8.3 Shear Force & Bending Moment Diagrams, Beam Bending & Shear Stresses

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

1. Shear Force ($V$): Positive (+ve) when left portion shears upward relative to right portion (or clockwise shearing couple).
2. 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

1. 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).
2. 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.
3. 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 $\kappa = \frac{d^2y}{dx^2} = \frac{M}{EI} = 0$.
4. 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.
5. 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

3. Pure Bending Mechanics & Euler-Bernoulli Beam Theory

Assumptions of Simple Bending Theory

1. 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: $\epsilon(y) = -y/R$).
2. Material is homogeneous, isotropic, and obeys Hooke's Law linearly in both tension and compression ($E_{\text{tension}} = E_{\text{compression}}$).
3. Beam is initially straight with a uniform constant cross-section throughout its length.
4. Radius of curvature $R$ of the neutral surface is large compared to cross-sectional dimensions.
5. 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:

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 ($y = 0$): $$\mathbf{\tau_{\max} = \frac{3}{2} \frac{V}{bh} = 1.5 \, \tau_{\text{avg}}}$$

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 ($y = 0$): $$\mathbf{\tau_{\max} = \frac{4}{3} \frac{V}{\pi R^2} = \frac{4}{3} \, \tau_{\text{avg}} \approx 1.333 \, \tau_{\text{avg}}}$$

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: $$\mathbf{\tau_{\text{NA}} = \frac{4}{3} \tau_{\text{avg}} = \frac{8V}{3bh}}$$
• Maximum Shear Stress ($\tau_{\max}$): Occurs at mid-height ($y = h/2$) from apex/base, NOT at the neutral axis! $$\mathbf{\tau_{\max} = \frac{3}{2} \tau_{\text{avg}} = 1.5 \, \tau_{\text{avg}} = \frac{3V}{bh}}$$

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: $$\mathbf{\tau_{\max} = \frac{9}{8} \tau_{\text{avg}} = 1.125 \, \tau_{\text{avg}}}$$

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:

1. Support reactions $R_A$ and $R_C$
2. Maximum sagging bending moment
3. Maximum hogging bending moment
4. Exact location of the point of contraflexure

Solution:

• Total downward load = $10 \times 6 = 60\text{ kN}$.
• Taking moments about $A$ ($\sum M_A = 0$): $$R_C \times 4 = (10 \times 6) \times 3 = 180 \implies \mathbf{R_C = 45\text{ kN}}$$
• Vertical equilibrium ($\sum F_y = 0$): $$R_A + R_C = 60 \implies R_A = 60 - 45 = \mathbf{15\text{ kN}}$$
• 1. Section in Span $AC$ ($0 \le x \le 4\text{ m}$): $$V(x) = R_A - w x = 15 - 10x$$ $$M(x) = R_A x - \frac{wx^2}{2} = 15x - 5x^2$$
  • Shear force is zero at $15 - 10x = 0 \implies x = 1.5\text{ m}$ from $A$.
  • Max Sagging Moment: $$M_{\max} = 15(1.5) - 5(1.5)^2 = 22.5 - 11.25 = \mathbf{+11.25\text{ kN}\cdot\text{m}}$$
• 2. Moment at Support $C$ ($x = 4\text{ m}$): $$M_C = 15(4) - 5(4)^2 = 60 - 80 = \mathbf{-20.0\text{ kN}\cdot\text{m}} \quad (\text{Max Hogging})$$
• 3. Point of Contraflexure: Setting $M(x) = 0$ in span $AC$: $$15x - 5x^2 = 0 \implies 5x(3 - x) = 0$$ 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: $$\sigma_{\max} = \frac{M_{\max}}{Z} = \frac{20 \times 10^6}{6.667 \times 10^5} = \mathbf{30.0\text{ MPa}}$$
• Maximum Transverse Shear Stress: $$\tau_{\max} = 1.5 \frac{V_{\max}}{b h} = 1.5 \times \frac{10,000}{100 \times 200} = 1.5 \times 0.50 = \mathbf{0.75\text{ MPa}}$$