12.4 Beam Deflections, Combined Stresses, and Mohr's Circle

12.4 Beam Deflections, Combined Stresses, and Mohr's Circle

Advanced structural design requires verifying both serviceability (deflection limitations under operational loads) and ultimate strength under multi-axial stress states. This section explores classical beam deflection methods, eccentric combined loading, analytical stress transformation, Mohr's Circle, and failure theories.

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1. Beam Deflection: Double Integration Method

From the differential geometry of beam curvature $\kappa = \frac{d^2 y / dx^2}{[1 + (dy/dx)^2]^{3/2}} \approx \frac{d^2 y}{dx^2}$ (valid for small slopes $dy/dx \ll 1$), equating curvature to flexural moment yields the Euler-Bernoulli beam differential equation: $\boxed{E I \frac{d^2 y}{dx^2} = M(x)}$ Integrating sequentially:

1. Slope Equation ($\theta$): $E I \frac{dy}{dx} = E I \theta(x) = \int M(x) \, dx + C_1$
2. Deflection Equation ($y$): $E I y(x) = \int \left[\int M(x) \, dx\right] dx + C_1 x + C_2$

Boundary Conditions for Integration Constants

• Simply Supported (Pin or Roller at $x = 0$): $y(0) = 0$.
• Fixed Support (Built-in at $x = 0$): $y(0) = 0$ and $\theta(0) = \frac{dy}{dx}\big|_{x=0} = 0$.
• Symmetry (Midspan $x = L/2$ of symmetric loading): $\theta(L/2) = 0$.
• Free Cantilever End (at $x = L$): Shear $V(L) = 0$, Moment $M(L) = 0$.

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2. Moment-Area Theorems (Mohr's Theorems)

Developed by Christian Otto Mohr, the Moment-Area Theorems provide powerful geometric shortcuts for evaluating slopes and deflections without full algebraic integration.

First Moment-Area Theorem

The change in slope between any two points $A$ and $B$ on the elastic curve equals the total area under the $M / (E I)$ diagram between those points: $\boxed{\theta_{B/A} = \theta_B - \theta_A = \int_A^B \frac{M(x)}{E I} \, dx = \text{Area of } \left(\frac{M}{E I}\right)_{A \to B}}$

Second Moment-Area Theorem

The vertical tangential deviation $t_{B/A}$ of point $B$ on the elastic curve relative to the tangent drawn from point $A$ equals the first moment of the area under the $M / (E I)$ diagram between $A$ and $B$, taken about point $B$: $\boxed{t_{B/A} = \int_A^B \frac{M(x)}{E I} (x_B - x) \, dx = \left[\text{Area of } \left(\frac{M}{E I}\right)_{A \to B}\right] \cdot \bar{x}_B}$ where $\bar{x}_B$ is the centroidal distance of the $M / (E I)$ area measured from reference point $B$.

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3. The Conjugate Beam Method

The Conjugate Beam Method transforms deflection analysis into an equivalent statics problem by establishing a fictitious beam of identical span loaded with the real beam's $M / (E I)$ diagram.

Fundamental Principles

• Slope in Real Beam: Equals vertical shear force in the conjugate beam: $\theta_{\text{real}} = V_{\text{conj}}$.
• Deflection in Real Beam: Equals bending moment in the conjugate beam: $y_{\text{real}} = M_{\text{conj}}$.

Real vs. Conjugate Support Conversions

Conjugate beam supports must satisfy boundary kinematic constraints:

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4. Superposition Method for Standard Beam Cases

Linear elasticity permits superimposing slopes and deflections from standard loading cases:

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5. Combined Stresses: Axial Load Plus Bending and the Kern

When a member is subjected to combined axial normal force $P$ and bending moment $M$ (or an eccentric axial force $P$ applied at eccentricity $e$ from the centroid where $M = P e$): $\sigma = \frac{P}{A} \pm \frac{M c}{I} = \frac{P}{A} \left(1 \pm \frac{e c}{r^2}\right)$ where $r = \sqrt{I/A}$ is the radius of gyration.

The Core / Kern of a Cross-Section

In masonry, unreinforced concrete, and precast post-tensioned members, materials possess negligible tensile strength. The kern is the central region of the cross-section within which an axial compressive force can be applied without generating tensile stress anywhere across the section.

