Shear stress distribution in beams

Why it matters

Bending stress usually governs a beam's size, but transverse shear governs short, heavily loaded members (brackets, pins, short cantilevers near a support) and thin webs of I-sections and extrusions. Shear stress also decides how far apart the bolts, rivets or welds of a built-up beam can be, and it is the stress that tends to split glued or layered parts such as laminated composites or bonded sensor mounts. Knowing where it peaks lets you put material, or fasteners, where they are needed.

Key ideas

Where beam shear stress comes from. When the bending moment changes along a beam (dM/dx = V ≠ 0), the bending stresses on the two faces of a short slice are unequal. For any part of the section above a chosen level to stay in equilibrium, a horizontal (longitudinal) shear force must act on that level. By complementary shear, an equal shear stress acts vertically on the cross-section. That is why a stack of loose planks slides at the interfaces when bent, while a glued block does not.

The shear formula. τ = V·Q/(I·b), where Q is the first moment about the neutral axis of the area between the level considered and the nearest free surface, and b is the section width at that level. Assumptions: shear stress is uniform across the width b and parallel to the vertical load direction; the beam obeys simple bending theory. It is accurate for narrow sections and approximate for wide or thin-walled flanges.

Distribution in common sections.

• Rectangle: parabolic over the depth, zero at the top and bottom, maximum at the NA: τ_max = 1.5 × V/A.
• Solid circle: maximum at the NA: τ_max = (4/3) × V/A.
• Thin-walled circular tube: τ_max = 2 × V/A.
• Triangle (apex up, base down): maximum at mid-height, τ_max = 1.5 × V/A, and at the NA (h/3 above the base) τ = (4/3) × V/A. This is a case where the maximum is not at the NA.
• I-section: in the web the stress is nearly uniform (parabolic but flat), with its maximum at the NA; at the flange-web junction the stress jumps because b changes from the flange width to the web thickness. The web carries most of the shear force (typically 90 % or more) while the flanges carry most of the bending moment. A quick estimate is τ_web ≈ V/(d·t_w).

Shear flow and fasteners. In a built-up beam, the longitudinal force per unit length that the joint between parts must transmit is the shear flow q = V·Q/I, where Q is for the attached part. Fastener spacing s follows from s = F_allow/q, where F_allow is the allowable load per fastener (or per row).

Where shear governs. For a beam of span L and depth h, bending stress scales with L/h while shear stress does not. So shear matters for short, deep beams, for thin webs, near supports where V is largest, and for materials weak in shear such as timber or laminates. In long slender beams bending governs almost always.

Shear deflection. Simple beam theory ignores shear deformation; it adds a small extra deflection, significant only for short, deep beams.

Formulas

τ = V·Q / (I·b)

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

τ = [V / (2I)]·(h²/4 − y²)

• Rectangle of depth h (m), at distance y from the NA (m).

τ_max = 1.5·V/A (rectangle), τ_max = (4/3)·V/A (solid circle), τ_max = 2·V/A (thin tube)

• A: whole cross-sectional area (m²); V/A is the average shear stress.

q = V·Q / I, s = F_allow / q

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

Worked examples

Example 1 (standard): rectangular bracket section. Given: a rectangular section 100 mm wide and 200 mm deep carries a shear force of 10 kN. Find the average and maximum shear stress, and the stress 50 mm from the NA.

1. Area: A = 100 × 200 = 20 000 mm². Average: V/A = 10 000/20 000 = 0.50 MPa.
2. τ_max = 1.5·V/A = 0.75 MPa, at the NA.
3. Check with the general formula: I = 100 × 200³/12 = 66.67 × 10⁶ mm⁴; Q at the NA = (100 × 100) × 50 = 500 000 mm³; τ = 10 000 × 500 000/(66.67 × 10⁶ × 100) = 0.75 MPa.
4. At y = 50 mm: τ = [V/(2I)]·(h²/4 − y²) = 10 000/(2 × 66.67 × 10⁶) × (10 000 − 2500) = 0.5625 MPa. Answer: τ_avg = 0.50 MPa, τ_max = 0.75 MPa at the NA, τ = 0.56 MPa at 50 mm from the NA.

Example 2 (GATE level): I-section. Given: a symmetric I-section, overall depth 200 mm, flanges 100 mm × 10 mm, web 10 mm thick (clear web depth 180 mm), carries V = 50 kN. Find the shear stress in the flange and in the web at the junction, and the maximum shear stress.

1. I = 100 × 200³/12 − 90 × 180³/12 = 66.667 × 10⁶ − 43.740 × 10⁶ = 22.93 × 10⁶ mm⁴.
2. Q at the junction (one flange): 100 × 10 × 95 = 95 000 mm³ (flange centroid 95 mm from the NA).
3. In the flange at the junction (b = 100 mm): τ = 50 000 × 95 000/(22.93 × 10⁶ × 100) = 2.07 MPa.
4. In the web at the junction (b = 10 mm): τ = 50 000 × 95 000/(22.93 × 10⁶ × 10) = 20.7 MPa.
5. Q at the NA: 95 000 + 10 × 90 × 45 = 135 500 mm³. τ_max = 50 000 × 135 500/(22.93 × 10⁶ × 10) = 29.6 MPa.
6. Quick estimate: V/(d·t_w) = 50 000/(200 × 10) = 25 MPa, close to the web average. Answer: flange 2.07 MPa, web at junction 20.7 MPa, τ_max ≈ 29.6 MPa at the NA. The tenfold jump at the junction comes purely from the change in width b.

Common mistakes

• Using the area of the whole section, or the area on the wrong side of the level, when computing Q.
• Measuring ȳ' from the top edge instead of from the NA.
• Using the flange width at a point in the web (or vice versa) for b.
• Assuming the maximum is always at the NA; check sections whose width changes, such as triangles and diamonds.
• Forgetting that shear stress is zero at a free top or bottom surface.
• Using V/A as the maximum; it is only the average.

For GATE ME

Expect ratios τ_max/τ_avg for rectangle and circle, shear stress at a given level in rectangular and I sections, the jump at the flange-web junction, the shape of the distribution for given sections, and shear flow or fastener spacing in built-up beams. Practise Q calculations for composite areas.

Quick check

1. What is τ_max/τ_avg for a rectangular section? For a solid circle?
2. Where is shear stress zero in a rectangular section?
3. A 150 mm × 300 mm rectangle carries V = 15 kN. Find τ 50 mm above the NA.
4. In an I-section, which part carries most of the shear force?
5. Shear flow is 20 N/mm and each bolt can carry 4 kN. What is the bolt spacing?

Answers: 1. 1.5; 4/3. 2. At the top and bottom surfaces. 3. 15 000/(2 × 337.5 × 10⁶) × (22 500 − 2500) = 0.444 MPa. 4. The web. 5. 4000/20 = 200 mm.