Find length of fillet welds shown in figure below to connect `<100100**10` to a gusset of `12\ mm` thickness. `f_y=250\ MPa` and `f_u=410\ MPa`

Solution:

`f_(wd)=f_u/(root()(3)**gamma_(MW))=189.371\ MPa`

`S_(m i n)=5\ mm`

`S_(m a x)` at side `=3/4**t=7.5\ mm`

We know, `S_(m a x)` at end side `=8.5\ mm`

Let, `S`  be the average of the `S_(m i n)`, `S_(m a x)` and `S_(m a x)` 

Thus, `S=7\ mm`

Effective throat thickness is, `t_t=4.9\ mm`

Let `l_e` be the effective length of weld, them

`F_(wd)=f_(wd)**A_e=f_(wd)**l_e**t_t`

or, `l_w=F_(wd)/(f_(wd)**t_t)=323.304\ mm`

The welding is provided here at the side and end edge and the length of weld along end edge is `100\ mm`. Let `l_1` and `l_2` be the length of the weld at toe and back of angle section, then the strength of the weld at end is `=100**4....

Solution:

`f_(wd)=f_u/(root()(3)**gamma_(MW))=189.371\ MPa`

`S_(m i n)=5\ mm`

`S_(m a x)` at side `=3/4**t=7.5\ mm`

We know, `S_(m a x)` at end side `=8.5\ mm`

Let, `S`  be the average of the `S_(m i n)`, `S_(m a x)` and `S_(m a x)` 

Thus, `S=7\ mm`

Effective throat thickness is, `t_t=4.9\ mm`

Let `l_e` be the effective length of weld, them

`F_(wd)=f_(wd)**A_e=f_(wd)**l_e**t_t`

or, `l_w=F_(wd)/(f_(wd)**t_t)=323.304\ mm`

The welding is provided here at the side and end edge and the length of weld along end edge is `100\ mm`. Let `l_1` and `l_2` be the length of the weld at toe and back of angle section, then the strength of the weld at end is `=100**4.9**189.371=92791.790\ N`

Strength of weld at the side is `207208.210\ N`. Thus,

`l_1**t_t**f_(wd)+l_2**t_t**f_(wd)=207208.210`

or, `l_1+l_2=223.304`.....(i)

Taking moment at the centre of gravity of the section, we get,

`l_1**27.6=l_2**72.4`

or, `l_1=2.623 l_2`....(ii)

Now, solving (i) and (ii), we get,

`l_1=61.636\ mm` and `l_2=161.669\ mm`

 But, the actual `l_1=61.636+2**4.9=71.436~=75\ mm`

Actual `l_2=161.669+2**4.9=171.469~=175\ mm`

Thus, the length of fillet weld is, `=100+175+75=350\ mm`