--- title: "Greatest bolt force, 100 kN at 500 mm" subject: "Design of Steel and Timber Structure" chapter: 4 key: "semester_6/design_of_steel_and_timber_structure:bolt-force-shaft-100kn-500mm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bolt-force-shaft-100kn-500mm site: BCT Engineering --- # Greatest bolt force, 100 kN at 500 mm Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures - Asked in: 2073 Magh, 8 marks - Video: [Forces in Bolt Group (Shear, Moment and Bracket Connection)](https://www.youtube.com/watch?v=PNYwCOBhWW4) — Civil Engineer Tonmoy Maity - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bolt-force-shaft-100kn-500mm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4 ## Question 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] ## 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. ## Answer **Data:** $P=100$ kN, $e=500$ mm, two rows of five bolts, gauge 100 mm, pitch 60 mm; the load lies in the plane of the bolt group (torsion). Take $x=\pm50$ mm, $y=0,\pm60,\pm120$ mm from the centroid. ```text P=100 kN o o y=+120 <------ 500 ------> o o +60 o o 0 o o -60 o o -120 ``` #### Step 1 - direct shear $F_d=\dfrac{P}{n}=\dfrac{100}{10}=10.0$ kN (vertical, same in every bolt) #### Step 2 - torsion $$ \Sigma(x^2+y^2)=10(50^2)+2\times2(60^2+120^2)=97000\ \text{mm}^2,\qquad T=100\times500=50000\ \text{kN·mm} $$ Critical bolt: farthest corner bolt, $x=50$, $y=120$ ($r=130$ mm). - Horizontal component: $F_x=\dfrac{Ty}{\Sigma r^2}=\dfrac{50000\times120}{97000}=61.86$ kN - Vertical component: $\dfrac{Tx}{\Sigma r^2}=25.77$ kN #### Step 3 - resultant $$ R=\sqrt{F_x^2+(F_d+F_{y,T})^2}=\sqrt{61.86^2+(10.0+25.77)^2}=71.46\ \text{kN} $$ **Answer: greatest force in a bolt $\approx71.5$ kN** (at the extreme corner bolt). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Ten bolts share 100 kN equally. What is the direct shear on each? A. 50 kN B. 5 kN C. 25 kN D. 10 kN Answer: D. F_d = P/n = 100/10 = 10 kN, vertical, in every bolt. (Key point: Direct shear) 2. What twisting moment T acts on the bolt group? A. 100 x 500 = 50000 kN·mm B. 100 x 100 = 10000 kN·mm C. 100 x 50 = 5000 kN·mm D. 100 x 60 = 6000 kN·mm Answer: A. The load acts 500 mm from the bolt line, so T = P x e = 50000 kN·mm. (Key point: Torsional moment) 3. Which bolt has the greatest force? A. The bolt at the centroid B. The middle bolt of a row C. The farthest corner bolt at x = 50, y = 120 D. A bolt on the horizontal axis Answer: C. Torsional force grows with distance from the centroid, so the extreme corner bolt (r = 130 mm) is critical. (Key point: Critical bolt) 4. Torsion gives F_x = 61.86 kN and a vertical part 25.77 kN; direct shear is 10 kN. What is the greatest bolt force? A. 61.86 + 25.77 + 10 = 97.6 kN B. sqrt(61.86^2 + 25.77^2) = 67.0 kN C. sqrt(61.86^2 + (10 + 25.77)^2) = 71.5 kN D. sqrt(61.86^2 + (25.77 - 10)^2) = 63.8 kN Answer: C. Direct and torsional vertical components act in the same direction at this bolt, so they add, then combine at right angles with F_x. 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