RRB JE Mechanical-SOM Questionbank 1

Which region of the stress-strain curve as shown below represents work hardening in ductile materials? RRB JE (Re-Exam) 04.06.2025

D-E
E-F
C-D
A-C

Explanation:

Elastic region (A-C), Plastic region (C-F), Work hardening region (C-D), Necking region (E-F).

Thermal stress in a composite bar DOES NOT depend on which of the following factors? RRB JE (Re-Exam) 04.06.2025

Coefficient of thermal expansion
Temperature change
Modulus of elasticity
Area of cross-section

Explanation:

Thermal stress (σ_T) = E × α × ΔT, where E = Young’s modulus, ΔT = change in temperature, α = coefficient of thermal expansion. It does not depend on area of cross-section.

The modulus of resilience is characterised by the area located under the stress-strain curve upto the —-? RRB JE (Re-Exam) 04.06.2025

ultimate point
proportional limit
point where strain hardening starts
fracture point

Explanation:

Modulus of resilience is the strain energy per unit volume up to the proportional limit. It is equal to ½ × σ × ε. Modulus of toughness is the total area under the stress-strain curve up to fracture point.

Determine the minimum thickness of the rectangular axial bar shown against yielding. Given Factor of Safety (FOS) = 2 and Yield stress = 310 MPa. RRB JE Stage 2 (22.04.2025 9:00 AM-11:00 AM)

25 mm
60 mm
155 mm
19.4 mm

Explanation:

σ = Yield stress/FOS = 310/2 = 155 MPa. P/A = 155, (120 × 10³)/(40 × t) = 155, t = (120 × 10³)/(155 × 40) = 19.354 mm ≈ 19.4 mm.

If elastic strength increases 3 times, then Proof Resilience —-? RRB JE Stage 2 (22.04.2025 2:30-4:30 PM)

increases 9 times
increases 3 times
decreases 9 times
decreases 3 times

Explanation:

Proof Resilience U = σₑ²/(2E). If σₙ = 3σₑ, then Uₙₑw = (3σₑ)²/(2E) = 9σₑ²/(2E) = 9U. So it increases 9 times.

A steel bar (E = 200 N/m², α = 12 × 10⁻⁶/°C) expands by 0.3 mm due to a temperature increase. If the original length of the bar was 15 cm, what was the temperature rise? RRB JE Stage 2 (22.04.2025 2:30-4:30 PM)

166.6°C
100°C
120.6°C
180°C

Explanation:

ΔL = α × L₀ × ΔT, ΔT = ΔL/(α × L₀) = 0.3/(12 × 10⁻⁶ × 150) = 0.3/(1.8 × 10⁻³) = 166.67°C.

When a body is subjected to two equal and opposite pulls, as a result of which the body tends to extend its length, the stress and strain induced are —-? RRB JE 29-08-2019

Tensile stress and compressive strain
Compressive stress and tensile strain
Tensile stress and tensile strain
Compressive stress and compressive strain

Explanation:

When a body is subjected to two equal and opposite pulls (tensile forces), the body tends to extend its length. The stress and strain induced are tensile stress and tensile strain.

____ is a beam with one end fixed and the other end simply supported. RRB JE 29-08-2019

Fixed beam
Continuous beam
Propped cantilever beam
Over-hanging beam

Explanation:

A propped cantilever beam is a beam with one end fixed and the other end simply supported.

The Hoop stress developed in the thin cylinders is given by —-? (where P = Internal pressure, d = Internal diameter and t = wall thickness) RRB JE 29-08-2019

Pd/3t
Pd/2t
Pd/t
Pd/4t

Explanation:

For thin cylinders: Hoop/Circumferential stress σ_t = Pd/(2t), Longitudinal stress σ_l = Pd/(4t). For thin spherical vessels: σ = Pd/(4t).

A carbon steel having a Brinell hardness number 100 should have ultimate tensile strength closer to —-? RRB JE 30-08-2019

220 N/mm²
150 N/mm²
800 N/mm²
350 N/mm²

Explanation:

Ultimate tensile strength ≈ 3.5 × BHN = 3.5 × 100 = 350 N/mm².

If ‘α’ is coefficient of thermal expansion, ‘ΔT’ is magnitude of change in temperature and ‘E’ is modulus of elasticity, then the expression for thermal stress induced in a rod of length ‘L’ that is fixed between two rigid ends is given as —-? RRB JE 31-08-2019

α E ΔT
α ΔT / E
α E L ΔT
α L ΔT

Explanation:

Thermal stress in a rod fixed between rigid ends is σ_th = α × E × ΔT.

