Unit 1 (Part B)

Part B: Mechanics of Materials

1. Derive the expression for the moment of inertia of a hollow and solid cylinder. Also discuss the application of it.

Moment of Inertia (I): A measure of an object's resistance to changes in its rotational motion. I = ∫ r² dm.

(i) Solid Cylinder:

Consider a solid cylinder of mass M, radius R, and length L. We want to find its moment of inertia about its central axis.

Imagine the cylinder is made of many thin cylindrical shells of radius r, thickness dr, and length L.

Mass density ρ = M/V = M/(π R² L).

Volume of a thin shell dV = (2π r dr) L.

Mass of the thin shell dm = ρ dV = (M/(π R² L)) (2π r L dr) = (2M/R²) r dr.

The moment of inertia of this thin shell (like a hollow cylinder) about the central axis is dI = dm · r².

dI = ((2M/R²) r dr) r² = (2M/R²) r³ dr

To find the total moment of inertia, integrate from r=0 to r=R:

I = ∫ dI = ∫₀ᴿ (2M/R²) r³ dr = (2M/R²) ∫₀ᴿ r³ dr

I = (2M/R²) [ r⁴/4 ]₀ᴿ = (2M/R²) ( R⁴/4 - 0 )

I_{solid cylinder} = ½ M R²

(ii) Hollow Cylinder (or Cylindrical Shell):**

Consider a thin hollow cylinder of mass M and radius R. All its mass is effectively at a distance R from the central axis of rotation.

Using the basic definition I = ∫ r² dm. Since r=R for all mass elements dm:

I = ∫ R² dm = R² ∫ dm

Since ∫ dm = M (total mass):

I_{hollow cylinder} = M R²

(For a thick hollow cylinder with inner radius R₁ and outer radius R₂, I = ½ M (R₁² + R₂²)).

Applications:**

Flywheels: Solid discs or cylinders store rotational energy. Their large moment of inertia helps maintain constant angular velocity, smoothing out fluctuations in engine speed.

Vehicle Wheels: The distribution of mass (affecting moment of inertia) impacts acceleration and handling. Lighter wheels with lower moment of inertia allow faster acceleration.

Gyroscopes: Rely on the principle of conservation of angular momentum, which is related to moment of inertia. Used in navigation systems.

Rotating Machinery: Understanding the moment of inertia of components like shafts, turbines, and rotors is crucial for designing and analyzing their dynamic behavior and energy requirements.

2. Find the expression for moment of inertia of a solid disc with respect to its tangent (i) Parallel to the disc surface and (ii) Perpendicular to the disc surface.

Let the solid disc have mass M and radius R.

Moment of Inertia about axes through the Center of Mass (CM):**

Axis perpendicular to the disc through CM (z-axis): I_z = ½ M R²

Axis along a diameter through CM (x or y-axis): By Perpendicular Axis Theorem (I_z = I_x + I_y), and since I_x = I_y by symmetry, 2I_diameter = I_z. Therefore, I_diameter = ½ I_z = ¼ M R².

We will use the **Parallel Axis Theorem**: I = I_CM + M d², where I_CM is the moment of inertia about a parallel axis through the center of mass, and d is the perpendicular distance between the two axes.

(i) Tangent Parallel to the disc surface:**

This tangent is parallel to a diameter. The axis passes through the edge of the disc.

The parallel axis through the CM is a diameter.

I_CM = I_diameter = ¼ M R².

The distance between the diameter and the parallel tangent is d = R.

Applying the Parallel Axis Theorem:

I_{tangent, parallel} = I_CM + M d² = ¼ M R² + M R²

I_{tangent, parallel} = (5/4) M R²

(ii) Tangent Perpendicular to the disc surface:**

This tangent is parallel to the axis passing perpendicularly through the center of the disc.

The parallel axis through the CM is the axis perpendicular to the disc through its center.

I_CM = I_z = ½ M R².

The distance between the central perpendicular axis and the parallel tangent at the rim is d = R.

