Mechanical
What is the condition for translational equilibrium of a body under coplanar forces?
- ΣM = 0 only
- ΣFx = 0 and ΣFy = 0
- ΣFx = 1 and ΣFy = 1
- ΣF = maximum
Explanation:
- For a body to be in translational equilibrium, the resultant force must be zero.
- For coplanar forces, this requires ΣFx = 0 and ΣFy = 0.
For complete equilibrium of a rigid body in a coplanar force system, which conditions must be satisfied?
- ΣFx = 0 only
- ΣFy = 0 only
- ΣM = 0 only
- ΣFx = 0, ΣFy = 0 and ΣM = 0
Explanation:
- A rigid body is in complete equilibrium when there is no resultant translation or rotation.
- Therefore, both force equilibrium and moment equilibrium conditions must be satisfied.
Which principle states that the algebraic sum of moments of all forces about a point is zero for a body in rotational equilibrium?
- Newton’s first law
- Principle of conservation of energy
- Moment equilibrium condition
- Hooke’s law
Explanation:
- For rotational equilibrium, the resultant moment about any reference point must be zero.
- Thus, ΣM = 0.
The moment of a force about a point is equal to:
- Force × area
- Force × perpendicular distance from the point to the line of action
- Force ÷ distance
- Mass × velocity
Explanation:
- The moment of force is M = F × d, where d is the perpendicular distance between the point and the line of action of the force.
- The SI unit is N·m.
According to Newton’s first law of motion, a body remains at rest or in uniform straight-line motion when:
- A force acts continuously on it
- The resultant external force is zero
- Its velocity is maximum
- Its acceleration is maximum
Explanation:
- Newton’s first law is also called the law of inertia.
- A body maintains its state of rest or uniform motion unless acted upon by a non-zero resultant external force.
Newton’s second law of motion gives the relationship between force and:
- Displacement only
- Rate of change of momentum
- Potential energy only
- Density
Explanation:
- Newton’s second law states that force is equal to the rate of change of momentum.
- For constant mass, F = ma.
Newton’s third law of motion states that:
- Every body remains at rest
- Force is always equal to mass × velocity
- For every action, there is an equal and opposite reaction
- Acceleration is always constant
Explanation:
- Newton’s third law states that forces between interacting bodies occur in equal-magnitude and opposite-direction pairs.
- The action and reaction forces act on different bodies.
What is the coefficient of friction?
- Normal reaction divided by friction
- Limiting friction divided by normal reaction
- Friction multiplied by normal reaction
- Weight divided by friction
Explanation:
- The coefficient of static friction is μ = Flim/N.
- At limiting equilibrium, friction reaches its maximum static value.
According to the laws of dry friction, limiting friction is proportional to:
- Area of contact only
- Velocity
- Normal reaction
- Temperature only
Explanation:
- Coulomb’s law of dry friction gives Flim = μN.
- Thus, limiting friction is proportional to normal reaction for a given pair of surfaces.
When a body is just about to move over a rough horizontal surface, friction is called:
- Rolling friction
- Kinetic friction
- Limiting friction
- Fluid friction
Explanation:
- When impending motion occurs, static friction reaches its maximum value.
- This maximum static friction is called limiting friction.
The angle of friction is the angle between:
- Weight and normal reaction
- Resultant reaction and normal reaction
- Friction and weight
- Applied force and displacement
Explanation:
- The resultant reaction is the vector sum of normal reaction N and limiting friction F.
- If φ is the angle of friction, tan φ = F/N = μ.
What is stress?
- Deformation per unit length
- Internal resisting force per unit area
- Force multiplied by area
- Change in volume per unit mass
Explanation:
- Stress is the internal resisting force developed per unit cross-sectional area due to an external load.
- Normal stress is σ = P/A.
- Its SI unit is Pa or N/m².
What is strain?
- Force per unit area
- Deformation per unit original dimension
- Force per unit volume
- Mass per unit volume
Explanation:
- Strain is the ratio of change in dimension to original dimension.
- Longitudinal strain = ΔL/L.
- Strain is dimensionless.
Which type of stress is produced when a body is subjected to an axial tensile load?
- Shear stress
- Bending stress only
- Tensile normal stress
- Torsional stress only
Explanation:
- An axial tensile load tends to elongate the member.
- The corresponding normal stress is tensile stress, σ = P/A.
Which type of stress is developed when two parallel forces act tangentially in opposite directions on a body?
- Compressive stress
- Tensile stress
- Shear stress
- Bearing stress only
Explanation:
- Shear stress acts tangentially to the area and tends to cause one layer of material to slide relative to another.
