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Mecklenburg-Vorpommern Abitur Construction Engineering - independent engineering foundations study materials
Mecklenburg-Vorpommern Abitur Construction Engineering - independent engineering foundations study resources

Mecklenburg-Vorpommern Abitur Construction Engineering - independent engineering foundations practice questions

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Question 1Materials, testing and selection

Which value is normally used as the modulus of elasticity (Elastizitätsmodul) of structural steel S235 in engineering calculations?

  1. 70 000 N/mm²
  2. 210 000 N/mm² Correct answer.
  3. 21 000 N/mm²
  4. 2 100 N/mm²

Explanation

E ≈ 210 000 N/mm² (210 GPa) for all common structural steels. It links stress and strain in Hooke's law σ = E · ε and is essentially independent of the steel grade - S235 and S355 have the same stiffness but different yield strengths.

Question 2Materials, testing and selection

A round steel tie bar with a diameter of 20 mm carries an axial tensile force of 62.8 kN. What is the normal stress in the bar?

  1. ≈ 100 N/mm²
  2. ≈ 157 N/mm²
  3. ≈ 200 N/mm² Correct answer.
  4. ≈ 3 140 N/mm²

Explanation

S = π/4 · d² = π/4 · 20² = 314.2 mm². σ = F/S = 62 800 N / 314.2 mm² ≈ 200 N/mm². For S235 (Re = 235 N/mm²) this bar is already close to the yield point.

Question 3Materials, testing and selection

A steel rod of length 2 500 mm and cross-section 200 mm² is loaded with 42 kN. With E = 210 000 N/mm², how much does it elongate elastically?

  1. 2.5 mm Correct answer.
  2. 0.25 mm
  3. 25 mm
  4. 1.0 mm

Explanation

σ = F/S = 42 000/200 = 210 N/mm². ε = σ/E = 210/210 000 = 0.001. ΔL = ε · L = 0.001 · 2 500 mm = 2.5 mm. Equivalently ΔL = F·L/(E·S).

Question 4Materials, testing and selection

A cube of an unknown metal measuring 40 mm × 40 mm × 40 mm has a mass of 172.8 g. Which metal is it most likely to be?

  1. Aluminium (ρ ≈ 2.7 g/cm³) Correct answer.
  2. Steel (ρ ≈ 7.85 g/cm³)
  3. Copper (ρ ≈ 8.9 g/cm³)
  4. Magnesium (ρ ≈ 1.74 g/cm³)

Explanation

V = 4 cm × 4 cm × 4 cm = 64 cm³. ρ = m/V = 172.8 g / 64 cm³ = 2.70 g/cm³, which matches aluminium.

Question 5Materials, testing and selection

A galvanised (zinc-coated) steel sheet is scratched down to the bare steel. Why does the steel still not rust at the scratch?

  1. The zinc layer flows into the scratch and seals it mechanically
  2. Zinc is less noble than iron, so it becomes the anode and corrodes preferentially, protecting the steel cathodically Correct answer.
  3. Zinc is more noble than iron and therefore protects it as a cathode
  4. The zinc raises the carbon content of the steel locally, making it corrosion resistant

Explanation

In the electrochemical series zinc (≈ −0.76 V) is less noble than iron (≈ −0.44 V). In the presence of an electrolyte the zinc dissolves as the sacrificial anode and the exposed steel becomes the cathode, where no metal dissolution occurs. This active (cathodic) protection is why galvanising still works at cut edges and scratches.

Question 6Materials, testing and selection

A square softwood post with a cross-section of 120 mm × 120 mm is loaded in compression parallel to the grain. If the permissible compressive stress is 8.5 N/mm², what is the maximum permissible axial load?

  1. ≈ 1.0 kN
  2. ≈ 61.2 kN
  3. ≈ 122.4 kN Correct answer.
  4. ≈ 1 224 kN

Explanation

A = 120 mm × 120 mm = 14 400 mm². F = σ · A = 8.5 N/mm² × 14 400 mm² = 122 400 N = 122.4 kN. (For a slender post a buckling check would also be required, which can reduce this value considerably.)

