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Junior Engineer practice questions

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Question 1Engineering fundamentals and applied mathematics

A gear train has a driving gear with 40 teeth meshing with a driven gear of 120 teeth. What is the gear ratio of the driven gear to the driving gear, and does it reduce or increase output speed?

  1. 1:3, and it reduces output speed
  2. 3:1, and it reduces output speed Correct answer.
  3. 3:1, and it increases output speed
  4. 1:3, and it increases output speed

Explanation

The ratio is driven teeth divided by driving teeth, 120/40 = 3, so the ratio is 3:1. Because the driven gear turns once for every three turns of the driver, output speed is reduced to one third and torque rises proportionally.

Question 2Engineering fundamentals and applied mathematics

In SI units, which combination of base units is equivalent to one newton?

  1. kg·m/s² Correct answer.
  2. kg·m²/s²
  3. kg·m/s
  4. kg·m²/s³

Explanation

Newton's second law gives force as mass times acceleration, so one newton is one kilogram-metre per second squared. The other combinations correspond to energy, momentum, and power respectively.

Question 3Engineering fundamentals and applied mathematics

A beam carries a point load of 12 kN at its midspan and is simply supported at both ends with a span of 4 m. Ignoring self-weight, what is the reaction at each support?

  1. 3 kN
  2. 24 kN
  3. 12 kN
  4. 6 kN Correct answer.

Explanation

A single central point load on a symmetric simply supported beam splits equally between the two supports. Half of 12 kN is 6 kN at each end, and the sum of reactions equals the applied load, satisfying vertical equilibrium.

Question 4Engineering fundamentals and applied mathematics

A material has a density of 2,700 kg/m³. What is the mass of a component with a volume of 0.004 m³?

  1. 0.108 kg
  2. 10.8 kg Correct answer.
  3. 1.08 kg
  4. 108 kg

Explanation

Mass equals density multiplied by volume, so 2,700 × 0.004 = 10.8 kg. Checking units, kg/m³ times m³ leaves kilograms, which confirms the operation.

Question 5Engineering fundamentals and applied mathematics

Which quantity is a scalar rather than a vector?

  1. Acceleration
  2. Velocity
  3. Kinetic energy Correct answer.
  4. Displacement

Explanation

Kinetic energy has magnitude only and no direction, so it is a scalar. Displacement, velocity, and acceleration all require a direction to be fully specified.

Question 6Engineering fundamentals and applied mathematics

A resistor of 220 Ω carries a current of 50 mA. What is the voltage across it?

  1. 110 V
  2. 4.4 V
  3. 11 V Correct answer.
  4. 44 V

Explanation

Ohm's law gives V = I × R = 0.050 A × 220 Ω = 11 V. Converting milliamperes to amperes before multiplying avoids a factor-of-1000 error.

Question 7Engineering fundamentals and applied mathematics

In a right triangle used to resolve a force, the side opposite an angle of 30° has length 5 m. What is the length of the hypotenuse?

  1. 5.0 m
  2. 2.5 m
  3. 8.7 m
  4. 10 m Correct answer.

Explanation

The sine of an angle equals opposite over hypotenuse, so hypotenuse = 5 / sin 30° = 5 / 0.5 = 10 m. The opposite side is half the hypotenuse whenever the angle is 30°.

Question 8Engineering fundamentals and applied mathematics

Which statement correctly describes the purpose of a factor of safety in design?

  1. It is the ratio of working stress to ultimate strength, always less than one.
  2. It is the ratio of ultimate strength to allowable working stress, providing margin against failure. Correct answer.
  3. It is the load at which a component first yields under test.
  4. It is the percentage by which a material's stiffness exceeds the design requirement.

Explanation

The factor of safety compares the material's failure strength with the stress actually permitted in service, so it is normally greater than one. It accounts for uncertainties in loading, material properties, and analysis.

