Bending of beams and Structural Application

Elastic Properties of Matter

17. Bending of Beams and Structural Applications

When a beam is supported at its two ends and loaded at the middle, it bends under the action of the applied load. During bending, the longitudinal layers of the beam undergo extension or compression.

For a simply supported beam subjected to a downward central load, the upper layers are compressed, whereas the lower layers are stretched. Between these two regions lies a neutral surface, where the longitudinal strain and bending stress are zero.

According to the theory of simple bending, the magnitude of longitudinal bending stress at a distance \(y\) from the neutral axis is

\[ \boxed{\sigma_b=\frac{M_b y}{I_A}} \]

where

• \(\sigma_b\) = longitudinal bending stress,
• \(M_b\) = bending moment at the section,
• \(y\) = perpendicular distance from the neutral axis,
• \(I_A\) = second moment of area of the cross-section about the neutral axis.

Thus, bending stress increases linearly with the distance from the neutral axis. It is maximum at the outermost upper and lower surfaces and zero at the neutral surface.

Applications of I-shaped and T-shaped Sections

Since the outer layers experience the greatest bending stress, material placed far from the neutral axis contributes more effectively to bending resistance. Consequently, beams can be designed with wider flanges and a relatively thin central web.

This principle is employed in the construction of I-shaped girders, T-shaped beams, and railway rails. Such sections provide a high bending stiffness with comparatively less material.

Similarly, hollow tubes are commonly used in bicycle frames and other lightweight structures because their material is distributed away from the central axis, providing efficient resistance to bending and torsion.

Physical significance: The bending stiffness of a uniform beam is proportional to \(YI_A\), where \(Y\) is Young's modulus and \(I_A\) is the second moment of area.

18. Determination of Young's Modulus by Bending Method

18.1 Principle

When a rectangular beam is supported horizontally on two knife edges and a load is applied at its midpoint, the beam undergoes non-uniform bending. The central depression depends on Young's modulus, the dimensions of the beam, the distance between the supports, and the applied load.

18.2 Experimental Arrangement

A rectangular beam AB, whose Young's modulus is to be determined, is placed symmetrically on two horizontal knife edges P and Q.

A weight hanger is suspended at the midpoint of the beam. Known masses are added to the hanger to produce a downward depression.

A spherometer or suitable vertical displacement measuring screw is mounted on a rigid support to measure the depression at the midpoint.

For electrical contact detection, the measuring screw and the conducting beam are connected in a circuit containing a Leclanché cell, high resistance, plug key, and galvanometer.

When the screw touches the beam, the circuit is completed and the galvanometer indicates contact. The corresponding screw reading is recorded.

The load is increased and decreased in steps, commonly using masses of \(0.5\,\mathrm{kg}\), and readings are taken during both loading and unloading. The mean readings are used to determine the depression.

18.3 Theory and Derivation

Consider a horizontal beam of rectangular cross-section supported at two points separated by a distance \(l\). A central load \(W=Mg\) is applied.

For a simply supported beam, the upward reaction at each support is

\[ R_P=R_Q=\frac{W}{2} \]

Let \(x\) be the distance from the left support. For the left half of the beam, the bending moment is

\[ M_b(x)=\frac{Wx}{2}, \qquad 0\leq x\leq\frac{l}{2}. \]

According to the Euler–Bernoulli theory of bending, for small deflections and constant flexural rigidity,

\[ YI_A\frac{d^2y}{dx^2}=M_b(x) \]

where the sign of the deflection coordinate is chosen consistently with the bending moment.

Therefore,

\[ YI_A\frac{d^2y}{dx^2}=\frac{Wx}{2}. \]

Integrating once,

\[ YI_A\frac{dy}{dx} =\frac{Wx^2}{4}+C_1. \]

At the midpoint, the slope is zero due to symmetry:

\[ \left.\frac{dy}{dx}\right|_{x=l/2}=0. \]

Hence,

\[ C_1=-\frac{Wl^2}{16}. \]

Integrating again,

\[ YI_Ay=\frac{Wx^3}{12} -\frac{Wl^2x}{16}+C_2. \]

At the left support, \(x=0\) and \(y=0\). Therefore, \(C_2=0\).

