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 Critical Section (Tdn) (Clause 6.3 of IS 800:2007). 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 = 2x896 = 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

- Edge distance of bolts (e) = 1.5dh = 1.5 x 22 = 33 ≈ 40mm
- End distance of bolts = 1.5dh = 1.5 x 22 = 33 ≈ 40mm

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

- Kb = least of
- e/(3dh) = 40/(3×22) = 0.606
- (f/(3dh)) – 0.25 = (60/(3×22)) – 0.25 = 0.659
- Fub/Fu = 410/400 = 0.975
- 1

Kb = 0.606

- 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.