Question 9Engineering fundamentals and applied mathematics

A pump delivers 0.012 m³/s through a pipe of internal diameter 0.1 m. What is the mean flow velocity?

  1. 1.53 m/s Correct answer.
  2. 15.3 m/s
  3. 3.06 m/s
  4. 0.15 m/s

Explanation

Area equals πd²/4 = π × 0.01/4 ≈ 0.00785 m². Dividing flow rate by area gives 0.012 / 0.00785 ≈ 1.53 m/s.

Question 10Engineering fundamentals and applied mathematics

A simply supported beam of span 6 m carries a uniformly distributed load of 5 kN/m over its full length. What is the maximum bending moment?

  1. 45 kN·m
  2. 22.5 kN·m Correct answer.
  3. 11.25 kN·m
  4. 7.5 kN·m

Explanation

Total load is 5 × 6 = 30 kN and each reaction is 15 kN. The maximum moment at midspan is wL²/8 = 5 × 36 / 8 = 22.5 kN·m.

Question 11Engineering fundamentals and applied mathematics

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance?

  1. 2 Ω Correct answer.
  2. 4.5 Ω
  3. 9 Ω
  4. 1.5 Ω

Explanation

For two resistors in parallel, R = (R₁R₂)/(R₁+R₂) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω. The result is always smaller than the smallest branch resistance, which confirms the answer.

Question 12Engineering fundamentals and applied mathematics

A body starts from rest and accelerates uniformly at 2.5 m/s². How far does it travel in 8 s?

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

Explanation

Using s = ut + ½at² with u = 0 gives s = 0.5 × 2.5 × 64 = 80 m. The squared time term dominates, so doubling the time quadruples the distance.

Question 13Engineering fundamentals and applied mathematics

A force of 200 N is applied at the end of a wrench 0.25 m from the centre of a bolt, perpendicular to the handle. What torque is produced?

  1. 25 N·m
  2. 8 N·m
  3. 50 N·m Correct answer.
  4. 800 N·m

Explanation

Torque equals force times perpendicular lever arm, 200 × 0.25 = 50 N·m. Because the force is perpendicular, no sine factor is needed.

Question 14Engineering fundamentals and applied mathematics

A steel bar of original length 2 m and cross-section 200 mm² carries a tensile load of 40 kN. Taking Young's modulus as 200 GPa, what is the elastic extension?

  1. 2 mm Correct answer.
  2. 0.5 mm
  3. 8 mm
  4. 4 mm

Explanation

Extension = FL/(AE) = (40,000 × 2)/(200 × 10⁻⁶ × 200 × 10⁹) = 80,000/40,000,000 = 0.002 m, or 2 mm. Consistent SI conversion of area and modulus is essential.

Question 15Engineering fundamentals and applied mathematics

In a series circuit of 12 V supply with resistors of 4 Ω and 8 Ω, what is the power dissipated in the 8 Ω resistor?

  1. 2.67 W
  2. 8 W Correct answer.
  3. 10.67 W
  4. 1.33 W

Explanation

Total resistance is 12 Ω, so current is 12/12 = 1 A. Power in the 8 Ω resistor is I²R = 1² × 8 = 8 W.

Question 16Engineering fundamentals and applied mathematics

A projectile is launched at 20 m/s at 30° above the horizontal on level ground. Neglecting air resistance and taking g as 10 m/s², what is the horizontal range?

  1. 17.3 m
  2. 20 m
  3. 34.6 m Correct answer.
  4. 40 m

Explanation

Range = v² sin 2θ / g = 400 × sin 60° / 10 = 400 × 0.866 / 10 ≈ 34.6 m. Alternatively, horizontal velocity is 17.32 m/s and flight time is 2 s.

Question 17Engineering fundamentals and applied mathematics

A hydraulic cylinder with a piston area of 0.005 m² must produce an output force of 15 kN. What pressure is required?

