Mechanics of Materials

Mechanics of Materials (also called Strength of Materials) studies the behavior of solid bodies subjected to various types of loading. This discipline bridges the gap between statics and structural design, enabling engineers to determine stresses, strains, and deformations in structural elements.

Stress and Strain

Normal Stress

Normal stress ($\sigma$) is the internal force per unit area acting perpendicular to a cross-section:

Where:

• $P$ = axial force (N or kN)
• $A$ = cross-sectional area (m$^2$ or mm$^2$)

Units: Pascal (Pa) = N/m$^2$, or more commonly MPa = N/mm$^2$

Tensile stress: Positive (stretching) Compressive stress: Negative (shortening)

Normal Strain

Normal strain ($\varepsilon$) is the change in length per unit original length:

Strain is dimensionless but often expressed in units of mm/mm or as a percentage.

Hooke's Law

For linear elastic materials:

Where $E$ is the modulus of elasticity (Young's modulus), typically:

• Steel: $E \approx 200$ GPa
• Concrete: $E \approx 25-35$ GPa
• Aluminum: $E \approx 70$ GPa

Deformation Under Axial Load

For members with varying cross-sections or loads:

Poisson's Ratio

When a material is stretched axially, it contracts laterally:

Typical values:

• Steel: $\nu \approx 0.3$
• Concrete: $\nu \approx 0.15-0.20$
• Rubber: $\nu \approx 0.5$

Shear Stress and Strain

Shear Stress

Where $V$ is the shear force and $A$ is the area over which it acts.

Shear Strain

Where $\delta_s$ is the shear displacement and $h$ is the height.

Shear Modulus

Relationship between elastic constants:

Stress-Strain Relationships

Key Points on Stress-Strain Curve

1. Proportional Limit: End of linear region
2. Elastic Limit: Maximum stress for elastic behavior
3. Yield Point ($\sigma_y$): Onset of plastic deformation
4. Ultimate Strength ($\sigma_u$): Maximum stress
5. Fracture Point: Failure

Offset Yield Strength

For materials without clear yield point, use 0.2% offset method:

Draw a line parallel to the elastic region, offset by $\varepsilon = 0.002$. The intersection with the stress-strain curve gives the yield strength.

Torsion of Circular Shafts

Shear Stress Distribution

For a circular shaft under torque $T$:

Where:

• $\rho$ = radial distance from center
• $J$ = polar moment of inertia

Maximum shear stress (at outer surface, $\rho = r$):

Where $Z_p = J/r$ is the polar section modulus.

Polar Moment of Inertia

Solid circular shaft:

Hollow circular shaft:

Angle of Twist

For varying torque or cross-section:

Power Transmission

Where $n$ is rotational speed in RPM.

Bending of Beams

Flexure Formula

For a beam in pure bending:

Where:

• $M$ = bending moment
• $y$ = distance from neutral axis
• $I$ = moment of inertia about neutral axis

Maximum bending stress:

Where $c$ is the distance to the extreme fiber and $S = I/c$ is the section modulus.

Section Modulus for Common Shapes

Shear Stress in Beams

Where:

• $V$ = shear force
• $Q$ = first moment of area above (or below) the point
• $I$ = moment of inertia
• $b$ = width at the point of interest

For a rectangular cross-section, maximum shear stress at neutral axis:

Beam Deflections

Differential Equation of the Elastic Curve

Successive integration gives slope and deflection:

Common Deflection Formulas

Simply supported beam, central point load $P$:

Simply supported beam, uniform load $w$:

Cantilever beam, end point load $P$:

Cantilever beam, uniform load $w$:

Superposition Method

For multiple loads, add individual deflections:

Combined Stresses

Principal Stresses

For a general 2D stress state ($\sigma_x$, $\sigma_y$, $\tau_{xy}$):

Maximum Shear Stress

Mohr's Circle

A graphical method for stress transformation:

• Center at $\left(\frac{\sigma_x + \sigma_y}{2}, 0\right)$
• Radius = $\sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$

Column Buckling

Euler's Critical Load

For a slender column with pinned ends:

Effective Length

For different end conditions, use effective length $L_e = KL$:

Slenderness Ratio

Where $r = \sqrt{I/A}$ is the radius of gyration.

Critical Stress

Real-World Application: Steel Beam Design

Designing a steel beam for a floor system.

Beam Analysis Example

import math

# Beam loading and geometry
beam_params = {
    'span': 8.0,              # meters
    'dead_load': 5.0,         # kN/m (self-weight + permanent loads)
    'live_load': 10.0,        # kN/m (occupancy load)
    'support_type': 'simply_supported',
    'steel_grade': 'A992',    # Fy = 345 MPa
    'E': 200000               # MPa (steel modulus)
}

# Load factors (LRFD)
factored_load = 1.2 * beam_params['dead_load'] + 1.6 * beam_params['live_load']

# Calculate maximum moment (simply supported, uniform load)
L = beam_params['span']
w = factored_load
M_max = w * L**2 / 8  # kN-m
M_max_Nmm = M_max * 1e6  # Convert to N-mm