- It is commonly represented by τ.
Hooke’s law states that, within the elastic limit:
- Stress is independent of strain
- Stress is directly proportional to strain
- Stress is inversely proportional to strain
- Strain is always zero
Explanation:
- Within the elastic limit, stress is proportional to strain.
- For direct loading, σ = Eε, where E is Young’s modulus.
What is the elastic limit of a material?
- The point at which the material melts
- The maximum stress up to which the material returns essentially to its original shape after unloading
- The point of fracture only
- The point of zero strain
Explanation:
- Within the elastic limit, removal of the load allows the material to recover its original dimensions, neglecting very small residual effects.
- Beyond the elastic range, permanent deformation may remain.
Young’s modulus is defined as the ratio of:
- Shear stress to shear strain
- Longitudinal stress to longitudinal strain
- Volumetric stress to longitudinal strain
- Load to volume
Explanation:
- Young’s modulus is E = σ/ε.
- It measures the stiffness of a material in axial loading.
Bulk modulus is defined as the ratio of:
- Shear stress to shear strain
- Volumetric stress to volumetric strain
- Longitudinal stress to longitudinal strain
- Torque to angular deformation
Explanation:
- Bulk modulus K measures resistance to uniform volumetric deformation.
- K = −Δp/(ΔV/V), with the negative sign accounting for compression.
Modulus of rigidity is the ratio of:
- Normal stress to longitudinal strain
- Volumetric stress to volumetric strain
- Shear stress to shear strain
- Load to displacement
Explanation:
- Modulus of rigidity or shear modulus is G = τ/γ.
- It measures resistance to shear deformation.
Which elastic constant represents resistance to change in volume?
- Young’s modulus
- Modulus of rigidity
- Bulk modulus
- Poisson’s ratio
Explanation:
- Bulk modulus measures the resistance of a material to uniform compression and change in volume.
Poisson’s ratio is the ratio of:
- Longitudinal strain to lateral strain
- Lateral strain to longitudinal strain in magnitude
- Stress to strain
- Shear stress to shear strain
Explanation:
- Poisson’s ratio is ν = − lateral strain/longitudinal strain.
- The negative sign reflects the opposite nature of lateral and longitudinal deformation under axial loading.
For a homogeneous beam subjected to transverse loading, the bending moment at a section is associated with:
- Axial deformation only
- Internal resistance to bending
- Fluid pressure
- Thermal expansion only
Explanation:
- Bending moment at a section represents the internal moment developed to resist the external loading.
- It produces bending stresses across the beam section.
What does a shear force diagram represent?
- Variation of bending stress only
- Variation of internal shear force along the length of a beam
- Variation of Young’s modulus
- Variation of density
Explanation:
- A shear force diagram, or SFD, shows how internal shear force varies from section to section along a beam.
What does a bending moment diagram represent?
- Variation of axial strain
- Variation of bending moment along the length of a beam
- Variation of fluid pressure
- Variation of density
Explanation:
- A bending moment diagram, or BMD, graphically represents the bending moment at different sections of a beam.
At a point load acting on a beam, the shear force diagram generally shows:
- A smooth curve without change
- A sudden jump equal to the magnitude of the point load
- Always zero
- An infinite slope only
Explanation:
- A concentrated point load causes a sudden change in shear force.
- The magnitude of the jump equals the applied point load, with sign determined by the adopted convention.
At a concentrated point load, the bending moment diagram generally:
- Has a sudden jump equal to the load
- Remains continuous, although its slope changes
- Becomes zero automatically
- Becomes infinite
Explanation:
- A point load causes a sudden change in shear force, which changes the slope of the BMD.
- The bending moment itself remains continuous at an ordinary point load.
At a point where the shear force changes sign from positive to negative, the bending moment is generally:
- Minimum only
- Maximum
- Always zero
- Infinite
Explanation:
- Since dM/dx = V, a change in shear force from positive to negative indicates that the bending moment changes from increasing to decreasing.
- Therefore, the bending moment is maximum at that section.
What is a composite bar?
- A bar made only from one material
- A bar consisting of two or more different materials acting together
- A bar with zero stress
- A bar without any deformation
Explanation:
- A composite bar contains members of different materials connected so that they act together under load.
- Examples include steel and concrete combinations.
For two materials rigidly connected in a composite bar and subjected to the same axial deformation over the same length, the compatibility condition is:
- Equal stress in both materials
- Equal strain in both materials
- Equal Young’s modulus
- Equal cross-sectional area
Explanation:
- When the members are rigidly connected and have the same original length, their elongations are compatible.