Question 7Materials, testing and selection

A tie rod of S235 steel (yield strength Re = 235 N/mm²) must carry a static tensile force of 90 kN with a safety factor of 1.5 against yielding. What is the minimum required cross-sectional area?

  1. ≈ 383 mm²
  2. ≈ 575 mm² Correct answer.
  3. ≈ 862 mm²
  4. ≈ 255 mm²

Explanation

σ_perm = Re / ν = 235 / 1.5 = 156.7 N/mm². A = F / σ_perm = 90 000 N / 156.7 N/mm² ≈ 574.5 mm². A round bar would need d = √(4A/π) ≈ 27.0 mm, so Ø28 mm would be selected.

Question 8Materials, testing and selection

A proportional tensile specimen has an original gauge length L0 = 100 mm and a diameter of 10 mm. After fracture the gauge length measures 117 mm, and the maximum force recorded during the test was 42.4 kN. What are the percentage elongation after fracture A and the tensile strength Rm?

  1. A = 17 %, Rm ≈ 540 N/mm² Correct answer.
  2. A = 17 %, Rm ≈ 424 N/mm²
  3. A = 117 %, Rm ≈ 540 N/mm²
  4. A = 14.5 %, Rm ≈ 540 N/mm²

Explanation

A = (Lu − L0)/L0 × 100 % = (117 − 100)/100 × 100 % = 17 %. S0 = π/4 · 10² = 78.54 mm², so Rm = 42 400 N / 78.54 mm² ≈ 540 N/mm². An elongation of 17 % indicates a ductile steel with substantial plastic reserve.

Question 9Statics, strength and mechanical transmission

For the same beam (L = 4.0 m, q = 3.0 kN/m), what is each end support reaction?

  1. 3.0 kN
  2. 6.0 kN Correct answer.
  3. 12.0 kN
  4. 1.5 kN

Explanation

Total load = qL = 12 kN; each simple support takes half → 6.0 kN.

Question 10Statics, strength and mechanical transmission

Required section modulus W_erf for M = 6.0 kNm if allowable bending stress σ_zul = 10 N/mm² is closest to which value? (Use consistent units: 6.0×10⁶ N·mm / 10 N/mm²)

  1. 0.6×10⁶ mm³ (600 cm³) Correct answer.
  2. 6.0×10⁶ mm³
  3. 60 mm³
  4. 10 mm³

Explanation

W = M/σ = 6.0e6 N·mm / 10 N/mm² = 6.0e5 mm³ = 600 cm³.

Question 11Statics, strength and mechanical transmission

A 120 × 200 mm rectangular beam (b=120, h=200). What is W about the strong axis?

  1. 800 000 mm³ Correct answer.
  2. 400 000 mm³
  3. 200 000 mm³
  4. 2 400 000 mm³

Explanation

W = b h² / 6 = 120 × 40 000 / 6 = 4 800 000 / 6 = 800 000 mm³.

Question 12Statics, strength and mechanical transmission

Increasing beam depth h while keeping width b constant is efficient for bending because W scales with...

  1. h (linear only)
  2. h² Correct answer.
  3. 1/h
  4. Independent of h

Explanation

W = b h² / 6 - depth enters squared, so deepening is highly efficient for bending capacity.

Question 13Building systems, load paths, heat and moisture

In an ideal steady one-dimensional series wall model with common heat flux and no internal heat generation, how is the temperature drop across each layer related to its thermal resistance?

  1. Its cost only
  2. Its thermal resistance R (larger R → larger ΔT share) Correct answer.
  3. Only its color
  4. Only its sd-value

Explanation

Stationary 1D profile: ΔT_i / ΔT_total = R_i / R_T - insulation takes most of the drop.

Question 14Building systems, load paths, heat and moisture

Wasserdampfdiffusion through building layers is driven primarily by...

  1. Differences in vapor pressure / concentration across the assembly Correct answer.
  2. Only liquid water pipes
  3. Only steel yield
  4. Only Ausbreitmaß

Explanation

Vapor moves from higher to lower partial pressure - basis of diffusion and Glaser analysis.