The ability of a material to resist deformation or deflection under stress is known as —-? RRB JE 31-08-2019

Ductility
Mechanical strength
Stiffness
Toughness

Explanation:

Stiffness is the ability of a material to resist deformation or deflection under stress. Stiffness (S) = Load (W)/Deformation (δ).

Young’s modulus of elasticity for a perfectly rigid body is —-? (RRB Bhubneshwar JE II 29.11.2008) (DMRC JE 22.09.2017)

zero
unity
infinity
cannot be known

Explanation:

For a perfectly rigid body, deformation = 0, strain = 0. E = σ/ε = σ/0 = ∞ (infinity).

The deformation per unit length is called —-? (RRB Bhopal TM SSE 25.10.2009)

Tensile stress
Compressive stress
Shear stress
Strain

Explanation:

Strain is the deformation per unit length. ε = δL/L. It is a dimensionless quantity.

The value of Poisson’s ratio is always less than —-? (RRB Allahabad JE 09.09.2012)

1
0.5
0.4
0.2

Explanation:

The value of Poisson’s ratio is always less than 0.5. For engineering materials: 0 ≤ μ ≤ 0.5. μ = 0 for cork, μ = 0.5 for rubber.

A metallic cube is subjected to equal pressure (P) on its all the six faces. If ε is volumetric strain produced, the ratio P/ε is called —-? (RRB Bangalore SSE 09.09.2012)

Elastic modulus
Shear modulus
Bulk modulus
Strain-Energy per unit volume

Explanation:

Bulk modulus K = σ/ε_v = P/ε_v. Also K = E/[3(1 – 2μ)].

The ratio of linear stress to linear strain is known as —-? (RRB Gorakhpur Design SSE 09.09.2012)

Poisson’s ratio
Bulk modulus
Modulus of rigidity
Modulus of elasticity

Explanation:

Modulus of elasticity (E) = Longitudinal stress/Longitudinal strain. Modulus of rigidity (G) = Shear stress/Shear strain. Bulk modulus (K) = Direct stress/Volumetric strain. Poisson’s ratio (μ) = Lateral strain/Longitudinal strain.

The relation between E (Modulus of elasticity) and K (bulk modulus of elasticity) is —-? (RRB Chandigarh SSE 09.09.2012)

E = K(1 – 2/m)
E = 2K(1 – 2/m)
E = 3K(1 – 2/m)
E = 4K(1 – 2/m)

Explanation:

E = 3K(1 – 2/m) where m = 1/μ, μ = Poisson’s ratio. Also E = 2G(1 + 1/m), E = 9KG/(3K + G).

Factor of safety is the ratio of —-? (BMRCL JE 24 Feb.2019)

breaking stress to working stress
endurance limit to yield stress
elastic limit to ultimate stress
ultimate stress to working stress

Explanation:

Factor of safety = Ultimate stress/Working stress.

If a material has numerically the same value for its modulus of rigidity and bulk modulus, then what is its Poisson’s ratio? (RRB Allahabad JE 19.12.2010)

0.25
0.2
0.15
0.125

Explanation:

E = 2G(1 + μ) and E = 3K(1 – 2μ). Given G = K, 2(1 + μ) = 3(1 – 2μ), 2 + 2μ = 3 – 6μ, 8μ = 1, μ = 1/8 = 0.125.

Which of the following is a dimensionless quantity? (RRB Mumbai JE 05.10.2008)

Shear stress
Poisson’s ratio
Torque
None of these

Explanation:

Poisson’s ratio is the ratio of lateral strain to longitudinal strain, hence it is a dimensionless quantity.

Young’s modulus of elasticity (E) of mild steel is —-? (RRB Mumbai JE 19.12.2010)

310 GPa
210 GPa
405 MPa
None of the above

Explanation:

Young’s modulus of mild steel is 210 GPa. According to Hooke’s Law, within elastic limit, stress is directly proportional to strain: σ = Eε.

The ratio of transverse displacement to the distance from the lower face is called as —-? DMRC JE 17.04.2018, 4:30-6:45pm

shear strain
longitudinal strain
lateral strain
tensile strain

Explanation:

The ratio of transverse displacement to the distance from the lower face is called shear strain.

Endurance limit of steel is associated with ____ number of cycles in fatigue loading. (DMRC JE 2013)

low
infinite
limited
1000

Explanation:

Endurance limit (Se) is the maximum stress that a standard specimen can sustain for infinite number of cycles (10⁶) without failure under completely reversed loading.

