke detection commonly uses magnetic reed or inductive proximity switches (digital). Load cells/thermocouples/pH are different measurands.

Question 47Measurement, feedback, PLC and automation

An AND of three inputs A,B,C is 1 only when:

  1. Any one of A,B,C is 1
  2. A≠B
  3. A=1 and B=1 and C=1 Correct answer.
  4. Exactly two inputs are 1

Explanation

Three-input AND requires all high. Any-one-high is OR; inequality is XOR-like; exactly-two is a different function.

Question 48Measurement, feedback, PLC and automation

Pneumatic holding of a vertical load when supply fails is best aided by:

  1. Removing all seals
  2. Venting both sides fully to atmosphere intentionally under load
  3. Using only larger silencers
  4. Check valves / locking units designed for the cylinder Correct answer.

Explanation

Non-return / locking concepts prevent uncontrolled dropping on pressure loss. Venting both sides or removing seals worsens holding.

Question 49Measurement, feedback, PLC and automation

A time-delay off (Ausschaltverzögerung) keeps the output true:

  1. Only before the input ever becomes true
  2. For a set time after the input goes false Correct answer.
  3. Forever without reset conditions
  4. Only at power-up for 1 µs always

Explanation

Off-delay timers extend the output after the enabling condition drops. That is distinct from on-delay (Einschaltverzögerung).

Question 50Measurement, feedback, PLC and automation

XOR of inputs A and B is 1 when:

  1. Both are 1
  2. Both are 0
  3. Exactly one of A or B is 1 Correct answer.
  4. A equals B

Explanation

XOR is exclusive OR: true for (1,0) or (0,1). Both 1 or both 0 yield 0; equality is XNOR territory.