3.2 Direct Shear, Bearing & Transverse Shear in Beams

3.2 Direct Shear, Bearing & Transverse Shear in Beams

Shear stresses develop whenever internal forces tend to make adjacent material planes slide relative to one another. Machine designers must separate direct (average) shear in pins, rivets, keys and bolts from transverse shear that accompanies bending in beams. (Torsional shear from twisting moments is treated separately in Section 3.3.) Both appear on nearly every connection and beam problem, usually alongside a bearing-stress check.

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1. Direct (Average) Shear in Fasteners & Joint Design

Direct shear occurs in bolted joints, rivets, cotter pins, and drive keys where parallel opposing forces act across a well-defined shear plane. The average direct shear stress is:

$$\tau_{\text{avg}} = \frac{V}{A}$$

where $V$ is the total internal shear force acting parallel to cross-sectional area $A$.

  Single Shear Joint:                    Double Shear Joint (Clevis & Pin):
        P                                       P/2
  [===========]                          [==============]  
        | Shear Plane                           | Shear Plane 1
        V                                P <--- [==============]
  [===========]                                 | Shear Plane 2
        P                                       P/2
                                         [==============]  

Single vs. Double Shear Configurations

• Single Shear: A single shear interface resists the external tensile load $P$. The internal shear force is $V = P$. For a cylindrical pin or bolt of diameter $d$: $\tau = \frac{P}{A} = \frac{4 P}{\pi d^2}$
• Double Shear: Two shear interfaces resist the load (such as a clevis-and-tongue connection). The applied load divides equally between both planes ($V = P/2$): $\tau = \frac{P / 2}{A} = \frac{P}{2 A} = \frac{2 P}{\pi d^2}$ Impact: Double shear reduces the shear stress on the pin by 50% relative to single shear for the same load and pin diameter, or permits a $29.3\%$ reduction in pin diameter for the same allowable shear stress.

Bearing Stress in Pinned Joints

Simultaneously, the contact interface between the cylindrical fastener and the plate hole wall experiences compressive bearing stress $\sigma_b$, evaluated over the projected rectangular area $A_b = d \cdot t$:

$$\sigma_b = \frac{P}{d \cdot t}$$

where $t$ is the plate thickness and $d$ is the nominal pin diameter.

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2. Transverse Shear Stress in Beams

When a beam is subjected to transverse loading, internal bending moments $M(x)$ and internal shear forces $V(x)$ coexist. The variation of bending stress along the span ($dM/dx = V$) creates horizontal interlaminar shear stresses. By complementary shear equilibrium ($\tau_{xy} = \tau_{yx}$), this horizontal shear stress is identical in magnitude to the vertical transverse shear stress acting across the beam cross-section.

The Transverse Shear Formula

$$\tau = \frac{V Q}{I b}$$

where:

• $V$ = internal transverse shear force at the cut section.
• $I$ = area moment of inertia of the entire cross-section about the neutral axis.
• $b$ = width of the cross-section at the specific depth $y$ where shear stress is evaluated.
• $Q$ = first moment of area of the portion of the cross-section $A'$ located above (or below) the level at which $\tau$ is evaluated, relative to the neutral axis:

$$Q = \int_{A'} y \, dA = \bar{y}' A'$$

Here, $\bar{y}'$ represents the distance from the cross-section's neutral axis to the centroid of the sub-area $A'$.

        +-----------------+  --- y = c (extreme fiber: Q = 0, tau = 0)
        |    Area A'      |   |
        +-----------------+  --- y1 (cut level: width = b, Q = y_bar' * A')
        |                 |   |
   - - -|- - - - - - - - -|- -|- - Neutral Axis (y = 0: Q_max, tau_max)
        |                 |
        +-----------------+

Shear Stress Distributions for Standard Sections

A. Solid Rectangular Cross-Section (Width $b$, Depth $h$)

For a rectangular section, $I = \frac{b h^3}{12}$. Evaluating $Q(y)$ at a distance $y$ from the neutral axis:

$A' = b \left(\frac{h}{2} - y\right), \quad \bar{y}' = y + \frac{1}{2}\left(\frac{h}{2} - y\right) = \frac{1}{2}\left(\frac{h}{2} + y\right)$

$Q(y) = \bar{y}' A' = \frac{b}{2} \left( \frac{h^2}{4} - y^2 \right)$

Substituting into the transverse shear formula:

$\tau(y) = \frac{V \left[ \frac{b}{2}\left(\frac{h^2}{4} - y^2\right) \right]}{\left(\frac{b h^3}{12}\right) b} = \frac{3 V}{2 b h} \left( 1 - \frac{4 y^2}{h^2} \right) = \frac{3 V}{2 A} \left( 1 - \frac{4 y^2}{h^2} \right)$