The magnitude of the maximum depression occurs at the midpoint \(x=l/2\):

\[ \delta=\frac{Wl^3}{48YI_A}. \]

For a rectangular beam of breadth \(b\) and vertical thickness \(d\), the second moment of area is

\[ I_A=\frac{bd^3}{12}. \]

Substituting,

\[ \delta= \frac{Wl^3}{48Y(bd^3/12)} \]

\[ \delta=\frac{Wl^3}{4Ybd^3}. \]

Since \(W=Mg\),

\[ \delta=\frac{Mgl^3}{4Ybd^3}. \]

Therefore, Young's modulus is

\[ \boxed{ Y=\frac{Mgl^3}{4bd^3\delta} } \tag{18.1} \]

18.4 Graphical Determination

A graph is plotted between the applied mass \(M\) along the horizontal axis and the corresponding depression \(\delta\) along the vertical axis.

Within the elastic limit, the graph is approximately a straight line.

The slope of the graph is

\[ S=\frac{\Delta\delta}{\Delta M}. \]

Using the bending equation,

\[ S=\frac{gl^3}{4Ybd^3}. \]

Therefore,

\[ \boxed{ Y=\frac{gl^3}{4bd^3S} } \tag{18.2} \]

Here, \(l\) is measured using a metre scale, while the breadth \(b\) and thickness \(d\) are measured using vernier calipers or a suitable micrometer.

Important: All measurements must be expressed in SI units. If the horizontal axis represents force \(W\) instead of mass \(M\), the corresponding slope is \(\Delta\delta/\Delta W\), and the factor \(g\) is omitted from the numerator.

19. Determination of Young's Modulus, Modulus of Rigidity and Poisson's Ratio by Searle's Method

19.1 Principle

Searle's dynamical method is based on the bending and torsional oscillations of a wire. By measuring the time periods of these oscillations, Young's modulus \(Y\), modulus of rigidity \(\eta\), and Poisson's ratio \(\sigma\) can be determined.

19.2 Experimental Arrangement

Two identical rigid rods AB and CD are connected at their midpoints by a thin experimental wire EF of length \(l\) and radius \(r\).

The rods are suspended horizontally by two fine, nearly torsionless threads from a rigid support. The rods remain parallel in their equilibrium positions.

19.3 Bending Oscillations

The corresponding ends of the two rods are brought slightly closer together and then released. The rods execute small angular oscillations in a horizontal plane due to the bending of the connecting wire.

These oscillations are called bending oscillations.

Let each rod rotate through an angle \(\theta\) in opposite directions. The relative angular deflection between the two ends of the wire is \(2\theta\).

If the wire bends into an approximately circular arc of radius of curvature \(R\), then

\[ l=R(2\theta). \]

Therefore,

\[ \boxed{ R=\frac{l}{2\theta} } \]

The bending moment developed in the wire is

\[ \tau_b=\frac{YI_A}{R}, \]

where \(I_A\) is the second moment of area of the circular cross-section of the wire.

\[ I_A=\frac{\pi r^4}{4}. \]

Hence,

\[ \tau_b= \frac{Y\pi r^4}{4R}. \]

Substituting \(R=l/(2\theta)\),

\[ \tau_b= \frac{Y\pi r^4}{4} \left(\frac{2\theta}{l}\right). \]

\[ \boxed{ \tau_b=\frac{Y\pi r^4}{2l}\theta } \]

This torque acts as a restoring torque on each rod.

Let \(I\) be the mass moment of inertia of each rod about the vertical axis through its centre of gravity.

According to rotational dynamics,

\[ I\frac{d^2\theta}{dt^2} =-\frac{Y\pi r^4}{2l}\theta. \]

Rearranging,

\[ \frac{d^2\theta}{dt^2} +\frac{Y\pi r^4}{2lI}\theta=0. \]

This is the differential equation of simple harmonic motion.