Abrupt change of cross section in a member subject to load can result in —-? (Konkan Railway TA 2017)

thermal stresses
creep
stress concentration
fatigue

Explanation:

Stress concentration is the localization of high stresses due to irregularities and abrupt changes in cross-section.

The ratio of Young’s modulus to modulus of rigidity for a material having Poisson’s ratio of 0.25 is —-? (RRB Kolkata Diesel JE 25.10.2009)

1.5
2
2.5
1

Explanation:

E = 2G(1 + μ). E/G = 2(1 + 0.25) = 2.5.

Young’s Modulus, Bulk Modulus (K) and Shear Modulus (G) are related by —-? (RRB Bangalore SSE 09.09.2012)

E = 9KG/(3K + G)
E = 9KG/(3K + G)
E = 9KG/(K + 3G)
E = 3KG/(9K + G)

Explanation:

E = 9KG/(3K + G). This is derived from E = 2G(1 + μ) and E = 3K(1 – 2μ).

A metal flat 40 mm wide, 10 mm thick section and 2 m length, is under axial compressive load of 40 kN. If contraction in length is 1 mm and increase in width is 0.006 mm. What is the value of Poisson’s ratio? (RRB Mumbai C&G JE 25.10.2009)

0.06
0.3
0.1
0.4

Explanation:

μ = Lateral strain/Longitudinal strain = (δb/b)/(δL/L) = (0.006/40)/(1/2000) = 0.00015/0.0005 = 0.3.

Relation between true strain (ε) and engineering strain (e) is —-? (RRB Bhopal SSE 09.09.2012)

ε = ln(e + 1)
ε = ln(e + 1)
ε = 1/ln(e + 1)
ε = 1/ln(e + 1)

Explanation:

True strain ε = ln(L/L₀) = ln(1 + e), where e is engineering strain.

The impact test is done to test ____ of a material. DMRC JE 17.04.2018 12:15-2:30 PM

ductility
toughness
strength
hardness

Explanation:

Impact test is done to test toughness of a material. Toughness is measured by Izod and Charpy impact testing machines.

The value of Poisson’s ratio for steel is between —-? (RRB Mumbai C&G SSE 25.10.2009)

0.01 to 0.1
0.2 to 0.25
0.25 to 0.33
0.4 to 0.7

Explanation:

Poisson’s ratio for steel varies between 0.25 to 0.33. Cast iron: 0.23-0.27, Aluminium: 0.33, Copper: 0.335, Cork: 0, Rubber: 0.5.

Which of the following materials generally exhibits a yield point? (RRB Bhubneshwar JE 11 29.11.2008)

Cast iron
Annealed and hot rolled mild steel
Soft brass
Glass

Explanation:

Annealed and hot rolled mild steel exhibits a yield point (upper and lower yield points). Cast iron, glass, and soft brass do not exhibit a clear yield point.

If the ratio G/E = 0.4, the Poisson’s ratio will be —-? (G – Rigidity modulus, E – Young’s modulus) (BMRCL JE 24 Feb. 2019) (RRB Allahabad JE 19.12.2010)

0.2
0.25
0.3
0.33

Explanation:

E = 2G(1 + μ). E/(2G) = 1 + μ. 1/(2 × 0.4) = 1 + μ, 1/0.8 = 1 + μ, 1.25 = 1 + μ, μ = 0.25.

The unit of modulus of elasticity is same as those of —-? RRB-JE 29.08.2019, Ist Shift (RRB JE [Exam Date: 27-08-2015 (Shift-1)] (RRB Allahabad JE 25.10.2009)

stress, strain and pressure
stress, force and modulus of rigidity
strain, force and pressure
stress, pressure and modulus of rigidity

Explanation:

The unit of modulus of elasticity is the same as stress, pressure, and modulus of rigidity (N/m²).

A uniform straight rod is subjected to axial load. Which of the following statement is correct? (RRB Mumbai JE 05.10.2008)

It induces maximum shear stress on the transverse plane
It induces maximum normal stress on the plane inclined at 45° to axis of rod
It induces maximum shear stress on the plane inclined at 45° to axis of rod
It induces zero shear stress on any plane inclined to axis of rod

Explanation:

A uniform straight rod subjected to axial load induces maximum shear stress on the plane inclined at 45° to the axis of the rod.