Question 15Building systems, load paths, heat and moisture

sd-value (diffusionsäquivalente Luftschichtdicke) equals...

  1. μ · d Correct answer.
  2. d / μ
  3. λ / d
  4. U only

Explanation

s_d = μ · d (with d in m) - used in Glaser / moisture design.

Question 16Building systems, load paths, heat and moisture

What does the thermal transmittance (Wärmedurchgangskoeffizient) U of a building component describe?

  1. The heat flow in watts through 1 m² of the component per 1 K temperature difference; lower values mean better insulation Correct answer.
  2. The heat flow through the component per hour in kWh; higher values mean better insulation
  3. The resistance of the component to heat flow, in m²K/W
  4. The temperature at the inner surface of the component

Explanation

U is measured in W/(m²·K) and states how many watts pass through one square metre for each kelvin of temperature difference between inside and outside. A modern insulated external wall reaches about 0.20 W/(m²·K), an uninsulated solid brick wall roughly 1.4 W/(m²·K).

Question 17Building systems, load paths, heat and moisture

An insulation layer 140 mm thick has λ = 0.035 W/(m·K). What is its thermal resistance R?

  1. 0.25 m²K/W
  2. 4.0 m²K/W Correct answer.
  3. 4 000 m²K/W
  4. 0.0049 m²K/W

Explanation

R = d/λ = 0.140 m / 0.035 W/(m·K) = 4.0 m²K/W. The thickness must be inserted in metres, matching the metre in the unit of λ.

Question 18Building systems, load paths, heat and moisture

An external wall consists of 15 mm internal plaster (λ = 0.70), 175 mm reinforced concrete (λ = 2.30) and 140 mm insulation (λ = 0.035). With Rsi = 0.13 and Rse = 0.04 m²K/W, what is the U-value?

  1. ≈ 0.23 W/(m²K) Correct answer.
  2. ≈ 0.25 W/(m²K)
  3. ≈ 4.27 W/(m²K)
  4. ≈ 0.34 W/(m²K)

Explanation

Layer resistances: plaster 0.015/0.70 = 0.021, concrete 0.175/2.30 = 0.076, insulation 0.140/0.035 = 4.000 m²K/W. RT = 0.13 + 0.021 + 0.076 + 4.000 + 0.04 = 4.267 m²K/W. U = 1/RT = 0.234 ≈ 0.23 W/(m²K).

Question 19Building systems, load paths, heat and moisture

Interior air at 20 °C and 60 % relative humidity has a dew point of about 12 °C. What does this mean for the wall surfaces of the room?

  1. Condensation forms wherever the inner surface temperature falls below about 12 °C Correct answer.
  2. Condensation forms wherever the surface is warmer than 12 °C
  3. The wall must be held at exactly 12 °C
  4. The dew point has no relation to the surface temperature

Explanation

At the dew point the air is saturated. Any surface colder than 12 °C chills the adjacent air below saturation, so the excess water vapour condenses out as Tauwasser. In practice thermal bridges at window reveals and external corners are the coldest spots and therefore fail first.

Question 20Building systems, load paths, heat and moisture

In a heated building in a cold climate, on which side of the insulation should the layer with the higher water-vapour diffusion resistance (vapour barrier or retarder) be placed?

  1. On the cold, outer side, to keep driving rain out
  2. On the warm, inner side, so that vapour cannot enter the construction and condense inside it Correct answer.
  3. In the middle of the insulation layer
  4. It makes no difference as long as the layer is continuous

Explanation

The rule is 'tighter inside than outside' (innen dichter als außen). In winter the higher indoor vapour pressure drives moisture outwards; it must be stopped before it reaches the cold zone where the temperature falls below the dew point. The Glaser method makes this visible by plotting the actual vapour pressure against the saturation pressure through the cross-section.

Question 21Building systems, load paths, heat and moisture

For a wall with RT = 4.27 m²K/W (U = 0.234 W/(m²K)) and Rsi = 0.13 m²K/W, with 20 °C inside and −10 °C outside, what is the internal surface temperature θsi?

