Mechanics of Materials · Chapter 7

Transverse Shear

Shear stress in beams, fastener design via shear flow, and the shear center of thin-walled sections.

Learning objectives

Explain why transverse shear is accompanied by longitudinal shear, and derive the shear formula τ = VQ/It.
Compute shear-stress distributions over beam cross sections and identify where they peak.
Use shear flow q = VQ/I to design fasteners and their spacing in built-up members.
Describe shear flow in thin-walled members and locate the shear center to prevent twisting.

7.1Shear in Straight Members

Transverse and Longitudinal Shear

A beam generally carries both an internal moment and an internal shear \(V\) — the resultant of a transverse shear-stress distribution over the cross section. By the complementary property of shear, an equal longitudinal shear stress acts along the beam's length. The classic demonstration: stack three unbonded boards and load them — they slide relative to one another; glue them and the longitudinal shear stress at the glue lines prevents the slip, making the boards act as one beam.

Warping of the Cross Section

Shear strain is not uniform over the depth, so shear warps the cross section — plane sections do not strictly remain plane when both bending and shear act. For slender beams the warping is small, which is what justifies still using the flexure formula alongside the shear analysis.

7.2The Shear Formula

Because the shear-strain distribution is not simple to postulate, the shear stress is found indirectly — from horizontal force equilibrium of a portion of a beam element. The bending stresses on the two faces of an element of length \(dx\) differ (moments \(M\) and \(M + dM\)), so a slice of the element cut at height \(y'\) can balance horizontally only if a longitudinal shear stress acts on the cut plane. Working this out with the flexure formula and \(V = dM/dx\):

The Shear Formula

\[ \tau = \frac{V\,Q}{I\,t} \]

\(\tau\) = shear stress at the point, assumed constant (averaged) across the width \(t\); equal on the cross section and the longitudinal plane (complementary property)
\(V\) = internal shear force; \(I\) = moment of inertia of the entire section about the neutral axis
\(t\) = width of the section at the point where τ is evaluated
\(Q = \bar{y}'A'\) — moment about the neutral axis of the area \(A'\) above (or below) the point

Calculating Q

\(Q\) is the variable most often done wrong. \(A'\) is the part of the cross section that the longitudinal shear stress "holds onto" the rest of the beam — everything above (or everything below) the level where τ is wanted — and \(\bar{y}'\) is the distance from the neutral axis to the centroid of \(A'\). Q is zero at the top and bottom fibers and maximum at the neutral axis; using the area below the point gives the same result (with \(\bar{y}'\) negative where applicable).

Limitations of the Shear Formula

The constant-across-the-width assumption is good for narrow sections: for a rectangle with width equal to its depth the true maximum edge stress is only ≈ 3 % above the formula's value, but for very wide, flat sections (b ≈ 4h) the error grows to ≈ 40 %. The formula also fails at points where the section boundary is not perpendicular to the stress direction (e.g. the sloped sides of a triangle) and at abrupt width changes such as flange–web junctions — at a free boundary the shear stress must act tangent to the boundary. Fortunately, engineers usually need only the maximum value, at the neutral axis, where the formula is at its best.

Shear-stress distribution τ = VQ/It over a rectangular beam section — parabolic with τmax = 1.5V/A at the neutral axis.

FIG. 7-1 (original diagram)

7.3Shear Flow in Built-Up Members

Shear Flow

Built-up members — boards nailed together, plates bolted or welded into girders — need fasteners to stop the parts sliding during bending. The load each fastener line must resist, measured as force per unit length of beam, is the shear flow \(q\). It follows from the same horizontal-equilibrium argument as the shear formula, without dividing by the width.

Shear Flow

\[ q = \frac{V\,Q}{I} \]

\(Q = \bar{y}'A'\), where \(A'\) is the area of the segment being held on at the fastened juncture
Units: N/m or lb/in — force per unit length along the beam.

Fastener Spacing

\[ F = q\,s \qquad\Rightarrow\qquad s = \frac{F}{q} \]

\(F\) = allowable shear force per fastener; \(s\) = spacing along the beam.
If the segment is attached along two or three fastener lines (e.g. two rows of nails), each line resists \(q/2\) or \(q/3\).

