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Pancyprian Appl. Mech. Sci. III study materials
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Pancyprian Appl. Mech. Sci. III practice questions

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Question 1Statics, strength and mechanical transmission

A steel rod carries an axial tensile force of 48 kN. The rod has a circular cross-section of diameter 20 mm. What is the average normal stress in the rod?

  1. 76.4 MPa
  2. 152.8 MPa Correct answer.
  3. 38.2 MPa
  4. 240 MPa

Explanation

σ = F/A. A = πd²/4 = π(0.020)²/4 = 3.142×10⁻⁴ m². With F = 48 000 N, σ = 48 000 / 3.142×10⁻⁴ = 1.528×10⁸ Pa = 152.8 MPa.

Question 2Statics, strength and mechanical transmission

A bolted joint carries a shear force of 18 kN on a single bolt of diameter 12 mm. Assuming single shear and uniform stress, the average shear stress in the bolt is closest to:

  1. 79.6 MPa
  2. 159.2 MPa Correct answer.
  3. 39.8 MPa
  4. 127 MPa

Explanation

τ = F/A with A = πd²/4 = π(0.012)²/4 = 1.131×10⁻⁴ m². τ = 18 000 / 1.131×10⁻⁴ = 1.591×10⁸ Pa ≈ 159.2 MPa.

Question 3Statics, strength and mechanical transmission

A rectangular bar 40 mm × 10 mm carries an axial compressive load of 20 kN. The compressive stress is:

  1. 50 MPa Correct answer.
  2. 5 MPa
  3. 500 MPa
  4. 20 MPa

Explanation

A = 0.040 × 0.010 = 4.0×10⁻⁴ m². σ = F/A = 20 000 / 4.0×10⁻⁴ = 5.0×10⁷ Pa = 50 MPa.

Question 4Statics, strength and mechanical transmission

A hollow circular shaft has outer diameter Do = 60 mm and inner diameter Di = 40 mm. It transmits a torque T = 1.0 kN·m. The maximum shear stress due to torsion is closest to:

  1. 29.5 MPa Correct answer.
  2. 51.5 MPa
  3. 15.0 MPa
  4. 73.0 MPa

Explanation

J = (π/32)(Do⁴ − Di⁴) = (π/32)(0.060⁴ − 0.040⁴) = (π/32)(1.296×10⁻⁵ − 2.56×10⁻⁶) = 1.021×10⁻⁶ m⁴. τ_max = T c / J = 1000 × 0.030 / 1.021×10⁻⁶ ≈ 2.94×10⁷ Pa ≈ 29.4 MPa, closest to 29.5 MPa.

Question 5Statics, strength and mechanical transmission

A simply supported beam of length L = 2.0 m carries a central concentrated load P = 4 kN. The maximum bending moment is:

  1. 1.0 kN·m
  2. 2.0 kN·m Correct answer.
  3. 4.0 kN·m
  4. 8.0 kN·m

Explanation

For a central point load on a simply supported beam, M_max = PL/4 = (4 kN)(2.0 m)/4 = 2.0 kN·m at mid-span.

Question 6Statics, strength and mechanical transmission

A beam cross-section has I = 2.0×10⁻⁶ m⁴. At a section where M = 1.5 kN·m, the bending stress at y = 40 mm from the neutral axis is:

  1. 30 MPa Correct answer.
  2. 7.5 MPa
  3. 75 MPa
  4. 3.0 MPa

Explanation

σ = My/I = (1500 N·m)(0.040 m) / 2.0×10⁻⁶ m⁴ = 3.0×10⁷ Pa = 30 MPa.

Question 7Statics, strength and mechanical transmission

A thin-walled cylindrical pressure vessel has internal diameter 200 mm and wall thickness 5 mm. For internal pressure p = 2.0 MPa, the hoop (circumferential) stress is:

  1. 20 MPa
  2. 40 MPa Correct answer.
  3. 10 MPa
  4. 80 MPa

Explanation

For a thin cylinder, σ_h = pD/(2t) = (2.0 MPa)(200 mm)/(2×5 mm) = 40 MPa.

