Mechanical Design & Fasteners
Bolted-Joint Stiffness Factor (C) and Separation Safety Formula
Calculate Shigley’s bolted-joint stiffness constant C = k_b / (k_b + k_m), bolt tension F_b = MAX(P, F_i + C*P), and joint-separation safety factor n_0 = F_i / (P*(1 - C)).
Stop adding 100% of external load P directly onto bolt preload F_i!
When external tensile load P pulls on a preloaded bolted flange (initial preload F_i), beginners often write F_b = F_i + P. In reality, the bolt (stiffness k_b) and clamped frustum members (stiffness k_m) deform in parallel! Prior to joint separation, the bolt absorbs only fraction C = k_b / (k_b + k_m) of P (F_b = F_i + C * P), while the remaining (1 - C) * P unloads member compression (F_m = F_i - (1 - C) * P).
Verify on a flange with bolt stiffness k_b = 500 kN/mm (B2), member stiffness k_m = 1,500 kN/mm (B3 -> C = 500 / 2000 = 0.2500), preload F_i = 45,000 N (B4), and external load P = 24,000 N (B5): bolt tension rises by only 0.25 * 24,000 = 6,000 N to 51,000 N (NOT 69,000 N!), residual clamping compression stays healthy at 27,000 N, and joint-separation safety factor is n_0 = 45,000 / (24,000 * 0.75) = 2.5000! Once P exceeds 60,000 N (e.g., P = 70,000 N), the flange gaps open and MAX(P, F_i + C * P) switches cleanly to full load 70,000 N.
| Bolted-Joint Quantity (C = 0.25, F_i = 45 kN) | Formula | Result at P = 24 kN |
|---|---|---|
| Joint Stiffness Constant (C) | k_b / (k_b + k_m) | 0.2500 (25% Bolt Share) |
| Active Bolt Tension (F_b) | MAX(P, F_i + C * P) | 51,000 N (vs Naive 69,000 N) |
| Residual Clamping Force (F_clamp) | MAX(0, F_i - (1 - C) * P) | 27,000 N (Gasket Sealed) |
| Separation Safety Factor (n_0) | F_i / (P * (1 - C)) | 2.5000 (> 1.0) |
Piecewise Pre- and Post-Separation Bolt Tension MAX(P, F_i + C * P) Formula
Wrapping F_i + C * P inside MAX(P, ...) handles both elastic load sharing below separation and 100% bolt load takeover after flange separation in one exact expression!
=LET(k_blt, B2, k_mem, B3, f_pre, B4, p_ext, B5, IF(OR(k_blt <= 0, k_mem <= 0, f_pre < 0, p_ext < 0), "Invalid Joint Input", LET(c_fac, k_blt / (k_blt + k_mem), ROUND(MAX(p_ext, f_pre + c_fac * p_ext), 2))))Joint-Separation Load Factor n_0 = F_i / (P * (1 - C)) Formula
Require P > 0 when computing n_0 (under zero external load, separation cannot occur).
=LET(k_blt, B2, k_mem, B3, f_pre, B4, p_ext, B5, IF(OR(k_blt <= 0, k_mem <= 0, f_pre <= 0, p_ext <= 0), "Invalid or Zero External Load", LET(c_fac, k_blt / (k_blt + k_mem), ROUND(f_pre / (p_ext * (1 - c_fac)), 4))))Try the example
Use the original values or change them to test your understanding.