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Tension member with lug angle, 400 kN

Short answer

Use ISA 100×100×12 main angle with lug ISA 75×75×10. Provide 5 M20 bolts for main angle to gusset, 6 for lug to gusset, and 7 for lug-to-main connection.

The bottom chord of a truss is subjected to an axial pull of 400 kN. The length of joint available is only 300 mm. Design the tension member using single equal angle section with requirements of lug angle if necessary (connected with 16 mm thick gusset plate). Use M 20 bolts of 4.6 grade (and steel as Fe410).

Given: axial pull (factored) T=400T=400 kN, joint length available 300 mm, gusset 16 mm, M20 bolts of grade 4.6, Fe410/E250 (fy=250f_y=250, fu=410f_u=410). A single angle connected by one leg would need 9 bolts (400/45.26400/45.26), i.e. about 480 mm of joint length, which is more than 300 mm, so a lug angle is used at the end (IS 800:2007 cl. 10.12).

Section

Required Ag≥400×103×1.10250=1760A_g\ge\dfrac{400\times10^3\times1.10}{250}=1760 mm². Try ISA 100×100×12: Ag=2257A_g=2257 mm².

  • Yielding: Tdg=2257×2501.10=513.0T_{dg}=\dfrac{2257\times250}{1.10}=513.0 kN >400>400 ✓
  • Rupture: with a lug angle both legs are effectively connected, so there is no shear-lag reduction. An=2257−12×22=1993A_n=2257-12\times22=1993 mm²; Tdn=0.9×1993×4101.25=588.3T_{dn}=\dfrac{0.9\times1993\times410}{1.25}=588.3 kN >400>400 ✓

Bolts (M20, 4.6)

  • Shear Vdsb=4003×2451.25×10−3=45.26V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26 kN
  • Bearing (e=40e=40, p=50p=50, t=12t=12 mm): kb=0.508k_b=0.508, Vdpb=99.9V_{dpb}=99.9 kN → bolt value 45.2645.26 kN

Forces with lug angle

Outstanding leg force ≈TAoutAg\approx T\dfrac{A_{out}}{A_g} with Aout≈(100−6)×12=1128A_{out}\approx(100-6)\times12=1128 mm²: Pout=199.9P_{out}=199.9 kN.

  • Connected leg force =400−199.9=200.1=400-199.9=200.1 kN → main angle to gusset: n=200.1/45.26=4.4n=200.1/45.26=4.4 → 5 bolts
  • Lug angle to gusset: 1.2 Pout=239.91.2\,P_{out}=239.9 kN → 6 bolts
  • Lug angle to main angle: 1.4 Pout=279.91.4\,P_{out}=279.9 kN → 7 bolts

Lug angle

The lug angle itself must develop 1.2Pout=239.91.2P_{out}=239.9 kN. ISA 65×65×8 (A=976A=976 mm², Tdg=221.8T_{dg}=221.8 kN) is not enough. Use ISA 75×75×10 (A=1402A=1402 mm²): Tdg=318.6T_{dg}=318.6 kN; net section with a hole in each leg Tdn=0.9(1402−2×10×22)×4101.25=284.0T_{dn}=\dfrac{0.9(1402-2\times10\times22)\times410}{1.25}=284.0 kN; both >239.9>239.9 kN ✓.

Layout

   gusset 16
  |==o==o==o==o==o==   main angle  (5 bolts @ 50)
  |==o==o==o==o==o==o  lug angle   (6 bolts @ 50)
        <- 300 mm ->

Pitch 50 mm (≥2.5d\ge2.5d), end distance 40 mm; length on the gusset 40+5×50=290≤30040+5\times50=290\le300 mm ✓. The lug angle is bolted to the gusset beyond the main angle and to the outstanding leg of the main angle (7 bolts).

Answer: ISA 100×100×12 with lug angle ISA 75×75×10; 5 M20 bolts for the main angle, 6 for the lug angle to the gusset and 7 for the lug-to-main-angle connection.

✓ Answer checked by Claude Opus 5.5 (high effort)

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