Design and detail a tension splice to connect `300\ mm12\ mm` flat with `300\ mm 16\ mm` flat using two cover plates to carry a tension of `400\ KN`. Use M20 high strength bolt of the property class `8.8` if
no slip is permitted at the ultimate load
slip is permitted at the ultimate load

Solution:

For M20 bolts of class 8.8,

`d=20 mm`

`d_0=20+2=22\ mm`

`f_(ub)=800 MPa`

`f_(yb)=640 MPa`

Also, Given,

`f_y=250 MPa` and `f_u=410\ MPa`

When no slip is permitted at the ultimate load

Nominal shear capacity of bolt is,

`V_(n s f)=mu_f n_eK_h gamma_(m f) F_0`

Where,

`mu_f=0.55` is the slip factor.

`n_e` is the number of effective interfaces offering frictional resistance to slip.

`K_h=1.0` is the clearances hole

`gamma_(mf)=1.25`

`F_0=A_b b F_0=0.78**(pi d^2)/4&&0.70=137.225\ KN` is the proof load.

thus, `V_(n s f)=0.55**1**1**137.225=75.471\ KN`

Thus, `V_(d s f)=75.379/1.25=60.379\ KN`

Now, the number of bolts is,

`n=400/60.379=6.625~=7`

Slip is permitted at ultimate load:

`gamma_(mf)=1.1`

`V_(d s f)=15.471/1.1=68.6\ KN`

Thus, the number of bolts is, `=400/68.6=5.831~=6`

The Design shearing stre....

🔗

Design and detail a tension splice to connect `300\ mm**12\ mm` flat with `300\ mm ** 16\ mm` flat using two cover plates to carry a tension of `400\ KN`. Use M20 high strength bolt of the property class `8.8` if no slip is permitted at the ultimate load slip is permitted at the ultimate load