Bolted Joint Stiffness Member⚠ unverified

Understanding the Parameters

• Grip length $t$ — the total thickness being clamped. A longer grip means a softer member (more material to compress), lowering $k_m$; this is why long-grip bolts are more forgiving of external load.

• Bolt diameter $d$ — sets the size of the pressure cone. A larger bolt spreads compression over a wider frustum, stiffening the members ($k_m$ rises roughly with $d$).

• Member modulus $E$ — steel members are ~3× stiffer than aluminium; $k_m$ scales linearly with $E$. If the two clamped plates differ, the frusta are taken in series.

• The $0.5774$ constant — this is $\tan 30^\circ$, the half-angle of the assumed pressure cone. It encodes the empirical finding that clamping pressure spreads outward from the bolt head/nut at roughly $30^\circ$ through the members.

Derivation (Approaching a Proof)

The compressed material under a bolt head is idealised as a frustum of a cone with a $30^\circ$ half-angle, radiating from the bolt-head bearing circle outward through the grip. For an incremental slice of thickness $dx$ at radius spread, the local area is that of an annulus and its compliance is $dx/(EA(x))$. Integrating the slice compliances in series along the frustum and inverting gives the stiffness of one frustum. Shigley evaluates the integral in closed form for a $30^\circ$ cone (half-angle $\alpha$ with $\tan\alpha = 0.5774$):

$$k_{\text{frustum}} = \frac{\pi E d \tan\alpha}{\ln\!\dfrac{(2t\tan\alpha + D - d)(D + d)}{(2t\tan\alpha + D + d)(D - d)}},$$

with $D$ the head bearing diameter (taken as $1.5d$). Substituting $\tan\alpha = 0.5774$, $D = 1.5d$, and combining the two frusta of a symmetric grip yields the compact registry form

$$k_m = \frac{0.5774\,\pi E d}{\ln\!\left[\dfrac{5\,(0.5774\,t + 0.5d)}{0.5774\,t + 2.5d}\right]}.$$

Dimensional check. The logarithm's argument is a ratio of lengths (dimensionless), so $k_m = \dfrac{(\text{–})\cdot\text{Pa}\cdot\text{m}}{(\text{–})} = (\text{N}/\text{m}^2)\cdot\text{m} = \text{N}/\text{m}$ — a stiffness, as required.

History and Development

Estimating member stiffness is the hard half of bolted-joint analysis: bolt stiffness is a simple bar ($k_b = A E/l$), but the members compress in a spreading three-dimensional field. The $30^\circ$ cone frustum model, refined by Shigley and by Wileman, Choudhury & Green (whose finite-element study produced an alternative exponential fit $k_m = E d\,A e^{Bd/t}$), became the textbook standard. The whole point is to get $C = k_b/(k_b+k_m)$ right, because $C$ decides how much of a fluctuating external load reaches the bolt — and therefore whether the joint survives fatigue.

Related Concepts: Joint Stiffness Ratio, Joint Stiffness, Bolt Shank Stiffness, Bolt Thread Stiffness, Bolt Load External, Bolt Fatigue Factor

Notes: $0.5774 = \tan 30^\circ$ (pressure-cone half-angle); head bearing diameter taken as $1.5d$. Longer grip → softer members → lower $k_m$. Dissimilar plates → frusta in series. Wileman's exponential fit is an alternative to this cone-integral form.