4.3 Beam Bending, Transverse Shear, Torsion & Mohr Circle Construction

4.2 Beam Bending, Torsion & Combined Stresses (Mohr's Circle)

Quick Answer: Beam bending generates normal flexural stresses defined by $\sigma = -My/I$, with maximum fiber stress $\sigma_{\max} = M/S$. Transverse shear produces internal shear stresses $\tau = VQ/(It)$, creating longitudinal shear flow $q = VQ/I$ that governs the sizing of web-to-flange welds in built-up girders. Torsion of circular members produces shear stresses $\tau = Tr/J$, whereas thin-walled closed tubes follow Bredt-Batho theory $\tau = T/(2A_m t)$. Under combined bending, torsion, and shear, 2D Mohr's Circle determines principal normal stresses $\sigma_{1,2} = \sigma_{\text{avg}} \pm R$ and maximum shear stress $\tau_{\max} = R$. For ductile structural metals, multiaxial yield is governed by the von Mises criterion ($\sigma_{vM} \le \sigma_y$), predicting pure shear yield at $\tau_y = 0.577\sigma_y$.

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1. Pure Bending & Euler-Bernoulli Beam Theory

Euler-Bernoulli beam theory (classical beam theory) describes the relationship between applied bending moments, cross-sectional geometry, and internal normal flexural stresses.

          y ^                                          Linear Normal Stress
            |       Top: Compression (σ < 0)           Distribution σ(y) = -My/I
       +----+----+  ------------------------           +---------------+ Compression (-)
       |    |    |             |                       |             /
       |    |    |             v                       |            /
-------+----+----+---------------------- Neutral Axis -+-----------/--- σ = 0
       |    |    |             ^                       |          /
       |    |    |             |                       |         /
       +----+----+  ------------------------           +--------+------> Tension (+)
            |       Bottom: Tension (σ > 0)
            +-----------------------------> x

Kinematic Assumptions

1. Plane Sections Remain Plane: Cross-sections perpendicular to the beam axis before bending remain planar and perpendicular to the deformed neutral axis after bending (neglecting transverse shear warping, valid for slender beams where span-to-depth ratio $L/h > 10$).
2. Homogeneous, Isotropic & Linearly Elastic: The material obeys Hooke's law ($\sigma = E \varepsilon$) and possesses identical elastic moduli in tension and compression.
3. Small Deflections: Displacements are small relative to cross-sectional dimensions, ensuring radius of curvature $\rho$ satisfies $1/\rho \approx d^2 v / dx^2$.

Derivation of the Flexure Formula

Consider an element of length $dx$ bent into an arc with radius of curvature $\rho$ measured from the center of curvature to the neutral surface. The longitudinal normal strain $\varepsilon_x$ at distance $y$ from the neutral axis is:

$$\varepsilon_x = -\frac{y}{\rho} = -\kappa y$$

where $\kappa = 1/\rho$ is the beam curvature. Applying Hooke's law ($\sigma_x = E \varepsilon_x$):

$$\sigma_x = -E \kappa y = -\frac{E y}{\rho}$$

Because no net axial force acts on the cross-section in pure bending ($\int_A \sigma_x dA = 0$):

$$-\frac{E}{\rho} \int_A y \, dA = 0 \implies \int_A y \, dA = 0$$

This proves that the neutral axis must pass through the centroid of the cross-sectional area.

The internal resultant bending moment $M$ resists the applied moment:

$$M = -\int_A y \, \sigma_x \, dA = \frac{E}{\rho} \int_A y^2 \, dA = \frac{E I}{\rho} = E I \kappa$$

where $I = \int_A y^2 dA$ is the second moment of area (area moment of inertia) about the neutral axis. Solving for curvature yields $\kappa = 1/\rho = M / (EI)$. Substituting $\kappa$ back into the stress equation produces the Flexure Formula:

$$\sigma_x(y) = -\frac{M y}{I}$$

where $y$ is positive upward from the neutral axis. Tension occurs where $y$ is negative (below neutral axis for positive bending moment), and compression occurs where $y$ is positive.

Maximum Flexural Stress & Section Modulus ($S$)

The extreme fiber stress occurs at maximum distance $c = y_{\max}$:

$$\sigma_{\max} = \frac{M c}{I} = \frac{M}{S}$$

where $S = I / c$ is the elastic section modulus. For standard cross-sections:

• Solid Rectangular ($b \times h$): $$I = \frac{b h^3}{12}, \qquad c = \frac{h}{2}, \qquad S = \frac{b h^2}{6}$$
• Solid Circular (Diameter $d$): $$I = \frac{\pi d^4}{64}, \qquad c = \frac{d}{2}, \qquad S = \frac{\pi d^3}{32}$$
• Hollow Circular Tube (Outer $D_o$, Inner $D_i$): $$I = \frac{\pi (D_o^4 - D_i^4)}{64}, \qquad S = \frac{\pi (D_o^4 - D_i^4)}{32 D_o}$$
• Parallel Axis Theorem (for Built-Up Welded Sections): $$I = \sum \left( \bar{I}_i + A_i d_i^2 \right)$$ where $\bar{I}_i$ is the local moment of inertia of component $i$, $A_i$ is its area, and $d_i$ is the perpendicular distance from its centroid to the composite neutral axis.

