10.6 Beam Deflections and Combined Axial/Bending Loading

10.6 Beam Deflections and Combined Axial/Bending Loading

Core Engineering Principle: Structural components in practice are rarely subjected to isolated single loads. Engineers must evaluate deflections using elastic differential equations and analyze combined normal and shear stress states produced by concurrent axial, flexural, torsional, and pressure loadings.

Elastic Beam Deflection: Double Integration Method

For a prismatic elastic beam with flexural rigidity $E I$ subjected to transverse loads, the relationship between internal bending moment $M(x)$ and transverse deflection $v(x)$ is governed by the second-order linear differential equation:

$E I \frac{d^2 v}{dx^2} = M(x)$

Integrating once yields the equation for beam slope $\theta(x) = \frac{dv}{dx}$:

$E I \frac{dv}{dx} = \int M(x) dx + C_1$

Integrating a second time yields the transverse deflection profile $v(x)$:

$E I v(x) = \iint M(x) dx^2 + C_1 x + C_2$

The integration constants $C_1$ and $C_2$ are evaluated using physical Boundary Conditions at support locations.

Superposition Method for Beam Deflections

Because the elastic governing differential equation is linear, the total deflection or slope caused by multiple transverse loads equals the algebraic sum of the deflections or slopes caused by each individual load acting separately.

Standard FE Exam Beam Deflection Formulas

Below are standard beam deflection equations compiled from the NCEES FE Reference Handbook:

Exam Tip: Ensure consistency of units when using beam formulas. If $E$ is in Pa ($\text{N/m}^2$), $I$ must be in $\text{m}^4$, $L$ in $\text{m}$, and $P$ in $\text{N}$, yielding deflection $v$ in meters.

Combined Axial and Bending Loading & Eccentricity

When a structural member is subjected simultaneously to an axial load $P$ and bending moments $M_y$ and $M_z$, normal stresses superimpose linearly:

$\sigma_{total} = \pm \frac{P}{A} \pm \frac{M_z y}{I_z} \pm \frac{M_y z}{I_y}$

Eccentric Axial Loading

An axial force $P$ applied at an eccentricity distance $e$ from the centroidal axis of a column cross-section is equivalent to a concentric axial force $P$ plus a couple moment $M = P \cdot e$.

The combined normal stress profile across the section depth is:

$\sigma = - \frac{P}{A} \pm \frac{(P e) c}{I} = - \frac{P}{A} \left( 1 \pm \frac{e c}{k^2} \right)$

where $k = \sqrt{I/A}$ is the Radius of Gyration of the cross-section.

Kern of a Cross-Section (Middle Third Rule)

For brittle or unreinforced materials (such as masonry, concrete, or soil foundations) that cannot resist tension, the eccentric load $P$ must remain within a central region called the Kern so that no tensile normal stresses develop anywhere on the cross-section.

For a solid rectangular cross-section of width $b$ and depth $h$, the load eccentricity must satisfy:

$e \le \frac{h}{6}$

This requirement is known as the Middle Third Rule (the resultant load must fall within the middle third of the cross-sectional depth).

Thin-Walled Pressure Vessels

A vessel is classified as thin-walled if the ratio of inner radius $r$ to wall thickness $t$ satisfies $\frac{r}{t} \ge 10$. Under internal gauge pressure $p$, radial stress through the wall thickness is negligible, creating a state of plane stress in the vessel wall.

Cylindrical Pressure Vessels

• Hoop (Circumferential) Stress ($\sigma_h$): Acts tangentially along the circumference: $\sigma_h = \frac{p r}{t}$

• Longitudinal (Axial) Stress ($\sigma_a$): Acts parallel to the longitudinal cylinder axis: $\sigma_a = \frac{p r}{2 t}$

Notice that hoop stress is exactly twice the magnitude of longitudinal stress ($\sigma_h = 2 \sigma_a$). Consequently, cylindrical pressure vessels under excess pressure tend to burst along longitudinal seams.

Spherical Pressure Vessels

Due to spherical symmetry, normal stress is identical in all tangential directions:

$\sigma_1 = \sigma_2 = \frac{p r}{2 t}$

| Vessel Geometry | Maximum In-Plane Stress | In-Plane Shear Stress | Absolute Max Shear Stress ($\tau_{abs, max}$) | | :--- | :---: | :---: | | Thin Cylinder | $\sigma_h = \frac{p r}{t}$ | $\tau_{in-plane} = \frac{\sigma_h - \sigma_a}{2} = \frac{p r}{4 t}$ | $\tau_{abs, max} = \frac{\sigma_h - 0}{2} = \frac{p r}{2 t}$ | | Thin Sphere | $\sigma = \frac{p r}{2 t}$ | $\tau_{in-plane} = 0$ | $\tau_{abs, max} = \frac{\sigma - 0}{2} = \frac{p r}{4 t}$ |

Worked Engineering Problems

Problem 1: Combined Eccentric Axial Load on a Column

Scenario: A short rectangular concrete column ($b = 300\text{ mm} = 0.30\text{ m}$, $h = 400\text{ mm} = 0.40\text{ m}$) carries a vertical compressive axial load $P = 480\text{ kN}$ applied at an eccentricity $e = 60\text{ mm} = 0.06\text{ m}$ from the geometric centroid along the principal $h$-axis. Calculate (a) the cross-sectional area $A$, (b) moment of inertia $I_x$, (c) equivalent bending moment $M$, and (d) maximum compressive and tensile normal stresses at the extreme cross-sectional fibers.

