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Mechanics
Bending Moment in Elastic Beams
The bending moment at any cross-section of a beam is the internal internal couple resulting from shearing stresses that resists external loads. It governs the curvature and structural deformation of beams under transverse loading.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of bending moment, relation with radius of curvature, internal elastic couples, and simple beam bending formulas.
PGT extension
Geometrical moment of inertia for various cross-sections, differential equation of the elastic curve, cantilever deflection, and loaded beams supported at ends.
Prerequisites
Hooke's law in one dimension
Young's modulus of elasticity
Shearing stress and strain
Definition and Nature of Bending Moment
The bending moment at a section of a beam is equal to the algebraic sum of the moments of all forces acting on either side of the section.
It acts as an internal couple balancing the moment of external loads to maintain static and rotational equilibrium.
In a symmetrically bent beam, the upper fibers experience longitudinal tension (or compression) while the lower fibers experience the opposite, with a neutral layer undergoing zero longitudinal strain.
Neutral Axis and Neutral Surface
The neutral surface is the longitudinal layer within the beam that undergoes neither extension nor compression during bending.
The intersection of the neutral surface with any transverse cross-section is called the neutral axis.
For beams of uniform symmetrical cross-section subjected to pure bending, the neutral axis passes through the centroid of the cross-section.
Geometrical Moment of Inertia (Flexural Rigidity)
The flexural rigidity of a beam is given by the product Y I, where Y is Young's modulus and I is the geometrical moment of inertia of the cross-section about the neutral axis.
For a rectangular cross-section of breadth b and depth d, the geometrical moment of inertia is $I = (bd^{3})/12$.
For a circular cross-section of radius r, the geometrical moment of inertia about the diametral axis is $I = (\pir^{4})/4$.
Internal Bending Moment and Radius of Curvature
The internal bending moment M developed in a beam is directly proportional to its geometrical moment of inertia and inversely proportional to the radius of curvature R of the neutral axis.
The relation is expressed as $M = (Y I)/R$.
Greater flexural rigidity (Y I) results in a smaller curvature (larger radius R) for a given applied bending moment.
Derivations
Relation Between Bending Moment and Radius of Curvature
Conditions: Pure bending of a beam of uniform cross-section, elastic limit not exceeded, plane cross-sections remain plane after bending.
Consider a small segment of length dx of a beam bent into an arc of a circle of radius R subtending an angle θ, such that d$x = R \theta$.
Consider a longitudinal filament at a distance y from the neutral axis. Its original length is d$x = R \theta$, and its stretched length is l' = (R + y)θ.
The longitudinal strain in this filament is $\epsilon = (l' - dx)/dx = ((R + y)\theta - R\theta)/(R\theta) = y/R$.
By Hooke's law, the longitudinal stress in the filament is $\sigma = Y \epsilon = (Y y)/R$.
The tensile or compressive force acting on an elemental area dA at distance y is d$F = \sigma dA = (Y / R) y dA$.
The moment of this force about the neutral axis is d$M = dF \cdot y = (Y / R) y^{2} dA$.
Integrating over the entire cross-section, the total internal bending moment is $M = \int dM = (Y / R) \int y^{2} dA$.
Recognizing that $I = \int y^{2} dA$ is the geometrical moment of inertia, we get $M = (Y I) / R$.
Result: $M = Y I / R$
Elevation of a Cantilever Loaded at the Free End
Conditions: Cantilever of length L, uniform rectangular cross-section (breadth b, depth d), loaded with a concentrated weight W at the free end.
Consider a cantilever fixed at x = 0 and loaded with weight W at the free end x = L.
At a distance x from the fixed end, the bending moment due to the load W is $M = W(L - x)$.
Equating this to the elastic bending moment, W(L - x) = (Y I)/R.
Using the small slope approximation where 1/R ≈ d²y/dx², the differential equation becomes Y I (d²y/dx²) = W(L - x).
Integrating once with respect to x using boundary conditions (dy/dx = 0 at x = 0), we find the slope dy/d$x = (W / Y I)(Lx - x^{2}/2)$.