- Thus, ε1 = ε2 when both experience the same elongation over the same length.
In a composite bar, the total load is equal to:
- Stress of one material only
- Sum of the loads carried by the individual materials
- Young’s modulus of all materials
- Total strain multiplied by length
Explanation:
- The applied load is shared by the component materials.
- Therefore, P = P1 + P2 + …
In a composite bar under axial load, the material with higher Young’s modulus generally carries:
- No load
- Always the same load as the softer material
- A greater share of load when the component strains are equal
- Only shear load
Explanation:
- For equal strain, σ = Eε.
- Therefore, a material with higher Young’s modulus develops greater stress and generally carries a greater load for the same area.
Torsion of a circular shaft primarily produces:
- Normal tensile stress only
- Normal compressive stress only
- Shear stress
- Hydrostatic stress
Explanation:
- A torque applied to a circular shaft produces twisting deformation.
- The principal stress associated directly with torsion is shear stress.
For a circular shaft subjected to pure torsion, the torsion equation is:
- P = ρAV
- T/J = τ/R = Gθ/L
- M/I = σ/y
- σ = Eε
Explanation:
- The torsion equation for a circular shaft is:
- T/J = τ/R = Gθ/L
- Here, T is torque, J is polar moment of inertia, τ is shear stress, R is outer radius, G is modulus of rigidity, θ is angle of twist and L is shaft length.
In a circular shaft under pure torsion, shear stress is maximum at the:
- Centre
- Neutral axis only
- Outer surface
- Centroid only
Explanation:
- From τ = Tr/J, shear stress varies linearly with radial distance r.
- Therefore, stress is zero at the centre and maximum at the outer surface.
For a solid circular shaft under pure torsion, shear stress at the centre is:
- Maximum
- Zero
- Equal to the surface stress
- Infinite
Explanation:
- From τ = Tr/J, when r = 0 at the centre, τ = 0.
- Shear stress increases linearly from zero at the centre to maximum at the outer surface.
What is the polar moment of inertia of a solid circular shaft of diameter d?
- πd4/64
- πd4/32
- πd3/32
- πd2/4
Explanation:
- For a solid circular section:
- J = πd4/32
- The second moment of area about a centroidal axis is I = πd4/64.
What is the torsional rigidity of a circular shaft?
- E/I
- GJ
- EI
- μ/A
Explanation:
- Torsional rigidity is the resistance offered by a shaft to angular twist.
- It is represented by GJ, where G is modulus of rigidity and J is polar moment of inertia.
What is buckling of a column?
- Uniform shortening under axial load
- Lateral instability of a slender column under compressive load
- Twisting due to torque
- Failure due only to shear
Explanation:
- A slender column subjected to sufficient axial compression may become unstable and bend laterally.
- This instability is called buckling.
Euler’s column theory is primarily applicable to:
- Short columns only
- Long slender columns
- Thick pressure vessels
- Short beams under shear
Explanation:
- Euler’s theory assumes an ideal long, slender column and elastic behavior.
- It is most suitable for long columns where buckling governs failure.
Euler’s critical buckling load for a column is proportional to:
- L²
- L
- 1/L²
- 1/L
Explanation:
- Euler’s critical load is:
- Pcr = π²EI/(Le)²
- Therefore, critical load decreases with the square of effective length.
For a column with both ends hinged, the effective length is:
- L/2
- L
- 2L
- √2L
Explanation:
- For a column hinged at both ends, the effective length factor K = 1.
- Therefore, Le = L.
For a column fixed at both ends, the effective length is:
- L
- L/√2
- L/2
- 2L
Explanation:
- For ideal fixed-fixed end conditions, the effective length factor is K = 0.5.
- Therefore, Le = L/2.
For a column fixed at one end and free at the other, the effective length is:
- L/2
- L
- 2L
- L/√2
Explanation:
- For an ideal fixed-free column, the effective length factor K = 2.
- Thus, Le = 2L.
According to Euler’s theory, the critical load depends on:
- Only the material density
- Young’s modulus, moment of inertia and effective length
- Only the cross-sectional area
- Only the column weight
Explanation:
- Euler’s critical load is Pcr = π²EI/Le2.
- Thus, it depends on Young’s modulus E, second moment of area I and effective length Le.
