Ultimate strength of ISA 100x100x10, M20 bolts

Data: ISA 100×100×10 ($A_g=1903$ mm²), gusset 12 mm, connected through one 100 mm leg, M20 (4.6) bolts in a single line, threads outside the shear plane. The number of bolts is not given: assumed 4 bolts, pitch 50 mm, end distance 35 mm, gauge 55 mm from the back; $d_0=22$ mm; $f_y=250$, $f_u=410$.

1. Yielding

$T_{dg}=\dfrac{1903\times250}{1.10}\times10^{-3}=432.5$ kN

2. Rupture with shear lag

• $A_{nc}=(100-5-22)\times10=730$ mm², $A_{go}=(100-5)\times10=950$ mm²
• $b_s=w+w_1-t=100+55-10=145$ mm ($w_1$ = gauge), $L_c=3\times50=150$ mm

$$\beta=1.4-0.076\times\frac{100}{10}\times\frac{250}{410}\times\frac{145}{150}=0.952\ \ (\text{limits }0.7\ \text{to }1.443)$$ $$T_{dn}=\frac{0.9\times730\times410}{1.25}+\frac{0.952\times950\times250}{1.10}=421.0\ \text{kN}$$

3. Block shear

$T_{db1}=343.1$ kN, $T_{db2}=286.3$ kN.

4. Bolts (threads outside the shear plane)

$V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{314}{1.25}\times10^{-3}=58.01$ kN; $V_{dpb}=83.2$ kN (10 mm angle). Four bolts: $232.0$ kN.

Answer: member design strength $=\min(432.5,\,421.0,\,286.3)=286.3$ kN (block shear governs); with the assumed 4 bolts the joint strength is $\min(286.3,\,232.0)=232.0$ kN.