Applying the Parallel Axis Theorem:

I_{tangent, perpendicular} = I_CM + M d² = ½ M R² + M R²

I_{tangent, perpendicular} = (3/2) M R²

3. Derive an expression for the torsional couple per unit angular twist when a cylinder is twisted.(12) A solid sphere of mass 2kg rotates with respect to its diameter with angular speed 5 rotation per second. The radius of the sphere is 25 cm. What is the (i) Moment of inertia and (ii) Kinetic energy associated with the rotation of the sphere with respect to its diameter?(4)

Part 1: Torsional Couple per Unit Angular Twist (Torsional Rigidity)**

Consider a solid cylinder (or wire) of length L and radius R. Let its lower end be twisted by an angle θ relative to its fixed upper end.

Imagine the cylinder is composed of numerous infinitesimally thin coaxial cylindrical shells. Consider one such shell of radius r and thickness dr.

When the cylinder is twisted, a line initially parallel to the axis on the surface of this shell (say, AB) gets twisted to AB'. The angle of shear is φ.

From the geometry, the arc length BB' = rθ. Also, BB' ≈ Lφ for small angles.

So, rθ = Lφ ⟹ φ = rθ/L.

Let G be the Shear Modulus (Modulus of Rigidity) of the material. Shear stress = G × Shear strain (φ).

Shear Stress = G φ = G rθ/L.

The area of the cross-section of the thin shell is dA = 2π r dr.

The tangential shearing force on this area is dF = Shear Stress × dA = (G rθ/L) (2π r dr) = (2π G θ/L) r² dr.

The torque (twisting couple) due to this force on the shell is dτ = dF × r.

dτ = ((2π G θ/L) r² dr) r = (2π G θ/L) r³ dr

To find the total torque required to twist the entire cylinder, integrate from r=0 to r=R:

τ = ∫ dτ = ∫₀ᴿ (2π G θ/L) r³ dr = (2π G θ/L) ∫₀ᴿ r³ dr

τ = (2π G θ/L) [ r⁴/4 ]₀ᴿ = (2π G θ/L) ( R⁴/4 )

τ = (π G R⁴ / (2L)) θ

The torque is proportional to the angle of twist θ. The constant of proportionality is the torsional couple per unit angular twist (also called torsional rigidity), denoted by C.

C = τ/θ = π G R⁴ / (2L)

For a hollow cylinder with inner radius R₁ and outer radius R₂, the integration limits change, and the result is C = π G (R₂⁴ - R₁⁴) / (2L).

Part 2: Rotating Solid Sphere Calculation**

Given:

Mass, M = 2 kg

Angular speed, f = 5 rotations per second. ω = 2π f = 2π(5) = 10π rad/s

Radius, R = 25 cm = 0.25 m

(i) Moment of Inertia (I):**

The moment of inertia of a solid sphere about its diameter is given by:

I = (2/5) M R²

I = (2/5) (2 kg) (0.25 m)²

I = (4/5) (0.0625) kg m²

I = 0.8 × 0.0625 kg m² = 0.05 kg m²

I = 0.05 kg m²

(ii) Rotational Kinetic Energy (KE):**

The kinetic energy of rotation is given by:

KE = ½ I ω²

KE = ½ (0.05 kg m²) (10π rad/s)²

KE = ½ (0.05) (100 π²) J

KE = 0.025 × 100 π² J = 2.5 π² J

Using π² ≈ 9.87:

KE ≈ 2.5 × 9.87 J ≈ 24.675 J

KE ≈ 24.68 J

4. Give the theory and experimental method of finding young’s modules of a cantilever.

Theory:**

A cantilever is a beam fixed horizontally at one end and loaded at the free end. When a load W is applied at the free end, the beam bends (depresses). We want to relate this depression y to the Young's Modulus E of the beam's material.

Consider a small segment dx of the beam at a distance x from the fixed end. The bending moment at this section is due to the load W acting at a distance (L-x) from the segment, where L is the length of the cantilever.

Bending Moment M = W(L-x).

The general formula relating bending moment M to the radius of curvature R of the bent beam is:

M = E I_g / R

Where I_g is the geometrical moment of inertia (or second moment of area) of the beam's cross-section about the neutral axis. For a rectangular beam of breadth b and thickness d, I_g = bd³/12.

The radius of curvature can be approximated as R ≈ 1 / (d²y/dx²) for small deflections.

Substituting these into the bending moment equation:

W(L-x) = E I_g (d²y/dx²)

Integrate with respect to x to find the slope dy/dx:

∫ W(L-x) dx = ∫ E I_g (d²y/dx²) dx

W(Lx - x²/2) + C₁ = E I_g (dy/dx)

Boundary condition: At the fixed end (x=0), the slope is zero (dy/dx = 0). So, C₁ = 0.