# Calculate maximum shear
V_max = w * L / 2  # kN

# Required section modulus
Fy = 345  # MPa
phi_b = 0.9  # Resistance factor for bending
S_required = M_max_Nmm / (phi_b * Fy)  # mm^3

# Required moment of inertia for deflection limit
deflection_limit = L * 1000 / 360  # L/360 in mm (for live load)
w_service = beam_params['live_load']  # kN/m (unfactored live load)
# delta = 5wL^4 / (384EI), solve for I
I_required = (5 * w_service * (L * 1000)**4) / (384 * beam_params['E'] * deflection_limit)

print(f"Steel Beam Design:")
print(f"  Span: {L} m")
print(f"  Factored load: {factored_load:.2f} kN/m")
print(f"  Maximum moment: {M_max:.2f} kN-m")
print(f"  Maximum shear: {V_max:.2f} kN")
print(f"\nSection Requirements:")
print(f"  Required section modulus: {S_required/1000:.0f} x 10^3 mm^3")
print(f"  Required moment of inertia: {I_required/1e6:.0f} x 10^6 mm^4")
print(f"  Deflection limit: {deflection_limit:.1f} mm (L/360)")

# Example W-section selection
W_section = {
    'designation': 'W310x52',
    'S': 748e3,      # mm^3
    'I': 119e6,      # mm^4
    'd': 317,        # mm (depth)
    'tw': 7.6,       # mm (web thickness)
}

# Check adequacy
S_ratio = W_section['S'] / S_required
I_ratio = W_section['I'] / I_required

print(f"\nSelected Section: {W_section['designation']}")
print(f"  Section modulus ratio: {S_ratio:.2f} (>1.0 OK)")
print(f"  Moment of inertia ratio: {I_ratio:.2f} (>1.0 OK)")

if S_ratio >= 1.0 and I_ratio >= 1.0:
    print(f"  Design: ADEQUATE")
else:
    print(f"  Design: SELECT LARGER SECTION")

Your Challenge: Shaft Design for Torsion

Design a solid circular shaft to transmit power while limiting stress and angle of twist.

Goal: Determine the required shaft diameter for given power transmission requirements.

Problem Setup

import math

# Shaft requirements
shaft_config = {
    'power': 75,              # kW
    'speed': 600,             # RPM
    'length': 2.0,            # meters
    'allowable_shear': 60,    # MPa
    'allowable_twist': 1.0,   # degrees per meter
    'G': 80000                # MPa (steel shear modulus)
}

# TODO: Calculate required shaft diameter
# Step 1: Calculate torque from power
omega = 2 * math.pi * shaft_config['speed'] / 60  # rad/s
T = shaft_config['power'] * 1000 / omega  # N-m
T_Nmm = T * 1000  # N-mm

# Step 2: Diameter based on shear stress
# tau_max = 16T / (pi * d^3)
# d^3 = 16T / (pi * tau_allow)
d_stress = (16 * T_Nmm / (math.pi * shaft_config['allowable_shear'])) ** (1/3)

# Step 3: Diameter based on angle of twist
# phi = TL / (GJ) = 32TL / (pi * G * d^4)
# Convert allowable twist to radians per mm
allowable_twist_rad = math.radians(shaft_config['allowable_twist'] / 1000)  # rad/mm
# d^4 = 32TL / (pi * G * phi_allow)
L_mm = shaft_config['length'] * 1000
d_twist = (32 * T_Nmm * L_mm / (math.pi * shaft_config['G'] * allowable_twist_rad * L_mm)) ** 0.25

# Step 4: Select governing diameter
d_required = max(d_stress, d_twist)

# Round up to nearest standard size (5mm increments)
d_selected = math.ceil(d_required / 5) * 5

print(f"Shaft Design Results:")
print(f"  Power: {shaft_config['power']} kW at {shaft_config['speed']} RPM")
print(f"  Torque: {T:.2f} N-m = {T_Nmm:.0f} N-mm")
print(f"\nDiameter Requirements:")
print(f"  Based on shear stress: {d_stress:.1f} mm")
print(f"  Based on angle of twist: {d_twist:.1f} mm")
print(f"  Governing diameter: {d_required:.1f} mm")
print(f"  Selected diameter: {d_selected} mm")

# Verify design
J_selected = math.pi * d_selected**4 / 32
tau_actual = T_Nmm * (d_selected/2) / J_selected
phi_actual = T_Nmm * L_mm / (shaft_config['G'] * J_selected)
phi_actual_deg = math.degrees(phi_actual) * 1000  # deg per meter

print(f"\nDesign Verification:")
print(f"  Actual shear stress: {tau_actual:.2f} MPa (limit: {shaft_config['allowable_shear']} MPa)")
print(f"  Actual twist: {phi_actual_deg:.3f} deg/m (limit: {shaft_config['allowable_twist']} deg/m)")
print(f"  Stress utilization: {tau_actual/shaft_config['allowable_shear']*100:.1f}%")
print(f"  Twist utilization: {phi_actual_deg/shaft_config['allowable_twist']*100:.1f}%")

What modifications would be needed if the shaft were hollow instead of solid, and how would that affect the weight?