Bolted joints under static and eccentric loading

Why it matters

Cylinder heads, wheel hubs, engine mounts, suspension brackets and gearbox housings are all held by bolts. A bolted joint rarely fails because the load was miscalculated in a single bolt; it fails because the preload was wrong, or because an eccentric load put far more force on one bolt of the group than the designer assumed. This topic gives the methods to find the worst-loaded bolt and size it.

Key ideas

Bolt size and stress area. Threads reduce the section. Tensile stress is calculated on the tensile stress area A_t (or, in many Indian textbooks, the core area based on the minor diameter d_c), not the nominal shank area. Take A_t and d_c for each metric size from your thread table (for example M12 coarse has d_c ≈ 9.85 mm and A_t ≈ 84.3 mm²).

Preload. Tightening stretches the bolt and compresses the clamped parts (members). When an external tensile load P is applied, only a fraction C of it goes into the bolt; the rest relieves member compression. The joint constant C = k_b/(k_b + k_m), where k_b and k_m are the bolt and member stiffnesses. Members are usually much stiffer than the bolt, so C is small (often 0.2–0.3) and the bolt sees little extra load, which is why a properly preloaded bolt resists fatigue well. The joint separates when the member compression reaches zero, at P = P_i/(1 − C).

Eccentric load in the plane of the joint (bolts in shear). A load P acting at distance e from the centroid G of a bolt group is replaced by:

1. a direct (primary) shear P/n on every bolt, parallel to P; and
2. a moment P·e that rotates the plate about G, giving each bolt a secondary shear proportional to its distance r_i from G and perpendicular to the line joining it to G. The bolt with the largest vector sum of the two is critical; it is usually the one farthest from G on the side where the two components point the same way.

Eccentric load perpendicular to the bolt axes (bracket bolted to a wall, bolts in tension). The bracket tends to tilt about its lower edge. Each bolt's tensile load is proportional to its distance l_i from the tilting edge. The bolts also share the direct shear P/n. The critical bolt (farthest from the edge) then carries combined tension and shear, checked with the maximum principal stress or maximum shear stress theory.

Good practice: place the bolt-group centroid on or near the line of the load, spread bolts as far from G as space allows (secondary loads fall as 1/r), use bolts of the same size, and use controlled tightening (torque-angle) so preload is predictable.

Formulas

Tensile stress: σ_t = P / A_t Resultant bolt load (preloaded): F_b = P_i + C·P Member load: F_m = P_i − (1 − C)·P Separation load: P_sep = P_i / (1 − C) Joint constant: C = k_b / (k_b + k_m) Primary shear: P₁ = P / n Secondary shear (in-plane moment): P₂,i = P·e·r_i / Σr² Resultant: R = √(P₁² + P₂² + 2·P₁·P₂·cos θ) Bracket on a wall: P_t,i = P·L·l_i / Σl², and for two bolts in each row P_t,max = P·L·l₂ / (2·(l₁² + l₂²)) Combined: τ_max = √((σ/2)² + τ²), σ₁ = σ/2 + τ_max

Symbols: P = external load (N); P_i = preload (N); A_t = tensile stress area (mm²); k_b, k_m = bolt and member stiffness (N/mm); n = number of bolts; e = eccentricity of the load from the group centroid (mm); r_i = distance of bolt i from the centroid (mm); θ = angle between the primary and secondary shear vectors; L = distance of the load from the wall (mm); l_i = distance of bolt i from the tilting edge (mm); σ, τ = tensile and shear stress in the bolt (MPa).

Worked examples

Example 1 (standard, bolts in shear). A bracket is fixed to a column by four bolts at the corners of a 100 mm × 100 mm square. A vertical load of 8 kN acts 200 mm from the centroid of the bolt group. Find the largest bolt shear force.

1. Primary shear on each bolt: P₁ = 8000/4 = 2000 N (vertical, downward).
2. Distance of each bolt from G: r = √(50² + 50²) = 70.71 mm; Σr² = 4 × 70.71² = 20 000 mm².
3. Moment: P·e = 8000 × 200 = 1.6 × 10⁶ N·mm.
4. Secondary shear: P₂ = 1.6 × 10⁶ × 70.71 / 20 000 = 5657 N, perpendicular to each radius.
5. For the two bolts nearer the load, the secondary force has a downward component, and its angle to the primary force is θ = 45°.
6. R = √(2000² + 5657² + 2 × 2000 × 5657 × cos 45°) = 7211 N ≈ 7.21 kN.
7. With a permissible shear stress of 60 MPa, the required core area is 7211/60 = 120.2 mm², so d_c ≥ 12.4 mm; choose the next size from the thread table.

Example 2 (GATE level, bolts in tension and shear). A bracket carries a vertical load of 10 kN at 300 mm from the wall. It is held by four bolts, two at 50 mm and two at 250 mm above the lower (tilting) edge. Permissible shear stress in the bolt = 60 MPa. Find the core diameter by maximum shear stress theory.

1. Maximum tensile load (upper bolts): P_t = 10 000 × 300 × 250 / (2 × (50² + 250²)) = 7.5 × 10⁸ / 130 000 = 5769 N.
2. Direct shear per bolt: P_s = 10 000/4 = 2500 N.
3. Maximum shear force equivalent: τ_max·A = √((5769/2)² + 2500²) = 3817 N.
4. Required area: A = 3817/60 = 63.6 mm².
5. Core diameter: d_c = √(4 × 63.6/π) = 9.0 mm, so M12 (d_c ≈ 9.85 mm) is adequate.

Example 3 (preload). A bolt has preload 20 kN and joint constant C = 0.25. An external load of 8 kN is applied. Bolt load = 20 + 0.25 × 8 = 22 kN; member compression = 20 − 0.75 × 8 = 14 kN; separation load = 20/0.75 = 26.7 kN.

Common mistakes

• Treating an eccentrically loaded joint as a single bolt in bending. The moment is resisted by the bolt group, through secondary forces proportional to r_i (in shear) or l_i (in tension).
• Adding primary and secondary forces as scalars instead of vectors, or using the wrong angle θ.
• Measuring l_i from the bolt centroid instead of the tilting edge for a wall bracket.
• Using the nominal diameter instead of the core or tensile-stress area.
• Assuming the whole external load adds to the preload; only C·P does while the joint stays closed.
• Forgetting that P·e must be in N·mm when r is in mm.

For GATE ME

Typical questions: the largest bolt force in a group under eccentric in-plane load; bolt tension in a wall bracket; bolt and member loads with preload and a given stiffness ratio; separation load; and choosing a bolt diameter from a permissible stress. Practise drawing the primary and secondary force vectors at each bolt before calculating.

Quick check

1. Three bolts share a 9 kN load acting through the group centroid. What is each bolt's shear?
2. In-plane eccentric loading: the secondary shear force on a bolt is proportional to what?
3. Preload 15 kN, C = 0.2, external load 10 kN. Find the bolt load.
4. For that joint, at what external load do the members separate?
5. Why is the tensile stress area used instead of the shank area?

Answers: 1. 3 kN. 2. Its distance from the bolt-group centroid. 3. 17 kN. 4. 18.75 kN. 5. The threaded section is smaller and is where the bolt fails.