Two angle section ISA 100 × 100 × 10 are connected to either side
Two angle section ISA 100 × 100 × 10 are connected to either side of a 14 mm gusset plate and is subjected to a working pull of 400 kN. If the welding in the workshop is performed by adopting the maximum size of the weld possible, what would be the suitable length of the weld l1 and l2 as shown in figure respectively? Assume fillet weld applied to section with rounded toes.
(Take Ultimate strength as 410 MPa)
A. 216 mm and 86 mm
B. 182 mm and 94 mm
C. 198 mm and 82 mm
D. 202 mm and 80 mm
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
\(Maximum\;size\;of\;the\;weld = \frac{3}{4} \times thickness\;of\;angle\;with\;rounded\;toe.\)
\( = \frac{3}{4} \times 10 = 7.5\;mm\)
Hence, provided weld size (w) = 7.5 mm
Length of weld = l1 + l2 = L (say)
∴ Design the strength of the weld
\( = \;L \times \left( {0.7 \times S} \right) \times \frac{{{f_u}}}{{\sqrt 3 \times 1.25}}\)
\( = L \times \left( {0.7 \times 7.5} \right) \times \frac{{410}}{{\sqrt 3 \times 1.25}}\)
= 994.2 L
Also, Force in each angle section will be = \(\frac{{Factor\;of\;safety \;\times \;working\;load}}{2}\)
\( = \frac{{1.5 \times 400\;}}{2} = 300\;kN\)
(Because of two angle section)
∴ 300 × 103 = 994.2 × L
⇒ L = 301.75 mm
∴ L = l1 + l2
To provide zero eccentricity
Also, \({\rm{l}}_1{\rm{\;}} \times {\rm{\;}}28.4{\rm{\;}} = {\rm{\;l}}_2{\rm{\;}} \times {\rm{\;}}\left( {100{\rm{\;}}-{\rm{\;}}28.4} \right)\)
\( \Rightarrow {l_2} = \frac{{28.4}}{{71.6}}{l_1}\)
\(\therefore {l_1} + \frac{{28.4}}{{71.6}}{l_1} = 301.75\)
⇒ l1 = 216.053 mm = 216 mm
l2 = 302 – 216
l2 = 86
Hence l1 = 216 mm and l2 = 86 mm