  1. ≈ 19.1 °C Correct answer.
  2. ≈ 16.1 °C
  3. ≈ 20.0 °C
  4. ≈ 13.0 °C

Explanation

The heat flux density is q = U · Δθ = 0.234 × 30 = 7.03 W/m². The same flux passes through the internal surface layer, dropping q · Rsi = 7.03 × 0.13 ≈ 0.91 K across it. Hence θsi = 20 − 0.91 ≈ 19.1 °C.

Question 22Building systems, load paths, heat and moisture

How much heat flows in one hour through 25 m² of a wall with U = 0.234 W/(m²K) at a temperature difference of 30 K?

  1. ≈ 0.18 kWh Correct answer.
  2. ≈ 5.85 kWh
  3. ≈ 7.0 kWh
  4. ≈ 175 kWh

Explanation

Φ = U · A · Δθ = 0.234 × 25 × 30 = 175.5 W. Over one hour that is 175.5 Wh = 0.176 ≈ 0.18 kWh - roughly the energy of a 175 W lamp burning for an hour.

Question 23Timber and concrete construction foundations

In macroscopic Holzaufbau, which tissue primarily conducts water in softwoods and strongly influences density and strength?

  1. Only bark cork
  2. Earlywood/latewood tracheids arranged in growth rings (with latewood usually denser) Correct answer.
  3. Only adhesive layers in CLT
  4. Concrete aggregate pockets

Explanation

Macroscopic softwood structure shows growth rings of earlywood and denser latewood tracheids that govern density and strength.

Question 24Timber and concrete construction foundations

Which Holzschutz strategy is primarily constructive (konstruktiver Holzschutz) rather than chemical?

  1. Flooding the section with creosote only
  2. Detailing to keep timber dry: overhangs, capillary breaks, ventilation, splash protection Correct answer.
  3. Ignoring gutters
  4. Painting concrete formwork oil onto the beam face only

Explanation

Konstruktiver Holzschutz keeps wood dry by geometry and detailing; chemical protection is additional when needed.

Question 25Timber and concrete construction foundations

A service-class / moisture-driven reduction of allowable stresses is applied because...

  1. Wet wood is always stronger
  2. Higher moisture generally reduces strength and stiffness and can enable decay if detailing fails Correct answer.
  3. Moisture only changes color
  4. EMC never affects design

Explanation

Elevated moisture reduces mechanical properties and raises durability risk - hence Nutzungsklasse / Holzschutz coupling.

Question 26Timber and concrete construction foundations

A simply supported RC beam span 5.0 m with uniform design load q_Ed = 20 kN/m: M_Ed,max is...

  1. 62.5 kNm Correct answer.
  2. 125 kNm
  3. 50 kNm
  4. 250 kNm

Explanation

M = q L² / 8 = 20 × 25 / 8 = 500/8 = 62.5 kNm.

Question 27Timber and concrete construction foundations

Longitudinal tensile reinforcement in an RC beam is placed primarily where?

  1. In the compression zone only for simple spans
  2. In the tension zone (bottom for midspan of simple spans; top over continuous supports) Correct answer.
  3. Only outside the formwork
  4. Only in the slab edge far from the beam

Explanation

Bewehrung follows tension: bottom bars midspan simple beam; top bars over supports in continuous beams - Bewehrungsplan logic.

Question 28Timber and concrete construction foundations

As = M / (z · σ_s) is a simplified estimate of required tensile steel. With M=62.5×10⁶ N·mm, z≈0.9×d, d=450 mm, σ_s=435 N/mm², As is closest to...

  1. About 355 mm² Correct answer.
  2. About 35 mm²
  3. About 3500 mm²
  4. About 10 mm²

Explanation

z ≈ 0.9×450 = 405 mm; As = 62.5e6 / (405×435) ≈ 62.5e6 / 176175 ≈ 355 mm².

Question 29Timber and concrete construction foundations

Why is curing (Nachbehandlung) critical after casting?

  1. It only cools steel
  2. It maintains moisture/temperature so cement hydration continues, improving strength and durability near the surface Correct answer.
  3. It replaces all cover
  4. It sets the Dachneigung

Explanation

Proper curing limits early drying and supports hydration - strength and durability of cover zone.