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2. Transverse Shear Stress in Beams & Flange-to-Web Welds

When a beam is subjected to transverse shear loads (non-uniform bending, $V = dM/dx$), transverse and longitudinal shear stresses are generated simultaneously to preserve equilibrium.

         Built-Up Welded I-Girder                  Shear Stress Distribution τ(y)

           |<----- b_f ----->|
       +---+-----------------+---+ t_f             +--+ Flange Shear (Small)
       |       Top Flange        |
       +---+--------+--------+---+                 |
           | /|     |     |\ | <--- Fillet Welds   |  Longitudinal Shear Flow: q = VQ/I
           |/ |     | t_w | \|      q_weld = q/2   | 
           |  +-----+-----+  |                     |  /-------------------------\
           |  |     |     |  |                     | |                           | Web carries
         h |  |     |     |  |                     | |                           | 90-95% of
           |  |     |     |  |                     | |                           | Shear Force
           |  +-----+-----+  |                     |  \-------------------------/  τ_max at N.A.
           |\ |     |     | /|
           | \|     |     |/ | <--- Fillet Welds   |
       +---+--------+--------+---+ t_f             |
       |      Bottom Flange      |                 +--+ Flange Shear
       +---+-----------------+---+

Jourawski's Shear Formula

By evaluating the horizontal force balance on a longitudinal slice of a beam element, Jourawski's Formula establishes transverse shear stress $\tau$ at any depth $y_1$:

$$\tau(y_1) = \frac{V Q(y_1)}{I t}$$

where:

• $V$ = Transverse shear force at the cross-section.
• $I$ = Total area moment of inertia of the entire cross-section about the neutral axis.
• $t$ = Thickness of the cross-section at the plane of the cut.
• $Q(y_1)$ = First moment of area of the cross-sectional portion above (or below) the level $y_1$ about the neutral axis: $$Q(y_1) = \int_{y_1}^c y \, dA = A' \bar{y}'$$ where $A'$ is the area of the section isolated beyond the cut, and $\bar{y}'$ is the distance from the neutral axis to the centroid of $A'$.

Characteristics of Beam Shear Distribution

• Rectangular Sections: Parabolic distribution with $\tau = 0$ at top and bottom surfaces, peaking at the neutral axis: $$\tau_{\max} = \frac{3 V}{2 A} = 1.5 \tau_{\text{avg}}$$
• Welded Wide-Flange (I-Girder) Sections: The flanges carry negligible vertical shear stress because $t = b_f$ is large and $Q$ is small. In the web, $t = t_w$ is thin, causing shear stress to concentrate heavily. The web carries $90\text{--}95\%$ of the total transverse shear force, typically approximated as: $$\tau_{\text{web, avg}} \approx \frac{V}{t_w h_w}$$

Longitudinal Shear Flow ($q$) & Sizing of Girder Welds

In built-up girders fabricated by welding two flange plates to a central web plate, horizontal shear forces tend to slip the flanges along the web. The longitudinal shear force per unit length is defined as shear flow ($q$, in $\text{N/mm}$ or $\text{lb/in}$):

$$q = \tau t = \frac{V Q_f}{I}$$

where $Q_f = A_{\text{flange}} \cdot d_f = (b_f t_f) d_f$, and $d_f$ is the distance from the composite neutral axis to the flange centroid.

Because two symmetrical fillet welds connect the flange to the web:

$$q_{\text{weld}} = \frac{q}{2} = \frac{V Q_f}{2 I}$$

Under AWS D1.1, the required effective weld throat $t_e$ for an allowable shear stress $\tau_{\text{allow}}$ (e.g., $0.30 \times \text{tensile strength}$ for matching filler metal) is:

$$t_e = \frac{q_{\text{weld}}}{\tau_{\text{allow}}} = \frac{V Q_f}{2 I \tau_{\text{allow}}}$$

For an equal-leg fillet weld, leg size is $w = t_e / 0.7071$.

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3. Torsion of Circular Shafts & Thin-Walled Tubes

Torsional moments ($T$) twist a member about its longitudinal axis, creating pure shear stresses.

Circular Solid and Tubular Shafts

In axisymmetric circular sections, cross-sections rotate as rigid planes without out-of-plane warping. Shear strain $\gamma$ varies linearly with radial distance $r$ from the centerline, reaching maximum at the outer surface ($r = c = D_o/2$):

$$\tau(r) = \frac{T r}{J}, \qquad \tau_{\max} = \frac{T c}{J} = \frac{T}{Z_p}$$

where:

• $J$ = Polar moment of inertia ($J = I_x + I_y$). For a solid shaft, $J = \frac{\pi d^4}{32}$. For a tubular shaft, $J = \frac{\pi (D_o^4 - D_i^4)}{32}$.
• $Z_p = J / c$ = Polar section modulus.
• The angle of twist $\phi$ over length $L$ is: $$\phi = \frac{T L}{G J}$$

Thin-Walled Closed Tubes (Bredt-Batho Theory)

For non-circular or circular thin-walled closed tubes with wall thickness $t$:

$$q = \tau t = \frac{T}{2 A_m} \implies \tau = \frac{T}{2 A_m t}$$

where $A_m$ is the area enclosed by the median line of the tube wall profile. The torsional constant is:

$$J = \frac{4 A_m^2}{\oint \frac{ds}{t}}$$

For constant thickness $t$, $J = \frac{4 A_m^2 t}{S_m}$, where $S_m$ is the perimeter of the median line.