Rankine’s formula is mainly used for:
- Only very short columns
- Columns of intermediate length by considering both crushing and buckling effects
- Only thin pressure vessels
- Only torsion members
Explanation:
- Rankine’s theory provides an empirical approach that accounts for both crushing and buckling tendencies.
- It is particularly useful for columns where Euler’s assumptions alone are not adequate.
Which column theory is more suitable for intermediate-length columns?
- Euler’s theory alone
- Rankine’s theory
- Bernoulli’s fluid theory
- Hooke’s law alone
Explanation:
- Rankine’s formula bridges the behavior between short-column crushing and long-column buckling.
- It is therefore commonly used for intermediate columns.
According to Rankine’s formula, the crippling load is generally expressed as:
- P = σA only
- P = σcA/[1 + α(Le/k)2]
- P = EI/L
- P = πEI/L
Explanation:
- A common Rankine form is:
- P = σcA/[1 + α(Le/k)2]
- Here, σc is crushing stress, A is area, Le is effective length, k is radius of gyration and α is the Rankine constant.
What is a thin-walled pressure vessel?
- A vessel with wall thickness greater than its diameter
- A pressure vessel having relatively small wall thickness compared with its diameter
- A vessel having zero internal pressure
- A solid cylinder without a cavity
Explanation:
- A thin-walled pressure vessel has a small wall thickness compared with its diameter.
- For such vessels, membrane-stress theory is commonly used.
For a thin cylindrical pressure vessel, the hoop stress is:
- pd/4t
- pd/2t
- pd/t
- pt/2d
Explanation:
- For a thin cylindrical vessel:
- Hoop stress, σh = pd/(2t)
- where p is internal pressure, d is internal diameter and t is wall thickness.
For a thin cylindrical pressure vessel, the longitudinal stress is:
- pd/2t
- pd/4t
- pd/t
- 2pd/t
Explanation:
- For a thin cylindrical vessel:
- Longitudinal stress, σl = pd/(4t)
- Thus, hoop stress is twice the longitudinal stress.
In a thin cylindrical pressure vessel, the hoop stress is how many times the longitudinal stress?
- Half
- Equal
- Twice
- Four times
Explanation:
- Hoop stress = pd/(2t).
- Longitudinal stress = pd/(4t).
- Therefore, σh/σl = 2.
Hoop stress in a thin cylindrical pressure vessel acts:
- Along the longitudinal axis of the cylinder
- Circumferentially around the cylinder
- Only through the thickness
- Radially inward only
Explanation:
- Hoop stress is also called circumferential stress.
- It acts around the circumference and tends to split the cylinder along its longitudinal direction.
Longitudinal stress in a thin cylindrical pressure vessel acts:
- Circumferentially
- Along the longitudinal axis of the cylinder
- Only radially
- Only through the wall thickness
Explanation:
- Longitudinal stress acts along the axis of the cylinder.
- It tends to separate the end caps from the cylindrical wall.
For a thin spherical pressure vessel, the membrane stress is:
- pd/t
- pd/2t
- pd/4t
- pd/8t
Explanation:
- For a thin spherical vessel:
- σ = pd/(4t)
- The stress is the same in all tangential directions at a point because of spherical symmetry.
Compared with a thin cylindrical vessel of the same diameter, pressure and thickness, the membrane stress in a thin spherical vessel is:
- Twice as high
- Equal to the hoop stress
- Half the hoop stress of the cylinder
- Four times as high
Explanation:
- Cylinder hoop stress = pd/(2t).
- Sphere stress = pd/(4t).
- Therefore, the spherical-vessel membrane stress is half the cylindrical hoop stress for the same p, d and t.
Which stress is generally critical in the design of a thin cylindrical pressure vessel?
- Longitudinal stress
- Hoop stress
- Radial stress only
- Shear stress due to torque
Explanation:
- For a thin cylinder, hoop stress is twice the longitudinal stress.
- Therefore, hoop stress generally governs the membrane-stress design.
If the internal pressure of a thin cylindrical vessel is doubled while diameter and thickness remain unchanged, the hoop stress will:
- Become half
- Remain unchanged
- Become double
- Become four times
Explanation:
- Hoop stress is σh = pd/(2t).
- Therefore, stress is directly proportional to internal pressure.
- Doubling pressure doubles hoop stress.
If the wall thickness of a thin cylindrical pressure vessel is doubled while pressure and diameter remain unchanged, the hoop stress will:
- Become double
- Become half
- Remain unchanged
- Become four times
Explanation:
- Hoop stress is inversely proportional to wall thickness:
- σh = pd/(2t).
- Therefore, doubling t reduces hoop stress to half.