8.3 Shear Force & Bending Moment Diagrams, Beam Bending & Shear Stresses
Key Takeaways
- The differential relations $\frac{dV}{dx} = -w(x)$ and $\frac{dM}{dx} = V(x)$ establish that bending moment attains an extremum (maximum or minimum) precisely where the shear force is zero or transitions through zero.
- A point of contraflexure occurs where the bending moment vanishes ($M = 0$) and changes sign, representing a physical reversal in beam curvature between sagging and hogging.
- Euler-Bernoulli pure bending theory ($\,\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}\, $) assumes plane sections remain plane, yielding a linear normal stress distribution across depth and maximum stress $\sigma_{\max} = \frac{M}{Z}$.
- Section modulus $Z$ defines flexural capacity; for solid rectangular sections $Z = \frac{bh^2}{6}$, and for solid circular sections $Z = \frac{\pi d^3}{32}$.
- Transverse shear stress follows Jourawski's formula $\tau = \frac{VQ}{Ib}$, producing parabolic distributions where $\tau_{\max} = 1.5\tau_{\text{avg}}$ for rectangular beams, $\tau_{\max} = \frac{4}{3}\tau_{\text{avg}}$ for circular beams, and $\tau_{\max} = 1.5\tau_{\text{avg}}$ at mid-height ($h/2$) for triangular sections.
Last updated: August 2026
1. Differential Equilibrium Relations & Sign Conventions
In structural analysis of mining conveyors, gantry girders, and haul truck chassis beams, determining internal shear force $V(x)$ and bending moment $M(x)$ profiles is essential for safe flexural and shear sizing.
Standard Sign Conventions
- Shear Force ($V$): Positive (+ve) when left portion shears upward relative to right portion (or clockwise shearing couple).
- Bending Moment ($M$): Positive (+ve) for Sagging (concave upward, producing compression in top fibers and tension in bottom fibers). Negative (-ve) for Hogging (convex upward, producing tension in top fibers and compression in bottom fibers).
Positive Shear (+V): Positive Bending / Sagging (+M):
^
| | Compressive Top (---)
+---+ \ /
| | \ /
v +---------------------+
Tensile Bottom (+++)
Differential Equilibrium Formulations
For a beam element subjected to distributed downward load intensity $w(x)$ (force per unit length):
Critical Analytical Deductions
- Slope of SFD: At any point, $\text{Slope of } V(x) = -w(x)$. Under a uniformly distributed load (UDL), SFD is a straight inclined line ($1^{\text{st}}$ degree polynomial).
- Slope of BMD: At any point, $\text{Slope of } M(x) = V(x)$. Where $V(x) = 0$, the slope of BMD is zero ($\frac{dM}{dx} = 0$), locating local maximum or minimum bending moments.
- Point of Contraflexure (Inflection Point): A section where the bending moment equals zero ($M = 0$) and changes sign (transitions between sagging and hogging). At this point, the beam curvature $\kappa = \frac{d^2y}{dx^2} = \frac{M}{EI} = 0$.
- Concentrated Loads: A point load causes an instantaneous vertical step jump in the SFD equal to the load magnitude, creating a sharp cusp/kink in the BMD.
- Concentrated Moments: An applied external couple $M_0$ produces no change in the SFD, but creates an instantaneous vertical step jump in the BMD equal to $M_0$.
2. Standard Beam Loadings, Maximum Moments & Contraflexure
| Support & Loading Configuration | SFD Shape | Maximum Shear Force $V_{\max}$ | BMD Shape | Maximum Bending Moment $M_{\max}$ |
|---|---|---|---|---|
| Cantilever: Point Load $W$ at free end | Uniform Rectangular | $W$ | Linear Triangle | $-W L$ (at fixed base) |
| Cantilever: Full UDL $w$ | Linear Triangle | $w L$ | Parabola ($2^{\text{nd}}$ deg) | $-\frac{w L^2}{2}$ (at fixed base) |
| Cantilever: UVL ($0$ at tip to $w$ at base) | Parabola ($2^{\text{nd}}$ deg) | $\frac{w L}{2}$ | Cubic ($3^{\text{rd}}$ deg) | $-\frac{w L^2}{6}$ (at fixed base) |
| Simply Supported: Central Point Load $W$ | Two Rectangles ($\pm W/2$) | $\frac{W}{2}$ | Symmetric Triangle | $+\frac{W L}{4}$ (at midspan) |
| Simply Supported: Full UDL $w$ | Linear Triangle ($\pm wL/2$) | $\frac{w L}{2}$ | Symmetric Parabola | $+\frac{w L^2}{8}$ (at midspan) |
| Simply Supported: Symmetric UVL ($w$ at center) | Parabolic | $\frac{w L}{4}$ | Cubic | $+\frac{w L^2}{12}$ (at midspan) |
| Simply Supported: End Couple $M_0$ | Uniform Rectangular ($-M_0/L$) | $\frac{M_0}{L}$ | Linear Triangle | $M_0$ (at applied end) |
3. Pure Bending Mechanics & Euler-Bernoulli Beam Theory
Assumptions of Simple Bending Theory
- Bernoulli's Hypothesis: Plane transverse sections before bending remain plane and perpendicular to the neutral axis after bending (longitudinal strain varies linearly with distance $y$ from the neutral axis: $\epsilon(y) = -y/R$).