Comparing with

\[ \frac{d^2\theta}{dt^2}+\omega_1^2\theta=0, \]

we obtain

\[ \omega_1^2=\frac{Y\pi r^4}{2lI}. \]

The time period of bending oscillations is

\[ T_1=\frac{2\pi}{\omega_1}. \]

Therefore,

\[ T_1=2\pi \sqrt{\frac{2lI}{Y\pi r^4}}. \]

Squaring and rearranging,

\[ T_1^2=\frac{8\pi lI}{Yr^4}. \]

Thus,

\[ \boxed{ Y=\frac{8\pi lI}{T_1^2r^4} } \tag{19.1} \]

19.4 Moment of Inertia of the Rods

For a thin rectangular rod of mass \(m\), length \(a\), and breadth \(b\), rotating about a perpendicular axis through its centre,

\[ \boxed{ I=\frac{m(a^2+b^2)}{12} } \]

For a uniform circular disc of mass \(m\) and radius \(R\), rotating about its central axis perpendicular to its plane,

\[ \boxed{ I=\frac{mR^2}{2} } \]

For a uniform solid circular rod of mass \(m\), length \(a\), and radius \(R\), rotating about an axis through its centre perpendicular to its length,

\[ \boxed{ I=\frac{m(a^2+3R^2)}{12} } \]

19.5 Torsional Oscillations

To study torsional oscillations, the auxiliary suspension threads are removed. One of the rods is rigidly fixed to a support so that the experimental wire becomes vertical and supports the second rod horizontally.

The suspended rod is rotated through a small angle about the vertical axis of the wire and then released.

The rod executes angular simple harmonic motion due to the restoring torque produced by twisting the wire. These oscillations are called torsional oscillations.

For a circular wire of length \(l\), radius \(r\), and modulus of rigidity \(\eta\), the torsional constant is

\[ C=\frac{\pi\eta r^4}{2l}. \]

The restoring torque for an angular displacement \(\theta\) is

\[ \tau=-C\theta. \]

Using the equation of rotational motion,

\[ I\frac{d^2\theta}{dt^2}=-C\theta. \]

Therefore,

\[ \frac{d^2\theta}{dt^2} +\frac{C}{I}\theta=0. \]

The time period of torsional oscillations is

\[ T_2=2\pi\sqrt{\frac{I}{C}}. \]

Substituting the value of \(C\),

\[ T_2=2\pi \sqrt{\frac{2lI}{\pi\eta r^4}}. \]

Squaring,

\[ T_2^2=\frac{8\pi lI}{\eta r^4}. \]

Therefore,

\[ \boxed{ \eta=\frac{8\pi lI}{T_2^2r^4} } \tag{19.2} \]

19.6 Determination of Poisson's Ratio

Dividing Equation (19.1) by Equation (19.2),

\[ \frac{Y}{\eta} = \frac{8\pi lI}{T_1^2r^4} \frac{T_2^2r^4}{8\pi lI}. \]

Hence,

\[ \boxed{ \frac{Y}{\eta}=\frac{T_2^2}{T_1^2} } \tag{19.3} \]

For a homogeneous, isotropic, linearly elastic material, the relation between Young's modulus, modulus of rigidity, and Poisson's ratio is

\[ Y=2\eta(1+\sigma). \]

Therefore,

\[ \sigma=\frac{Y}{2\eta}-1. \]

Using Equation (19.3),

\[ \sigma=\frac{T_2^2}{2T_1^2}-1. \]

Thus,

\[ \boxed{ \sigma=\frac{T_2^2-2T_1^2}{2T_1^2} } \tag{19.4} \]

Therefore, by measuring the time periods \(T_1\) and \(T_2\), together with the dimensions of the wire and the moment of inertia of the rod, the elastic constants \(Y\), \(\eta\), and \(\sigma\) can be determined.

Assumptions: The wire is homogeneous and isotropic, the oscillations are small, the rods are sufficiently rigid, and the effects of damping, suspension threads, and wire inertia are negligible.

