--- title: "Check eccentric fillet weld, 8 mm, P=100 kN" subject: "Design of Steel and Timber Structure" chapter: 4 key: "semester_6/design_of_steel_and_timber_structure:rivet-eccentric-weld-p100" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/rivet-eccentric-weld-p100 site: BCT Engineering --- # Check eccentric fillet weld, 8 mm, P=100 kN Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures - Asked in: 2068 Bhadra (old course), 14 marks - Video: [Problem on Eccentrically loaded welded joints, DMM -1](https://www.youtube.com/watch?v=LOi_cmJwQyM) — Reference Book - 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/rivet-eccentric-weld-p100 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4 ## Question For the electric (eccentric) connection as shown, determine whether the joint is safe or not. Size of the fillet weld is 8mm and load P is equal to 100KN. Assume that permissible shear stress in the weld is 108 MPa. [Figure: 14 mm thick gusset plate welded to a column flange with 20 cm weld width at top and bottom and 20 cm depth; P = 100 kN acts at 40 cm from the weld face; 20 cm dimension marked] ## Short answer The weld experiences direct shear and bending moment, giving resultant stress 155.9 MPa. This exceeds the permissible shear stress of 108 MPa. Hence the joint is not safe. ## Answer **Basis:** working stress method (permissible shear stress in weld 108 N/mm²). The gusset (14 mm) is welded to the column flange by a C-shaped fillet weld: top and bottom 200 mm and vertical 200 mm. $P=100$ kN acts parallel to the flange at $e=400$ mm from the weld plane, giving direct shear and bending. Weld size $s=8$ mm, throat $t_t=0.7\times8=5.6$ mm. #### Weld group (per mm of throat) - $L=2\times200+200=600$ mm - $I_{xx}=\dfrac{200^3}{12}+2\times200\times100^2=4666667$ mm³ - $M=100\times10^3\times400=40\times10^6$ N·mm #### Force per mm of weld at the extreme weld (y = 100 mm) - Shear: $q_s=\dfrac{P}{L}=\dfrac{100000}{600}=166.7$ N/mm - Bending: $q_b=\dfrac{My}{I_{xx}}=\dfrac{40\times10^6\times100}{4666667}=857.1$ N/mm - Resultant $q=\sqrt{q_s^2+q_b^2}=873.2$ N/mm #### Stress in the weld $$ \tau=\frac{q}{t_t}=\frac{873.2}{5.6}=155.9\ \text{N/mm}^2>108\ \text{N/mm}^2 $$ **Answer: the joint is NOT safe** (stress is about 156 N/mm² against 108 N/mm² permissible). It needs a larger weld (about 12 mm) or a deeper weld. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What throat thickness is used for an 8 mm fillet weld? A. 8 mm B. 0.7 x 8 = 5.6 mm C. 0.5 x 8 = 4 mm D. 1.5 x 8 = 12 mm Answer: B. Effective throat is 0.7 times the weld size. (Key point: Throat thickness) 2. Total weld length is 2 x 200 + 200 = 600 mm. What is the direct shear per mm? A. 100000/600 = 166.7 N/mm B. 40 x 10^6/600 C. 100000/200 = 500 N/mm D. 100000 x 600 Answer: A. q_s = P/L = 166.7 N/mm. (Key point: Direct shear) 3. Why is the bending force (857.1 N/mm) much bigger than the shear? A. Welds are weak in bending B. The eccentricity e = 400 mm gives a large moment M = 40 x 10^6 N·mm C. Shear is ignored D. The weld is short Answer: B. M = Pe is large because of the 400 mm eccentricity, so the bending force dominates. (Key point: Dominant action) 4. Stress is 155.9 N/mm^2 and permissible is 108 N/mm^2. What is concluded? A. The joint is safe B. Only shear needs checking C. The load can be doubled D. Not safe; a larger weld (about 12 mm) is needed Answer: D. Stress exceeds the permissible value, so the weld must be enlarged or deepened. 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