E I_g (dy/dx) = W(Lx - x²/2)

Integrate again to find the depression y:

∫ E I_g (dy/dx) dx = ∫ W(Lx - x²/2) dx

E I_g y = W(L x²/2 - x³/6) + C₂

Boundary condition: At the fixed end (x=0), the depression is zero (y = 0). So, C₂ = 0.

E I_g y = W((Lx²/2) - (x³/6))

The maximum depression occurs at the free end (x=L):

E I_g y_max = W(L³/2 - L³/6) = W ((3L³ - L³)/6) = WL³/3

Rearranging for Young's Modulus E, and substituting I_g = bd³/12 for a rectangular beam:

E = WL³ / (3 y_max I_g) = WL³ / (3 y_max (bd³/12))

E = 4WL³ / (bd³ y_max)

(Often y_max is simply written as y).

Experimental Method:**

Setup: Clamp one end of the rectangular beam (e.g., a wooden meter scale) horizontally to a rigid support (like a table edge).

Measurement Point: Place a pin vertically at the free end of the beam. Position a travelling microscope focused on the tip of the pin to measure its vertical position accurately.

Loading: Hang a weight hanger from the free end, near the pin.

Initial Reading: Note the reading on the microscope's vertical scale for the unloaded beam (hanger only, mass m₀).

Loading Steps: Add slotted weights (mass m) to the hanger in equal increments (e.g., 50g). For each load W = (m₀+m)g, record the corresponding reading on the microscope.

Unloading Steps: Remove the weights in the same increments, recording the readings again.

Calculate Depression: Calculate the mean reading for each load. The depression y for an added mass m is the difference between the mean reading for that load and the initial reading (hanger only).

Beam Dimensions: Measure the length L of the cantilever from the fixed point to the loading point. Measure the breadth b and thickness d of the beam at several points using vernier calipers and a screw gauge, respectively, and calculate the average values.

Calculation: Calculate E using the formula E = (4L³ / (bd³)) (W/y). The term W/y (or mg/y) can be found from the slope of a graph plotting Load (W or m) vs. Depression (y).

5. What is uniform bending? Derive an expression for the elevation at the centre of a beam which is loaded at both end. Describe an experiment to determine Young’s modulus of a beam by uniform bending.

Uniform Bending:**

Uniform bending occurs when a beam bends into the arc of a circle. This happens when the bending moment is constant along the length of the beam segment being considered. A common way to achieve this is by supporting the beam symmetrically on two knife edges and applying equal loads at its ends.

Theory for Elevation:**

Consider a beam supported symmetrically on two knife edges A and B, separated by a distance l. Equal weights W are hung from the ends C and D, such that AC = BD = a.

The reaction forces at the supports A and B are each equal to W.

Consider the portion of the beam between the knife edges A and B. The external bending moment at any point x between A and B is constant:

M = W × a. (The upward force W at A creates a moment W(x-a) and the downward force W at C creates a moment -Wx. Total moment W(x-a) - Wx = -Wa. Considering the magnitude, M=Wa).

Since the bending moment M is constant between A and B, the beam bends into a circular arc of constant radius R.

The relation between bending moment M, Young's Modulus E, geometrical moment of inertia I_g, and radius of curvature R is:

M = E I_g / R

Wa = E I_g / R ⟹ R = E I_g / (Wa)

Let O be the center of the circular arc and y be the elevation of the midpoint of AB above the supports. From the geometry of the circle (property of intersecting chords, or Pythagoras on triangle OAE where E is midpoint of AB):

(2R-y)y = (l/2)² (Approximation for small elevation y compared to R)

2Ry - y² ≈ (l/2)². Since y is small, y² can be neglected.

2Ry ≈ l²/4 ⟹ y = l² / (8R)

Substitute the expression for R:

y = l² / (8 (E I_g / (Wa))) = Wal² / (8 E I_g)

For a rectangular beam, I_g = bd³/12.

y = Wal² / (8 E (bd³/12)) = 12Wal² / (8 E bd³)

y = 3Wal² / (2Ebd³)

Rearranging for Young's Modulus E:

E = 3Wal² / (2bd³y)

Experimental Method:**

Setup: Place the beam symmetrically on two knife edges separated by a distance l.

