12.3 Beam Stresses: Flexural and Transverse Shear

12.3 Beam Stresses: Flexural and Transverse Shear

Flexural members (beams and girders) represent fundamental load-carrying elements in buildings, bridges, and offshore platforms. Designing a beam requires analyzing two concurrent internal stress distributions: longitudinal normal stresses generated by bending moments ($\sigma_b$), and transverse/horizontal shearing stresses generated by vertical shear forces ($\tau$).

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1. Internal Forces and Differential Equilibrium in Beams

Consider an infinitesimal beam segment of length $dx$ subjected to a downward distributed load $w(x)$, shear force $V$, and bending moment $M$:

Differential Equilibrium Relationships

1. Vertical Force Equilibrium ($\sum F_y = 0$): $V - (V + dV) - w(x) \, dx = 0 \implies \boxed{\frac{dV}{dx} = -w(x)}$ $\Delta V = V_2 - V_1 = - \int_{x_1}^{x_2} w(x) \, dx = - (\text{Area under distributed load curve})$
2. Moment Equilibrium ($\sum M_O = 0$): $-M + (M + dM) - V \, dx - w(x) \frac{(dx)^2}{2} = 0 \implies \boxed{\frac{dM}{dx} = V(x)}$ $\Delta M = M_2 - M_1 = \int_{x_1}^{x_2} V(x) \, dx = (\text{Area under shear diagram})$

Critical Curve Properties for Shear and Moment Diagrams

• Slope of Shear Diagram: Equals the negative intensity of the distributed load ($\text{Slope } = -w$).
• Slope of Moment Diagram: Equals the local vertical shear force ($\text{Slope } = V$).
• Peak Bending Moments: Occur where the shear diagram crosses zero ($V = 0$) or passes through a sharp discontinuity.
• Point of Inflection: A point along the beam where $M(x) = 0$, indicating a reversal in curvature (from sagging/positive to hogging/negative bending).

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2. Pure Bending and the Flexure Formula (Navier's Hypothesis)

Euler-Bernoulli Assumptions

1. Navier's Hypothesis: Cross-sections originally plane and perpendicular to the longitudinal axis remain plane and perpendicular to the neutral axis after bending (shear deformations are neglected).
2. Material Linearity: The beam is homogeneous, isotropic, and obeys Hooke's Law ($\sigma = E \epsilon$).
3. Prismatic Beam: The cross-section is uniform along the length with a vertical longitudinal plane of symmetry containing the loads.

Longitudinal Strain and the Neutral Axis

Due to beam curvature $\kappa = 1/\rho$, longitudinal strain $\epsilon_x$ varies linearly with vertical distance $y$ from the neutral surface: $\epsilon_x = - \frac{y}{\rho} = - \kappa y$ At the neutral axis ($y = 0$), longitudinal strain and normal stress are identically zero. Equilibrium of axial forces ($\int_A \sigma_x \, dA = 0$) proves that the neutral axis must pass directly through the centroid of the cross-section.

The Flexure Formula

Equating internal resisting moment to applied bending moment $M$: $M = - \int_A y \sigma_x \, dA = \frac{E}{\rho} \int_A y^2 \, dA = \frac{E I}{\rho}$ $\boxed{\sigma_b = - \frac{M y}{I}}$ $\boxed{\sigma_{\max} = \frac{M c}{I} = \frac{M}{S}}$ where:

• $M$ is internal bending moment.
• $y$ is vertical coordinate measured from the neutral axis (positive upward).
• $c$ is distance from neutral axis to the outermost fiber ($c = y_{\max}$).
• $I$ is centroidal second moment of area (moment of inertia) about the bending axis: $I = \int_A y^2 \, dA$.
• $S = \frac{I}{c}$ is the elastic section modulus ($\text{mm}^3$ or $\text{m}^3$).

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3. Section Modulus and Cross-Sectional Geometry

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4. Unsymmetrical (Biaxial) Bending About Principal Axes

When applied bending moments act in a plane inclined relative to the principal centroidal axes $(y, z)$ of the cross-section: $\sigma(y, z) = - \frac{M_z y}{I_z} + \frac{M_y z}{I_y}$ where $M_z$ and $M_y$ are moment vector components along the principal axes. The orientation of the neutral axis (line of zero stress, $\sigma = 0$) forms an angle $\alpha$ with the $z$-axis: $\tan \alpha = \left(\frac{I_z}{I_y}\right) \tan \theta$ where $\theta$ is the inclination angle of the applied moment vector relative to the $z$-axis. Unless $I_y = I_z$ (such as for square or circular sections), the neutral axis is NOT perpendicular to the applied moment plane.