21. Determination of Modulus of Rigidity by Statical Method

21.1 Principle

The statical method is based on the relationship between the applied twisting couple and the resulting angle of twist of a cylindrical wire or rod.

For a circular wire of length \(l\), radius \(r\), and modulus of rigidity \(\eta\), the applied torque required to produce an angle of twist \(\theta\) is

\[ \boxed{ \tau=\frac{\pi\eta r^4}{2l}\theta } \tag{21.1} \]

21.2 Horizontal Twisting Apparatus

A cylindrical rod MN is mounted horizontally, with one end rigidly clamped to a fixed support. The other end is connected to the axle of a pulley of radius \(R\).

A cord is wound around the pulley, and a known mass \(M\) is suspended from its free end. The suspended mass applies a twisting torque to the rod.

Two pointers \(P_1\) and \(P_2\) are attached to the rod at two positions separated by a known distance \(l\). The pointers move over circular scales \(S_1\) and \(S_2\), allowing the angular twist between the two sections to be determined.

This arrangement has two important limitations:

1. The single applied force produces a lateral pull on the rod, which may increase friction at the bearings.
2. Errors in the alignment of the pointers and circular scales may introduce inaccuracies in the measured angles of twist.

21.3 Vertical Twisting Apparatus

To reduce these difficulties, a vertical twisting apparatus is employed.

The experimental wire MN is suspended vertically from a rigid support and clamped at its upper end N. Its lower end M is connected to a heavy cylindrical drum of radius \(R\).

Two cords are wound around the drum in opposite directions. They leave the drum tangentially and pass over frictionless pulleys.

Equal masses \(M\) are suspended from the free ends of the cords. The two tangential forces form a couple that twists the wire without producing a resultant lateral force on the drum.

Two pointers are attached to the wire at different positions and their angular deflections are measured using circular scales.

21.4 Theory and Derivation

Let each suspended mass be \(M\). The magnitude of each tangential force is

\[ F=Mg. \]

Since the perpendicular distance between the two parallel forces is \(2R\), the applied couple is

\[ \tau=F(2R). \]

Therefore,

\[ \boxed{ \tau=2MgR } \]

Let the two pointers be situated at distances \(l_1\) and \(l_2\) from the fixed end of the wire, with \(l_2>l_1\).

If their respective angular deflections are \(\theta_1\) and \(\theta_2\), then the angular twist between the pointers is

\[ \theta=\theta_2-\theta_1. \]

The length of the wire between the pointers is

\[ l=l_2-l_1. \]

From the torsion equation,

\[ \tau=\frac{\pi\eta r^4}{2l}\theta. \]

Substituting the experimental quantities,

\[ 2MgR= \frac{\pi\eta r^4} {2(l_2-l_1)} (\theta_2-\theta_1). \]

Rearranging,

\[ \boxed{ \eta= \frac{4MgR(l_2-l_1)} {\pi r^4(\theta_2-\theta_1)} } \tag{21.2} \]

This expression gives the modulus of rigidity of the experimental wire.

Note: The angular deflections must be expressed in radians. The wire must remain within its elastic limit, and the applied couple should be transmitted without appreciable friction.

22. Determination of Modulus of Rigidity by Dynamical Method (Inertia Table)

22.1 Principle

The dynamical method is based on the torsional oscillations of a body suspended from a cylindrical wire.

When the body is rotated through a small angle about the vertical axis and released, the restoring torque of the wire produces angular simple harmonic motion.

The time period of these oscillations depends on the moment of inertia of the suspended body and the torsional constant of the wire.

22.2 Experimental Arrangement

A thin cylindrical wire of length \(l\) and radius \(r\) is suspended vertically from a rigid support.

An inertia table or a rigid body of known moment of inertia \(I\) is attached to the lower end of the wire.

The body is rotated through a small angle about the vertical axis and then released. It executes torsional oscillations.

The time taken for a known number of complete oscillations is measured using a stopwatch. The mean time period is calculated.

22.3 Theory and Derivation

Let \(C\) be the restoring torque per unit angular twist of the wire.

