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 (T_dg) (Clause 6.2 of IS 800:2007).
2. Design Strength Due to Rupture of Critical Section (T_dn) (Clause 6.3 of IS 800:2007).
3. Design Strength Due to Block Shear (T_db) (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, T_u = T_dg = 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.

T_{dg =} A_gF_y/ꙋ_m0 → A_{g =} T_dgꙋ_m0/F_y

A_g = (300 x 10³ x 1.1)/250 = 1320 mm²

The total gross area of tension member (A_g) required is 1320 mm². Remember this is the area of two angle sections. Therefore,

Gross area of single angle section (A_g1) = (A_g)/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, A_g0 = 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,

t_a = 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, d_h = 20 + 2 = 22mm (Table 19 of IS 800:2007)

F_u = 410 N/mm²

F_ub = 400 N/mm²

F_y = 250 N/mm²

1. Edge distance of bolts (e) = 1.5d_h = 1.5 x 22 = 33 ≈ 40mm
2. End distance of bolts = 1.5d_h = 1.5 x 22 = 33 ≈ 40mm

• Minimum pitch (p) = 2.5d = 2.5 x 20 = 50 ≈ 60mm

1. K_b = least of
2. e/(3d_h) = 40/(3×22) = 0.606
3. (f/(3d_h)) – 0.25 = (60/(3×22)) – 0.25 = 0.659
4. F_ub/F_u = 410/400 = 0.975
5. 1

K_b = 0.606

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

= 2 x (F_ub/√3) x (A_nb/ꙋ_mb) where, A_nb = 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 K_b d t F_ub)/ ꙋ_mb

= (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 = (T_u/Bolt Value) = (300/90.545) = 3.31 ≈ 4 nos

STEP 3:

Check of Strength due to rupture of critical section,

The design strength,

T_dn = 0.9A_ncf_u/ꙋ_m1 + βA_gof_y/ ꙋ_m0

Where,

β = 1.4 – 0.076(w/t)( f_y/f_u)( b_s/L_c) ≤ (f_u ꙋ_m0/f_y ꙋ_m1)

≥ 0.7

w = outstand leg width = 60mm

t = total thickness of angles = 16mm

w₁ = end distance = 40mm

b_s = shear lag distance = w + w₁ – t = 60 + 40 – 16 = 84mm

L_c = 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)

A_nc = (60+60-2 x 22) x 8 = 608mm²

A_go = (60 x 16) = 960mm²

T_dn = 0.9A_ncf_u/ꙋ_m1 + βA_gof_y/ ꙋ_m0

= ((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 (T_db),

T_db = [(A_vgf_y)/(_√3ꙋ_m0) + (0.9A_tnf_u)/( ꙋ_m1)] or [(0.9A_vnf_u)/(_√3ꙋ_m1) + (A_tgf_y)/(ꙋ_m0)]

A_vg = 220 x 16 = 3520 mm²

A_vn = (220-3×22-(22/2)) x 16 = 2288 mm²

A_tg = 40 x 16 = 640 mm²

A_tn = (40-(22/2)) x 16 = 464 mm²

T_db = [((3520×250)/ _(√3x1.1)) + ((0.9x464x410/1.25)]

= 461880.22 + 136972.8

= 598853.02 N

= 598.85 KN

T_db = [((0.9x2288x410)/ _(√3x1.25)) + ((640×250/1.1)]

= 389952.53 + 145454.54

= 535407.07 N

= 535.41 KN

T_db = 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.