--- title: "Gusseted base for ISHB 350, 2000 kN" subject: "Design of Steel and Timber Structure" chapter: 6 key: "semester_6/design_of_steel_and_timber_structure:gusseted-base-ishb350-2000kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/gusseted-base-ishb350-2000kn site: BCT Engineering --- # Gusseted base for ISHB 350, 2000 kN Design of Steel and Timber Structure · Chapter 6: Compression Members - Asked in: 2077 Chaitra, 10 marks - Video: [easiest way to solve gusset base designing problem](https://www.youtube.com/watch?v=mTw9vQgPZ3E) — Civil Engineering YouTube Channel - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/gusseted-base-ishb350-2000kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6 ## Question Design a gusseted base for a column ISHB 350 @ 710 N/m with two plates 450mm×20mm carrying factored load 2000KN. The column is to be supported on Concrete pedestal with M20 grade concrete. ## Short answer Bearing strength9 MPa. Base plate500×520×16 mm. Shaft: ISHB350 with two450×20 cover plates. Gusset angles ISA100×100×10,450 mm long.10 mm fillet weld;4 M24 anchor bolts. ## Answer **Data and assumptions:** ISHB 350 @ 710 N/m ($h=350$, $b_f=250$, $t_f=11.6$ mm) with **two 450 × 20 mm plates** (one on each flange) forming the column shaft: overall shaft size $D=350+2\times20=390$ mm by $B_s=450$ mm. Factored load $P=2000$ kN; pedestal M20; E250 steel. Gusset angles (ISA 100×100×10) are used on the long sides. ```text |==450x20 plate==| | ISHB 350 | shaft 390 x 450 |==450x20 plate==| ISA cleats ---> |_| |_| <--- ISA cleats =================================== base plate L x B x t ``` #### 1. Bearing and plate area $0.45f_{ck}=9.0$ N/mm²; $A_{req}=\dfrac{2000\times10^3}{9.0}=222222$ mm². Provide base plate $L\times B=500\times520$ mm (area $260000$ mm² $>A_{req}$): $$ w=\frac{2000\times10^3}{500\times520}=7.69\ \text{N/mm}^2<9.0\ \text{N/mm}^2\ \checkmark $$ #### 2. Thickness of base plate Projections beyond the shaft: $a=\dfrac{500-390}{2}=55$ mm, $b=\dfrac{520-450}{2}=35$ mm. $$ t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.69\times(55^2-0.3\times35^2)\times1.10}{250}}=15.0\ \text{mm} $$ Because the plate is also loaded between the gusset angles, **provide $t_s=16$ mm** (minimum 16 mm for a gusseted base). #### 3. Transfer of load from the shaft to the base plate The column end is not machined, so the full 2000 kN is carried by fillet welds (shop welds, $s=10$ mm): $$ q=\frac{f_u}{\sqrt3\gamma_{mw}}\times0.7s=189.4\times0.7\times10=1326\ \text{N/mm},\qquad L_w=\frac{2000\times10^3}{1326}=1509\ \text{mm} $$ Weld all round the shaft and the cover plates: perimeter $2(390+450)=1680$ mm $\ge$ required; in addition provide the two gusset angles ISA 100×100×10, 450 mm long, welded to the cover plates and to the base plate with 8 mm fillets, which carry the balance. If the column is milled to bear, the welds need carry only a part. #### 4. Anchor bolts 4 nos. M24 (or more) near the corners, embedded 600 mm with plate washers. **Answer: base plate 500 × 520 × 16 mm; shaft ISHB 350 + 2 PL 450×20; ISA 100×100×10 gusset angles; 10 mm fillet weld; 4 M24 anchor bolts.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the M20 pedestal, what permissible bearing stress is used to find the required base plate area? A. 0.45 x 20 = 9.0 N/mm^2 B. 0.67 x 20 = 13.4 N/mm^2 C. 20/1.5 = 13.3 N/mm^2 D. 0.36 x 20 = 7.2 N/mm^2 Answer: A. IS 800 cl. 7.4.1 limits bearing stress to 0.45 f_ck, which is 9.0 N/mm^2 for M20. Dividing 2000 kN by it gives A_req = 222222 mm^2. (Key point: Bearing strength of concrete) 2. The shaft is 390 x 450 mm and the base plate is 500 x 520 mm. What projections a and b are used in the thickness formula? A. a = 110 mm, b = 70 mm B. a = 55 mm, b = 35 mm C. a = 55 mm, b = 70 mm D. a = 35 mm, b = 55 mm Answer: B. Each projection is half of the difference: (500-390)/2 = 55 mm and (520-450)/2 = 35 mm. (Key point: Plate projections) 3. Why must the fillet welds carry the full 2000 kN in this gusseted base? A. Anchor bolts are designed to carry the compression B. Because the concrete cannot take more than 9 N/mm^2 C. Because the column end is not machined, so there is no direct bearing D. Because the gusset angles are made of weaker steel Answer: C. Without machining, contact bearing cannot be relied on, so the welds transmit the whole load to the base plate. If the end is milled to bear, welds need carry only part. 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