--- title: "Max tension in 220x12 plate with rivets" subject: "Design of Steel and Timber Structure" chapter: 4 key: "semester_6/design_of_steel_and_timber_structure:rivet-flat-tension-220x12" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/rivet-flat-tension-220x12 site: BCT Engineering --- # Max tension in 220x12 plate with rivets Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures - Asked in: 2068 Bhadra (old course), 14 marks - 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/rivet-flat-tension-220x12 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4 ## Question A flat plate (220mm × 12mm) is loaded in tension and connected with 10mm thick gusset plate as shown. If the rivets are 20mm dia power driven shop rivets, calculate the maximum tension the flat can carry. [Figure: 220 mm wide, 12 mm thick plate with 6 rivets in staggered arrangement; vertical spacings 30 mm; horizontal 50, 60, 60, 50 mm] ## Short answer Rivet shear capacity is 36.31 kN per rivet; six rivets provide 217.8 kN. Net section strength is 293.4 kN. Maximum tension the flat can carry is 217.8 kN (governed by rivet shear). ## Answer **Basis:** working stress method (IS 800:1984), Fe410 ($f_y=250$, permissible tension $0.6f_y=150$ N/mm²). 20 mm power-driven shop rivets: hole $d_0=21.5$ mm, $\tau_{vf}=100$, $\sigma_{pf}=300$ N/mm². The six rivets lie in the lap and act in single shear. The gusset (10 mm) is thinner than the flat (12 mm), so the gusset governs bearing. **Reading of the figure:** the dimensions 50 + 60 + 60 + 50 = 220 mm are across the width, so the rivets lie on three gauge lines 60 mm apart (at 50, 110 and 170 mm from one edge), staggered at 30 mm along the load. #### Rivet value - Shear: $R_s=\tau_{vf}\dfrac{\pi}{4}d_0^2=100\times363.1\times10^{-3}=36.31$ kN - Bearing: $R_b=\sigma_{pf}d_0t=300\times21.5\times10\times10^{-3}=64.50$ kN - Rivet value $=36.31$ kN → six rivets: $217.8$ kN #### Strength of the flat (on the net area) - Solid plate (for reference): $150\times220\times12\times10^{-3}=396.0$ kN - Straight section through two holes (outer gauge lines): $A_n=(220-2\times21.5)\times12=2124$ mm² → $P=318.6$ kN - Zig-zag section through three holes (one on each gauge line, $g=60$, $s=30$): $A_n=\left(220-3\times21.5+2\times\dfrac{30^2}{4\times60}\right)\times12=1956$ mm² → $P=293.4$ kN **Answer: maximum tension the flat can carry $=\min(217.8,\ 318.6,\ 293.4)=217.8$ kN**, governed by shearing of the rivets. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Bearing is calculated using 10 mm thickness. Why? A. The gusset plate is thinner than the flat B. The flat is 12 mm C. 10 mm is the minimum D. Gusset is stronger Answer: A. Bearing uses the thinner connected plate, here the 10 mm gusset. (Key point: Bearing thickness) 2. Shear value is 36.31 kN and bearing value is 64.50 kN. What is the rivet value? A. 64.50 kN B. 100.8 kN C. 50.4 kN D. 36.31 kN Answer: D. The smaller value governs, so rivet value = 36.31 kN. (Key point: Rivet value) 3. The zig-zag section through two holes gives a larger net area (2484 mm^2) than the single-hole section (2382 mm^2). Why? A. Added stagger term p^2/(4g) increases net width B. The rivets are smaller C. Gauge is subtracted D. The plate is thicker Answer: A. Along a staggered path the net width is reduced by hole diameters but increased by p^2/(4g), so the single-hole section is critical. (Key point: Net section) 4. Which of the strengths governs the maximum tension? A. Gross section, 396.0 kN B. Net section, 357.3 kN C. Zig-zag section, 372.6 kN D. Six rivets in shear, 217.8 kN Answer: D. The least value 217.8 kN, rivet shear, governs. 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