7.4Shear Flow in Thin-Walled Members

Shear Flow Around a Thin-Walled Section

For members whose wall thickness is small compared to their height and width, \(q = VQ/I\) describes how the shear flow is distributed around the cross section. Two simplifications follow from the thin walls: q is taken constant across the thickness, and only the component parallel to the wall is considered (the flange faces are stress-free, so the perpendicular component must vanish).

Pattern for a Wide-Flange Beam

In the flanges, Q grows linearly from the free edges toward the web, so q varies linearly, peaking at the web. In the web, q varies parabolically, with its maximum at the neutral axis. Integrating the distributions shows the flange forces balance horizontally, while the web carries nearly the entire vertical shear V (typically ~95 % or more) — which is why design often simply checks τ ≈ V/A_web. The shear flow "flows" continuously around the section, like water in a channel: inward along the top flanges, down the web, outward along the bottom flanges.

7.5Shear Center for Open Thin-Walled Members

Why a Channel Twists

If a transverse load on a channel section passes through its centroid, the channel bends and twists. The reason: the shear-flow distribution integrates to a force \(F_f\) in each flange and \(V = P\) in the web, and the flange forces form an unbalanced couple about the web. To bend without twisting, the load must act through a point O — the shear center (flexural center) — offset from the web so its moment cancels the flange couple.

Shear Center Eccentricity (Channel)

\[ e = \frac{F_f\,d}{P} \]

\(F_f\) = resultant shear-flow force in one flange; \(d\) = distance between flange centerlines
The shear center always lies on an axis of symmetry; for sections with two axes of symmetry it coincides with the centroid. Its location is a property of geometry only — independent of P.
For sections made of intersecting thin elements (angles, tees), all the wall shear flows pass through the corner point, so the shear center is at that intersection.

Physical meaning

Internal resultants create local stress through member geometry; use a cut free-body and stated sign convention before superposition.

Problem-solving strategy

Obtain signed internal resultants from equilibrium.
Choose axes and identify the precise material point.
Apply only relations whose assumptions fit the geometry.
Audit units, direction, and the governing failure mode.

EXAMPLE 1Neutral-axis shear in a rail

An 80 mm by 160 mm rectangular timber rail has internal shear \(V=12\,000\ \text{N}\). Determine its neutral-axis transverse shear stress and compare it with average shear stress.

Recognition strategy

A known transverse internal resultant and a requested point stress identify the beam-shear relation, rather than direct average shear.

Given information

Use \(b=80\ \text{mm}\), \(h=160\ \text{mm}\), and a prismatic homogeneous beam away from the applied load.

Model and interpretation

Take positive \(V\) upward on the cut. The internal force \(V\) is a resultant; \(\tau\) is a local traction.

Step-by-step solution

1. Plan. A section free-body gives the specified internal shear, which must be established before converting it into a stress distribution.

2. Formulate. For the complete rectangle, \(I=bh^3/12=27.31\times10^6\ \text{mm}^4\); at the neutral axis, \(Q=80\cdot80\cdot40=256\,000\ \text{mm}^3\).

3. Calculate. With local width \(t=80\ \text{mm}\), \(\tau=VQ/(It)=1.41\ \text{MPa}\).

4. Resolve. The average is \(V/A=0.938\ \text{MPa}\), and \(1.5V/A=1.41\ \text{MPa}\) confirms the rectangular neutral-axis maximum.

Verification

Units: N and mm give \(\text{N}/\text{mm}^2\), or MPa, after the stated section-property substitution; the calculation is dimensionally consistent.

Signs and direction: The stated direction follows the cut-face convention. Reversing the internal resultant reverses the traction direction, but not the reported magnitude.

Plausibility: The computed magnitude has the expected limiting behavior and is independently consistent with the physical location of the maximum response.

Final result: \(\boxed{\tau_{\max}=1.41\ \text{MPa}}\)

Engineering interpretation

This idealized elastic result is not a timber allowable-stress rating. Grain direction, knots, moisture, notches, and the actual shear diagram can control splitting near supports.

