6.3 Bending Stresses, Torsion & Column Buckling Analysis

6.3 Bending Stresses, Torsion & Column Buckling Analysis

1. Shear Force and Bending Moment Diagrams (SFD/BMD)

Beams are structural members supporting transverse loads. Internal load resultants at any cross-section consist of a Shear Force ($V$) and a Bending Moment ($M$).

1.1 Differential Load-Shear-Moment Relationships

For a beam loaded by distributed load $w(x)$ (positive downward):

$$\frac{dV}{dx} = -w(x) \implies V(x) = V_0 - \int w(x) dx$$ $$\frac{dM}{dx} = V(x) \implies M(x) = M_0 + \int V(x) dx$$

Key rules for constructing SFD and BMD:

1. Peak bending moments ($M_{\text{max}}$) occur precisely where the shear force passes through zero ($V = 0$).
2. Concentrated point loads cause instantaneous vertical jumps in the shear diagram equal to load magnitude.
3. Concentrated couples cause instantaneous vertical jumps in the bending moment diagram.

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2. Flexure Formula and Transverse Shear Stress in Beams

2.1 The Elastic Flexure Formula

Under pure bending, cross-sections remain plane (Euler-Bernoulli beam theory). Bending stress $\sigma$ varies linearly with distance $y$ from the neutral axis (NA):

$$\sigma = - \frac{M y}{I}$$

where $M$ is internal bending moment, $I$ is area moment of inertia about NA, and $y$ is distance from NA. Maximum flexural stress at extreme fiber distance $y = c$ is:

$$\sigma_{\text{max}} = \frac{M c}{I} = \frac{M}{Z}$$

where $Z = \frac{I}{c}$ is the Elastic Section Modulus:

• Rectangular Section ($b \times h$): $Z = \frac{b h^2}{6}$
• Solid Circular Section (diameter $d$): $Z = \frac{\pi d^3}{32}$

2.2 Transverse Shear Stress Formula

Transverse loads generate horizontal and vertical shear stresses $\tau$ across beam cross-sections:

$$\tau = \frac{V Q}{I b}$$

where $V$ is shear force, $I$ is moment of inertia, $b$ is section width at specified depth, and $Q$ is first moment of area above (or below) the point of interest:

$$Q = A' \bar{y}'$$

For a rectangular section ($b \times h$), shear stress varies parabolically with depth, reaching its maximum at the neutral axis ($y = 0$):

$$\tau_{\text{max}} = 1.5 \frac{V}{A} = \frac{3 V}{2 b h}$$

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3. Torsion of Circular Shafts

When a circular shaft is subjected to torque $T$, internal shear stress acts tangentially across the cross-section.

3.1 Torsional Shear Stress Formula

Shear stress $\tau$ varies linearly with radial distance $r$ from the central axis:

$$\tau = \frac{T r}{J}$$

Maximum shear stress $\tau_{\text{max}}$ occurs at the outer boundary ($r = R = d/2$):

$$\tau_{\text{max}} = \frac{T d}{2 J}$$

where $J$ is the Polar Moment of Inertia:

• Solid Shaft (diameter $d$): $J = \frac{\pi d^4}{32} \implies \tau_{\text{max}} = \frac{16 T}{\pi d^3}$
• Hollow Shaft (outer $D$, inner $d$): $J = \frac{\pi (D^4 - d^4)}{32}$

3.2 Angle of Twist ($\theta$)

Total angular twist $\theta$ (in radians) over length $L$ of a uniform circular shaft under torque $T$ is:

$$\theta = \frac{T L}{G J}$$

where $G J$ represents torsional rigidity.

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4. Combined Axial, Bending, and Torsional Loading

In real machine components (e.g., drive shafts under torque, axial thrust, and gear bending), combined stresses act simultaneously on critical surface elements.

4.1 Superposition of Stresses

1. Total Normal Stress ($\sigma_x$): Sum of axial and flexural stresses: $$\sigma_x = \pm \frac{P}{A} \pm \frac{M y}{I}$$
2. Total Torsional Shear Stress ($\tau_{xy}$): $$\tau_{xy} = \frac{T r}{J}$$
3. Combined Principal Stresses: $$\sigma_{1,2} = \frac{\sigma_x}{2} \pm \sqrt{ \left( \frac{\sigma_x}{2} \right)^2 + \tau_{xy}^2 }$$

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5. Euler Column Buckling and Johnson Parabola

Columns are long, slender structural members subjected to axial compressive loads. Failure occurs by elastic instability (buckling) at stresses significantly below material yield strength.

5.1 Euler Buckling Critical Load ($P_{\text{cr}}$)

Derived by Leonhard Euler, the critical load $P_{\text{cr}}$ for a pin-ended slender column is:

$$P_{\text{cr}} = \frac{\pi^2 E I}{(K L)^2}$$

where $E$ is Young's modulus, $I$ is minimum area moment of inertia, $L$ is column length, and $K$ is the Effective Length Factor reflecting end support conditions:

• Pinned-Pinned: $K = 1.0 \implies L_e = L$
• Fixed-Fixed: $K = 0.5 \implies L_e = 0.5 L$
• Fixed-Free (Cantilever): $K = 2.0 \implies L_e = 2.0 L$
• Fixed-Pinned: $K = 0.7 \implies L_e = 0.7 L$

5.2 Critical Stress and Slenderness Ratio ($KL/r$)

Dividing $P_{\text{cr}}$ by cross-sectional area $A$ yields Euler critical stress $\sigma_{\text{cr}}$:

$$\sigma_{\text{cr}} = \frac{P_{\text{cr}}}{A} = \frac{\pi^2 E}{(K L / r)^2}$$

where $r = \sqrt{I / A}$ is the Radius of Gyration and $\frac{K L}{r}$ is the dimensionless Slenderness Ratio.