  1. 300 kPa
  2. 75 MPa
  3. 3 MPa Correct answer.
  4. 75 kPa

Explanation

Pressure equals force divided by area: 15,000 N / 0.005 m² = 3,000,000 Pa, or 3 MPa. Pascal's principle allows this pressure to be transmitted to other parts of a closed system.

Question 18Engineering fundamentals and applied mathematics

A simply supported beam of rectangular cross-section 50 mm wide and 100 mm deep is loaded in bending. If the depth is doubled to 200 mm while the width stays the same, how does the second moment of area change?

  1. It doubles
  2. It stays the same
  3. It quadruples
  4. It increases eightfold Correct answer.

Explanation

For a rectangle, I = bd³/12, so the second moment of area varies with the cube of depth. Doubling the depth multiplies I by 2³ = 8. This is why deep sections are efficient in bending.

Question 19Engineering fundamentals and applied mathematics

A 3-phase induction motor has 4 poles and is supplied at 50 Hz. What is its synchronous speed?

  1. 750 rpm
  2. 1,500 rpm Correct answer.
  3. 1,000 rpm
  4. 3,000 rpm

Explanation

Synchronous speed is 120f/p = 120 × 50 / 4 = 1,500 rpm. The rotor runs slightly slower than this, with the slip producing torque.

Question 20Engineering fundamentals and applied mathematics

A component is loaded such that the principal stresses are 80 MPa and 20 MPa in tension. What is the maximum shear stress in the plane of these stresses?

  1. 30 MPa Correct answer.
  2. 50 MPa
  3. 20 MPa
  4. 60 MPa

Explanation

Maximum in-plane shear stress equals half the difference of the principal stresses: (80 − 20)/2 = 30 MPa. This is the radius of Mohr's circle for the given state of stress.

Question 21Engineering fundamentals and applied mathematics

A conveyor drum of radius 0.3 m rotates at 120 revolutions per minute. What is the linear speed of the belt on the drum surface?

  1. 37.7 m/s
  2. 3.77 m/s Correct answer.
  3. 0.38 m/s
  4. 0.75 m/s

Explanation

Angular velocity is 120 × 2π / 60 ≈ 12.57 rad/s. Belt speed is ωr = 12.57 × 0.3 ≈ 3.77 m/s.

Question 22Engineering fundamentals and applied mathematics

A cantilever of length 2 m carries a point load of 10 kN at its free end. What are the bending moment and shear force at the fixed support?

  1. 20 kN·m moment and 10 kN shear Correct answer.
  2. 10 kN·m moment and 20 kN shear
  3. 20 kN·m moment and 20 kN shear
  4. 10 kN·m moment and 10 kN shear

Explanation

The fixed-end moment equals load times lever arm, 10 × 2 = 20 kN·m, hogging. The shear force at the support equals the full applied load, 10 kN, since no other vertical forces act.

Question 23Engineering fundamentals and applied mathematics

A closed-loop control system has a forward path gain of 20 and a negative feedback factor of 0.2. What is the closed-loop gain?

  1. 20
  2. 2.5
  3. 100
  4. 4 Correct answer.

Explanation

For negative feedback, closed-loop gain is G/(1 + GH) = 20/(1 + 20 × 0.2) = 20/5 = 4. Feedback trades raw gain for reduced sensitivity to variations in the forward path.

Question 24Engineering fundamentals and applied mathematics

Water flows through a horizontal pipe that narrows from 100 mm to 50 mm diameter. If the pressure in the wide section is 200 kPa and velocity there is 1 m/s, what is the approximate pressure in the narrow section? Take density as 1,000 kg/m³.

  1. 185 kPa Correct answer.
  2. 215 kPa
  3. 192.5 kPa
  4. 50 kPa

Explanation

By continuity the narrow-section velocity is 4 m/s, since area falls by a factor of four. Bernoulli then gives p₂ = 200,000 − ½ × 1,000 × (16 − 1) = 200,000 − 7,500 ≈ 192.5 kPa, so the closest listed value is about 192.5 kPa; the drop is 7.5 kPa.

