What is the procedure to design the double angle tension member in steel structures with formulas? Kindly elaborate on it?

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## 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,

_{dg}) (Clause 6.2 of IS 800:2007)._{dn}) (Clause 6.3 of IS 800:2007)._{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/mm

^{2}STEP 1:_{u }= T_{dg}= 1.5 x 200 = 300 N/mm^{2}T

_{dg = }A_{g}F_{y}/ꙋ_{m0 }→ A_{g = }T_{dg}ꙋ_{m0}/F_{y}A

_{g}= (300 x 10^{3}x 1.1)/250 = 1320 mm^{2}The total gross area of tension member (A

_{g}) required is 1320 mm^{2}. Remember this is the area of two angle sections. Therefore,Gross area of single angle section (A

_{g1}) = (A_{g})/2 = 1320/2 = 660 mm^{2}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= 896mm

^{2}Total Gross Area, A

_{g0 }= 2×896 = 1792 mm^{2 }> 1320 mm^{2 }(O.K)STEP 2:Total thickness of angles having outstanding legs placed back to back,

t

_{a}= 8+8 =16mmLet 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/mm^{2}F

_{ub}= 400 N/mm^{2}F

_{y}= 250 N/mm^{2}_{h}= 1.5 x 22 = 33 ≈ 40mm_{h}= 1.5 x 22 = 33 ≈ 40mm_{b}= least of_{h}) = 40/(3×22) = 0.606_{h})) – 0.25 = (60/(3×22)) – 0.25 = 0.659_{ub}/F_{u}= 410/400 = 0.975K

_{b}= 0.606= 2 x (F

_{ub}/√3) x (A_{nb}/ꙋ_{mb}) where, A_{nb}= 0.78 x (πd^{2}/4)= 2 x (400/√3) x ((0.78 x (π(20)

^{2}/4))/1.25)= 90.545 KN

= (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 nosSTEP 3:Check of Strength due to rupture of critical section,

The design strength,

T

_{dn}= 0.9A_{nc}f_{u}/ꙋ_{m1}+ βA_{go}f_{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

w

_{1}= end distance = 40mmb

_{s}= shear lag distance = w + w_{1}– t = 60 + 40 – 16 = 84mmL

_{c}= length of end connection = 3 x 60 = 180mmβ = 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 = 608mm^{2}A

_{go}= (60 x 16) = 960mm^{2}^{ }T_{dn}= 0.9A_{nc}f_{u}/ꙋ_{m1}+ βA_{go}f_{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_{vg}f_{y})/(_{√3}ꙋ_{m0}) + (0.9A_{tn}f_{u})/( ꙋ_{m1})] or [(0.9A_{vn}f_{u})/(_{√3}ꙋ_{m1}) + (A_{tg}f_{y})/(ꙋ_{m0})]A

_{vg}= 220 x 16 = 3520 mm^{2}A

_{vn}= (220-3×22-(22/2)) x 16 = 2288 mm^{2}A

_{tg}= 40 x 16 = 640 mm^{2}A

_{tn}= (40-(22/2)) x 16 = 464 mm^{2}T

_{db}= [((3520×250)/_{ (√3}x1.1)) + ((0.9x464x410/1.25)]= 461880.22 + 136972.8

= 598853.02 N

= 598.85 KN

T

_{db}= [((0.9x2288x410)/_{ (√3}x1.25)) + ((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.