What is the procedure to design the double angle tension member in steel structures with formulas? Kindly elaborate on it?
Both have different properties & a separate cost which describes below. steel As compared aluminium,steel have less corrosion resistance Steel are chiper than aluminium Steel give lesser malleability and elastic properties than aluminium In weight steel have higher weight Steel is tough & haRead more
Both have different properties & a separate cost which describes below.
steel
- As compared aluminium,steel have less corrosion resistance
- Steel are chiper than aluminium
- Steel give lesser malleability and elastic properties than aluminium
- In weight steel have higher weight
- Steel is tough & harder than alluminium
- Generally we use steel as a reinforced member in concrete structures
- Esay to weld with other structural components.
Aluminum
- Higher corrosion resistance
- Higher cost
- Higher
- malleability and elastic properties
- Lesser weight
- Generally we not use in concrete
- Difficult to weld
- Aluminum is soft metal compare to steel.
We are generally use steel as a reinforcement.
We doesn’t use aluminium because of soft metal property
See less
Amit Bhuriya
As per Section 6 (Design of Tension Members) of IS 800:2007 – Code of Practice for Construction in Steel, The design strength of the tension member is the minimum of following, Design strength due to yielding of the gross section (Tdg) (Clause 6.2 of IS 800:2007). Design Strength Due to Rupture of CRead more
As per Section 6 (Design of Tension Members) of IS 800:2007 – Code of Practice for Construction in Steel,
The design strength of the tension member is the minimum of following,
Let’s take an example to understand the designing of double angle tension member:
Data Known:
Service Load, T = 200 N/mm2
STEP 1:
Factored Load, Tu = Tdg = 1.5 x 200 = 300 N/mm2
Considering the tension member fails due to yielding of gross section, determine the gross area of angles required.
Tdg = AgFy/ꙋm0 → Ag = Tdgꙋm0/Fy
Ag = (300 x 103 x 1.1)/250 = 1320 mm2
The total gross area of tension member (Ag) required is 1320 mm2. Remember this is the area of two angle sections. Therefore,
Gross area of single angle section (Ag1) = (Ag)/2 = 1320/2 = 660 mm2
From steel table (SP 6-1), choose an angle having gross area of single angle about 25% to 40% more than computed above.
Taking Rolled Steel Equal Angle ISA 60x60x8 having following properties,
Sectional Area, A= 896mm2
Total Gross Area, Ag0 = 2×896 = 1792 mm2 > 1320 mm2 (O.K)
STEP 2:
Designing Connections: – We can provide either bolted or welded connections, so let us provide bolted connections.
Total thickness of angles having outstanding legs placed back to back,
ta = 8+8 =16mm
Let us provide 20mm diameter bolts of grade 4.6 and Steel of grade Fe415,
Diameter of bolt, d = 20mm
Diameter of bolt hole, dh = 20 + 2 = 22mm (Table 19 of IS 800:2007)
Fu = 410 N/mm2
Fub = 400 N/mm2
Fy = 250 N/mm2
Kb = 0.606
= 2 x (Fub/√3) x (Anb/ꙋmb) where, Anb = 0.78 x (πd2/4)
= 2 x (400/√3) x ((0.78 x (π(20)2/4))/1.25)
= 90.545 KN
= (2.5 Kb d t Fub)/ ꙋmb
= (2.5 x 0.606 x 20 x 16 x 400)/1.25
= 155.136 KN
Therefore, Bolt Value = Least of (90.545, 155.136) = 90.545 KN
No. of bolts required, N = (Tu/Bolt Value) = (300/90.545) = 3.31 ≈ 4 nos
STEP 3:
Check of Strength due to rupture of critical section,
The design strength,
Tdn = 0.9Ancfu/ꙋm1 + βAgofy/ ꙋm0
Where,
β = 1.4 – 0.076(w/t)( fy/fu)( bs/Lc) ≤ (fu ꙋm0/fy ꙋm1)
≥ 0.7
w = outstand leg width = 60mm
t = total thickness of angles = 16mm
w1 = end distance = 40mm
bs = shear lag distance = w + w1 – t = 60 + 40 – 16 = 84mm
Lc = length of end connection = 3 x 60 = 180mm
β = 1.4 – 0.076(60/16)(250/410)(84/180)
= 1.4 – 0.081
= 1.319 ≤ (410×1.1/250×1.25) = 1.4432 (OK)
≥ (0.7) (OK)
Anc = (60+60-2 x 22) x 8 = 608mm2
Ago = (60 x 16) = 960mm2
Tdn = 0.9Ancfu/ꙋm1 + βAgofy/ ꙋm0
= ((0.9 x 608 x 410)/1.25) + ((1.319 x 960 x 250)/1.1)
= 179481.6 + 287781.81
= 467343.6 N
= 467.34 KN > 300 KN (O.K)
STEP 4:
Check for Strength Due to Block Shear (Tdb),
Tdb = [(Avgfy)/(√3ꙋm0) + (0.9Atnfu)/( ꙋm1)] or [(0.9Avnfu)/(√3ꙋm1) + (Atgfy)/(ꙋm0)]
Avg = 220 x 16 = 3520 mm2
Avn = (220-3×22-(22/2)) x 16 = 2288 mm2
Atg = 40 x 16 = 640 mm2
Atn = (40-(22/2)) x 16 = 464 mm2
Tdb = [((3520×250)/ (√3x1.1)) + ((0.9x464x410/1.25)]
= 461880.22 + 136972.8
= 598853.02 N
= 598.85 KN
Tdb = [((0.9x2288x410)/ (√3x1.25)) + ((640×250/1.1)]
= 389952.53 + 145454.54
= 535407.07 N
= 535.41 KN
Tdb = min (535.41, 598.85) = 535.41 KN > 300 KN (O.K)
Therefore, the selected section is safe.
So, Provide 2 ISA 60x60x8 angles placed in such a way that outstanding legs are placed back to back and attached with 20mm diameter bolts of grade 4.6. Edge distance and End distance is 40mm and pitch is 60mm.
See less