--- title: "Splice plate for ISHB 350 column" subject: "Design of Steel and Timber Structure" chapter: 6 key: "semester_6/design_of_steel_and_timber_structure:column-splice-ishb350-1000kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/column-splice-ishb350-1000kn site: BCT Engineering --- # Splice plate for ISHB 350 column Design of Steel and Timber Structure · Chapter 6: Compression Members - Asked in: 2080 Chaitra, 8 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/column-splice-ishb350-1000kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6 ## Question Design splice plate and the connection for a column section ISHB 350 to support a factored axial load of 1000 kN. Use M20 bolts of grade 4.6 and steel of grade E250. Assume the column ends are machined. The connection is to be spliced at a height of 3 meters. ## Short answer Design splice for 50% of factored load, 500 kN. Flange splice plates 250 x 10 mm with 4 M20 bolts each side per flange; web splice plates 2 x (200 x 8 mm) with 2 M20 bolts each side. ## Answer **Data:** ISHB 350 @ 72.4 kg/m ($h=350$, $b_f=250$, $t_f=11.6$, $t_w=10.1$ mm, $A=9221$ mm²), factored axial load $1000$ kN at the splice, ends machined for bearing. M20 bolts grade 4.6 ($d_0=22$ mm); E250. #### Design basis Because the column ends are machined and in full contact, most of the load passes by direct bearing. The splice plates and bolts are designed to carry **at least 50 % of the factored load** ($P_s$) and to hold the two lengths in line (against bending from accidental eccentricity and erection loads). $$ P_s=0.5\times1000=500\ \text{kN} $$ The load is shared in proportion to area: flanges $A_f=2\times250\times11.6=5800$ mm², web $A_w=(350-2\times11.6)\times10.1=3301$ mm². - Flanges: $P_f=P_s\dfrac{A_f}{A}=314.5$ kN → 157.2 kN per flange - Web: $P_w=P_s\dfrac{A_w}{A}=179.0$ kN #### Bolt values (M20, 4.6) - Single shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN; double shear (threads conservatively in both shear planes): $2\times45.26=90.5$ kN - Bearing ($e=40$, $p=60$): $k_b=0.606$; on flange 11.6 mm $=115.3$ kN, on web 10.1 mm $=100.4$ kN - Flange bolt value $45.26$ kN (single shear, outer splice plate); web bolt value $90.5$ kN (double shear, two web plates). #### Flange splice $n=\dfrac{157.2}{45.26}=3.5$ → **4 bolts on each side of the joint per flange** (2 rows of 2, pitch 60, gauge 120, end 40 mm). Flange splice plate $250\times10$ mm, length $=2\times(40+60+40)+$ gap $\approx300$ mm. Stress in the plate $=\dfrac{157.2\times10^3}{250\times10}=62.9$ N/mm² $179$ kN ✓); web plates (two, one on each face) $200\times8$ mm, 250 mm long, with 2 rows of bolts. #### Notes - The splice is at 3 m height, so bending from lateral loads is small; the plates and bolts also hold the two lengths in line and resist erection loads. - Packing plates are provided if the two lengths differ in size. **Answer: flange splice plates 250 × 10 mm with 4 M20 bolts each side per flange; web splice plates 2 × (200 × 8 mm) with 2 M20 bolts each side.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The column ends are machined and in full contact. What load must the splice plates and bolts carry? A. Nothing, since bearing carries all B. At least 50% of the factored load (500 kN) C. The full 1000 kN D. 10% of the load only Answer: B. Most load goes by direct bearing, but the splice is designed for at least 50% of the factored load, P_s = 500 kN, and to hold the lengths in line. (Key point: Splice design load) 2. How is the splice load shared between flanges and web? A. Equally between flanges and web B. In proportion to their areas C. All by the web D. In proportion to the bolt diameters Answer: B. P_f = P_s A_f/A = 314.5 kN (157.2 kN per flange) and P_w = P_s A_w/A = 179.0 kN. (Key point: Distribution to flanges and web) 3. How many bolts are needed per flange per side, with value 45.26 kN and load 157.2 kN per flange? A. 2 B. 4 C. 6 D. 8 Answer: B. n = 157.2/45.26 = 3.5, rounded up to 4 bolts, arranged in 2 rows of 2. (Key point: Flange bolts) 4. Why is the web bolt value 100.4 kN rather than 45.26 kN? A. Web bolts are larger B. Two web plates, one on each face, put the bolts in double shear, governed by bearing on the 10.1 mm web C. Web steel is higher grade D. Bolt value is simply doubled for any web Answer: B. Double shear capacity is 103.3 kN, but bearing on the 10.1 mm web is 100.4 kN, which governs. 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