Skip to main content

Greatest bolt force, 100 kN at 500 mm

Short answer

100 kN load at 500 mm eccentricity, ten bolts, gauge 100 mm, pitch 60 mm. Direct shear 10 kN, torsional components yield resultant 71.5 kN in the extreme corner bolt.

A shaft transmits load of 100 kN at an eccentricity of 500 mm across a bracket plate bolted to a stanchion. Two rows of bolts 100 mm apart are provided with five bolts per row. The pitch of bolts in each row is 60 mm. Find the greatest force induced in bolt. [Figure: two vertical rows of 5 bolts at 60 mm pitch, rows 50 mm either side of the line of action's mid-line (50 | 50); 100 kN acts 500 mm from the bolt line]

Data: P=100P=100 kN, e=500e=500 mm, two rows of five bolts, gauge 100 mm, pitch 60 mm; the load lies in the plane of the bolt group (torsion). Take x=±50x=\pm50 mm, y=0,±60,±120y=0,\pm60,\pm120 mm from the centroid.

   P=100 kN                  o   o   y=+120
   <------ 500 ------>       o   o   +60
                             o   o   0
                             o   o   -60
                             o   o   -120

Step 1 - direct shear

Fd=Pn=10010=10.0F_d=\dfrac{P}{n}=\dfrac{100}{10}=10.0 kN (vertical, same in every bolt)

Step 2 - torsion

Σ(x2+y2)=10(502)+2×2(602+1202)=97000 mm2,T=100×500=50000 kN⋅mm\Sigma(x^2+y^2)=10(50^2)+2\times2(60^2+120^2)=97000\ \text{mm}^2,\qquad T=100\times500=50000\ \text{kN·mm}

Critical bolt: farthest corner bolt, x=50x=50, y=120y=120 (r=130r=130 mm).

  • Horizontal component: Fx=TyΣr2=50000×12097000=61.86F_x=\dfrac{Ty}{\Sigma r^2}=\dfrac{50000\times120}{97000}=61.86 kN
  • Vertical component: TxΣr2=25.77\dfrac{Tx}{\Sigma r^2}=25.77 kN

Step 3 - resultant

R=Fx2+(Fd+Fy,T)2=61.862+(10.0+25.77)2=71.46 kNR=\sqrt{F_x^2+(F_d+F_{y,T})^2}=\sqrt{61.86^2+(10.0+25.77)^2}=71.46\ \text{kN}

Answer: greatest force in a bolt ≈71.5\approx71.5 kN (at the extreme corner bolt).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Design of Steel and Timber Structure chapter 4 questions →