--- title: "Slab base for SC 220, 1000 kN, M30" subject: "Design of Steel and Timber Structure" chapter: 6 key: "semester_6/design_of_steel_and_timber_structure:slab-base-sc220-m30" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/slab-base-sc220-m30 site: BCT Engineering --- # Slab base for SC 220, 1000 kN, M30 Design of Steel and Timber Structure · Chapter 6: Compression Members - Asked in: 2071 Bhadra, 6 marks - Video: [Design of Slab Base for Column | Question-01 | IS:800 Based | CSVTU | Other Universities](https://www.youtube.com/watch?v=2l_vGcKqj7o) — PRABHAT SOLUTION - 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/slab-base-sc220-m30 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6 ## Question Design a slab base for a column SC220 to transfer an axial load of 1000 kN. Take Fe410 grade steel and M30 for concrete. ## Short answer Bearing strength13.5 MPa. Plate320×320 mm, pressure9.77 MPa. Thickness16 mm (≥ flange).4 M20 anchor bolts.,6 mm weld. ## Answer **Assumption:** SC 220 (IS 808 column section) is taken as $h=220$ mm, $b_f=220$ mm, $t_f\approx16$ mm. The load of 1000 kN is taken as the factored axial load. Concrete M30 ($f_{ck}=30$), steel Fe410/E250 ($f_y=250$). #### 1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1) $0.45f_{ck}=0.45\times30=13.50$ N/mm² #### 2. Plate size $$ A_{req}=\frac{P}{0.45f_{ck}}=\frac{1000\times10^3}{13.50}=74074\ \text{mm}^2 $$ Projection of about 50 mm all round the column (not less than 50 mm): $L=220+2\times50=320$ mm, $B=220+2\times50=320$ mm. Area provided $=102400$ mm² $>A_{req}$ ✓ $$ w=\frac{P}{BL}=\frac{1000\times10^3}{320\times320}=9.77\ \text{N/mm}^2<13.50\ \text{N/mm}^2 $$ #### 3. Thickness (cl. 7.4.3.1) Projections $a=50$ mm and $b=50$ mm (larger and smaller): $$ t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times9.77\times(50^2-0.3\times50^2)\times1.10}{250}}=13.71\ \text{mm} $$ It must also be not less than the flange thickness $t_f=16$ mm. **Provide $t_s=16$ mm.** #### 4. Connections - The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (the load is transferred mainly by contact). - Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift). **Answer: base plate 320 × 320 × 16 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M30 concrete.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For M30 concrete, what is the permissible bearing stress and the required plate area for 1000 kN? A. 9.0 N/mm^2, 111111 mm^2 B. 13.5 N/mm^2, 74074 mm^2 C. 20.1 N/mm^2, 49751 mm^2 D. 13.5 N/mm^2, 102400 mm^2 Answer: B. 0.45 x 30 = 13.5 N/mm^2, so A_req = 1000x10^3/13.5 = 74074 mm^2. 102400 mm^2 is the area actually provided. (Key point: Bearing strength of concrete) 2. The plate is 320 x 320 mm for a 220 x 220 mm column. How is this size fixed? A. By adding 50 mm projection all round the column, then checking the area exceeds A_req B. By making the plate area exactly equal to A_req C. By using twice the flange width D. By adding 100 mm on one side only Answer: A. A projection of at least 50 mm is added on every side (220 + 2x50 = 320). The area of 102400 mm^2 then exceeds 74074 mm^2, so w = 9.77 < 13.5 N/mm^2. (Key point: Plate size and projection) 3. The formula gives t_s = 13.71 mm, yet 14 mm is provided. Why? A. Plate thickness must always be even B. The thickness must not be less than the column flange thickness (14 mm) C. To resist the anchor bolt tension D. Because the weld size is 6 mm Answer: B. The slab thickness must be at least the flange thickness of the column, here about 14 mm, so 14 mm governs. 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