Question 8Statics, strength and mechanical transmission

For the thin cylinder with p = 2.0 MPa, D = 200 mm and t = 5 mm, the longitudinal stress is:

  1. 40 MPa
  2. 20 MPa Correct answer.
  3. 10 MPa
  4. 80 MPa

Explanation

σ_l = pD/(4t) = (2.0)(200)/(4×5) = 20 MPa, half the hoop stress for a closed thin cylinder.

Question 9Statics, strength and mechanical transmission

A pin of diameter 10 mm is in double shear and carries a load of 25 kN. The average shear stress in the pin is closest to:

  1. 159 MPa Correct answer.
  2. 318 MPa
  3. 79.6 MPa
  4. 398 MPa

Explanation

In double shear, resisting area = 2 × (πd²/4) = 1.571×10⁻⁴ m². τ = 25 000 / 1.571×10⁻⁴ ≈ 159 MPa.

Question 10Statics, strength and mechanical transmission

A short column of cross-sectional area 2500 mm² carries a concentric compressive load of 150 kN. The compressive stress is:

  1. 60 MPa Correct answer.
  2. 6 MPa
  3. 600 MPa
  4. 15 MPa

Explanation

A = 2500 mm² = 2.5×10⁻³ m². σ = 150 000 / 2.5×10⁻³ = 60 MPa.

Question 11Statics, strength and mechanical transmission

A shaft of diameter 40 mm is subjected to a torque of 500 N·m. The maximum torsional shear stress is closest to:

  1. 39.8 MPa Correct answer.
  2. 19.9 MPa
  3. 79.6 MPa
  4. 9.95 MPa

Explanation

For a solid circular shaft, τ_max = 16T/(πd³) = 16×500 / (π × 0.040³) ≈ 39.8 MPa.

Question 12Statics, strength and mechanical transmission

A plate of thickness 8 mm is punched with a hole of diameter 20 mm. If the ultimate shear strength of the plate is 300 MPa, the force required to punch the hole is closest to:

  1. 151 kN Correct answer.
  2. 75 kN
  3. 302 kN
  4. 37.7 kN

Explanation

Sheared area A = π d t = π(0.020)(0.008) = 5.027×10⁻⁴ m². F = τ A = (300×10⁶)(5.027×10⁻⁴) ≈ 151 kN.

Question 13Statics, strength and mechanical transmission

A welded joint has a throat thickness of 5 mm and length 80 mm. If the allowable shear stress in the weld is 80 MPa, the allowable shear load on a single fillet (throat area = t×L) is:

  1. 32 kN Correct answer.
  2. 16 kN
  3. 64 kN
  4. 8 kN

Explanation

A = t × L = 0.005 × 0.080 = 4.0×10⁻⁴ m². F_allow = τ A = (80×10⁶)(4.0×10⁻⁴) = 32 kN.

Question 14Statics, strength and mechanical transmission

At a point in a plane stress state, σ_x = 80 MPa, σ_y = 20 MPa and τ_xy = 0. The maximum principal stress is:

  1. 80 MPa Correct answer.
  2. 20 MPa
  3. 50 MPa
  4. 100 MPa

Explanation

With zero shear, the principal stresses are σ_x and σ_y. The algebraically larger is σ_1 = 80 MPa.

Question 15Statics, strength and mechanical transmission

A timber post 100 mm × 100 mm carries a compressive load of 80 kN. If the allowable compressive stress is 8 MPa, the post is:

  1. at the allowable stress (σ = 8.0 MPa) Correct answer.
  2. overstressed at 80 MPa
  3. understressed at 0.8 MPa
  4. overstressed at 16 MPa

Explanation

The supplied average compression model gives80000N/10000mm²=8N/mm²=8MPa, equal to the stated allowable stress. This answers only that comparison; buckling, connections, load factors and other real timber design checks are not established by the average-stress calculation.

Question 16Statics, strength and mechanical transmission

A circular bar of diameter 16 mm is subjected to an axial tensile load of 12 kN. The normal stress is closest to:

  1. 59.7 MPa Correct answer.
  2. 29.8 MPa
  3. 119 MPa
  4. 15 MPa

Explanation

A = π(0.016)²/4 = 2.011×10⁻⁴ m². σ = 12 000 / 2.011×10⁻⁴ ≈ 59.7 MPa.