Loading: Suspend two weight hangers of equal mass (m₀) from the ends of the beam at equal distances a from the knife edges.

Measurement Point: Place a pin vertically at the midpoint of the beam (between the knife edges). Focus a travelling microscope on the tip of the pin.

Initial Reading: Note the microscope reading for the unloaded beam (hangers only).

Loading Steps: Add equal slotted weights (mass m) simultaneously to both hangers in increments. For each total load W = (m₀+m)g on each hanger, record the microscope reading.

Unloading Steps: Remove the weights simultaneously in steps, recording the readings again.

Calculate Elevation: Find the mean reading for each load. The elevation y for an added mass m is the difference between the initial reading and the mean reading for that load.

Beam Dimensions: Measure the distance l between the knife edges, the distance a from knife edge to load point, the average breadth b, and average thickness d of the beam.

Calculation: Calculate E using the formula E = (3al² / (2bd³)) (W/y). The term W/y (or mg/y) is found from the slope of a graph plotting Load (W or m) vs. Elevation (y).

6. Explain with necessary theory the determination of young’s modulus of elasticity of the material of the beam supported at its ends and loaded in the middle. Describe an experiment to determine the young’s modulus of the material using this method.

This describes **Non-Uniform Bending** (specifically, center loading).

Theory:**

Consider a beam of length l supported on two knife edges at its ends A and B. A load W is applied at the center C.

The reaction force at each support is R_A = R_B = W/2.

Consider a point P on the beam at a distance x from end A (0 ≤ x ≤ l/2). The bending moment at P is:

M = R_A × x = (W/2) x. (This applies for the left half of the beam).

The relation between bending moment M, Young's Modulus E, geometrical moment of inertia I_g, and curvature is:

M = E I_g (d²y/dx²)

Where y is the depression of the beam at position x.

E I_g (d²y/dx²) = (W/2) x

Integrate with respect to x to find the slope dy/dx:

E I_g (dy/dx) = (W/2) (x²/2) + C₁ = Wx²/4 + C₁

Boundary condition: Due to symmetry, the slope at the center (x=l/2) must be zero (dy/dx = 0).

0 = W(l/2)²/4 + C₁ ⟹ C₁ = -Wl²/16

E I_g (dy/dx) = Wx²/4 - Wl²/16

Integrate again to find the depression y:

E I_g y = (W/4) (x³/3) - (Wl²/16) x + C₂ = Wx³/12 - Wl²x/16 + C₂

Boundary condition: At the support A (x=0), the depression is zero (y = 0). So, C₂ = 0.

E I_g y = Wx³/12 - Wl²x/16

The maximum depression occurs at the center (x=l/2):

E I_g y_max = W(l/2)³/12 - Wl²(l/2)/16 = Wl³/96 - Wl³/32

E I_g y_max = Wl³ (1/96 - 3/96) = Wl³ (-2/96) = -Wl³/48

The negative sign indicates depression (downward displacement). Taking the magnitude:

y_max = Wl³ / (48 E I_g)

For a rectangular beam, I_g = bd³/12.

y_max = Wl³ / (48 E (bd³/12)) = Wl³ / (4 E bd³)

Rearranging for Young's Modulus E:

E = Wl³ / (4bd³ y_max)

(Often y_max is simply written as y).

Experimental Method:**

Setup: Place the beam on two knife edges separated by a distance l.

Loading: Suspend a weight hanger from the exact midpoint of the beam.

Measurement Point: Place a pin vertically at the midpoint, aligned with the load application point. Focus a travelling microscope on the tip of the pin.

Initial Reading: Note the microscope reading with only the hanger (mass m₀) attached.

Loading Steps: Add slotted weights (mass m) to the hanger in increments. For each total load W = (m₀+m)g, record the microscope reading.

Unloading Steps: Remove the weights in steps, recording the readings again.

Calculate Depression: Find the mean reading for each load. The depression y for an added mass m is the difference between the mean reading for that load and the initial reading.

Beam Dimensions: Measure the distance l between the knife edges, the average breadth b, and average thickness d of the beam.

Calculation: Calculate E using the formula E = (l³ / (4bd³)) (W/y). The term W/y (or mg/y) is found from the slope of a graph plotting Load (W or m) vs. Depression (y).