Question 25Engineering fundamentals and applied mathematics

A simply supported beam of span L carries a uniformly distributed load w, and a second identical beam is fixed at both ends under the same load. What is the ratio of the maximum bending moment in the simply supported beam to that in the fixed-ended beam?

  1. 1:2
  2. 3:2 Correct answer.
  3. 2:3
  4. 1:1

Explanation

The simply supported maximum moment is wL²/8, while the fixed-ended maximum (at the supports) is wL²/12. The ratio (wL²/8)/(wL²/12) = 12/8 = 3:2, showing that end fixity reduces peak moment.

Question 26Engineering fundamentals and applied mathematics

A simply supported beam of length 4 m carries a point load of 20 kN at its midspan. What is the maximum bending moment in the beam?

  1. 20 kN·m Correct answer.
  2. 40 kN·m
  3. 10 kN·m
  4. 80 kN·m

Explanation

For a simply supported beam with a central point load, the maximum bending moment occurs at midspan and equals P·L/4. Substituting P = 20 kN and L = 4 m gives M = 20 × 4 / 4 = 20 kN·m. The reaction at each support is 10 kN, and the moment at midspan is 10 kN × 2 m = 20 kN·m.

Question 27Engineering fundamentals and applied mathematics

A 10 kg block is pulled across a horizontal floor by a horizontal force of 50 N. If the coefficient of kinetic friction between the block and the floor is 0.3 and g = 9.81 m/s², what is the acceleration of the block?

  1. 2.06 m/s² Correct answer.
  2. 5.00 m/s²
  3. 2.94 m/s²
  4. 7.94 m/s²

Explanation

The net horizontal force equals the applied force minus kinetic friction. Friction is μ·N = 0.3 × (10 × 9.81) = 29.43 N, so the net force is 50 − 29.43 = 20.57 N. Dividing by mass gives a = 20.57 / 10 = 2.057 m/s², approximately 2.06 m/s².

Question 28Engineering fundamentals and applied mathematics

A 2 kg mass attached to a spring of stiffness 200 N/m oscillates on a frictionless horizontal surface. What is the natural frequency of the system?

  1. 0.159 Hz
  2. 3.18 Hz
  3. 10.0 Hz
  4. 1.59 Hz Correct answer.

Explanation

The natural angular frequency is ω = √(k/m) = √(200/2) = √100 = 10 rad/s. The natural frequency in hertz is f = ω/(2π) = 10/(2π) ≈ 1.59 Hz. This follows directly from the standard formula for a simple harmonic oscillator.

Question 29Engineering fundamentals and applied mathematics

A 100 Ω strain gauge with a gauge factor of 2 is bonded to a steel member. When the member is loaded, the resistance of the gauge increases by 0.4 Ω. What is the strain experienced by the member?

  1. 0.008
  2. 0.002 Correct answer.
  3. 0.001
  4. 0.004

Explanation

The gauge factor GF relates fractional resistance change to strain: GF = (ΔR/R)/ε. Rearranging gives ε = (ΔR/R)/GF = (0.4/100)/2 = 0.004/2 = 0.002. This dimensionless strain corresponds to 2000 microstrain.

Question 30Engineering fundamentals and applied mathematics

A closed-loop control system has a forward-path transfer function G(s) = K/(s(s+2)) and unity negative feedback. Determine the range of K for which the closed-loop system is stable.

  1. K > 2
  2. K > 0 Correct answer.
  3. K < 0
  4. 0 < K < 2

Explanation

The closed-loop characteristic equation is 1 + G(s) = 0, giving s(s+2) + K = 0, or s² + 2s + K = 0. For a second-order polynomial with positive leading coefficient, Routh-Hurwitz stability requires all coefficients to be positive: 2 > 0 and K > 0. Thus the system is stable for any K > 0.

