Shear stress distribution in beams

Why it matters

Bending stress usually governs long beams, but short, heavily loaded members (axle stubs near bearings, spring hangers, chassis cross-members near mounts, gear teeth roots) and thin webs can fail in shear. Shear stress distribution also decides how built-up beams are welded or bolted together and why the web of an I-section, not its flanges, carries the shear. Knowing the shape of the distribution lets you find the peak quickly.

Key ideas

Where shear stress comes from. When the bending moment changes along a beam (dM/dx = V ≠ 0), the bending stresses on two neighbouring sections differ. The part of the section above any level would slide relative to the part below unless a horizontal (longitudinal) shear stress acts on that level. By complementary shear, an equal vertical shear stress acts on the cross-section at that level. Stack two planks and bend them: they slip. Glue them: the glue carries exactly this longitudinal shear.

The shear formula. At a level y in the section, τ = V·Q/(I·b).

• V is the shear force at the section.
• Q is the first moment, about the neutral axis, of the area between that level and the nearer extreme fibre: Q = A′·ȳ′.
• I is the second moment of area of the whole section about the NA.
• b is the width of the section at that level. Assumptions: τ is uniform across the width b (good for narrow sections), the material is linear elastic, and the bending stresses follow the flexure formula.

Rectangular section. τ = (V / 2I)·(d²/4 − y²): a parabola, zero at the top and bottom fibres (free surfaces cannot carry shear), maximum at the NA: τ_max = 1.5·V/A = 1.5 τ_avg.

Circular section. Also parabolic in y: τ = (4V / 3A)·(1 − y²/R²), with τ_max = (4/3)·τ_avg at the NA. For a thin-walled circular tube, τ_max = 2·τ_avg.

Triangular section (base b, height h, apex up). The maximum is not at the NA but at mid-height: τ_max = 1.5·τ_avg at h/2 from the apex, while τ at the NA (h/3 from the base) is (4/3)·τ_avg.

I-sections and T-sections. Q grows continuously from the extreme fibre to the NA, but the width b jumps at the flange–web junction. So the stress jumps there: small in the wide flange, then much larger in the narrow web. In the web the distribution is a shallow parabola with its maximum at the NA. The web carries nearly all the shear force (typically 90 % or more), so a quick estimate is τ ≈ V/(d·t_w), using the overall depth and web thickness.

Shear flow. q = V·Q/I (N/m) is the longitudinal shear force per unit length of beam at a level. It sets the spacing of rivets, bolts or welds connecting flanges to webs: spacing s = (force capacity of one fastener) / q. In thin-walled open sections the shear flow also defines the shear centre, the point through which loads must act to avoid twisting (channels have it outside the web).

Limits. The formula is poor for very wide, shallow sections and near concentrated loads and supports; there, detailed analysis or design-code rules apply.

Formulas

τ = V·Q / (I·b)

• τ: shear stress at the level considered (Pa); V: shear force (N); Q = A′·ȳ′: first moment of the area beyond that level about the NA (m³); I: second moment of area of the whole section (m⁴); b: width at that level (m).

τ = (V / (2·I))·(d²/4 − y²), τ_max = 1.5·V / (b·d)

• Rectangle b × d; y measured from the NA (m).

τ = (4·V / (3·A))·(1 − y²/R²), τ_max = (4/3)·V / A

• Solid circle of radius R.

τ_max = 2·V / A

• Thin-walled circular tube.

τ_max = 1.5·V / A at mid-height; τ_NA = (4/3)·V / A (triangle)

q = V·Q / I, s = F_fastener / q

• q: shear flow (N/m); F_fastener: allowable force per fastener (N); s: spacing (m).

Worked examples

Example 1 (standard): rectangular beam. Given: a rectangular section 100 mm wide and 200 mm deep carries a shear force V = 40 kN. Find τ_avg, τ_max and τ at 50 mm above the NA.

1. A = 100 × 200 = 20 000 mm². τ_avg = V/A = 40 000 / 20 000 = 2.0 MPa.
2. τ_max = 1.5·τ_avg = 3.0 MPa at the NA.
3. I = 100 × 200³/12 = 66.67 × 10⁶ mm⁴. At y = 50 mm: τ = [40 000 / (2 × 66.67 × 10⁶)] × (100² − 50²) = 3.0 × 10⁻⁴ × 7500 = 2.25 MPa.
4. Check with V·Q/(I·b): Q = 100 × 50 × 75 = 375 000 mm³, so τ = 40 000 × 375 000 / (66.67 × 10⁶ × 100) = 2.25 MPa. Answer: τ_avg = 2.0 MPa, τ_max = 3.0 MPa, τ(50 mm) = 2.25 MPa.

Example 2 (GATE level): I-section. Given: an I-section with two flanges 150 mm × 20 mm and a web 260 mm × 10 mm (overall depth 300 mm) carries V = 100 kN. Find the shear stress in the flange and in the web at the junction, and the maximum shear stress.

1. I = (150 × 300³ − 140 × 260³)/12 = (4.050 × 10⁹ − 2.461 × 10⁹)/12 = 132.45 × 10⁶ mm⁴.
2. Q at the junction (one flange): 150 × 20 × (150 − 10) = 420 000 mm³.
3. In the flange, b = 150 mm: τ = 100 000 × 420 000 / (132.45 × 10⁶ × 150) = 2.11 MPa.
4. In the web, b = 10 mm: τ = 2.11 × 150/10 = 31.71 MPa.
5. Q at the NA: 420 000 + 10 × 130 × 65 = 504 500 mm³, so τ_max = 100 000 × 504 500 / (132.45 × 10⁶ × 10) = 38.09 MPa.
6. Quick check: V/(web depth × thickness) = 100 000 / (260 × 10) = 38.5 MPa, close to the exact maximum. Answer: flange 2.11 MPa, web at junction 31.7 MPa, maximum 38.1 MPa at the NA.

Common mistakes

• Using the full section's first moment (which is zero about the NA) instead of the area beyond the level.
• Using the flange width in the web, or vice versa, at the junction.
• Assuming the maximum shear stress always occurs at the NA (not for triangles and some other shapes).
• Applying 1.5·τ_avg to a circular or I-section.
• Unit slips: with N and mm, τ comes out directly in MPa.
• Forgetting that shear stress is zero at free top and bottom surfaces.

For GATE ME

Expect the ratio τ_max/τ_avg for rectangle, circle and triangle, shear stress at a level in a rectangle or at an I- or T-section junction, the location of maximum shear in a triangular section, and shear flow for fastener spacing. Practise computing Q quickly for composite sections.

Quick check

1. What is τ_max/τ_avg for a solid circular section?
2. Where is the shear stress maximum in a triangular section?
3. Rectangle 150 mm × 300 mm, V = 12 kN. Find τ_max.
4. Why does the shear stress jump at an I-section's flange–web junction?

Answers: 1. 4/3. 2. At mid-height. 3. 1.5 × 12 000 / 45 000 = 0.4 MPa. 4. Q is continuous there but the width b changes suddenly.