EXAMPLE 2Rivet pitch for a cover plate

A cover plate of area \(2160\ \text{mm}^2\) lies 132 mm from a beam neutral axis. At a cut \(V=90\ \text{kN}\), \(I=186\times10^6\ \text{mm}^4\), and one rivet line has 9.0 kN allowable shear. Find maximum pitch.

Recognition strategy

An interface force per unit length identifies shear flow, then \(F=qs\) converts flow to demand on one rivet.

Given information

Use one connector line, full composite action, and the plate as the area held onto by the connection.

Model and interpretation

Isolate the plate beyond the interface. Its longitudinal force change is resisted by the rivets even though \(V\) is transverse.

Step-by-step solution

1. Plan. The cut free-body establishes \(V=90\,000\ \text{N}\), the internal resultant that drives the interface shear flow.

2. Formulate. The held-on area gives \(Q=A\bar y=2160(132)=285\,120\ \text{mm}^3\), not the first moment of the whole section.

3. Calculate. The flow is \(q=VQ/I=90\,000(285\,120)/(186\times10^6)=138.0\ \text{N/mm}\).

4. Resolve. Setting \(F=qs=9000\ \text{N}\) gives \(s=65.2\ \text{mm}\), so installed pitch cannot exceed this value.

Verification

Units: N and mm give \(\text{N}/\text{mm}^2\), or MPa, after the stated section-property substitution; the calculation is dimensionally consistent.

Signs and direction: The stated direction follows the cut-face convention. Reversing the internal resultant reverses the traction direction, but not the reported magnitude.

Plausibility: The computed magnitude has the expected limiting behavior and is independently consistent with the physical location of the maximum response.

Final result: \(\boxed{s_{\max}=65\ \text{mm}}\)

Engineering interpretation

Connection design also requires code pitch and edge-distance limits, bearing, slip, fatigue, corrosion protection, and the complete shear-force diagram.

EXAMPLE 3Channel shear center

A thin channel has centerline web depth \(d=120\ \text{mm}\), equal flange length \(b=45\ \text{mm}\), and thickness \(t=4\ \text{mm}\). Locate its shear-center offset from the web centerline.

Recognition strategy

An open singly symmetric section under transverse shear needs a shear-center calculation to prevent elastic twist.

Given information

Use thin-wall centerline dimensions, neglect corner radii, and assume elastic tangential wall shear flow.

Model and interpretation

Positive offset is toward the open side. The flange-flow couple is balanced by the transverse load through the shear center.

Step-by-step solution

1. Plan. Thin-wall inertia is \(I_x=td^3/12+2bt(d/2)^2=1.872\times10^6\ \text{mm}^4\).

2. Formulate. Flange-flow integration gives \(F_f=Vtdb^2/(4I_x)\), a force resultant rather than a stress.

3. Calculate. Moment equilibrium gives \(e=F_fd/V=td^2b^2/(4I_x)\) for a load through the shear center.

4. Resolve. Substitution gives \(e=4(120)^2(45)^2/[4(1.872\times10^6)]=15.6\ \text{mm}\) toward the open side.

Verification

Units: N and mm give \(\text{N}/\text{mm}^2\), or MPa, after the stated section-property substitution; the calculation is dimensionally consistent.

Signs and direction: The stated direction follows the cut-face convention. Reversing the internal resultant reverses the traction direction, but not the reported magnitude.

Plausibility: The 15.6 mm offset is below the 45 mm flange length and independent of load magnitude, as expected for a geometry-defined shear center.

Final result: \(\boxed{e=15.6\ \text{mm}\text{ toward the open side}}\)

Engineering interpretation

The 15.6 mm load line eliminates ideal elastic twist only. Restrained warping, torsional stiffness, dynamic loading, connections, local plate buckling, and fabrication tolerances still require review.

Chapter recap

Use (VQ/(It)) for local beam shear, (VQ/I) for interface flow, and the shear center to prevent twist.
Before moving on, draw the cut free-body, declare signs and units, and state the applicable failure or serviceability check.