Integrating again with respect to x using the boundary condition (y = 0 at x = 0), we obtain the deflection equation $y = (W / Y I)(Lx^{2}/2 - x^{3}/6)$.
The maximum depression at the free end (x = L) is obtained by substituting x = L.
Result: $\delta = W L^{3} / (3 Y I)$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Bending Moment Equation
$$M = \frac{Y I}{R}$$
M = Bending moment, Y = Young's modulus, I = Geometrical moment of inertia, R = Radius of curvature N·m
Pure bending within elastic limits
Geometrical Moment of Inertia (Rectangular)
$$I = \frac{b d^3}{12}$$
$b = Breadth of cross-section$, $d = Depth of cross-section$ m&sup4;
Axis passing through centroid parallel to breadth
Geometrical Moment of Inertia (Circular)
$$I = \frac{\pi r^4}{4}$$
$r = Radius of circular cross-section$ m&sup4;
Axis along diameter
Flexural Rigidity
$$FR = Y I$$
Y = Young's modulus, I = Geometrical moment of inertia N·m²
Isotropic homogeneous material
Depression of Cantilever (Loaded at Free End)
$$\delta = \frac{W L^3}{3 Y I}$$
W = Load, L = Length of cantilever, Y = Young's modulus, I = Geometrical moment of inertia m
Cantilever of uniform cross-section, small deflections
Depression of Beam Supported at Both Ends (Central Load)
$$\delta = \frac{W L^3}{48 Y I}$$
W = Central load, L = Distance between supports, Y = Young's modulus, I = Geometrical moment of inertia m
Beam supported freely at both ends, load applied at midpoint
Bending Stress
$$\sigma = \frac{M y}{I}$$
\sigma = Bending stress, M = Bending moment, y = Distance from neutral axis, I = Moment of inertia N/m²
Linear elastic deformation
Curvature of Benched Beam
$$\frac{1}{R} = \frac{M}{Y I}$$
R = Radius of curvature, M = Bending moment, Y = Young's modulus, I = Moment of inertia m&sup-1;
Valid for small slopes
Graphs and experiments
Bending of a Cantilever (Depression vs. Load)
Draw: A straight line plot of depression (δ) on the vertical axis versus applied load (W) on the horizontal axis.
Interpret: The linear relationship confirms Hooke's law in bending; the slope of the line determines the experimental value of Young's modulus (Y).
Exam use: Frequently tested in viva and numerical problems to calculate Young's modulus or depression.
Bending Stress Distribution Across Cross-Section
Draw: A triangular stress distribution diagram across the depth of a rectangular beam showing zero stress at the neutral axis and maximum tensile/compressive stress at the outer fibers.
Interpret: Stress varies linearly with distance from the neutral axis, justifying the use of outer material to bear maximum structural loads.
Exam use: Conceptual multiple-choice questions regarding stress magnitude at the neutral axis versus outer edges.
Elastic Curve of a Cantilever
Draw: A curved profile starting horizontally at the fixed end and curving downwards parabolically to the free end.
Interpret: Represents the vertical displacement y as a function of horizontal position x along the cantilever span.
Exam use: Used in PGT derivations for slope and deflection differential equations.
Solved numericals
TGT core
A metal rod of rectangular cross-section with breadth 2 cm and depth 3 cm is bent by a couple. If Young's modulus of the material is 2 x 10^11 N/m^2 and the radius of curvature is 100 m, calculate the internal bending moment.
Identify given parameters: b = 0.02 m, d = 0.03 m, $Y = 2 \times 10^11 N/m^2$, R = 100 m.
A cantilever of length 1.5 m has a rectangular cross-section of breadth 0.04 m and depth 0.02 m. A mass of 2 kg is suspended at its free end. Given $g = 9.8 m/s^2$ and Young's modulus $Y = 1.0 x 10^11 N/m^2$, calculate the depression at the free end.
Identify given parameters: L = 1.5 m, b = 0.04 m, d = 0.02 m, $W = m g = 2 \times 9.8 = 19.6 N$, $Y = 1.0 \times 10^11 N/m^2$.
Two beams of the same material and mass have identical lengths. Beam A has a square cross-section of side a, while Beam B has a rectangular cross-section with depth 2a and breadth a/2. Compare their flexural rigidities.
