--- title: "Tension member with lug angle, 400 kN" subject: "Design of Steel and Timber Structure" chapter: 5 key: "semester_6/design_of_steel_and_timber_structure:tension-lug-angle-400kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-lug-angle-400kn site: BCT Engineering --- # Tension member with lug angle, 400 kN Design of Steel and Timber Structure · Chapter 5: Tension Members - Asked in: 2081 Chaitra, 8 marks; 2072 Magh, 12 marks (asked 2 times) - Video: [Design of lug Angles problems](https://www.youtube.com/watch?v=ECxaZ1WkuKU) — civilsofttech - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-lug-angle-400kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5 ## Question 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). ## 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. ## Answer **Given:** axial pull (factored) $T=400$ kN, joint length available 300 mm, gusset 16 mm, M20 bolts of grade 4.6, Fe410/E250 ($f_y=250$, $f_u=410$). A single angle connected by one leg would need 9 bolts ($400/45.26$), 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 $A_g\ge\dfrac{400\times10^3\times1.10}{250}=1760$ mm². Try **ISA 100×100×12**: $A_g=2257$ mm². - Yielding: $T_{dg}=\dfrac{2257\times250}{1.10}=513.0$ kN $>400$ ✓ - Rupture: with a lug angle both legs are effectively connected, so there is no shear-lag reduction. $A_n=2257-12\times22=1993$ mm²; $T_{dn}=\dfrac{0.9\times1993\times410}{1.25}=588.3$ kN $>400$ ✓ #### Bolts (M20, 4.6) - Shear $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN - Bearing ($e=40$, $p=50$, $t=12$ mm): $k_b=0.508$, $V_{dpb}=99.9$ kN → bolt value $45.26$ kN #### Forces with lug angle Outstanding leg force $\approx T\dfrac{A_{out}}{A_g}$ with $A_{out}\approx(100-6)\times12=1128$ mm²: $P_{out}=199.9$ kN. - Connected leg force $=400-199.9=200.1$ kN → main angle to gusset: $n=200.1/45.26=4.4$ → **5 bolts** - Lug angle to gusset: $1.2\,P_{out}=239.9$ kN → **6 bolts** - Lug angle to main angle: $1.4\,P_{out}=279.9$ kN → **7 bolts** #### Lug angle The lug angle itself must develop $1.2P_{out}=239.9$ kN. ISA 65×65×8 ($A=976$ mm², $T_{dg}=221.8$ kN) is not enough. Use **ISA 75×75×10** ($A=1402$ mm²): $T_{dg}=318.6$ kN; net section with a hole in each leg $T_{dn}=\dfrac{0.9(1402-2\times10\times22)\times410}{1.25}=284.0$ kN; both $>239.9$ kN ✓. #### Layout ```text 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 ($\ge2.5d$), end distance 40 mm; length on the gusset $40+5\times50=290\le300$ 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is a lug angle used here? A. The member is in compression B. The joint length of 300 mm is too short to connect the main angle alone, and the lug connects the outstanding leg C. To reduce the angle size D. To avoid bolts Answer: B. A lug angle shortens the connection and brings the outstanding leg into action. (Key point: Need for lug angle) 2. Why is no shear lag reduction applied in the rupture check? A. The member is short B. Bolts are high strength C. Net area is small D. With a lug angle the whole angle is effectively connected Answer: D. The outstanding leg is connected by the lug, so the full section is effective. (Key point: Rupture check) 3. How is the outstanding leg force estimated? A. T x A_out/A_g B. T/2 C. T x A_g/A_out D. Bolt value x n Answer: A. The force is shared in proportion to area: 400 x 1128/2257 = 199.9 kN. (Key point: Outstanding leg force) 4. What forces are the lug connections designed for? A. Equal to P_out B. 1.2 P_out lug-to-gusset and 1.4 P_out lug-to-angle C. 0.5 P_out D. T total Answer: B. Extra eccentricity is covered by 1.2 P_out for the lug to gusset and 1.4 P_out for lug to main angle. 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