When the suspended body is displaced through an angle \(\theta\), the restoring torque is

\[ \tau=-C\theta. \]

The negative sign indicates that the torque acts opposite to the angular displacement.

According to Newton's second law for rotational motion,

\[ \tau=I\frac{d^2\theta}{dt^2}. \]

Therefore,

\[ I\frac{d^2\theta}{dt^2}=-C\theta. \]

Rearranging,

\[ \frac{d^2\theta}{dt^2} +\frac{C}{I}\theta=0. \]

This is the differential equation of angular simple harmonic motion.

Hence, the angular frequency is

\[ \omega=\sqrt{\frac{C}{I}}. \]

The time period is

\[ T=\frac{2\pi}{\omega}. \]

Therefore,

\[ \boxed{ T=2\pi\sqrt{\frac{I}{C}} } \]

For a circular wire,

\[ C=\frac{\pi\eta r^4}{2l}. \]

Substituting,

\[ T=2\pi \sqrt{\frac{2lI}{\pi\eta r^4}}. \]

Squaring both sides,

\[ T^2= 4\pi^2 \left(\frac{2lI}{\pi\eta r^4}\right). \]

Simplifying,

\[ T^2=\frac{8\pi lI}{\eta r^4}. \]

Rearranging,

\[ \boxed{ \eta=\frac{8\pi lI}{r^4T^2} } \tag{22.1} \]

Thus, knowing the length \(l\), radius \(r\), moment of inertia \(I\), and time period \(T\), the modulus of rigidity of the suspension wire can be determined.

22.4 Determination of Time Period

If the time required for \(N\) complete oscillations is \(t\), then the time period is

\[ \boxed{T=\frac{t}{N}} \]

For greater accuracy, the experiment is repeated several times and the mean time period is calculated.

22.5 Precautions

1. The suspension wire should be straight, uniform, and free from kinks.
2. The angular displacement should be small so that the motion remains approximately simple harmonic.
3. The suspended body should oscillate about the vertical axis without appreciable lateral motion.
4. The wire length and diameter should be measured carefully.
5. The time should be measured over a sufficiently large number of oscillations.
6. The moment of inertia of the suspended system should be known accurately, including any significant attachments.

23. Summary of Important Formulae

Method	Elastic Constant	Formula
Central Bending	Young's Modulus	\(Y=\dfrac{Mgl^3}{4bd^3\delta}\)
Searle's Bending Oscillations	Young's Modulus	\(Y=\dfrac{8\pi lI}{T_1^2r^4}\)
Searle's Torsional Oscillations	Modulus of Rigidity	\(\eta=\dfrac{8\pi lI}{T_2^2r^4}\)
Searle's Method	Poisson's Ratio	\(\sigma=\dfrac{T_2^2-2T_1^2}{2T_1^2}\)
Statical Torsion	Modulus of Rigidity	\(\eta=\dfrac{4MgR(l_2-l_1)} {\pi r^4(\theta_2-\theta_1)}\)
Dynamical Torsion	Modulus of Rigidity	\(\eta=\dfrac{8\pi lI}{r^4T^2}\)

24. Physical Significance of Elastic Constants

Young's modulus (\(Y\)): It measures the resistance of a material to longitudinal deformation under tensile or compressive stress.

\[ Y=\frac{\text{Longitudinal stress}} {\text{Longitudinal strain}} \]

Modulus of rigidity (\(\eta\)): It measures the resistance of a material to shear deformation.

\[ \eta=\frac{\text{Shear stress}} {\text{Shear strain}} \]

Poisson's ratio (\(\sigma\)): It is the negative ratio of transverse strain to longitudinal strain under uniaxial loading.

\[ \sigma= -\frac{\text{Transverse strain}} {\text{Longitudinal strain}} \]

For an isotropic, linearly elastic material, the relationship between Young's modulus, modulus of rigidity, and Poisson's ratio is

\[ \boxed{Y=2\eta(1+\sigma)} \]

Young's modulus and modulus of rigidity have the SI unit pascal (\(\mathrm{Pa}\)), whereas Poisson's ratio is dimensionless.