When a rod of circular cross section is fixed at one end and subjected to an axial load of 500 N, the deflection under the load is found to be 2.4 mm. What will be the amount of deflection under the same load if the dia of rod is doubled and length is reduced to half of the original length? (Konkan Railway TA 2017)

1.2 mm
0.6 mm
0.3 mm
0.15 mm

Explanation:

δ = PL/(AE). For rod 1: δ₁ = PL/(πd²E/4). For rod 2: δ₂ = P(L/2)/(π(2d)²E/4). δ₂/δ₁ = 1/8. δ₂ = 2.4/8 = 0.3 mm.

Toughness of a material means —-? (RRB Kolkata Diesel JE 25.10.2009)

strength
machine ability
stress reliving
softening

Explanation:

Toughness is the ability of a material to withstand stress (resist fracture due to high impact loads) without fracture. It represents the strength of the material.

The percentage reduction in area in case of cast iron when it is subjected to tensile test is of the order of —-? (RRB Chandigarh SSE 25.10.2009) (RRB Gorakhpur RDSO SSE 25.10.2009) (DMRC JE 20.04.2018)

Explanation:

Cast iron is a brittle material and does not show any yielding before failure. Percentage reduction in area is nearly 0% (less than 2%, generally 0.5%).

The impact strength of a material is an index of its —-? (RRB Allahabad JE 25.10.2009)

toughness
tensile strength
hardness
fatigue strength

Explanation:

The impact strength of a material is an index of its toughness. Toughness is measured by Izod and Charpy impact testing machines.

In the case of fatigue loading of materials, the stress level corresponding to infinite life is known as —-? (RRB Allahabad JE 09.09.2012)

Loading limit
Fatigue limit
Tensile limit
Endurance limit

Explanation:

The fatigue limit or endurance limit is the stress level below which an infinite number of loading cycles can be applied without causing fatigue failure.

The Poisson’s ratio for rubber is —-? (RRB Bhubneshwar JE II 29.11.2008)

0.4
0
0.5
0.2

Explanation:

The Poisson’s ratio for rubber is 0.5.

The capacity of a material to absorb energy when it is elastically deformed is —-? (RRB Jammu SSE 09.09.2012)

Resilience
Ductility
Toughness
Yield stress

Explanation:

Resilience is the capacity of a material to absorb energy when elastically deformed. Toughness is the ability to absorb energy and plastically deform without fracturing.

Discontinuity in geometry causes stress concentration in machine components. Which of the following is NOT a method of alleviating stress concentration at corner of a stepped shaft? (RRB Gorakhpur RDSO SSE 25.10.2009) RRB Chandigarh Section Eng. Mech. 15.03.2009

Providing fillet
Providing groove
Providing washer
Providing fastener

Explanation:

Providing fastener is not a method of alleviating stress concentration. Methods include providing fillet, groove, washer, and undercutting.

To which of the following is the proof stress related? (RRB Bhubneshwar JE-II 19.12.2010) RRB Bangalore Material Engg. 21.11.2004

Elongation
Necking
Yielding
Fracture

Explanation:

Proof stress is related to yielding. In ductile materials where yield point is not clearly defined, proof stress is used to determine the yield point.

A bar of 12 mm diameter is tested on a universal testing machine. Gauge length = 200 mm, Load at proportional limit = 20 kN, Change in length at proportional limit = 0.2 mm, Change in diameter at proportional limit = 0.0025 mm. Find the value of Poisson’s ratio. (Konkan Railway SSE 2015)

ν = 0.557
ν = 0.208
ν = 0.355
ν = 0.456

Explanation:

μ = Lateral strain/Longitudinal strain = (δd/d)/(δL/L) = (0.0025/12)/(0.2/200) = 0.0002083/0.001 = 0.208.

Which of the following has the highest lateral strain when there is a longitudinal strain because of longitudinal force? (RRB Jammu JE 25.10.2009) (RRB Bangalore SSE 09.09.2012)

Iron
Rubber
Copper
Bronze

Explanation:

Lateral strain = μ × Longitudinal strain. Since rubber has the highest Poisson’s ratio (0.5), it has the highest lateral strain.

The Young’s modulus and thermal stress developed in a steel rod of diameter 2 cm and length 2 m is 200 GPa and 288 MPa respectively. This experiences heating from temperature 30°C to 150°C and the rod has been restricted in its original position. Calculate the value of coefficient of thermal expansion. (RRB Bangalore SSE 09.09.2012)

1.2 × 10⁻⁵ /°C
12 × 10⁻⁴ /°C
12 × 10⁻⁵ /°C
None of these

Explanation:

σ_th = E × α × ΔT. 288 × 10⁶ = 200 × 10⁹ × α × 120. α = 288 × 10⁶/(200 × 10⁹ × 120) = 1.2 × 10⁻⁵ /°C.