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5. Horizontal and Transverse Shear Stress (Jourawski's Formula)

In a beam subjected to transverse loading, the variation in bending moment along $x$ produces differing normal compressive and tensile forces on vertical sections. To maintain equilibrium of an elemental slice, horizontal shearing stresses must develop along longitudinal planes. By the complementary property of shear, horizontal shear stress $\tau_{xy}$ is identically equal to vertical transverse shear stress $\tau_{yx}$.

Jourawski's Shear Formula

$\boxed{\tau = \frac{V Q}{I b}}$ where:

• $V$ is vertical shear force at the cross-section.
• $I$ is moment of inertia of the entire cross-section about the neutral axis.
• $b$ is width of the cross-section at the specific horizontal plane where shear stress is evaluated.
• $Q$ is the first moment of area of that portion of the cross-section located beyond the cut plane (above or below), taken about the centroidal neutral axis: $\boxed{Q = \int_{y_1}^c y \, dA = A' \bar{y}'}$ where $A'$ is the partial area above level $y_1$, and $\bar{y}'$ is the vertical distance from the neutral axis to the centroid of $A'$.

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6. Shear Stress Distributions: Rectangular, Circular, and Flanged Profiles

Loading diagram...

1. Rectangular Cross-Section ($b \times h$)

At distance $y$ from the neutral axis, the area above $y$ is $A' = b (h/2 - y)$ with centroid at $\bar{y}' = y + \frac{1}{2}(h/2 - y) = \frac{1}{2}(h/2 + y)$. Thus: $Q(y) = A' \bar{y}' = \frac{b}{2} \left(\frac{h^2}{4} - y^2\right)$ Substituting into $\tau = \frac{V Q}{I b}$ with $I = \frac{b h^3}{12}$: $\tau(y) = \frac{V [\frac{b}{2}(h^2/4 - y^2)]}{(\frac{b h^3}{12}) b} = \frac{6 V}{b h^3} \left(\frac{h^2}{4} - y^2\right) = \frac{3 V}{2 A} \left[1 - \left(\frac{2y}{h}\right)^2\right]$

• Parabolic distribution with $\tau = 0$ at top and bottom surfaces ($y = \pm h/2$).
• Maximum shear stress occurs at the neutral axis ($y = 0$): $\boxed{\tau_{\max} = \frac{3 V}{2 A} = 1.50 \frac{V}{A}}$

2. Solid Circular Cross-Section ($d$)

Integration across a circular boundary yields a parabolic variation peaking at the neutral axis: $\boxed{\tau_{\max} = \frac{4 V}{3 A} = 1.333 \frac{V}{\pi R^2}}$

3. Structural Wide-Flange / I-Beams

• Flanges: Because flange width $b_f$ is large, horizontal shear stress $\tau = \frac{V Q}{I b_f}$ is very small.
• Web: At the flange-web junction, width drops abruptly from $b_f$ to $t_w$, causing a sudden discontinuity jump in shear stress.
• Web Shear Contribution: The web carries $90\%\text{ to }98\%$ of the total vertical shear force. In structural steel practice, average web shear stress is commonly approximated as: $\tau_{\text{web, avg}} \approx \frac{V}{A_{\text{web}}} = \frac{V}{d \cdot t_w}$

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7. Shear Flow and Built-Up Beam Fastener Spacing

When a beam is fabricated by fastening multiple plates or planks together (e.g., wooden box beams, built-up steel girders with cover plates), the fasteners must resist the longitudinal shear force attempting to slide the layers past one another.

Shear Flow ($q$)

Shear flow is the shear force per unit length along the longitudinal axis of the beam: $\boxed{q = \tau b = \frac{V Q}{I}} \quad (\text{N/mm or kN/m})$

Fastener Pitch Spacing ($s$)

Let $R_v$ be the total shear capacity of all fasteners positioned at a single longitudinal cross-section (for $m$ bolts or shear planes per station, $R_v = m R_{\text{bolt}}$). Over a longitudinal pitch spacing $s$, the total resisting capacity is $R_v = q \cdot s$: $\boxed{s = \frac{R_v}{q} = \frac{R_v I}{V Q}}$

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8. Worked Example: Built-Up T-Beam Flexural Capacity and Nail Pitch

Problem Statement: A built-up timber T-beam is constructed by nailing a horizontal flange plank ($200\text{ mm}$ wide $\times 50\text{ mm}$ thick) to a vertical stem plank ($50\text{ mm}$ wide $\times 200\text{ mm}$ deep). The beam supports a simply supported span of $L = 4.0\text{ m}$ carrying a concentrated load $P = 18\text{ kN}$ at midspan.