Open vs. Closed Sections in Welded Construction

Critical Structural Principle: Closed hollow structural sections (HSS) and welded box girders possess immense torsional resistance because continuous shear flow $q$ circulates around the closed perimeter. If an open structural shape is used (e.g., a C-channel, I-beam, or a slit pipe), continuous shear circulation is impossible. Shear stress is governed by open thin-walled torsion:

$$J_{\text{open}} = \frac{1}{3} \sum b_i t_i^3$$

Because $J_{\text{open}} \propto t^3$ while $J_{\text{closed}} \propto A_m^2 t$, a closed welded box girder typically exhibits $100\text{ to }1000\text{ times}$ greater torsional stiffness than an open I-beam of identical steel mass. Open sections subjected to twisting warp severely and suffer premature lateral-torsional buckling.

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4. Combined Stresses & Two-Dimensional Mohr's Circle

Real weldments rarely experience pure uniaxial tension or pure torsion alone. They endure combined axial tension, bending, and shear loads. Evaluating structural integrity requires transforming these combined stresses into principal stresses.

                                  τ (Shear Stress)
                                         ^
                                         |             Point X (σ_x, -τ_xy)
                                         |                 *
                                         |                /|
                                         |               / |
                                         |           R  /  |
                                         |             /   | τ_xy
                        σ_2              | C(σ_avg, 0)/    |
                         *---------------+-----------*-----+------*---------> σ (Normal)
                                         |            \    |     σ_1
                                         |             \   |
                                         |              \  |
                                         |               \ |
                                         |                \|
                                         |                 *
                                         |             Point Y (σ_y, +τ_xy)

Plane Stress Transformation Equations

Given in-plane normal stresses $\sigma_x, \sigma_y$ and shear stress $\tau_{xy}$ acting on an infinitesimal material element, the stresses on an inclined plane rotated by angle $\theta$ counterclockwise are:

$$\sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \left(\frac{\sigma_x - \sigma_y}{2}\right) \cos(2\theta) + \tau_{xy} \sin(2\theta)$$ $$\tau_{x'y'} = -\left(\frac{\sigma_x - \sigma_y}{2}\right) \sin(2\theta) + \tau_{xy} \cos(2\theta)$$

Graphical Construction of Mohr's Circle

Eliminating parameter $2\theta$ from the transformation equations reveals the equation of a circle: $(\sigma_{x'} - \sigma_{\text{avg}})^2 + \tau_{x'y'}^2 = R^2$.

1. Center Coordinates ($C$): $$C = (\sigma_{\text{avg}}, 0) = \left( \frac{\sigma_x + \sigma_y}{2}, 0 \right)$$
2. Circle Radius ($R$): $$R = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }$$
3. Principal Normal Stresses ($\sigma_1, \sigma_2$): The intersections of the circle with the horizontal normal stress axis (where shear stress $\tau = 0$): $$\sigma_1 = \sigma_{\text{avg}} + R = \frac{\sigma_x + \sigma_y}{2} + \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }$$ $$\sigma_2 = \sigma_{\text{avg}} - R = \frac{\sigma_x + \sigma_y}{2} - \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }$$
4. Principal Planes Orientation ($\theta_p$): $$\tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}$$
5. Maximum In-Plane Shear Stress ($\tau_{\max}$): The apex of the circle: $$\tau_{\max} = R = \frac{\sigma_1 - \sigma_2}{2}$$ The planes of maximum shear stress are oriented at $45^\circ$ to the principal planes.

Three-Dimensional Mohr's Circle & Absolute Maximum Shear Stress

In plane stress ($\sigma_z = 0$), the third principal stress is $\sigma_3 = 0$. Plotting all three principal stresses $(\sigma_1, \sigma_2, \sigma_3)$ yields three Mohr's circles. The absolute maximum shear stress is:

$$\tau_{\text{abs, max}} = \frac{\sigma_{\max} - \sigma_{\min}}{2}$$

• If $\sigma_1 > 0$ and $\sigma_2 < 0$: $\sigma_{\max} = \sigma_1$ and $\sigma_{\min} = \sigma_2$, so $\tau_{\text{abs, max}} = \frac{\sigma_1 - \sigma_2}{2}$.
• If $\sigma_1 > 0$ and $\sigma_2 > 0$: $\sigma_{\max} = \sigma_1$ and $\sigma_{\min} = \sigma_3 = 0$, so $\tau_{\text{abs, max}} = \frac{\sigma_1 - 0}{2} = \frac{\sigma_1}{2}$.

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