A $60 m m$ long and $6 m m$ thick fillet weld carries a steady load of $15 k N$ along the weld. The shear strength of the weld material is equal to $200 M P a$ . The factor of safety is

6.8

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4.8

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2.4

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3.4

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Solution

The correct option is D $3.4$
$P_{P F W} = 0.707 t \times L_{e} \times τ_{s}$
{here PFW $\to$ parallel filter weld}
$\frac{15 \times 10^{3}}{0.707 \times 6 \times 60} = τ_{s}$
or $τ_{s} = 58.9344 M P a$
Given shear strength of material
$= 200 M P a$
$F O S = \frac{200}{58.9344} = 3.4$

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A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN along the weld. The shear strength of the weld material is equal to 200 MPa. The factor of safety is

The correct option is D 3.4PPFW=0.707 t×Le×τs {here PFW → parallel filter weld} 15×1030.707×6×60=τs or τs=58.9344 MPa Given shear strength of material =200 MPa FOS=20058.9344=3.4

A $60 m m$ long and $6 m m$ thick fillet weld carries a steady load of $15 k N$ along the weld. The shear strength of the weld material is equal to $200 M P a$ . The factor of safety is

The correct option is D $3.4$
$P_{P F W} = 0.707 t \times L_{e} \times τ_{s}$
{here PFW $\to$ parallel filter weld}
$\frac{15 \times 10^{3}}{0.707 \times 6 \times 60} = τ_{s}$
or $τ_{s} = 58.9344 M P a$
Given shear strength of material
$= 200 M P a$
$F O S = \frac{200}{58.9344} = 3.4$