• Rectangular Cross-Section ($b \times h$): $\sigma_{\min} = -\frac{P}{b h} + \frac{P e_y (h/2)}{b h^3 / 12} = -\frac{P}{A}\left(1 - \frac{6 e_y}{h}\right) \ge 0 \implies \boxed{e_y \le \frac{h}{6}}$ This is the famous Middle-Third Rule: to prevent tension, the resultant compressive force must fall within the middle third of the depth (total kern width $= 2(h/6) = h/3$). For biaxial eccentricities, the kern forms a diamond (rhombus) with diagonals $h/3$ and $b/3$.
• Solid Circular Cross-Section ($d$): $\sigma_{\min} = -\frac{P}{A} + \frac{P e (d/2)}{\pi d^4 / 64} = -\frac{P}{A}\left(1 - \frac{8 e}{d}\right) \ge 0 \implies \boxed{e \le \frac{d}{8} = \frac{r}{4}}$ This constitutes the Middle-Quarter Rule: the kern is a concentric circle of diameter $d/4$.

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6. Plane Stress Transformation Equations

Consider an element in a state of plane stress subjected to normal stresses $(\sigma_x, \sigma_y)$ and shear stress $\tau_{xy}$. On an inclined plane oriented at angle $\theta$ counterclockwise from the positive $x$-axis: $\boxed{\sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \left(\frac{\sigma_x - \sigma_y}{2}\right) \cos 2\theta + \tau_{xy} \sin 2\theta}$ $\boxed{\tau_{x'y'} = - \left(\frac{\sigma_x - \sigma_y}{2}\right) \sin 2\theta + \tau_{xy} \cos 2\theta}$ $\sigma_{y'} = \frac{\sigma_x + \sigma_y}{2} - \left(\frac{\sigma_x - \sigma_y}{2}\right) \cos 2\theta - \tau_{xy} \sin 2\theta$

Stress Invariant

The sum of normal stresses on any two mutually perpendicular planes is constant: $\sigma_{x'} + \sigma_{y'} = \sigma_x + \sigma_y = I_1$

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7. Principal Stresses and Maximum Shear Stress

Principal Planes and Principal Stresses

Differentiating $\sigma_{x'}$ with respect to $\theta$ and setting to zero yields the principal planes where normal stresses reach their extrema: $\boxed{\tan 2\theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}}$

Substituting $\theta_p$ into the transformation equations yields the principal normal stresses: $\boxed{\sigma_{1, 2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}}$

Maximum In-Plane Shear Stress

Differentiating $\tau_{x'y'}$ with respect to $\theta$ yields the planes of maximum in-plane shear, oriented at $45^\circ$ to the principal planes: $\tan 2\theta_s = -\frac{\sigma_x - \sigma_y}{2 \tau_{xy}} = -\frac{1}{\tan 2\theta_p}$ $\boxed{\tau_{\max, \text{in-plane}} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \frac{\sigma_1 - \sigma_2}{2}}$ On these maximum shear planes, a uniform normal stress acts: $\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2}$.

Absolute Maximum Shear Stress (3D Analysis)

In plane stress, the third principal stress perpendicular to the plane is zero: $\sigma_3 = 0$.

• Opposite Signs ($\sigma_1 > 0$ and $\sigma_2 < 0$): $\tau_{\max, \text{abs}} = \frac{\sigma_1 - \sigma_2}{2} = \tau_{\max, \text{in-plane}}$
• Same Signs (Both Tensile: $\sigma_1 \ge \sigma_2 > 0$): $\boxed{\tau_{\max, \text{abs}} = \frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma_1 - 0}{2} = \frac{\sigma_1}{2}}$
• Same Signs (Both Compressive: $\sigma_2 \le \sigma_1 < 0$): $\tau_{\max, \text{abs}} = \frac{\sigma_3 - \sigma_2}{2} = \frac{0 - \sigma_2}{2} = \frac{|\sigma_2|}{2}$

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8. Mohr's Circle for Plane Stress

Mohr's Circle graphically represents plane stress transformation. Squaring and adding the transformation equations yields the equation of a circle: $(\sigma_{x'} - \sigma_{\text{avg}})^2 + \tau_{x'y'}^2 = R^2$.

Loading diagram...

• Center of Circle: $C = (\sigma_{\text{avg}}, 0) = \left(\frac{\sigma_x + \sigma_y}{2}, 0\right)$.
• Radius of Circle: $R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \tau_{\max, \text{in-plane}}$.
• Angular Relation: A physical rotation of angle $\theta$ on the stress element corresponds to a rotation of $2\theta$ in the same direction on Mohr's circle.

