According to Hooke’s law, $F=kx$ → deflection (x) ∝ load (F) for elastic springs.

Toughness = area under stress-strain curve → represents total energy absorbed before failure.

Resilience is the ability to absorb energy within elastic limit — important for materials under impact loads like springs.

The area under the load–deflection graph gives total work done by the load, part of which is stored as strain energy.

For both ends fixed, effective length Le=L/2. The fixing at both ends increases stability.

Since strain energy is a form of energy, its SI unit is Joule (J).

Fixed beams resist rotation and deflection, storing more strain energy under the same load compared to other supports.

Buckling happens when compressive axial load causes lateral instability, even if stress is below the material’s yield point.

Impact loading produces higher stresses because the load acts quickly and generates additional kinetic energy

Sudden loads cause dynamic stress twice that of gradually applied static loads, as energy is absorbed instantaneously.

Effective length Le​=0.7L for a column with one end fixed and the other hinged, as per Euler’s conditions.

Euler’s theory assumes elastic buckling and is valid only for long, slender columns where failure occurs by buckling, not by crushing.

Critical (Euler’s) load is the load that causes sudden lateral deflection (buckling) in a long column before material failure.

Cross-section remains circular since torsional deformation preserves the shape but twists the cross-section.

Higher modulus (E) means stiffer material, resulting in less deflection.

At an inflection point, bending moment changes sign and slope becomes zero — beam shape changes curvature direction.

Strain energy U=1/2(TθU), representing elastic energy stored due to twisting.

Spring constant (k) measures stiffness — the force required per unit extension or compression.

For simply supported beams, deflection is zero at supports and maximum at mid-span.

Failure occurs when shear stress exceeds the shear yield limit of the material.

It represents the energy per unit volume stored in a material up to elastic limit — important for springs and shock absorbers.

Outer fibers experience maximum tensile and compressive stresses during bending, while stress is zero at the neutral axis

This relation connects bending moment (M), moment of inertia (I), stress (σ), radius of curvature (R), and distance from neutral axis (y).

The neutral axis passes through the centroid of the beam’s cross-section, where bending stress is zero.

Hollow shafts are lighter and resist torque efficiently since maximum shear stress acts at the outer surface.

J=πd^4/32​​. It defines the shaft’s capacity to resist torsional shear.

Torsional rigidity is the product of modulus of rigidity (G) and polar moment of inertia (J). It represents the shaft’s resistance to twist.

In solid circular shafts, shear stress starts at zero at the center and increases linearly toward the outer surface.

Angle of twist θ=TL/JG . It increases with torque and length, while higher rigidity or polar moment reduces twist.

In a simply supported beam loaded downward, top fibers shorten (compression) while bottom fibers elongate (tension).
This is fundamental for stress distribution in bending.

From flexure formula: σ=My​/I.
So, bending stress increases directly with bending moment and distance from neutral axis.

Under pure bending (no shear), neutral axis passes through the centroid of the cross-section, where bending stress is zero.

From differential relation, dM/dx=V
Thus, the slope of bending moment curve = shear force at that section.

In a UDL case, shear decreases linearly from maximum positive at one support to maximum negative at the other, forming a triangle shape.

For a UDL, shear varies linearly (V = –w × x).
Hence, the shear force diagram forms a triangle, starting maximum at fixed end and zero at free end.

Where shear force is zero, bending moment reaches a maximum or minimum value.
This point is critical in beam design since bending stress peaks there.

The load acts exactly at mid-span, causing shear force to shift from +ve to –ve side with equal magnitude.
This indicates zero shear at that point.

For simply supported beam,
Maximum moment at midspan = WL/4.
This formula is used frequently in basic bending stress calculations.

A beam with two supports and load at the center is simply supported.
Maximum bending moment occurs at mid-span (M = WL/4).

For a cantilever beam with uniformly distributed load (w):
Bending moment at distance x, M = –(w x²)/2 → a parabolic curve.
Maximum moment at the fixed end = –(wL²)/2.

A point load creates a sudden change (jump) in shear force value at that section, as load acts at a single point.

Fundamental beam relations:

• dMdx=V\frac{dM}{dx} = VdxdM​=V

• dVdx=−w\frac{dV}{dx} = -wdxdV​=−w
  These differential equations form the base for shear and bending diagrams.

For a cantilever of length L with load W:
Maximum bending moment = WL at the fixed end, since it resists both load and moment.

At supports of a simply supported beam, no moment can be developed since the beam can freely rotate — hence M = 0.

For a symmetrically loaded beam, shear force changes sign at the midpoint — meaning zero shear force and maximum bending moment.

For a UDL over the entire span (L):
Mmax = wL²/8 at the mid-span.
This point experiences maximum bending stress — crucial for beam section design.

At the point of contraflexure (inflection), the bending moment changes sign — from positive (sagging) to negative (hogging).
This helps in designing continuous beams.