____ will exhibit viscoelastic behavior. (RRB Gorakhpur Design SSE 09.09.2012)

Steel
Diamond
Organic polymers
Neoprene

Explanation:

Neoprene will exhibit viscoelastic behavior. Materials with viscoelastic effects include synthetic polymers (PVC), wood, and human tissues, as well as metals at high temperature.

Which of the following material is most elastic? (RRB Jammu SSE 09.09.2012) (RRB Bhopal TM SSE 25.10.2009) (RRB Bangalore SSE 09.09.2012)

Rubber
Brass
Steel
Glass

Explanation:

Steel is more elastic than rubber because steel returns to its original shape faster than rubber when deforming forces are removed.

What is the effect on the Young’s modulus of a wire, if the radius of a wire subjected to a load P is doubled? (RRB Bhubneshwar JE II 29.11.2008)

Doubled
Halved
Become one-fourth
Remains unaffected

Explanation:

Young’s modulus (E) is a property of the material and does not depend on the dimensions of the material. It remains constant.

Which of the following shows the CORRECT graph for the stress-strain curve for an ideal elastic strain hardening material? (JMRC JE 10.06.2017)

Linear Elastic
Rigid Material
Ideal elastic strain hardening material
Elasto plastic or visco plastic

Explanation:

Ideal elastic strain hardening material shows a stress-strain curve with elastic region followed by strain hardening (work hardening) region.

The elongation produced in a bar due to its self-weight is given by —-? (RRB Mumbai SSE 05.10.2008)

9.81ρL²/E
9.81ρL²/(2E)
9.81ρL/E
9.81ρL²/(2E)

Explanation:

Elongation due to self-weight = γL²/(2E) = ρgL²/(2E) = 9.81ρL²/(2E).

A steel rod of original length 200 mm and final length of 200.2 mm after application of an axial tensile load of 20 kN. What will be the strain developed in the rod? (BMRCL JE 24 Feb.2019)

0.01
0.1
0.001
0.0001

Explanation:

Strain = (L_f – L_i)/L_i = (200.2 – 200)/200 = 0.2/200 = 0.001.

A bar having a cross-sectional area of 700 mm² is subjected to axial loads at the positions indicated. The value of stress in the segment QR is —-? (RRB Allahabad JE 19.12.2010)

40 MPa
50 MPa
70 MPa
120 MPa

Explanation:

σ_QR = F_QR/A = 28 × 10³/700 = 40 MPa.

What will be the change in length (mm) of a steel bar having a square cross section of dimension 40 mm × 40 mm, which is subjected to an axial compressive load of 250 kN. If the length of the bar is 4 m and modulus of elasticity is E = 250 GPa? (RRB Allahabad SSE 19.12.2010)

2.5
1.25
2
1.5

Explanation:

δL = PL/(AE) = (250 × 10³ × 4)/(0.04 × 0.04 × 250 × 10⁹) = 1000 × 10³/(0.0016 × 250 × 10⁹) = 2.5 mm.

A steel rod whose diameter is 2 cm and is 2 m long, experiences heating of temperature 30°C to 150°C. The coefficient of thermal expansion is α = 12 × 10⁻⁶/°C and Young’s modulus is 200 GPa. If the rod has been restricted to its original position, then the thermal stress (MPa) developed will be —-? (Konkan Railway TA 2017)

234
256
288
300

Explanation:

σ_th = E × α × ΔT = 200 × 10³ × 12 × 10⁻⁶ × 120 = 288 MPa.

Which formula correctly depicts the elongation in a tapered rod? (Konkan Railway SSE 2015)

δ = PL/(AE)
δ = WL²/(2AE)
δ = 4PL/(πEd₁d₂)
δ = 4PL/(πEd₁d₂)

Explanation:

Elongation in a tapered rod δ = 4PL/(πEd₁d₂), where d₁ and d₂ are diameters at the two ends.

A steel rod whose diameter is 6 cm and is 1 m long, experiences heating from temperature 40°C to 200°C. The coefficient of thermal expansion is α = 12 × 10⁻⁶/°C and Young’s modulus is 300 GPa. If the rod has not been restricted in its original position, the thermal stress (MPa) developed is —-? (RRB Kolkata Diesel JE 25.10.2009)

Explanation:

If the rod is not restricted (free to expand), no thermal stress is developed. Thermal stress = 0.