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9. Classical Failure Theories

To predict yielding or fracture under multiaxial stress, engineers apply failure criteria calibrated against uniaxial yield strength ($F_y = \sigma_y$) or ultimate strength ($\sigma_u$):

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10. Worked Example: Combined Stresses, Principal Planes, and Mohr's Circle Analysis

Problem Statement: A solid cylindrical steel shaft of diameter $d = 80\text{ mm}$ is subjected to a simultaneous axial tensile force of $P = 150\text{ kN}$, a bending moment $M = 3.20\text{ kN}\cdot\text{m}$, and a torsional torque $T = 4.00\text{ kN}\cdot\text{m}$.

1. Determine the state of stress $(\sigma_x, \sigma_y, \tau_{xy})$ at the critical point on the top surface of the shaft.
2. Calculate the principal stresses and orientation of the principal planes.
3. Calculate the maximum in-plane shear stress and absolute maximum shear stress.
4. Evaluate safety against yielding under the von Mises criterion if yield strength is $\sigma_y = 300\text{ MPa}$.

Step-by-Step Solution:

1. Cross-Section Properties: $A = \frac{\pi d^2}{4} = \frac{\pi (80)^2}{4} = 5,026.55\text{ mm}^2$ $I = \frac{\pi d^4}{64} = \frac{\pi (80)^4}{64} = 2.0106 \times 10^6\text{ mm}^4$ $J = 2 I = 4.0212 \times 10^6\text{ mm}^4$ $c = r = 40\text{ mm}$

2. Stresses at Top Surface Point:

   • Axial Tensile Stress: $\sigma_{\text{axial}} = \frac{P}{A} = \frac{150 \times 10^3\text{ N}}{5,026.55\text{ mm}^2} = 29.84\text{ MPa}$
   • Bending Tensile Stress (top fiber under positive moment): $\sigma_{\text{bend}} = \frac{M c}{I} = \frac{(3.20 \times 10^6\text{ N}\cdot\text{mm})(40\text{ mm})}{2.0106 \times 10^6\text{ mm}^4} = 63.66\text{ MPa}$
   • Total Normal Stress ($\sigma_x$): $\sigma_x = \sigma_{\text{axial}} + \sigma_{\text{bend}} = 29.84 + 63.66 = \boxed{93.50\text{ MPa}}$
   • Transverse Normal Stress ($\sigma_y$): No transverse normal pressure acts: $\sigma_y = 0$.
   • Torsional Shear Stress ($\tau_{xy}$): $\tau_{xy} = \frac{T c}{J} = \frac{(4.00 \times 10^6\text{ N}\cdot\text{mm})(40\text{ mm})}{4.0212 \times 10^6\text{ mm}^4} = \boxed{39.79\text{ MPa}}$

3. Principal Stresses and Principal Planes: $\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{93.50 + 0}{2} = 46.75\text{ MPa}$ $R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{(46.75)^2 + (39.79)^2} = \sqrt{2185.56 + 1583.24} = \sqrt{3768.80} = 61.39\text{ MPa}$ $\boxed{\sigma_1 = \sigma_{\text{avg}} + R = 46.75 + 61.39 = 108.14\text{ MPa}}$ $\boxed{\sigma_2 = \sigma_{\text{avg}} - R = 46.75 - 61.39 = -14.64\text{ MPa}}$

   • Principal Plane Angle: $\tan 2\theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y} = \frac{2 (39.79)}{93.50} = \frac{79.58}{93.50} = 0.8511$ $2\theta_p = 40.40^\circ \implies \boxed{\theta_p = 20.20^\circ}$

4. Maximum Shear Stresses:

   • Maximum In-Plane Shear Stress: $\tau_{\max, \text{in-plane}} = R = \boxed{61.39\text{ MPa}}$
   • Absolute Maximum Shear Stress: Since $\sigma_1 = +108.14\text{ MPa}$ is tensile and $\sigma_2 = -14.64\text{ MPa}$ is compressive (opposite signs), the 3D principal stress $\sigma_3 = 0$ lies between $\sigma_1$ and $\sigma_2$: $\boxed{\tau_{\max, \text{abs}} = \frac{\sigma_1 - \sigma_2}{2} = \frac{108.14 - (-14.64)}{2} = 61.39\text{ MPa}}$

5. von Mises Failure Criterion Check: $\sigma_v = \sqrt{\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2} = \sqrt{(108.14)^2 - (108.14)(-14.64) + (-14.64)^2}$ $\sigma_v = \sqrt{11,694.3 + 1,583.2 + 214.3} = \sqrt{13,491.8} = 116.15\text{ MPa}$ $\text{Factor of Safety } FS = \frac{\sigma_y}{\sigma_v} = \frac{300\text{ MPa}}{116.15\text{ MPa}} = \boxed{2.58}$ The shaft is fully safe against ductile yielding.

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11. CELE Board Exam Traps & Common Computational Errors