Question 31Materials, mechanics, and structural behavior

A steel bar of original length 2.0 m is loaded in tension and elongates by 1.2 mm. What is the engineering strain?

  1. 0.0012
  2. 0.0024
  3. 0.0006 Correct answer.
  4. 0.0003

Explanation

Engineering strain equals change in length divided by original length, so 1.2 mm divided by 2000 mm gives 0.0006. The units cancel because both quantities are lengths. Using the deformed length in the denominator is the true-strain definition and gives a negligibly different value here.

Question 32Materials, mechanics, and structural behavior

Which statement correctly describes the behaviour of a typical mild steel specimen just beyond the upper yield point?

  1. Stress rises steadily until the ultimate tensile strength is reached.
  2. Stress drops to a lower yield plateau while strain continues to increase. Correct answer.
  3. Stress falls to zero as the specimen fractures immediately.
  4. Stress remains constant while strain decreases elastically.

Explanation

After yielding, mild steel shows a yield drop followed by a plateau of roughly constant stress, the lower yield stress, during which Lüders bands propagate and strain increases. Strain hardening only begins after this plateau ends. The behaviour is a materials characteristic visible on the tensile stress-strain curve.

Question 33Materials, mechanics, and structural behavior

A simply supported beam of span 4 m carries a central point load of 12 kN. What is the maximum bending moment?

  1. 24 kN·m
  2. 6 kN·m
  3. 48 kN·m
  4. 12 kN·m Correct answer.

Explanation

For a simply supported beam with a central point load, the maximum moment is PL/4. Substituting 12 kN and 4 m gives 12 kN·m at midspan. Reactions are each 6 kN, and multiplying a reaction by half the span confirms the value.

Question 34Materials, mechanics, and structural behavior

What does the modulus of elasticity of a material represent?

  1. The slope of the linear elastic portion of its stress-strain curve Correct answer.
  2. The stress at which permanent deformation begins
  3. The area under the entire stress-strain curve up to fracture
  4. The maximum stress the material can sustain before necking

Explanation

The modulus of elasticity, or Young's modulus, is the ratio of stress to strain in the linear elastic region, which is the slope of that straight segment. It measures stiffness rather than strength. A steeper slope indicates a stiffer material that deforms less for a given stress.

Question 35Materials, mechanics, and structural behavior

Which property is most directly associated with the area under a stress-strain curve up to the point of fracture?

  1. Creep resistance
  2. Hardness
  3. Toughness Correct answer.
  4. Poisson's ratio

Explanation

The total area under the stress-strain curve to fracture represents the energy absorbed per unit volume before failure, which defines toughness. A tough material combines reasonable strength with substantial ductility. Hardness, creep resistance and Poisson's ratio describe different characteristics.

Question 36Materials, mechanics, and structural behavior

A structural member is loaded such that the principal stresses are 100 MPa and 40 MPa, both tensile. What is the maximum shear stress in the plane of these stresses?

  1. 20 MPa
  2. 140 MPa
  3. 70 MPa
  4. 30 MPa Correct answer.

Explanation

The maximum in-plane shear stress equals half the difference of the principal stresses, that is (100 − 40)/2 = 30 MPa. It corresponds to the radius of Mohr's circle for the in-plane state. The average normal stress on that plane is 70 MPa.

Question 37Materials, mechanics, and structural behavior

In a reinforced concrete beam designed by the ultimate strength method, what is the primary role of the longitudinal steel reinforcement in the tension zone?

  1. To prevent shrinkage cracking in the concrete cover
  2. To resist tensile stresses that concrete cannot carry reliably Correct answer.
  3. To increase the compressive strength of the concrete
  4. To reduce the self-weight of the beam

Explanation

Concrete is strong in compression but weak and brittle in tension, so steel bars are placed where flexural tension occurs to carry those forces. The two materials act together through bond, with the steel providing ductility before failure. This is the basis of reinforced concrete design.