5.3 Intermediate Columns and Johnson Parabola

Euler's equation applies only when $\sigma_{\text{cr}} \le \sigma_y / 2$, which defines the Transition Slenderness Ratio ($C_c$):

$$C_c = \sqrt{ \frac{2 \pi^2 E}{\sigma_y} }$$

• Slender Columns ($KL/r \ge C_c$): Fail by elastic Euler buckling (valid $\sigma_{\text{cr}}$ formula).
• Intermediate Columns ($KL/r < C_c$): Fail by inelastic buckling governed by the empirical Johnson Parabola Formula:

$$\sigma_{\text{cr}} = \sigma_y \left[ 1 - \frac{\sigma_y (K L / r)^2}{4 \pi^2 E} \right]$$

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6. Worked Numerical Examples

Step-by-Step Example 1: Maximum Bending Stress in Beam

Problem: A simply supported rectangular beam of span $L = 6.0\text{ m}$ carries a uniform distributed load $w = 12.0\text{ kN/m}$. Cross-section dimensions: width $b = 150\text{ mm}$, height $h = 300\text{ mm}$. Calculate maximum bending moment $M_{\text{max}}$, section modulus $Z$, and maximum bending stress $\sigma_{\text{max}}$.

Solution:

1. Maximum bending moment for simply supported beam under uniform load: $$M_{\text{max}} = \frac{w L^2}{8} = \frac{12.0\text{ kN/m} \times (6.0\text{ m})^2}{8} = \frac{12.0 \times 36.0}{8} = 54.0\text{ kN}\cdot\text{m} = 54.0 \times 10^6\text{ N}\cdot\text{mm}$$
2. Section modulus $Z$: $$Z = \frac{b h^2}{6} = \frac{150\text{ mm} \times (300\text{ mm})^2}{6} = \frac{150 \times 90,000}{6} = 2,250,000\text{ mm}^3 = 2.25 \times 10^6\text{ mm}^3$$
3. Maximum flexural stress $\sigma_{\text{max}}$: $$\sigma_{\text{max}} = \frac{M_{\text{max}}}{Z} = \frac{54.0 \times 10^6\text{ N}\cdot\text{mm}}{2.25 \times 10^6\text{ mm}^3} = 24.0\text{ MPa}$$

Step-by-Step Example 2: Torsion and Twist Angle of Solid Shaft

Problem: A solid drive shaft of diameter $d = 50\text{ mm}$ and length $L = 2.0\text{ m}$ transmits torque $T = 1.5\text{ kN}\cdot\text{m}$. Shear Modulus $G = 77\text{ GPa}$. Determine maximum shear stress $\tau_{\text{max}}$ and total angle of twist $\theta$ in degrees.

Solution:

1. Polar moment of inertia $J$: $$J = \frac{\pi d^4}{32} = \frac{\pi (0.050\text{ m})^4}{32} = \frac{\pi \times 6.25 \times 10^{-6}}{32} = 6.13592 \times 10^{-7}\text{ m}^4$$
2. Maximum torsional shear stress $\tau_{\text{max}}$: $$\tau_{\text{max}} = \frac{T c}{J} = \frac{1500\text{ N}\cdot\text{m} \times 0.025\text{ m}}{6.13592 \times 10^{-7}\text{ m}^4} = \frac{37.5}{6.13592 \times 10^{-7}} = 61,115,550\text{ Pa} = 61.12\text{ MPa}$$
3. Angle of twist $\theta$ in radians: $$\theta = \frac{T L}{G J} = \frac{1500 \times 2.0}{(77 \times 10^9) \times (6.13592 \times 10^{-7})} = \frac{3000}{47,246.58} = 0.063496\text{ rad}$$
4. Convert $\theta$ to degrees: $$\theta_{\text{deg}} = 0.063496 \times \frac{180^\circ}{\pi} = 3.638^\circ \approx 3.64^\circ$$

Step-by-Step Example 3: Euler Column Buckling Load

Problem: A structural steel column of length $L = 4.5\text{ m}$ has pinned ends ($K = 1.0$). Properties: $E = 200\text{ GPa}$, minimum moment of inertia $I_{\text{min}} = 8.5 \times 10^6\text{ mm}^4 = 8.5 \times 10^{-6}\text{ m}^4$. Calculate Euler critical buckling load $P_{\text{cr}}$.

Solution:

1. Calculate effective length $L_e = K L = 1.0 \times 4.5\text{ m} = 4.5\text{ m}$.
2. Apply Euler buckling formula: $$P_{\text{cr}} = \frac{\pi^2 E I}{(K L)^2} = \frac{\pi^2 \times (200 \times 10^9\text{ N/m}^2) \times (8.5 \times 10^{-6}\text{ m}^4)}{(4.5\text{ m})^2}$$
3. Evaluate numerator and denominator: $$\text{Numerator} = 9.8696044 \times 200 \times 10^9 \times 8.5 \times 10^{-6} = 16,778,327\text{ N}\cdot\text{m}^2$$ $$\text{Denominator} = 20.25\text{ m}^2$$
4. Calculate $P_{\text{cr}}$: $$P_{\text{cr}} = \frac{16,778,327}{20.25} = 828,559\text{ N} = 828.56\text{ kN}$$