- Material is homogeneous, isotropic, and obeys Hooke's Law linearly in both tension and compression ($E_{\text{tension}} = E_{\text{compression}}$).
- Beam is initially straight with a uniform constant cross-section throughout its length.
- Radius of curvature $R$ of the neutral surface is large compared to cross-sectional dimensions.
- Transverse shear deformation is neglected (valid for slender beams with span-to-depth ratio $L/h > 10$).
Bending Strain ε(y): Bending Stress σ(y):
-ε_top (Compression) -σ_top (Compression)
<---| <---|
| |
-----------+----------- (NA) -----------+----------- (Neutral Axis, σ=0)
| |
|---> |--->
+ε_bot (Tension) +σ_bot (Tension)
The Governing Bending Equation
- $M$ = Applied bending moment at the cross-section ($\text{N}\cdot\text{mm}$)
- $I$ = Area moment of inertia about the neutral centroidal axis ($\text{mm}^4$)
- $\sigma$ = Normal bending stress at fiber distance $y$ from neutral axis ($\text{MPa}$)
- $y$ = Perpendicular distance from neutral axis to the fiber ($\text{mm}$)
- $E$ = Young's Modulus of the beam material ($\text{MPa}$)
- $R$ = Radius of curvature of the bent neutral axis ($\text{mm}$)
- Flexural Rigidity: The product $EI$ represents the beam's resistance to bending deflection.
Section Modulus ($Z$)
The maximum bending stress occurring at the extreme fibers ($y = y_{\max}$) is given by:
| Cross-Section Geometry | Moment of Inertia $I_{\text{NA}}$ | Extreme Fiber $y_{\max}$ | Section Modulus $Z$ |
|---|---|---|---|
| Solid Rectangle ($b \times h$) | $\frac{b h^3}{12}$ | $\frac{h}{2}$ | $\mathbf{\frac{b h^2}{6}}$ |
| Hollow Rectangle ($B\times H$ outer, $b\times h$ inner) | $\frac{B H^3 - b h^3}{12}$ | $\frac{H}{2}$ | $\mathbf{\frac{B H^3 - b h^3}{6 H}}$ |
| Solid Circle (Diameter $d$) | $\frac{\pi d^4}{64}$ | $\frac{d}{2}$ | $\mathbf{\frac{\pi d^3}{32}}$ |
| Hollow Circle ($D_o, D_i$) | $\frac{\pi(D_o^4 - D_i^4)}{64}$ | $\frac{D_o}{2}$ | $\mathbf{\frac{\pi(D_o^4 - D_i^4)}{32 D_o}}$ |
| Solid Triangle (Base $b$, Height $h$) | $\frac{b h^3}{36}$ | $\frac{2h}{3}$ (Apex fiber) | $\mathbf{\frac{b h^2}{24}}$ (at apex) |
Beams of Uniform Strength
A beam of uniform strength is designed such that the maximum bending stress $\sigma_{\max}$ is constant along its entire span ($Z(x) \propto M(x)$). This is achieved by either:
- Varying beam width with constant depth: $b(x) = \frac{6 M(x)}{\sigma_{\max} h^2}$.
- Varying beam depth with constant width: $h(x) = \sqrt{\frac{6 M(x)}{\sigma_{\max} b}}$ (parabolic profile).
4. Transverse Shear Stress in Beams (Jourawski's Formula)
When a beam is subjected to transverse loading causing shear force $V$, transverse shear stresses $\tau$ develop across the cross-section in addition to bending stresses.
Jourawski Transverse Shear Formula
- $V$ = Total vertical shear force acting at the section ($\text{N}$)
- $Q(y) = A' \bar{y}$ = First moment of area of the portion above distance $y$ about the neutral axis ($\text{mm}^3$)
- $I$ = Second moment of area of the entire cross-section about the neutral axis ($\text{mm}^4$)
- $b(y)$ = Width of the cross-section at height $y$ where $\tau$ is evaluated ($\text{mm}$)
5. Shear Stress Distribution Across Standard Cross-Sections
Rectangular Section: Circular Section: Triangular Section:
+-------+ --- /\ (Apex, τ=0)
| | τ=0 / \ τ=0 / \
-----+-------+----- (NA) ----+-----+---- (NA) /----+---\ (h/2, τ_max = 1.5 τ_avg)
| (Parabolic) \ / /------+-----\ (NA: h/3, τ = 1.33 τ_avg)
+-------+ τ=0 --- τ=0 +-------------+ (Base, τ=0)
τ_max = 1.5 τ_avg τ_max = 1.33 τ_avg
1. Solid Rectangular Cross-Section ($b \times h$)
At distance $y$ from the neutral axis:
- Distribution: Parabolic across depth; $\tau = 0$ at outer top and bottom fibers ($y = \pm h/2$).