Question 17Statics, strength and mechanical transmission

A cantilever beam of length 1.2 m carries an end load of 2.5 kN. The maximum bending moment (at the fixed end) is:

  1. 3.0 kN·m Correct answer.
  2. 2.5 kN·m
  3. 1.5 kN·m
  4. 6.0 kN·m

Explanation

For a cantilever with end load, M_max = P L = 2.5 × 1.2 = 3.0 kN·m at the support.

Question 18Statics, strength and mechanical transmission

A solid circular shaft must transmit T = 1.2 kN·m at an allowable shear stress of 60 MPa. The minimum required diameter is closest to:

  1. 46.7 mm Correct answer.
  2. 37.1 mm
  3. 58.8 mm
  4. 23.4 mm

Explanation

From τ = 16T/(πd³), d³ = 16T/(πτ) = 16×1200/(π×60×10⁶) = 1.018×10⁻⁴; d = ∛(1.018×10⁻⁴) ≈ 0.0467 m = 46.7 mm.

Question 19Statics, strength and mechanical transmission

A bolt of diameter 16 mm bears on a plate of thickness 10 mm under a load of 20 kN. The average bearing stress is:

  1. 125 MPa Correct answer.
  2. 62.5 MPa
  3. 250 MPa
  4. 99.5 MPa

Explanation

Bearing area A_b = d × t = 0.016 × 0.010 = 1.6×10⁻⁴ m². σ_b = 20 000 / 1.6×10⁻⁴ = 125 MPa.

Question 20Statics, strength and mechanical transmission

A bar of length 1.5 m and cross-section 200 mm² is aluminium with E = 70 GPa. Under an axial tensile load of 14 kN, the elongation is closest to:

  1. 1.50 mm Correct answer.
  2. 0.75 mm
  3. 3.00 mm
  4. 0.15 mm

Explanation

δ = FL/(AE) = (14 000)(1.5) / [(200×10⁻⁶)(70×10⁹)] = 21 000 / 1.4×10⁷ = 1.5×10⁻³ m = 1.50 mm.

Question 21Statics, strength and mechanical transmission

Poisson's ratio ν = 0.30. If the axial strain is 1.0×10⁻³, the lateral strain magnitude is:

  1. 3.0×10⁻⁴ Correct answer.
  2. 1.0×10⁻³
  3. 3.0×10⁻³
  4. 7.0×10⁻⁴

Explanation

|ε_lat| = ν |ε_axial| = 0.30 × 1.0×10⁻³ = 3.0×10⁻⁴ (opposite sign to axial strain).

Question 22Statics, strength and mechanical transmission

A shear strain γ = 0.002 develops under shear stress τ = 160 MPa. The shear modulus G is:

  1. 80 GPa Correct answer.
  2. 40 GPa
  3. 160 GPa
  4. 320 GPa

Explanation

G = τ/γ = (160×10⁶)/0.002 = 80 GPa. Choose the option that matches this evaluated result and reject values from misapplied formulas.

Question 23Statics, strength and mechanical transmission

A concrete cylinder shortens by 0.30 mm under compression. Original gauge length was 150 mm. The compressive strain magnitude is:

  1. 0.0020 Correct answer.
  2. 0.0025
  3. 0.00030
  4. 0.020

Explanation

ε = 0.30/150 = 0.0020. Choose the option that matches this evaluated result and reject values from misapplied formulas.

Question 24Statics, strength and mechanical transmission

Within the elastic range, stress is proportional to strain. If stress doubles while E is constant, strain:

  1. doubles Correct answer.
  2. halves
  3. quadruples
  4. remains unchanged

Explanation

From ε = σ/E with E fixed, strain is directly proportional to stress, so doubling stress doubles strain.

