Two 12 mm thick steel flats are spliced by two 6 mm thick plates with four M16 bolts of product grade C and property class `4.6` as shown below. Determine ultimate load carrying capacity of connection. `f_y=250 MPa` and `f_u=410\ MPa`.

Solution:

For M16 bolts of class 4.6,

`d=16 mm`

`d_0=16+2=18mm`

`f_(ub)=400 MPa`

`f_(yb)=240 MPa`

Also, Given,

 `f_y=250 MPa` and `f_u=410\ MPa`

The Design shearing strength of bolts is,

`V_(dsb)=V_(nsb)/gamma_(mb)`

`=(f_(ub)/(root()(3) gamma_(mb)))(n_n*A_(nb)+n_s*A_(sb))`

Here,

`n_n=1`

`n_s=1`

Now, `A_s=(pi d^2)/4=(pi*16^2)/4=201.06\ mm^2`

And, `A_n=0.78**201.06=156.82\ mm^2`

Thus,

`V_(dsb)=400/(root()(3)**1.25)**(156.82+201.06)`

`=66119\ N`

`=66.119\ KN`

Again, the design bearing she....

Solution:

For M16 bolts of class 4.6,

`d=16 mm`

`d_0=16+2=18mm`

`f_(ub)=400 MPa`

`f_(yb)=240 MPa`

Also, Given,

 `f_y=250 MPa` and `f_u=410\ MPa`

The Design shearing strength of bolts is,

`V_(dsb)=V_(nsb)/gamma_(mb)`

`=(f_(ub)/(root()(3) gamma_(mb)))(n_n*A_(nb)+n_s*A_(sb))`

Here,

`n_n=1`

`n_s=1`

Now, `A_s=(pi d^2)/4=(pi*16^2)/4=201.06\ mm^2`

And, `A_n=0.78**201.06=156.82\ mm^2`

Thus,

`V_(dsb)=400/(root()(3)**1.25)**(156.82+201.06)`

`=66119\ N`

`=66.119\ KN`

Again, the design bearing shear strength of a single bolt is,

`V_(dpb)=(2.5 K_b d tf_u)/gamma_(mb)`

Where, `K_b` is the smaller value of `e/(3d_0)`, `p/(3d_0)-0.25`, `f_(ub)/f_u`, and `1`

Here, `e/(3d_0)=40/(3**18)=0.7407`

`p/(3d_0)-0.25=80/(3**18)-0.25=1.23`

`f_(ub)/f_u=400/410=0.975`

Thus, `K_b=0.7407`

Hence, the design bearing shear strength of a single bolt is,

`V_(dpb)=(2.5**0.7407**16**12**410)/1.25`

`=116615.8 N`

`=116.6\ KN`

Thus, the design strength of a single bolt is `66.119\ KN`

The design strength of the connection of 4 bolts is `4**66.119=264.476\ KN`

Now, the design strength of plates governed by rupture is given by,

`T_(dn)=(0.9 f_u A_n)/gamma_(m1)`

`=(0.9**410**(200-2**18)**12)/1.25`

`=580953\ N`

`=580.953\ KN`

Again, for the design strength due to block shear,

`A_(vg)=2(40+80)**12=2880\ mm^2`

`A_(vn)=2(40+80-18-18/2)=2232\ mm^2`

`A_(tg)=(100**12)=1200\ mm^2`

`A_(tn)=(100-18/2-18/2)=984\ mm^2`

Now,

`T_(db1)=[(2880**250)/(root()(3)**1.1)+(0.9**984**410)/1.25]` 

`=668.379\ KN`

Also, `T_(db2)=[(0.9**2232**410)/(root()(3)**1.25)+(1200**250)/1.10]`

`=653.136\ KN`

Thus, `T_(db)=653.136\ KN`

Thus, the ultimate design load carrying capacity of the connection is `264.484\ KN`