Question 38Materials, mechanics, and structural behavior

What is the main purpose of a tensile test on a steel coupon?

  1. To evaluate the permeability of the material to water
  2. To assess the fatigue life under cyclic loading
  3. To measure the hardness profile through the thickness
  4. To determine strength and ductility properties such as yield stress, ultimate strength and elongation Correct answer.

Explanation

A tensile test loads a specimen axially to failure while recording load and extension, giving a stress-strain curve. From it, yield strength, ultimate tensile strength, elastic modulus and elongation at fracture are obtained. These are the fundamental mechanical properties used in design.

Question 39Materials, mechanics, and structural behavior

A cantilever beam of length 3 m carries a uniformly distributed load of 5 kN/m over its full length. What is the magnitude of the maximum bending moment at the fixed support?

  1. 45 kN·m
  2. 22.5 kN·m Correct answer.
  3. 7.5 kN·m
  4. 15 kN·m

Explanation

For a cantilever with a uniform load, the fixed-end moment is wL²/2. Substituting 5 kN/m and 3 m gives 5 × 9 / 2 = 22.5 kN·m. The hogging moment is largest at the support and decreases to zero at the free end.

Question 40Materials, mechanics, and structural behavior

A steel rod with a cross-sectional area of 200 mm² carries an axial tensile load of 50 kN. If the elastic modulus is 200 GPa, what is the axial stress?

  1. 100 MPa
  2. 250 MPa Correct answer.
  3. 25 MPa
  4. 500 MPa

Explanation

Axial stress is force divided by area, so 50 000 N divided by 200 mm² equals 250 N/mm², which is 250 MPa. The elastic modulus is not needed for the stress and would only be required for deformation. The result is well below typical steel yield strength.

Question 41Materials, mechanics, and structural behavior

A simply supported beam has a rectangular cross-section 100 mm wide and 200 mm deep. What is its second moment of area about the horizontal centroidal axis?

  1. 66.7 × 10⁶ mm⁴ Correct answer.
  2. 200 × 10⁶ mm⁴
  3. 33.3 × 10⁶ mm⁴
  4. 100 × 10⁶ mm⁴

Explanation

For a rectangle, I equals bh³/12 about the centroidal axis parallel to the base. Using b = 100 mm and h = 200 mm gives 100 × 8 000 000 / 12, which is 66.7 × 10⁶ mm⁴. The depth enters cubed, so it dominates the stiffness.

Question 42Materials, mechanics, and structural behavior

Which statement best explains why a short, stocky steel column fails by yielding rather than by elastic buckling?

  1. Its elastic modulus is higher than that of a slender column.
  2. Its yield strength increases as the length decreases.
  3. Its cross-section has a larger radius of gyration than a slender column.
  4. Its slenderness ratio is low, so the buckling load exceeds the squash load. Correct answer.

Explanation

Buckling capacity falls as the effective length increases relative to the radius of gyration. For a short member the Euler buckling load is very high, so the section reaches its yield or squash load first. Failure is then governed by material strength rather than instability.

Question 43Materials, mechanics, and structural behavior

What is the purpose of applying a partial safety factor to material strength in limit state design?

  1. To account for variability and uncertainty in material properties and workmanship Correct answer.
  2. To convert working loads into factored loads
  3. To reduce the self-weight of the structure
  4. To increase the ductility of the structural material

Explanation

Partial safety factors on material strength reduce the characteristic value to a design value that reflects possible weaknesses, variability and construction tolerances. Load factors are applied separately to actions. Together they ensure the design resistance exceeds the design effect of actions with an accepted reliability.

Question 44Materials, mechanics, and structural behavior

A material has a Poisson's ratio of 0.3. If it is subjected to an axial tensile strain of 0.001, what is the magnitude of the lateral strain?