- Maximum Shear Stress: Occurs at the neutral axis ($y = 0$):
2. Solid Circular Cross-Section (Diameter $d$, Radius $R$)
- Distribution: Parabolic across depth; $\tau = 0$ at outer fibers ($y = \pm R$).
- Maximum Shear Stress: Occurs at the neutral axis ($y = 0$):
3. Triangular Cross-Section (Base $b$, Height $h$)
- Neutral axis is located at distance $h/3$ from the base ($2h/3$ from apex).
- Shear stress at the Neutral Axis:
- Maximum Shear Stress ($\tau_{\max}$): Occurs at mid-height ($y = h/2$) from apex/base, NOT at the neutral axis!
4. Diamond / Rhombus Cross-Section (Diagonals $b, h$)
- At Neutral Axis: $\tau_{\text{NA}} = \tau_{\text{avg}}$.
- Maximum Shear Stress occurs at distance $y = \pm 3h/8$ from the apex:
5. I-Beam Cross-Section
- Flanges: Carry $>85 - 90%$ of the total bending moment; shear stress is very small due to large flange width $B$.
- Web: Carries $>90 - 95%$ of the vertical shear force $V$.
- Shear Stress Step Jump: At the flange-web junction ($y = h_w/2$), width drops abruptly from $B_{\text{flange}}$ to $b_{\text{web}}$, causing a sudden surge in shear stress by the exact ratio $\frac{B_{\text{flange}}}{b_{\text{web}}}$.
6. Worked Step-by-Step Engineering Calculations
Example 1: Overhanging Beam Analysis & Point of Contraflexure
Problem: A beam $AB$ of length $L = 6\text{ m}$ is simply supported at $A$ and at $C$ ($4\text{ m}$ from $A$), with an overhang $CB = 2\text{ m}$. It carries a UDL of $w = 10\text{ kN/m}$ over its entire length $AB$. Find:
- Support reactions $R_A$ and $R_C$
- Maximum sagging bending moment
- Maximum hogging bending moment
- Exact location of the point of contraflexure
Solution:
- Total downward load = $10 \times 6 = 60\text{ kN}$.
- Taking moments about $A$ ($\sum M_A = 0$):
- Vertical equilibrium ($\sum F_y = 0$):
- 1. Section in Span $AC$ ($0 \le x \le 4\text{ m}$):
- Shear force is zero at $15 - 10x = 0 \implies x = 1.5\text{ m}$ from $A$.
- Max Sagging Moment:
- 2. Moment at Support $C$ ($x = 4\text{ m}$):
- 3. Point of Contraflexure: Setting $M(x) = 0$ in span $AC$: Excluding $x = 0$, the point of contraflexure is located at $\mathbf{x = 3.0\text{ m}}$ from support $A$ (sagging transitions to hogging).
Example 2: Transverse Shear vs Bending Stress in Timber Beam
Problem: A simply supported rectangular beam ($b = 100\text{ mm}, h = 200\text{ mm}$) of span $L = 4\text{ m}$ carries a central point load $W = 20\text{ kN}$. Calculate the maximum bending stress $\sigma_{\max}$ and maximum transverse shear stress $\tau_{\max}$.
Solution:
- Reactions: $R_A = R_B = W/2 = 10\text{ kN}$.
- Maximum shear force: $V_{\max} = 10\text{ kN} = 10,000\text{ N}$.
- Maximum bending moment: $M_{\max} = \frac{WL}{4} = \frac{20 \times 10^3 \times 4}{4} = 20,000\text{ N}\cdot\text{m} = 20 \times 10^6\text{ N}\cdot\text{mm}$.
- Section Modulus: $Z = \frac{b h^2}{6} = \frac{100 \times (200)^2}{6} = \frac{4,000,000}{6} = 6.667 \times 10^5\text{ mm}^3$.
- Maximum Bending Stress:
- Maximum Transverse Shear Stress:
Test Your Knowledge
For a beam of triangular cross-section (base b, height h) subjected to a vertical shear force V, where does the maximum transverse shear stress occur and what is its magnitude?
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Test Your Knowledge
A simply supported beam of length L carries a uniformly distributed load w (N/m) over its entire span. At what point does the bending moment attain its maximum value, and what is its magnitude?
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B
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Test Your Knowledge
Two beams of equal cross-sectional area are subjected to bending: Beam A has a solid square cross-section (a × a), while Beam B has a solid circular cross-section (diameter d). What is the ratio of their section moduli Z_square / Z_circle?
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