• At the outer free surfaces ($y = \pm h/2$): $\tau = 0$.
• At the neutral axis ($y = 0$): shear stress reaches its parabolic maximum:

$$\tau_{\max} = 1.5 \frac{V}{A} = \frac{3 V}{2 A}$$

B. Solid Circular Cross-Section (Radius $R$, Diameter $d$)

For a solid circle, $I = \frac{\pi R^4}{4}$. At the neutral axis ($y = 0$), $A'$ is a semicircle of area $\frac{\pi R^2}{2}$, with centroid location $\bar{y}' = \frac{4 R}{3 \pi}$. The width at the neutral axis is $b = 2 R = d$. Evaluating $Q$:

$Q_{\max} = \bar{y}' A' = \left(\frac{4 R}{3 \pi}\right) \left(\frac{\pi R^2}{2}\right) = \frac{2}{3} R^3 = \frac{d^3}{12}$

$$\tau_{\max} = \frac{V Q_{\max}}{I b} = \frac{V \left(\frac{2}{3} R^3\right)}{\left(\frac{\pi R^4}{4}\right)(2 R)} = \frac{4 V}{3 \pi R^2} = \frac{4 V}{3 A} \approx 1.333 \frac{V}{A}$$

C. Wide-Flange (I-Beam) Sections & Shear in the Web

In standard structural W-shapes and I-beams:

• Flanges have large width $b_f$, resulting in negligible vertical shear stress ($\tau = VQ/Ib_f$). Flanges carry $>85-90\%$ of bending moments.
• The web has a thin width $t_w$. Across the web, $Q(y)$ remains large, causing the web to carry between 90% and 98% of the total vertical transverse shear force $V$.

In practical PE exam engineering design, average shear stress in the web is evaluated as:

$$\tau_{\text{web, avg}} \approx \frac{V}{d \cdot t_w}$$

where $d$ is overall beam depth and $t_w$ is web thickness.

Shear Flow in Built-Up Beams

For composite timber or steel beams fabricated with nails, bolts, or welds, shear flow $q$ is defined as shear force per unit length:

$$q = \tau b = \frac{V Q}{I} \quad [\text{N/m} \text{ or } \text{lb/in}]$$

If fasteners have an allowable shear capacity $V_{\text{fastener}}$ and are installed with $n$ fasteners across each transverse joint line, the required longitudinal pitch spacing $s$ is:

$$s = \frac{n \cdot V_{\text{fastener}}}{q} = \frac{n \cdot V_{\text{fastener}} I}{V Q}$$

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3. Worked Example: Pin Joint and Built-Up Beam

Part A: Clevis Pin in Double Shear

A $12\,\text{mm}$ steel pin carries $P = 20\,\text{kN}$ through a $10\,\text{mm}$-thick tang held between two fork plates.

• Pin shear (double shear, $V = P/2$ on each plane):

$$\tau = \frac{2P}{\pi d^2} = \frac{2(20{,}000)}{\pi (12)^2} = \frac{40{,}000}{452.4} = 88.4\,\text{MPa}$$

• Bearing on the tang (projected area $d \cdot t$):

$$\sigma_b = \frac{P}{d\,t} = \frac{20{,}000}{(12)(10)} = 166.7\,\text{MPa}$$

If the same pin were loaded in single shear, the shear stress would double to $176.8\,\text{MPa}$; the bearing stress on the tang would not change.

Part B: Fastener Spacing in a Built-Up Beam

Two identical steel bars, each $100\,\text{mm}$ wide and $50\,\text{mm}$ deep, are stacked to form a $100 \times 100\,\text{mm}$ beam and joined by a single row of bolts along the interface. The maximum vertical shear is $V = 10\,\text{kN}$, and each bolt may carry $V_{\text{bolt}} = 6\,\text{kN}$ in shear.

1. Moment of inertia of the full section:

$$I = \frac{b h^3}{12} = \frac{100 (100)^3}{12} = 8.333 \times 10^6\,\text{mm}^4$$

1. First moment of the top bar about the neutral axis (the interface is the neutral axis):

$$Q = A' \bar{y}' = (100 \times 50)(25) = 125{,}000\,\text{mm}^3$$

1. Shear flow that the bolts must transfer:

$$q = \frac{V Q}{I} = \frac{(10{,}000)(125{,}000)}{8.333 \times 10^6} = 150\,\text{N/mm}$$

1. Maximum bolt pitch with one bolt per row ($n = 1$):

$$s = \frac{n V_{\text{bolt}}}{q} = \frac{6{,}000}{150} = 40\,\text{mm}$$

1. Check: the peak transverse shear stress in the solid section is $\tau_{\max} = 1.5 V/A = 1.5(10{,}000)/10{,}000 = 1.5\,\text{MPa}$, the same value $V Q/(I b)$ gives at the interface. The bolts, not the steel, govern this joint.