Stress Concentration Factors for Simple Tubular K Joints

Unbalanced out of plane bending

K image
$$ \beta = \frac{d}{D} $$
$$ \alpha = \frac{2L}{D} $$
$$ \gamma = \frac{D}{2T} $$
$$ \tau = \frac{t}{T} $$
$$ \zeta = \frac{g}{D} $$
Location Equation Short chord correction
Chord Saddle $$ \begin{align*} \mbox{Eqn 10A} &= \tau_A^{-0.54}\gamma^{-0.05}\left(0.99-0.47\beta_A+0.08\beta_A^4\right)\left(\gamma \tau_A \beta_A \left(1.7 - 1.05 \beta_A^3\right)\left(\sin\left(\theta_A\right)\right)^{1.6}\right)\\ \mbox{Eqn 10B} &= \tau_B^{-0.54}\gamma^{-0.05}\left(0.99-0.47\beta_B+0.08\beta_B^4\right)\left(\gamma \tau_B \beta_B \left(1.7 - 1.05 \beta_B^3\right)\left(\sin\left(\theta_B\right)\right)^{1.6}\right) \end{align*} $$ $$ \left(\mbox{Eqn 10A}\right)\left(1-0.08\left(\beta_B\gamma\right)^{0.5}\exp\left(-0.8x\right)\right)+\left(\mbox{Eqn 10B}\right)\left(1-0.08\left(\beta_B\gamma\right)^{0.5}\exp\left(-0.8x\right)\right)\left(2.05\beta_{max}^{0.5}\exp\left(-1.3x\right)\right)\tag{Eqn 23} $$ where $$ x=1+\frac{\zeta\sin\theta}{\beta} $$ F4
Brace saddle $\tau^{-0.54}\gamma^{-0.05}\left(0.99-0.47\beta+0.08\beta^4\right)\left(\mbox{Eqn 23}\right)$ F4

Short chord correction factor $\left(\alpha \lt 12 \right)$

$F4 = 1 - 1.07\beta^{1.88}\exp\left(-0.16 \gamma^{-1.06} \alpha^{2.4}\right)$

Note that the designation of braces A and B is not geometry dependent. It is nominated by the user.