--- title: "Splice connection for ISHB 250, 900 kN" subject: "Design of Steel and Timber Structure" chapter: 6 key: "semester_6/design_of_steel_and_timber_structure:column-splice-ishb250-900kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/column-splice-ishb250-900kn site: BCT Engineering --- # Splice connection for ISHB 250, 900 kN Design of Steel and Timber Structure · Chapter 6: Compression Members - Asked in: 2074 Bhadra, 8 marks - Video: [#Design of column splice#Design of steel structures#Welding Connection#](https://www.youtube.com/watch?v=KFndzy8iTzE) — CIVIL ENGINEERING COURSE Theory & Practical - 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/column-splice-ishb250-900kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6 ## Question An ISHB 250 @ 536 N/m column carrying a factored axial load of 900KN. The column ends machined. Design the splice connection. Use M16 bolts. ## Short answer Splice designed for 50% load (450 kN) sharing by area. Flange splice: two plates 250×10 mm, each side per flange 6 M16 bolts. Web splice: two plates 170×8 mm, each side 3 M16 bolts. ## Answer **Data:** ISHB 250 @ 536 N/m ($h=250$, $b_f=250$, $t_f=9.7$, $t_w=8.8$ mm, $A=6971$ mm²), factored axial load 900 kN, ends machined for bearing; M16 bolts (grade 4.6, $d_0=18$ mm); E250. #### Design basis With machined ends in contact, the load is mostly transferred by bearing; the splice plates and bolts are designed for **50 %** of the factored load, shared by area: - $P_s=0.5\times900=450$ kN - Flanges $A_f=2\times250\times9.7=4850$ mm²; web $A_w=(250-2\times9.7)\times8.8=2029$ mm² - Flange share $P_f=P_s\dfrac{A_f}{A}=313.1$ kN (156.5 kN per flange); web share $P_w=131.0$ kN #### Bolt values (M16, 4.6) - Single shear $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=29.01$ kN; double shear (threads conservatively in both shear planes) $=2\times29.01=58.0$ kN - Bearing ($e=35$, $p=50$): $k_b=0.648$; flange 9.7 mm: $82.5$ kN; web 8.8 mm: $74.8$ kN - Flange bolt value $29.01$ kN; web bolt value $58.0$ kN #### Flange splice $n=\dfrac{156.5}{29.01}=5.4$ → **6 bolts each side per flange** (2 rows of 3, pitch 50 mm, gauge 100 mm, end 35 mm). Outer flange plate $250\times10$ mm, about 340 mm long. Plate stress $=\dfrac{156.5\times10^3}{250\times10}=62.6$ N/mm² $131$ kN ✓); web plates $170\times8$ mm, 200 mm long. **Answer: flange plates 250 × 10 mm with 6 M16 bolts each side per flange; two web plates 170 × 8 mm with 3 M16 bolts each side.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are the splice plates and bolts designed for only 50% of the factored load (450 kN)? A. Because bolts are twice as strong in bearing B. Because the column is assumed to carry only half the load C. Because machined ends in contact transfer most of the load by bearing D. Because the flanges alone carry the load Answer: C. With ends machined and in contact, bearing transfers most of the load, so the splice carries 50% as a minimum design requirement. (Key point: Design load for splice with machined ends) 2. How is the splice load of 450 kN divided between flanges and web? A. Equally between flanges and web B. In proportion to the areas of flanges and web C. Entirely to the web D. In proportion to the bolt diameters Answer: B. The share is by area: flange share = 450 x A_f/A = 313.1 kN and the web takes 131.0 kN. (Key point: Sharing of load by area) 3. The load per flange is 156.5 kN and the bolt value in single shear is 29.01 kN. How many bolts are provided each side per flange? A. 6, since 5.4 is rounded up B. 5, since 5.4 is rounded down C. 3, as two rows share the load D. 2, because of double shear Answer: A. n = 156.5/29.01 = 5.4, rounded up to 6 bolts (2 rows of 3). The flange plate is on one face only, so single shear governs. 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