Question 25Statics, strength and mechanical transmission

A rod has stiffness k = AE/L = 50 kN/mm. Under a 10 kN tensile load the elongation is:

  1. 0.20 mm Correct answer.
  2. 0.50 mm
  3. 5.0 mm
  4. 0.02 mm

Explanation

δ = F/k = 10 / 50 = 0.20 mm. Choose the option that matches this evaluated result and reject values from misapplied formulas.

Question 26Statics, strength and mechanical transmission

Two bars in series carry the same axial load F. Bar 1: A1 = 200 mm², L1 = 400 mm, E1 = 200 GPa. Bar 2: A2 = 100 mm², L2 = 200 mm, E2 = 100 GPa. The ratio of elongations δ1/δ2 equals:

  1. 0.5 Correct answer.
  2. 1.0
  3. 2.0
  4. 4.0

Explanation

δ ∝ L/(AE) at constant F. δ1/δ2 = (L1/L2)×(A2/A1)×(E2/E1) = (400/200)×(100/200)×(100/200) = 2 × 0.5 × 0.5 = 0.5.

Question 27Thermal Physics

An ideal gas occupies 0.050 m³ at absolute pressure 200 kPa and temperature 300 K. Using R = 287 J/(kg·K) for air, the mass is closest to:

  1. 0.116 kg Correct answer.
  2. 0.232 kg
  3. 0.058 kg
  4. 1.16 kg

Explanation

From PV = mRT, m = PV/(RT) = (200 000×0.050)/(287×300) = 10 000 / 86 100 ≈ 0.116 kg.

Question 28Thermal Physics

Air in a rigid closed tank is heated from 300 K to 450 K. Absolute pressure starts at 100 kPa. The final pressure is:

  1. 150 kPa Correct answer.
  2. 100 kPa
  3. 67 kPa
  4. 200 kPa

Explanation

For constant volume, P/T is constant, so P2 = P1 (T2/T1) = 100 × (450/300) = 150 kPa.

Question 29Thermal Physics

A closed system receives 40 kJ of heat and performs 15 kJ of work. The change in internal energy ΔU is:

  1. +25 kJ Correct answer.
  2. +55 kJ
  3. −25 kJ
  4. +15 kJ

Explanation

First law (heat in positive, work out positive): ΔU = Q − W = 40 − 15 = +25 kJ.

Question 30Thermal Physics

Water of mass 2.0 kg is heated from 20°C to 80°C. Taking c = 4.18 kJ/(kg·K), the heat required (no phase change) is closest to:

  1. 502 kJ Correct answer.
  2. 251 kJ
  3. 1004 kJ
  4. 83.6 kJ

Explanation

Q = m c ΔT = 2.0 × 4.18 × (80−20) = 2.0 × 4.18 × 60 = 501.6 kJ ≈ 502 kJ.

Question 31Thermal Physics

An ideal gas expands isothermally. For an ideal gas isothermal process, the change in internal energy ΔU is:

  1. zero Correct answer.
  2. equal to Q
  3. equal to W
  4. equal to PV

Explanation

For an ideal gas, U depends only on temperature. In an isothermal process ΔT = 0, so ΔU = 0 and Q = W.

Question 32Thermal Physics

Air (cp = 1.005 kJ/(kg·K)) of mass 0.5 kg is heated at constant pressure from 300 K to 400 K. The heat transfer is closest to:

  1. 50.3 kJ Correct answer.
  2. 100.5 kJ
  3. 25.1 kJ
  4. 201 kJ

Explanation

At constant pressure Q = m cp ΔT = 0.5 × 1.005 × 100 = 50.25 kJ ≈ 50.3 kJ.

Question 33Thermal Physics

A piston–cylinder device contains gas that expands and does 30 kJ of boundary work while receiving 10 kJ of heat. The change in internal energy is:

  1. −20 kJ Correct answer.
  2. +20 kJ
  3. +40 kJ
  4. −40 kJ

Explanation

ΔU = Q − W = 10 − 30 = −20 kJ. Choose the option that matches this evaluated result and reject values from misapplied formulas.

