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Failure mechanisms of bolted connection

Short answer

Bolted connections fail by bolt shear, bearing, tension, plate shear-out, net section rupture, gross yielding, block shear, or prying. For eccentric load in-plane, bolts resist direct shear plus torsional shear; resultant checked against bolt shear strength. For load out-of-plane, bolts resist direct shear plus tension, checked with (V/V_db)^2 + (T/T_db)^2 ≤ 1.0.

Explain different failure mechanisms of bolted connection. Also discuss about combination of stresses developed in different type of eccentrically loaded bolted connection.

Failure mechanisms of a bolted connection

  1. Shear failure of bolt: the bolt shears across one plane (single shear) or two planes (double shear). Strength Vdsb=fub3γmb(nnAnb+nsAsb)V_{dsb}=\dfrac{f_{ub}}{\sqrt3\gamma_{mb}}(n_nA_{nb}+n_sA_{sb}).
  2. Bearing failure of bolt/plate: the plate elongates or tears in front of the bolt. Vdpb=2.5kbdtfu/γmbV_{dpb}=2.5k_bdtf_u/\gamma_{mb}. Prevented by adequate end and pitch distances.
  3. Tension failure of bolt: the bolt breaks in the threaded part under direct tension, Tdb=Tnb/γmbT_{db}=T_{nb}/\gamma_{mb} with Tnb=0.9fubAnb≤fybAsbγmb/γm0T_{nb}=0.9f_{ub}A_{nb}\le f_{yb}A_{sb}\gamma_{mb}/\gamma_{m0}.
  4. Tearing of plate at the edge (shear-out): if the end distance is too small, the plate shears out behind the bolt.
  5. Tension (net-section) rupture of plate: the plate fails across the net section through the bolt holes, Tdn=0.9Anfu/γm1T_{dn}=0.9A_nf_u/\gamma_{m1}.
  6. Yielding of gross section of the plate: Tdg=Agfy/γm0T_{dg}=A_gf_y/\gamma_{m0}.
  7. Block shear failure: a block of plate tears out along shear and tension planes at the bolt group.
  8. Bolt bending (long grip): prevented by limiting the grip to 8d8d (with reduced shear strength when the grip exceeds 5d5d).
  9. Prying failure: extra tension in bolts due to flexible connected plates.
  10. Slip in HSFG joints when the friction is exceeded.

Combined stresses in eccentric connections

  • Load in the plane of the bolt group (eccentricity in the plane): bolts get direct shear P/nP/n plus shear due to torsion Tr/Σr2T r/\Sigma r^2; the vector sum of both is the resultant shear on the critical bolt, R=(Fx)2+(Fy)2R=\sqrt{(F_x)^2+(F_y)^2}, checked against Vdb=min⁡(Vdsb,Vdpb)V_{db}=\min(V_{dsb},V_{dpb}).
  • Load perpendicular to the plane of the bolt group (bracket on column flange): bolts get direct shear V=P/nV=P/n and tension due to moment T=My/Σy2T=My/\Sigma y^2. Combined check for bearing-type bolts (IS 800:2007 cl. 10.3.6):
(VVdb)2+(TTdb)2≤1.0\left(\frac{V}{V_{db}}\right)^2+\left(\frac{T}{T_{db}}\right)^2\le1.0

For HSFG (friction-type) bolts (cl. 10.4.6) the interaction is (VsfVdf)2+(TfTdf)2≤1.0\left(\dfrac{V_{sf}}{V_{df}}\right)^2+\left(\dfrac{T_f}{T_{df}}\right)^2\le1.0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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