Question 30Engineering 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 31Engineering 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.

Question 32Engineering models, drawings, evidence and project decisions

A drawing is made to a scale of 1 : 5. A feature measures 34 mm on the drawing. What is its true size on the workpiece?

  1. 6.8 mm
  2. 170 mm Correct answer.
  3. 34 mm
  4. 39 mm

Explanation

In the notation 1 : 5 the drawing is the first figure and reality the second, so the object is drawn five times smaller. The true size is 34 mm × 5 = 170 mm.

Question 33Engineering models, drawings, evidence and project decisions

In network planning (Netzplantechnik), what characterises the critical path?

  1. It is the path containing the greatest number of activities
  2. It consists of the activities with zero total float, and its length determines the total project duration Correct answer.
  3. It is the cheapest sequence of activities
  4. It is the path that can be delayed without consequence

Explanation

The critical path is the longest chain of activities through the network in terms of DURATION. Every activity on it has a total float of zero, so any delay there pushes the project completion date back by the same amount. It is therefore where management attention and any acceleration measures must be concentrated.

Question 34Engineering models, drawings, evidence and project decisions

An activity in a precedence network (Vorgangsknotennetz) has an earliest start FAZ = 12 days and a latest start SAZ = 17 days. What is its total float, and is the activity critical?

  1. GP = 5 days; the activity is not critical Correct answer.
  2. GP = 5 days; the activity is critical
  3. GP = 29 days; the activity is not critical
  4. GP = 0 days; the activity is critical

Explanation

The total float is GP = SAZ − FAZ = 17 − 12 = 5 days (equivalently SEZ − FEZ). The activity may therefore start up to five days late without delaying the project end. Since GP > 0 it does not lie on the critical path.

Question 35Engineering models, drawings, evidence and project decisions

What is the central aim of ergonomic workplace design (Ergonomie) in production?

  1. To adapt the workplace, tools and work process to human physical and mental characteristics, reducing strain and health risk Correct answer.
  2. To adapt the worker to the machine by training alone
  3. To maximise machine utilisation regardless of operator load
  4. To reduce the number of workers on a production line

Explanation

The guiding principle is that the work is adapted to the human being, not the other way round. In practice this means working heights matched to body dimensions, avoidance of forced postures and repeated heavy lifting, adequate lighting and noise limits, and job enlargement against monotony. In the REFA work-system model this concerns the interaction of the element 'Mensch' with the equipment and the environmental influences.

Question 36Engineering models, drawings, evidence and project decisions

Which phases does a life-cycle assessment of a technical product cover?

  1. Only the manufacturing phase in the factory
  2. Raw-material extraction, manufacture, distribution, use, and disposal or recycling Correct answer.
  3. Only the use phase, since it lasts longest
  4. Design and manufacture only, since these are the phases the engineer controls

Explanation

A life-cycle view runs from cradle to grave, or to cradle again where material loops are closed. The reason for the whole-life perspective is that the dominant phase varies by product: for a car or a refrigerator the use phase dominates the environmental burden, while for a window frame or a smartphone manufacture does. Optimising a single phase in isolation often merely shifts the burden elsewhere.

Question 37Engineering models, drawings, evidence and project decisions

A PV module requires 1 200 kWh of primary energy to manufacture and generates 320 kWh of electricity per year at the intended site. Assuming that 2.5 kWh of primary energy is saved per kWh of electricity generated, what is the energy payback time (energetische Amortisationszeit)?

  1. 1.5 years Correct answer.
  2. 3.75 years
  3. 0.67 years
  4. 9.4 years

Explanation

The primary energy saved each year is 320 kWh × 2.5 = 800 kWh. The payback time is the manufacturing energy divided by the annual saving: 1 200 / 800 = 1.5 years. Since modules are warranted for 25 years or more, they return their manufacturing energy many times over.

Question 38Statics, 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 39Statics, 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 40Statics, 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 41Statics, 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 42Statics, 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 43Statics, 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 44Statics, 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 45Statics, 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 46Statics, 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 47Statics, 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 48Statics, 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 49Statics, 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 50Statics, 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.