Stress Concentration Factors for Simple Tubular K Joints

Out of plane bending on one brace only

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 $$ \begin{align*} \mbox{Eqn 10A} &= \tau_A^{-0.54}\gamma^{-0.05}\left(0.99-0.47\beta_A+0.08\beta_A^4\right)\left(\gamma \tau_A \beta_A \left(1.7 - 1.05 \beta_A^3\right)\left(\sin\left(\theta_A\right)\right)^{1.6}\right) \end{align*} $$ $$ \left(\mbox{Eqn 10A}\right)\left(1-0.08\left(\beta_B\gamma\right)^{0.5}\exp\left(-0.8x\right)\right)\tag{Eqn 25} $$ where $$ x=1+\frac{\zeta\sin\theta}{\beta} $$ F3
Brace saddle $\tau^{-0.54}\gamma^{-0.05}\left(0.99-0.47\beta+0.08\beta^4\right)\left(\mbox{Eqn 25}\right)$ F3

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

$F3 = 1 - 0.55\beta^{1.8}\gamma^{0.16}\exp\left(-0.49 \gamma^{-0.89} \alpha^{1.8}\right)$

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