For Beam A (square cross-section of side a): Moment of inertia $I_A = (a \times a^3) / 12 = a^4 / 12$.
For Beam B (breadth $b = a/2$, depth $d = 2a): Moment of inertia I_B = ((a/2) \times (2a)^3) / 12 = ((a/2) \times 8a^3) / 12 = (4a^4) / 12 = a^4 / 3$.
Flexural rigidity ratio is given by (FR)_A / (FR)_$B = (Y I_A) / (Y I_B) = I_A / I_B$.
Answer: Flexural rigidity of Beam B is 4 times that of Beam A (FR_A : F$R_B = 1 : 4)$
Assumptions and limitations
The material of the beam is homogeneous, isotropic, and obeys Hooke's law strictly.
Plane transverse sections before bending remain plane after bending (Bernoulli-Euler hypothesis).
The radius of curvature of the bent beam is very large compared to the transverse dimensions of the cross-section.
Deformations are small, permitting the approximation 1/$R = d^{2}y/dx^{2}$.
Common misconceptions
Common error: The neutral axis experiences maximum tensile stress during bending.
Correct: The neutral axis experiences zero stress and zero strain during bending.
The neutral layer lies at the center where longitudinal stretching and compression transition into one another.
Common error: Bending moment depends only on the applied force, not on the shape of the cross-section.
Correct: Internal bending moment balances the external couple and depends directly on the geometrical moment of inertia of the cross-section.
Cross-sectional geometry (such as I-beams) significantly dictates how a beam resists bending for a given load.
Common error: Depression of a cantilever is directly proportional to the fourth power of its length.
Correct: Depression is directly proportional to the cube of its length (δ ∝ L³).
Dimensional analysis and exact integration of the differential equation yield an L³ dependence.
Quick revision
Bending moment M is related to radius of curvature R by $M = Y I / R$.
Geometrical moment of inertia for a rectangular beam is $I = bd^{3}/12$.
Flexural rigidity is the product of Young's modulus and geometrical moment of inertia (Y I).
Depression of a loaded cantilever varies directly as L³ and inversely as Y I.
Must remember
The bending moment $M$ in a beam is related to the radius of curvature $R$ and flexural rigidity $YI$ by the equation $M = \frac{YI}{R}$.
The neutral surface is the longitudinal layer within the beam that undergoes neither extension nor compression during bending, experiencing zero strain and zero stress.
For a rectangular cross-section of breadth $b$ and depth $d$, the geometrical moment of inertia is $I = \frac{bd^3}{12}$.
For a circular cross-section of radius $r$, the geometrical moment of inertia about the diametral axis is $I = \frac{\pi r^4}{4}$.
The depression of a cantilever loaded at its free end is given by $\delta = \frac{WL^3}{3YI}$.
The depression of a beam supported at both ends with a central load is given by $\delta = \frac{WL^3}{48YI}$.
Glossary
Bending Moment ($M$)The algebraic sum of the moments of all forces acting on either side of a section of a beam, acting as an internal couple balancing external loads.
Neutral AxisThe intersection of the neutral surface with any transverse cross-section of the beam, passing through the centroid for symmetrical sections.
Flexural Rigidity ($YI$)The product of Young's modulus $Y$ and the geometrical moment of inertia $I$, representing the resistance of a beam to bending.
Geometrical Moment of Inertia ($I$)A property of the cross-sectional shape of a beam that determines its capacity to resist bending about an axis.
Radius of Curvature ($R$)The radius of the circular arc into which the neutral axis of the beam is bent under an applied bending moment.
Geometrical Moment of Inertia for Different Shapes
Cross-Section Shape
Formula for Moment of Inertia ($I$)
Key Parameters
Rectangular
$I = \frac{bd^3}{12}$
$b$ = breadth, $d$ = depth
Circular
$I = \frac{\pi r^4}{4}$
$r$ = radius of circular cross-section
Depression in Beams under Different End Conditions
Beam Configuration
Load Position
Depression Formula ($\delta$)
Cantilever
Loaded at free end
$\delta = \frac{WL^3}{3YI}$
Supported at Both Ends
Central load
$\delta = \frac{WL^3}{48YI}$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the state of stress and strain at the neutral axis of a bent elastic beam?