Question 34Thermal Physics

Steam enters a turbine at enthalpy h1 = 3200 kJ/kg and leaves at h2 = 2600 kJ/kg. Neglecting KE, PE and heat loss, the work output per kilogram is:

  1. 600 kJ/kg Correct answer.
  2. 5800 kJ/kg
  3. 3200 kJ/kg
  4. 2600 kJ/kg

Explanation

Steady-flow energy balance with negligible KE/PE/Q: w = h1 − h2 = 3200 − 2600 = 600 kJ/kg.

Question 35Thermal Physics

An ideal-gas polytropic process follows PV^n = constant with n = 1.4. If P1 = 100 kPa, V1 = 0.2 m³ and V2 = 0.1 m³, P2 is closest to:

  1. 264 kPa Correct answer.
  2. 200 kPa
  3. 50 kPa
  4. 140 kPa

Explanation

P2 = P1 (V1/V2)^n = 100 × (2)^1.4 = 100 × 2.639 ≈ 264 kPa.

Question 36Thermal Physics

The thermal efficiency of a heat engine that receives 500 kJ of heat and rejects 350 kJ to a sink is:

  1. 30% Correct answer.
  2. 70%
  3. 43%
  4. 14%

Explanation

η = 1 − Q_out/Q_in = 1 − 350/500 = 0.30 = 30%. Equivalently W_net/Q_in = 150/500 = 30%.

Question 37Thermal Physics

A gas is compressed adiabatically and reversibly (isentropic). For an ideal gas with γ = 1.4, if T1 = 290 K and the pressure ratio P2/P1 = 8, T2 is closest to:

  1. 525 K Correct answer.
  2. 290 K
  3. 2320 K
  4. 410 K

Explanation

For the supplied reversible adiabatic ideal-gas model, T2=T1(P2/P1)^((gamma−1)/gamma). Thus290×8^(0.4/1.4)≈525K. Absolute temperature and the stated constant heat-capacity-ratio assumption are required; this is not an actual compressor discharge-temperature guarantee.

Question 38Thermal Physics

Enthalpy for an ideal gas may be written h = u + Pv. If u = 200 kJ/kg and Pv = 80 kJ/kg, then h is:

  1. 280 kJ/kg Correct answer.
  2. 120 kJ/kg
  3. 200 kJ/kg
  4. 80 kJ/kg

Explanation

h = u + Pv = 200 + 80 = 280 kJ/kg. Choose the option that matches this evaluated result and reject values from misapplied formulas.

Question 39Thermal Physics

A refrigerator has coefficient of performance COP = Q_C/W_in = 3.0. If it removes 9.0 kW of heat from the cold space, the power input is:

  1. 3.0 kW Correct answer.
  2. 27 kW
  3. 6.0 kW
  4. 12 kW

Explanation

W_in = Q_C / COP = 9.0 / 3.0 = 3.0 kW. Choose the option that matches this evaluated result and reject values from misapplied formulas.

Question 40Thermal Physics

In a constant-pressure process, 2 kg of air (cp = 1.00 kJ/(kg·K)) is heated so that temperature rises by 50 K. The enthalpy change of the air is:

  1. 100 kJ Correct answer.
  2. 50 kJ
  3. 200 kJ
  4. 25 kJ

Explanation

For an ideal gas ΔH = m cp ΔT = 2 × 1.00 × 50 = 100 kJ (and Q_p = ΔH).

Question 41Thermal Physics

Absolute zero on the Celsius scale corresponds to:

  1. −273.15°C Correct answer.
  2. 0°C
  3. −100°C
  4. 273.15°C

Explanation

The absolute (Kelvin) zero point is −273.15°C. Thermodynamic absolute temperature T(K) = t(°C) + 273.15.

Question 42Thermal Physics

A closed rigid tank contains 0.2 kg of air. Heat addition of 10 kJ raises temperature. With cv = 0.718 kJ/(kg·K), the temperature rise is closest to:

  1. 70 K Correct answer.
  2. 50 K
  3. 14 K
  4. 140 K

Explanation

For constant volume, Q = m cv ΔT, so ΔT = Q/(m cv) = 10 / (0.2 × 0.718) = 10 / 0.1436 ≈ 69.6 K ≈ 70 K.