Solution:

1. Calculate sectional properties: $A = b h = (0.30\text{ m})(0.40\text{ m}) = 0.12\text{ m}^2$ $I_x = \frac{b h^3}{12} = \frac{(0.30)(0.40)^3}{12} = \frac{(0.30)(0.064)}{12} = 0.0016\text{ m}^4$ $c = \frac{h}{2} = 0.20\text{ m}$ $S_x = \frac{I_x}{c} = \frac{0.0016}{0.20} = 0.008\text{ m}^3$

2. Calculate direct axial stress component (compressive): $\sigma_{axial} = - \frac{P}{A} = - \frac{480,000\text{ N}}{0.12\text{ m}^2} = -4,000,000\text{ Pa} = -4.0\text{ MPa}$

3. Calculate equivalent bending moment $M$ and flexural stress component: $M = P \cdot e = (480,000\text{ N})(0.06\text{ m}) = 28,800\text{ N}\cdot\text{m} = 28.8\text{ kN}\cdot\text{m}$ $\sigma_{bending} = \pm \frac{M}{S_x} = \pm \frac{28,800\text{ N}\cdot\text{m}}{0.008\text{ m}^3} = \pm 3,600,000\text{ Pa} = \pm 3.6\text{ MPa}$

4. Superimpose stresses at extreme fibers:

• Highly Compressed Edge (same side as load eccentricity): $\sigma_{max, comp} = -4.0\text{ MPa} - 3.6\text{ MPa} = -7.6\text{ MPa (compressive)}$
• Opposite Edge: $\sigma_{opp} = -4.0\text{ MPa} + 3.6\text{ MPa} = -0.4\text{ MPa (compressive)}$ Since $\sigma_{opp} = -0.4\text{ MPa} < 0$, the entire cross-section remains under compression, consistent with $e = 60\text{ mm} \le \frac{h}{6} = \frac{400}{6} = 66.67\text{ mm}$.

Problem 2: Cylindrical Pressure Vessel under Combined Pressure and Torque

Scenario: A thin-walled cylindrical steel pressure tank with inner radius $r = 0.50\text{ m}$ and wall thickness $t = 10\text{ mm} = 0.01\text{ m}$ is pressurized to internal gauge pressure $p = 2.0\text{ MPa}$. Simultaneously, a torque $T = 50\text{ kN}\cdot\text{m}$ is applied to the tank. Determine (a) hoop stress $\sigma_h$, (b) longitudinal stress $\sigma_a$, (c) torsional shear stress $\tau_{xy}$, and (d) maximum principal stress $\sigma_1$.

Solution:

1. Calculate pressure stress components: $\sigma_h = \frac{p r}{t} = \frac{(2.0 \times 10^6\text{ Pa})(0.50\text{ m})}{0.01\text{ m}} = 100.0\text{ MPa}$ $\sigma_a = \frac{p r}{2 t} = \frac{100.0}{2} = 50.0\text{ MPa}$

2. Calculate torsional shear stress $\tau_{xy}$ using thin-walled tube formula $\tau = \frac{T}{2 \pi r^2 t}$: $\tau_{xy} = \frac{50,000\text{ N}\cdot\text{m}}{2 \pi (0.50\text{ m})^2 (0.01\text{ m})} = \frac{50,000}{2 \pi (0.25)(0.01)} = \frac{50,000}{0.015708} = 3,183,099\text{ Pa} \approx 3.18\text{ MPa}$

3. Calculate principal stress $\sigma_1$ via 2D transformation with $\sigma_x = \sigma_a = 50$, $\sigma_y = \sigma_h = 100$, $\tau_{xy} = 3.18$: $\sigma_{avg} = \frac{50 + 100}{2} = 75.0\text{ MPa}$ $R = \sqrt{\left(\frac{50 - 100}{2}\right)^2 + (3.18)^2} = \sqrt{(-25)^2 + 10.11} = \sqrt{625 + 10.11} = \sqrt{635.11} = 25.20\text{ MPa}$ $\sigma_1 = \sigma_{avg} + R = 75.0 + 25.20 = 100.20\text{ MPa}$