Skip to main content

Safe load P on bolted bracket, M20 4.6

Short answer

Ten M20 grade 4.6 bolts, gauge 120 mm, pitch 80 mm. Critical bolt force = 0.2609P, capacity 45.26 kN. Factored load capacity 173.5 kN, safe working load 115.7 kN.

Determine the safe load P that can be carried by the joint shown in figure below. Use M20 bolts of grade 4.6. The thickness of the flange of I-section is 9.1 mm and that of bracket plate 10 mm thick. [Figure: bracket plate bolted to I-section flange with two vertical lines of 6 bolts each (bolt gauge 120 mm); load P acts at 200 mm from the column face; vertical spacings 40, 80, 80, 80, 80, 40 mm]

Assumptions: the spacing 40-80-80-80-80-40 mm gives 5 bolts in each vertical line (end 40 mm, four pitches of 80 mm), so 10 bolts M20 (grade 4.6) in all, gauge 120 mm. PP acts in the plane of the bolts at e=200e=200 mm from the bolt centroid. Plates: flange 9.1 mm, bracket 10 mm; fu=410f_u=410 N/mm².

Bolt value

  • Shear: Vdsb=4003×2451.25×10−3=45.26V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26 kN
  • Bearing on the thinner plate (9.1 mm): kb=min⁡(0.606,0.962,0.976,1)=0.606k_b=\min(0.606,0.962,0.976,1)=0.606; Vdpb=90.4V_{dpb}=90.4 kN
  • Bolt value Vdb=45.26V_{db}=45.26 kN

Force in the critical bolt per unit load PP (kN)

  • Σ(x2+y2)=10(602)+2×2(802+1602)=164000\Sigma(x^2+y^2)=10(60^2)+2\times2(80^2+160^2)=164000 mm²
  • Corner bolt (x=60, y=160x=60,\ y=160): Fx=PeyΣr2=0.1951PF_x=\dfrac{Pey}{\Sigma r^2}=0.1951P; Fy=P10+PexΣr2=0.1732PF_y=\dfrac{P}{10}+\dfrac{Pex}{\Sigma r^2}=0.1732P
  • R=Fx2+Fy2=0.2609PR=\sqrt{F_x^2+F_y^2}=0.2609P

Safe load

0.2609Pu=45.26⇒Pu=173.5 kN0.2609P_u=45.26\Rightarrow P_u=173.5\ \text{kN}

Answer: design (factored) load Pu≈173.5P_u\approx173.5 kN; the safe working load P≈Pu/1.5=115.7P\approx P_u/1.5=115.7 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Design of Steel and Timber Structure chapter 4 questions →