At the free end, no load can resist bending, hence the bending moment is zero.
Maximum moment occurs at the fixed end due to the reaction moment.

Shear force (V) = dM/dx.
It represents the internal transverse force resisting shear. A sudden change in load causes a sudden change in shear force but not in bending moment.

For structural steel, Poisson’s ratio typically ranges from 0.25–0.3.
It ensures predictable deformation under loads and is used in stress–strain and deflection calculations.

A rigid body doesn’t deform under any amount of applied force, meaning strain = 0.
This is an idealization used in theoretical mechanics but doesn’t exist in real materials

Bulk modulus (K) = Volumetric stress / Volumetric strain.
It determines how compressible a material is. Higher K = lower compressibility.
Important for materials used under hydrostatic or internal pressure, like cylinders and vessels.

An elastic body returns to its original shape after load removal and follows Hooke’s law (stress ∝ strain).
Such materials are predictable and widely used in engineering analysis.

The correct relation is E = 2G(1 + ν).
It shows how shear and tensile properties are connected for isotropic materials — essential for analyzing both bending and torsional stresses.

Due to Poisson’s effect, when a material is stretched longitudinally, it contracts laterally.
The extent of this contraction depends on Poisson’s ratio (ν).

Typical Young’s Modulus values:

• Rubber ≈ 0.01 GPa

• Aluminum ≈ 70 GPa

• Steel ≈ 200 GPa ✅
  Hence, steel is the stiffest material among the listed — ideal for load-bearing components.

In the linear (elastic) region, the slope of the stress-strain curve represents the Elastic Modulus (E).
It defines the stiffness of materials — steeper slope = stiffer material.

Using relation E = 2G(1 + ν):
200 = 2G(1 + 0.3) → G = 76.9 GPa ≈ 80 GPa.
This shows how elastic constants are interdependent and crucial for determining shaft rigidity and deformation.

According to Hooke’s Law, strain ∝ stress within the elastic limit.
Beyond this limit, the material undergoes permanent deformation. This concept is fundamental in designing springs and load-bearing structures to stay within safe stress limits.

Elasticity means the ability to return to original shape after deformation.
Materials with high Young’s modulus (like steel) are more elastic because they deform less under the same load compared to soft materials like rubber.

A Poisson’s ratio > 0.5 violates the law of conservation of volume — physically impossible.
Most metals have values between 0.25 and 0.35, while rubber approaches 0.5.

For isotropic materials, E = 9KG / (3K + G).
This formula links the three primary elastic constants, helping engineers calculate missing properties for structural analysis.

Modulus of rigidity (G) represents a material’s resistance to shear deformation.
For steel, G ≈ 80 GPa.
It’s crucial for torsional design in shafts and couplings.

Since Young’s modulus is stress divided by strain (dimensionless), its unit is the same as stress = N/m² or Pascal (Pa).
It indicates the magnitude of stress needed for a given strain.

Bulk modulus (K) = Hydrostatic stress / Volumetric strain.
It measures the resistance to uniform compression — important in pressure vessels, hydraulic components, and submarines.

Young’s modulus (E) = Stress / Strain in the same direction.
It measures stiffness or the ability to resist deformation.
Typical values:

• Steel ≈ 200 GPa

• Aluminum ≈ 70 GPa
  Used in tensile and bending calculations.

For incompressible materials like rubber, the volume remains constant even under load, giving ν = 0.5.
It means any elongation in one direction causes equal contraction in the other two — used for rubber seals and elastomeric materials.

Poisson’s ratio (ν) = (Lateral strain) / (Longitudinal strain).
When a material is stretched in one direction, it contracts in the perpendicular direction. Most engineering materials (like steel) have ν ≈ 0.3. It helps in calculating deformation in multi-axial loading systems.

Hooke’s Law states that stress is directly proportional to strain within the elastic limit — i.e., stress ∝ strain or σ = E × ε. Beyond this point, permanent deformation begins. It’s the base for linear elasticity used in structures, beams, and mechanical design.

Elastic materials return to their original shape once the load is removed.

Perfect elasticity means no permanent deformation.

Hooke’s Law: Stress ∝ Strain within elastic limit.

Tangential forces produce shear stress.

E = Stress / Strain in the linear (elastic) region.

Tensile stress tries to elongate the material.

Elastic materials regain original shape after deformation.

Working stress ensures a safety margin below ultimate stress

Axial load acts perpendicular to cross-section, causing normal stress.

Resilience = energy stored per unit volume within elastic range.

Strain energy = ½ × Stress × Strain.

μ = Lateral Strain / Longitudinal Strain.

Yield stress indicates the onset of plastic behavior.

E = Stress / Strain; it measures stiffness of a material.

Ultimate stress represents maximum strength before fracture.

Beyond the yield point, material experiences plastic deformation

Hooke’s law applies while material obeys linear stress–strain relation.

1 Pascal = 1 N/m².

Strain = ΔL / L; it is dimensionless.

Stress = Force / Area; it defines internal resistance to deformation