Question 43Thermal Physics

Work done by an ideal gas in a reversible isothermal expansion from V1 to V2 is W = mRT ln(V2/V1). For mRT = 5 kJ and V2/V1 = e (≈2.718), W is:

  1. 5 kJ Correct answer.
  2. 5e kJ
  3. 0 kJ
  4. 2.5 kJ

Explanation

ln(e) = 1, so W = 5 × 1 = 5 kJ. Choose the option that matches this evaluated result and reject values from misapplied formulas.

Question 44Pneumatic and hydraulic model reasoning

The hydrostatic pressure increase at a depth of 5.0 m in fresh water (ρ = 1000 kg/m³, g = 9.81 m/s²) is closest to:

  1. 49.1 kPa Correct answer.
  2. 5.0 kPa
  3. 98.1 kPa
  4. 9.81 kPa

Explanation

Δp = ρ g h = 1000 × 9.81 × 5.0 = 49 050 Pa ≈ 49.1 kPa.

Question 45Pneumatic and hydraulic model reasoning

A mercury manometer (ρ_Hg = 13 600 kg/m³) shows a column difference of 250 mm. The gauge pressure difference (g = 9.81 m/s²) is closest to:

  1. 33.4 kPa Correct answer.
  2. 3.34 kPa
  3. 24.5 kPa
  4. 2.45 kPa

Explanation

Δp = ρ g h = 13 600 × 9.81 × 0.250 ≈ 33 350 Pa ≈ 33.4 kPa.

Question 46Pneumatic and hydraulic model reasoning

Oil of specific gravity 0.85 flows in a pipe of diameter 100 mm at mean velocity 2.0 m/s. The volume flow rate is closest to:

  1. 0.0157 m³/s Correct answer.
  2. 0.0628 m³/s
  3. 0.0314 m³/s
  4. 0.200 m³/s

Explanation

A = πd²/4 = π(0.10)²/4 = 7.854×10⁻³ m². Q = A V = 7.854×10⁻³ × 2.0 ≈ 0.0157 m³/s. (SG is not needed for volume flow.)

Question 47Pneumatic and hydraulic model reasoning

Continuity for incompressible flow: a pipe reduces from diameter 200 mm to 100 mm. If upstream velocity is 1.5 m/s, the downstream velocity is:

  1. 6.0 m/s Correct answer.
  2. 3.0 m/s
  3. 1.5 m/s
  4. 0.375 m/s

Explanation

A1 V1 = A2 V2. Area ratio = (D1/D2)² = 4, so V2 = 4 × 1.5 = 6.0 m/s.

Question 48Pneumatic and hydraulic model reasoning

Water flows in a horizontal pipe. Upstream velocity is 2.0 m/s and downstream (narrower section) velocity is 6.0 m/s. Neglecting losses, the pressure drop from upstream to downstream (ρ = 1000 kg/m³) is:

  1. 16 kPa Correct answer.
  2. 18 kPa
  3. 32 kPa
  4. 4 kPa

Explanation

Bernoulli for horizontal frictionless flow: Δp = ρ(V2² − V1²)/2 = 1000 × (36 − 4)/2 = 1000 × 16 = 16 000 Pa = 16 kPa.

Question 49Pneumatic and hydraulic model reasoning

A wooden block of volume 0.020 m³ floats in fresh water with 75% of its volume submerged. The mass of the block is:

  1. 15 kg Correct answer.
  2. 20 kg
  3. 10 kg
  4. 7.5 kg

Explanation

Buoyant force = weight. Displaced volume = 0.75 × 0.020 = 0.015 m³. m = ρ V_disp = 1000 × 0.015 = 15 kg.

Question 50Pneumatic and hydraulic model reasoning

Dynamic viscosity μ = 0.001 Pa·s. Between two parallel plates 0.5 mm apart, the top plate moves at 2.0 m/s and the bottom is fixed (linear Couette profile). The shear stress is:

  1. 4.0 Pa Correct answer.
  2. 0.001 Pa
  3. 4000 Pa
  4. 1.0 Pa

Explanation

τ = μ (du/dy) = 0.001 × (2.0 / 0.0005) = 0.001 × 4000 = 4.0 Pa.