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Module 1: Column Bases

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Video1
In this lecture I will be discussing about a design example of the gusset base. So in last lecture we have discussed the design procedure of the gusset base and which includes the determination of the gusset plate dimensions and the base plate dimensions and the number of bolts. So how to find out step by step that will be clear if we go through this example. Example. A column section ISHB 350 @ 710.2 N/m carries a factor axial compressive load of 1700 kN and factored bending moment of 85 kN-m. Design the base plate and its connections. Assume concrete pedestal of M-20 grade. Solution: Assume Fe 410 grade of steel: fu=410 MPa, f y=250 MPaFor M20 grade of concrete: Bearing strength of concrete¿0.45f ck=0.45× 20=9 N/mm2 Partial safety factor: (Table 5, IS 800: 2007)γm0=1.1 γmb=1.25 Properties of ISHB 350 @ 710.2 N/m: [table 1, SP-6(1)-1964] tf=11.6 mm tw=10.1 mm D=350 mm bf=250 mm A=9221 mm2 Design compressive load, P=1700 kN Design bending moment, M=85 kN-m Eccentricity, e= M P = 85×106 1700× 103=50 mm. Let us provide 16 mm thick gusset plate, one on each side of the column flanges and two cleat angles ISA 200×150×15 mm.Provide a length of base plate, L=550 mm. Now, eL=50 550 = 1 11 <16 Hence, the base plate is in compression throughout. To limit the bearing pressure from concrete to 0.45 f ck, the width required,B= 2 P L× 0.45f ck = 2×1700×103 550×0.45× 20=686.87 mm ≃690 mm Projection of base plate beyond leg angle toe¿ 690−(350+2×16+2×150) 2 =4 mm Let us provide a base plate 690×550 mm in size. Area provided, A=690× 550=379.5× 103 mm2Section modulus of the base plate, Ze= 550 ×6902 6 =43.64 ×106 mm3 Maximum pressure, fmax= PA + MZe = 1700× 103 379.5 ×103 + 85×106 43.64 ×106 ¿6.43 N/mm2 < 9 N/mm2 fmin= P A− MZe =1700×103 379.5×103− 85×106 43.64 ×106 ¿2.53 N/mm2 < 9  N/mm2 Calculation for the thickness of base plate: Base pressure at section x-x, ¿2.53+(6.43−2.53)× 551 690 =5.64 N/mm2 Moment at critical section x-x, ¿5.64×139 × 139 2 + 12 ×139 ×(6.43−5.64 )×(32 ×139) ¿59573.1 N-mm Moment capacity of base plate, M d=1.2 f y γm0 Ze ¿1.2× 250 1.1 ×(1 6 ×1×t2a)=45.45t2a N-mm ⇒ 45.45t 2a=59573.1 ⇒ t a=36.2 mm (aggregate thickness of base plate and cleat angle) Therefore, thickness of base plate, tb=36.2−15=21.2 mm ≃22 mm > 11.6 mm Provide a base plate 690×550×22 mm in size. 
Video 2  
Bolted connection:Provide 24-mm diameter bolts of grade 4.6. The bolts will be in single shear. Strength bolt in single shear (cl. 10.3.3, IS 800:2007) ¿ A nb(f√ub 3) γmb = (0.78 × π ×4242)×(400 √3 ) 1.25 ×10−3= 65.2 kN Strength of bolt in bearing ¿ 2.5kb dt f u /γ mb (cl. 10.3.4, IS 800:2007) For 24 mm diameter bolts the minimum edge distance,e=1.5× d 0=1.5× (24+2)=39 mm The minimum pitch, p=2.5 ×24=60 mm Let us provide an edge distance of 40 mm and pitch of 65 mm.k b is smaller of (3ed 0 = 3×4026 =0.51),(3pd 0 −0.25= 3×6526 −0.25=0.58), (ffub u = 400 410 =0.98) and 1.0 Hence kb=0.51 ∴ Strength in bearing ¿ 2.5×0.51×24 ×11.6 × 410 1.25 ×10−3 ¿ 116.43 kN Hence, the strength of bolt ¿ 65.2 kN Assuming column end and gusset material to have complete bearing, 50% of the load will beassumed to pass directly and 50% of the load will pass through the connections. Number of bolts required to connect column flanges with gusset plates, n 1= 0.5 ×1700 65.2 =13.03≃16 Provide 8, 24 mm diameter bolts on each flange in two rows as shown in the figure.The number of bolts required to connect the cleat angle with gusset plate will be the same. Dimension of gusset plate: Height of gusset plate ¿200+2 ×40+65=345 mm Length of gusset plate ¿ length of base plate ¿550 mm Provide gusset plate 550×345×16 mm in size. So this is how one can design a gusset plate when the column is under compression and moment, okay. So this is all about the todays lecture and before going to finish this entire lecture I would like to suggest one thing that please go through the IS code 800-2007 IS: 800-2007, this is not a mathematical course or some structural analysis course that everything you will be understanding through logic, here logic is definitely there but numerical equations are there expressions are there which is derived from some experimental observations like some coefficients are obtained from experimental observations, right. So through this lecture means entire lecture means I hope that you will get a concept of design of limit state method for steel structure. So how to design a steel member using limit state method that concepts I hope it will be clear however it will be much clear if you work out some example then the concept will be much clear, I thank for your patients and I hope you will enjoy this lecture, thank you.