Design a welded connection for member at a joint of truss shown in figure below. `f_y=250\ MPa` and `f_u=410\ MPa`. The forces shown are at services load.

Solution:

From Steel table, We have,

For `<100100**10`, `C_(x x)=28.4\ mm=C_(y y)`

For `<9090**8`, `C_(x x)=25.1\ mm=C_(y y)`

For `<7575**10`, `C_(x x)=21.00\ mm=C_(y y)`

For `<6565**8`, `C_(x x)=18.9\ mm=C_(y y)`

For the design of member carrying `50\ KN` tension load,

Design load `=1.5**50=75\ KN`

`S_(m i n)=3\ mm`

Considering only the side weld, `S_(m a x)=3/4**8=6\ mm`

Let the size of the fillet weld id `5\ mm`

Now, the effective throat thickness, `t_t=KS=0.7**5=3.5\ mm`

Let, `l_1`  and `l_2` be the length of the weld along the back face and toe face and let the given tension is balanced by the strength weld provided a....

Solution:

From Steel table, We have,

For `<100100**10`, `C_(x x)=28.4\ mm=C_(y y)`

For `<9090**8`, `C_(x x)=25.1\ mm=C_(y y)`

For `<7575**10`, `C_(x x)=21.00\ mm=C_(y y)`

For `<6565**8`, `C_(x x)=18.9\ mm=C_(y y)`

For the design of member carrying `50\ KN` tension load,

Design load `=1.5**50=75\ KN`

`S_(m i n)=3\ mm`

Considering only the side weld, `S_(m a x)=3/4**8=6\ mm`

Let the size of the fillet weld id `5\ mm`

Now, the effective throat thickness, `t_t=KS=0.7**5=3.5\ mm`

Let, `l_1`  and `l_2` be the length of the weld along the back face and toe face and let the given tension is balanced by the strength weld provided at toe and back face of angle section `F_1` and `F_2`  at back and toe face of given angle section. Then,

`F_1=l_1**t_t**f_(wd)`

Where, `f_(wd)=f_u/(root()(3)**gamma_(MW))=189.371\ MPa` is the strength of the fillet weld.

`F_2=l_2**t_t**f_(wd)`

Now, `l_1t_1f_(wd)+l_2t_2 f_(wd)=P(N)`

or, `l_1+l_2=P/(t_t**f_(wd))`.....(i)

Taking moment at the centre of gravity of section,

`l_1t_1f_(wd)**C_(x x)=l_2t_2 f_(wd)**(b-C_(x x))`

`l_1=(l_2(b-C_(x x)))/(C_(x x))` .....(ii)

Now, Solving the equations (i) and (ii), we get,

`l_2=P/(t_t**f_(wd))**C_(x x)/b`

`l_1=P/(t_t**f_(wd))**(b-C_(x x))/b`

Thus, `l_1=32.902\ mm` and `l_2=80.254\ mm`

Now,

The actual `l_1=32.902+2**3.5=40\ mm`

The actual `l_2=80.254+2**3.5=87.254\ mm~=90\ mm`

Thus, provide the size of fillet weld `5\ mm` and length of weld along toe side and along back side `40\ mm` and `90\ mm` respectively.

Design of Member carrying Compressive load,

Solution:

From Steel table, We have,

For `<100100**10`, `C_(x x)=28.4\ mm=C_(y y)`

For `<9090**8`, `C_(x x)=25.1\ mm=C_(y y)`

For `<7575**10`, `C_(x x)=21.00\ mm=C_(y y)`

For `<6565**8`, `C_(x x)=18.9\ mm=C_(y y)`

For the design of member carrying `50\ KN` tension load,

Design load `=1.5**50=75\ KN`

`S_(m i n)=3\ mm`

Considering only the side weld, `S_(m a x)=3/4**8=6\ mm`

Let the size of the fillet weld id `5\ mm`

Now, the effective throat thickness, `t_t=KS=0.7**5=3.5\ mm`

Let, `l_1`  and `l_2` be the length of the weld along the back face and toe face and let the given tension is balanced by the strength weld provided at toe and back face of angle section `F_1` and `F_2`  at back and toe face of given angle section. Then,

`F_1=l_1**t_t**f_(wd)`

Where, `f_(wd)=f_u/(root()(3)**gamma_(MW))=189.371\ MPa` is the strength of the fillet weld.

`F_2=l_2**t_t**f_(wd)`

Now, `l_1t_1f_(wd)+l_2t_2 f_(wd)=P(N)`

or, `l_1+l_2=P/(t_t**f_(wd))`.....(i)

Taking moment at the centre of gravity of section,

`l_1t_1f_(wd)**C_(x x)=l_2t_2 f_(wd)**(b-C_(x x))`

`l_1=(l_2(b-C_(x x)))/(C_(x x))` .....(ii)

Now, Solving the equations (i) and (ii), we get,

`l_2=P/(t_t**f_(wd))**C_(x x)/b`

`l_1=P/(t_t**f_(wd))**(b-C_(x x))/b`

Thus, `l_1=32.902\ mm` and `l_2=80.254\ mm`

Now,

The actual `l_1=32.902+2**3.5=40\ mm`

The actual `l_2=80.254+2**3.5=87.254\ mm~=90\ mm`

Thus, provide the size of fillet weld `5\ mm` and length of weld along toe side and along back side `40\ mm` and `90\ mm` respectively.

Design of Member carrying `78**1.5=117\ KN` Compressive load,