1. Locate the centroid and determine the moment of inertia about the neutral axis ($I_{\text{NA}}$).
2. Calculate the maximum flexural tensile and compressive stresses.
3. If the flange and stem are joined by nails having an allowable lateral shear capacity of $R_{\text{nail}} = 800\text{ N}$, determine the maximum longitudinal nail spacing $s$.

Step-by-Step Solution:

1. Centroid Location (measured from bottom of stem):

   • Flange ($1$): $A_1 = 200 \times 50 = 10,000\text{ mm}^2$, $y_1 = 200 + 25 = 225\text{ mm}$.
   • Stem ($2$): $A_2 = 50 \times 200 = 10,000\text{ mm}^2$, $y_2 = 100\text{ mm}$.
   • Total Area: $A = A_1 + A_2 = 20,000\text{ mm}^2$. $\bar{y} = \frac{A_1 y_1 + A_2 y_2}{A} = \frac{(10,000)(225) + (10,000)(100)}{20,000} = \frac{2,250,000 + 1,000,000}{20,000} = 162.5\text{ mm}$
   • Distance to top fiber: $c_{\text{top}} = (200 + 50) - 162.5 = 87.5\text{ mm}$.
   • Distance to bottom fiber: $c_{\text{bottom}} = 162.5\text{ mm}$.

2. Moment of Inertia via Parallel Axis Theorem ($I_{\text{NA}}$):

   • Flange: $I_1 = \frac{b_1 h_1^3}{12} + A_1 d_1^2 = \frac{200 (50)^3}{12} + 10,000 (225 - 162.5)^2 = 2.083 \times 10^6 + 10,000 (62.5)^2$ $I_1 = 2.083 \times 10^6 + 39.063 \times 10^6 = 41.146 \times 10^6\text{ mm}^4$
   • Stem: $I_2 = \frac{b_2 h_2^3}{12} + A_2 d_2^2 = \frac{50 (200)^3}{12} + 10,000 (162.5 - 100)^2 = 33.333 \times 10^6 + 10,000 (62.5)^2$ $I_2 = 33.333 \times 10^6 + 39.063 \times 10^6 = 72.396 \times 10^6\text{ mm}^4$
   • Total Moment of Inertia: $\boxed{I_{\text{NA}} = I_1 + I_2 = 113.542 \times 10^6\text{ mm}^4}$

3. Maximum Bending Moment and Maximum Stresses:

   • Simply supported beam with center load $P = 18\text{ kN}$ on span $L = 4.0\text{ m}$: $V_{\max} = \frac{P}{2} = 9.0\text{ kN} = 9,000\text{ N}$ $M_{\max} = \frac{P L}{4} = \frac{(18\text{ kN})(4.0\text{ m})}{4} = 18.0\text{ kN}\cdot\text{m} = 18.0 \times 10^6\text{ N}\cdot\text{mm}$
   • Maximum compressive stress (at top fiber, $y = +c_{\text{top}} = +87.5\text{ mm}$): $\sigma_{\text{comp}} = \frac{M c_{\text{top}}}{I} = \frac{(18.0 \times 10^6)(87.5)}{113.542 \times 10^6} = \boxed{13.87\text{ MPa}}$
   • Maximum tensile stress (at bottom fiber, $y = -c_{\text{bottom}} = -162.5\text{ mm}$): $\sigma_{\text{tens}} = \frac{M c_{\text{bottom}}}{I} = \frac{(18.0 \times 10^6)(162.5)}{113.542 \times 10^6} = \boxed{25.76\text{ MPa}}$

4. Nail Spacing ($s$):

   • To prevent the flange from sliding off the stem, evaluate $Q$ of the flange alone: $Q_{\text{flange}} = A_1 d_1 = (10,000\text{ mm}^2)(62.5\text{ mm}) = 625,000\text{ mm}^3$
   • Shear flow at the flange-stem interface under maximum shear $V = 9,000\text{ N}$: $q = \frac{V Q}{I} = \frac{(9,000\text{ N})(625,000\text{ mm}^3)}{113.542 \times 10^6\text{ mm}^4} = \frac{5.625 \times 10^9}{113.542 \times 10^6} = 49.54\text{ N/mm}$
   • Required nail pitch spacing (single line of nails, $R_v = 800\text{ N}$): $\boxed{s = \frac{R_{\text{nail}}}{q} = \frac{800\text{ N}}{49.54\text{ N/mm}} = 16.15\text{ mm} \approx 16\text{ mm}}$

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9. CELE Board Exam Traps & Common Computational Errors