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

## 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,

_{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.

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