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    What is the Procedure to Design the Double Angle Tension Member in Steel Structures with formulas?

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    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,

    1. Design strength due to yielding of the gross section (Tdg) (Clause 6.2 of IS 800:2007).
    2. Design Strength Due to Rupture of Critical Section (Tdn) (Clause 6.3 of IS 800:2007).
    3. Design Strength Due to Block Shear (Tdb) (Clause 6.4 of IS 800:2007).

    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

    1. Edge distance of bolts (e) = 1.5dh = 1.5 x 22 = 33 ≈ 40mm
    2. End distance of bolts = 1.5dh = 1.5 x 22 = 33 ≈ 40mm
    • Minimum pitch (p) = 2.5d = 2.5 x 20 = 50 ≈ 60mm
    1. Kb = least of
    2. e/(3dh) = 40/(3×22) = 0.606
    3. (f/(3dh)) – 0.25 = (60/(3×22)) – 0.25 = 0.659
    4. Fub/Fu = 410/400 = 0.975
    5. 1

    Kb = 0.606

    1. Design strength of Bolt (i.e Bolt Value)
      • Design shearing strength of bolt in double shear

    = 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

      • Design bearing capacity of bolt

    = (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.

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