Explanation: The neutral axis is the longitudinal layer that undergoes neither extension nor compression; hence, both its longitudinal stress and strain are zero.
2. What is the formula for the geometrical moment of inertia $I$ of a rectangular cross-section of breadth $b$ and depth $d$ about an axis passing through its centroid parallel to the breadth?
Explanation: For a rectangular cross-section of breadth $b$ and depth $d$, the geometrical moment of inertia about the centroidal axis parallel to the breadth is $I = \frac{bd^3}{12}$.
3. The internal bending moment $M$ developed in a beam is related to the radius of curvature $R$ by which of the following expressions?
Explanation: The bending moment equation states that $M = \frac{YI}{R}$, where $YI$ is the flexural rigidity and $R$ is the radius of curvature.
4. What is the expression for the geometrical moment of inertia of a circular cross-section of radius $r$ about its diametral axis?
Explanation: For a circular cross-section of radius $r$, the geometrical moment of inertia about a diametral axis is given by $I = \frac{\pi r^4}{4}$.
5. What is the flexural rigidity of a beam made of an isotropic homogeneous material?
Explanation: Flexural rigidity is defined as the product of Young's modulus $Y$ and the geometrical moment of inertia $I$, i.e., $FR = YI$.
6. How does the depression $\delta$ of a cantilever loaded at its free end depend on its length $L$?
Explanation: The depression of a cantilever loaded at the free end is $\delta = \frac{WL^3}{3YI}$, showing that depression is directly proportional to the cube of its length ($L^3$).
7. For a beam supported freely at both ends and loaded at the midpoint, what is the formula for central depression $\delta$?
Explanation: The central depression of a beam supported at both ends with a load $W$ at the midpoint is $\delta = \frac{WL^3}{48YI}$.
8. What happens to the upper fibers of a symmetrically bent beam under normal bending conditions?
Explanation: In a symmetrically bent beam, upper fibers experience either longitudinal tension or compression depending on the direction of the bending moment, while lower fibers experience the opposite.
9. What is the SI unit of flexural rigidity ($YI$)?
Explanation: Young's modulus $Y$ has units of $\text{N}/\text{m}^2$ and geometrical moment of inertia $I$ has units of $\text{m}^4$, making the unit of flexural rigidity $YI$ equal to $(\text{N}/\text{m}^2) \cdot \text{m}^4 = \text{N}\cdot\text{m}^2$.
10. What is the geometrical locus called where the neutral surface intersects any transverse cross-section of a beam?
Explanation: The intersection of the neutral surface with any transverse cross-section is defined as the neutral axis.
11. For a beam of uniform symmetrical cross-section subjected to pure bending, where does the neutral axis pass through?
Explanation: For beams of uniform symmetrical cross-section subjected to pure bending, the neutral axis passes directly through the centroid of the cross-section.
12. If the flexural rigidity of a beam is increased while keeping the applied bending moment constant, what happens to the radius of curvature $R$?
Explanation: Since $M = \frac{YI}{R}$, rearranging gives $R = \frac{YI}{M}$. Thus, a greater flexural rigidity ($YI$) results in a larger radius of curvature $R$ (smaller curvature) for a given bending moment.
13. What is the expression for bending stress $\sigma$ at a distance $y$ from the neutral axis?
Explanation: The bending stress formula is $\sigma = \frac{My}{I}$, where $M$ is the bending moment, $y$ is the distance from the neutral axis, and $I$ is the moment of inertia.
14. What physical condition must hold true for the bending moment equation $M = \frac{YI}{R}$ to be strictly valid?
Explanation: The relation $M = \frac{YI}{R}$ is derived under the assumption of pure bending within the elastic limits of the material.