  1. 0.0033
  2. 0.0003 Correct answer.
  3. 0.0013
  4. 0.0007

Explanation

Poisson's ratio is the negative ratio of lateral strain to axial strain, so the lateral strain magnitude equals 0.3 × 0.001 = 0.0003. The lateral strain is compressive when the axial strain is tensile. This contraction arises from conservation of volume in the elastic range.

Question 45Materials, mechanics, and structural behavior

A beam of span 6 m is simply supported and carries a uniformly distributed load of 8 kN/m. What is the maximum bending moment?

  1. 18 kN·m
  2. 36 kN·m Correct answer.
  3. 72 kN·m
  4. 48 kN·m

Explanation

For a simply supported beam with a uniform load, the maximum moment at midspan is wL²/8. Substituting 8 kN/m and 6 m gives 8 × 36 / 8 = 36 kN·m. Each reaction is wL/2 = 24 kN, and multiplying by 3 m minus the load moment confirms the result.

Question 46Materials, mechanics, and structural behavior

Which mechanism most commonly causes fatigue failure in a steel bridge detail subjected to repeated traffic loading?

  1. Gradual creep deformation at ambient temperature
  2. Corrosion of the concrete cover
  3. Elastic recovery of the member after each load cycle
  4. Initiation and growth of a crack from a stress concentration under cyclic stress Correct answer.

Explanation

Fatigue cracks usually start at geometric discontinuities such as welds, holes or sharp changes of section, where local stress is amplified. Repeated cycles then propagate the crack until the remaining section fractures. Fatigue strength therefore depends on detail category and stress range.

Question 47Materials, mechanics, and structural behavior

In a reinforced concrete section at the ultimate limit state, which assumption underlies the flexural design?

  1. Steel and concrete have identical elastic moduli
  2. Plane sections remain plane, with strain compatible between steel and concrete Correct answer.
  3. The neutral axis coincides with the centroid of the section
  4. Concrete carries all tensile stress after cracking

Explanation

Flexural theory assumes that strains vary linearly through the depth, so the strain in the steel is related to the strain in the concrete by similar triangles. Stress is then obtained from the material constitutive laws, with concrete tensile strength neglected. This compatibility assumption is central to ultimate strength design.

Question 48Materials, mechanics, and structural behavior

A short column of cross-sectional area 25 000 mm² carries an axial load of 1500 kN. What is the average compressive stress?

  1. 0.06 MPa
  2. 600 MPa
  3. 60 MPa Correct answer.
  4. 6 MPa

Explanation

Average compressive stress is load divided by area, so 1 500 000 N divided by 25 000 mm² gives 60 N/mm², which is 60 MPa. This value is within the range of concrete strength, so the section would need checking against design resistance. Uniform stress is assumed for a short axially loaded column.

Question 49Materials, mechanics, and structural behavior

Why is a shear force diagram for a beam with only point loads drawn as a series of horizontal steps?

  1. Shear varies linearly along each segment between point loads.
  2. The bending moment is constant between point loads.
  3. Each point load causes an abrupt jump in shear, while between loads the shear is constant. Correct answer.
  4. The beam stiffness changes at each point load.

Explanation

Shear equals the algebraic sum of vertical forces to one side of a section. A point load introduces a concentrated force, so the shear changes discontinuously by that amount at its location and stays constant in between. A distributed load, by contrast, produces a sloping shear diagram.

Question 50Materials, mechanics, and structural behavior

A steel bar is constrained between two rigid supports and is heated uniformly. Which statement describes the resulting stress state?

  1. A compressive stress develops because free thermal expansion is prevented. Correct answer.
  2. A bending moment develops about the bar's centroidal axis.
  3. No stress develops because steel has a low thermal expansion coefficient.
  4. A tensile stress develops because the bar tries to contract.

Explanation

If expansion is fully restrained, the thermal strain cannot occur, so an elastic compressive strain of equal magnitude is induced. The stress equals the product of the elastic modulus, the thermal expansion coefficient and the temperature change. Restraint, not the magnitude of expansion alone, governs the stress.