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GATE 2023 CE (CE2) – Question 40

Steel Structures · Connections: simple, eccentric, beam-column · 2 marks · Multiple choice

Two plates are connected by fillet welds of size 10 mm and subjected to tension, as shown in the figure. The thickness of each plate is 12 mm. The yield stress and the ultimate stress of steel under tension are 250 MPa and 410 MPa, respectively. The welding is done in the workshop (partial safety factor, $\gamma_{mw}=1.25$). As per the Limit State Method of IS 800: 2007, what is the minimum length (in mm, rounded off to the nearest higher multiple of 5 mm) required of each weld to transmit a factored force P equal to 275 kN?

100 mm wide plate overlaps a 150 mm wide plate; two longitudinal side fillet welds transfer P.
  1. 100
  2. 105
  3. 110
  4. 115

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Correct answer: (B) 105

Explanation

**Step 1: design strength of the fillet weld.** For shop welds ($\gamma_{mw}=1.25$) the design shear strength per unit throat area is
$$f_{wd}=\frac{f_u/\sqrt3}{\gamma_{mw}}=\frac{410/\sqrt3}{1.25}=189.4\text{ MPa}.$$

**Step 2: effective throat thickness.** For a 10 mm fillet weld with equal legs, $t_t=0.7\times10=7$ mm.

**Step 3: required effective length.** Two longitudinal welds share the load $P=275$ kN, so
$$2\,L_w\,t_t\,f_{wd}\geq P\;\Rightarrow\;L_w\geq\frac{275\,000}{2\times7\times189.4}=103.7\text{ mm}.$$

**Step 4: round up** to the next multiple of 5: **105 mm** (option B).

(IS 800 also requires end returns and a minimum length of 4 times the weld size, i.e. 40 mm, which this easily meets.)

Official GATE 2023 answer key: https://gate2026.iitg.ac.in/doc/download/Answer_keys2023/CE2_ANS_GATE2023.pdf