15. What is the dimension of the geometrical moment of inertia $I$?
Explanation: Geometrical moment of inertia involves integrating area elements multiplied by distance squared ($\int y^2 dA$), giving dimensions of length to the fourth power, $[L^4]$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A cantilever of length $L$ and uniform cross-section is loaded at its free end with a load $W$. If Young's modulus is $Y$ and geometrical moment of inertia is $I$, the depression $\delta$ at the free end is given by:
Explanation: The standard derivation for a cantilever loaded at its free end yields a depression of $\delta = \frac{WL^3}{3YI}$.
2. The geometrical moment of inertia of a circular cross-section of radius $r$ about its diametral axis is:
Explanation: The geometrical moment of inertia of a circular cross-section about its diametral axis is $I = \frac{\pi r^4}{4}$ (whereas polar moment of inertia is $\frac{\pi r^4}{2}$).
3. A beam of length $L$ is supported freely at both ends and carries a load $W$ at its midpoint. The depression at the center is:
Explanation: For a beam supported at both ends with a central load $W$, the depression at the midpoint is $\delta = \frac{WL^3}{48YI}$.
4. What is the relationship between the internal bending moment $M$, flexural rigidity $YI$, and the radius of curvature $R$?
Explanation: The bending moment equation states that $M = \frac{YI}{R}$, connecting the internal bending couple to the material/geometrical rigidity and the curvature radius.
5. For a rectangular beam of breadth $b$ and depth $d$, the geometrical moment of inertia about the centroidal axis parallel to breadth is:
Explanation: The standard geometrical moment of inertia for a rectangular cross-section about its neutral (centroidal) axis parallel to breadth $b$ is $I = \frac{bd^3}{12}$.
6. Which layer in a symmetrically bent elastic beam experiences neither extension nor compression?
Explanation: The neutral surface is defined as the longitudinal layer within the beam that undergoes neither extension nor compression during bending.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. If the depth $d$ of a rectangular beam is doubled while keeping its breadth $b$ constant, how does its geometrical moment of inertia $I$ change?
Explanation: Since $I = \frac{bd^3}{12}$, doubling the depth $d$ changes $I$ by a factor of $2^3 = 8$ times.
2. Which of the following represents the correct expression for curvature $\frac{1}{R}$ of a bent beam?
Explanation: From $M = \frac{YI}{R}$, the curvature is given by $\frac{1}{R} = \frac{M}{YI}$.
3. In a beam supported at both ends with a central load, if the length of the beam is doubled while keeping all other parameters constant, the depression at the center becomes:
Explanation: Since $\delta = \frac{WL^3}{48YI}$, depression is proportional to $L^3$. Doubling the length increases the depression by $2^3 = 8$ times.
4. What role does the internal bending moment play in a statically balanced bent beam?
Explanation: The internal bending moment acts as an internal couple balancing the moment of external loads to maintain static and rotational equilibrium.
5. What is the bending stress at the neutral axis of a beam subjected to pure bending?
Explanation: Since the neutral axis experiences zero longitudinal strain, the bending stress ($\sigma = \frac{My}{I}$, where $y = 0$ at the neutral axis) is zero.
6. If two beams of the same material and same length have circular cross-sections of radii $r$ and $2r$ respectively, what is the ratio of their geometrical moments of inertia?
Explanation: Since $I = \frac{\pi r^4}{4}$, $I$ is proportional to $r^4$. For radius $2r$, the moment of inertia increases by $2^4 = 16$ times, giving a ratio of $1 : 16$.
7. Which parameter directly measures a beam's resistance to bending for a given material and cross-section?
Explanation: Flexural rigidity ($YI$) is the direct measure of how stiff a beam is against bending under external moments.
8. Why are girders for construction typically manufactured with an 'I' shaped cross-section?
Explanation: An 'I' beam concentrates material where the distance $y$ from the neutral axis is greatest, maximizing the geometrical moment of inertia $I$ and flexural rigidity for a given weight of material.
9. If a cantilever of length $L$ has its length halved ($L/2$) while keeping the load and cross-section identical, how does its free-end depression change?
Explanation: Depression $\delta \propto L^3$. Halving the length results in $(1/2)^3 = 1/8$ of the original depression.
10. What is the SI unit of bending stress $\sigma$?
Explanation: Stress is force per unit area, so its SI unit is $\text{N}/\text{m}^2$ (or Pascal).