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 property class `8.8` if

• no slip is permitted at ultimate load
• slip is permitted at 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 Desi....

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 strength of bolts is,

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

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

`=206.628\ 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`

let `e=40\ mm>1.5d_0` and `p=100 mm >2.5d`

Here, `e/(3d_0)=40/(3**22)=0.606`

`p/(3d_0)-0.25=100/(3**22)-0.25=1.265`

`f_(ub)/f_u=800/410=1.951`

Thus, `K_b=0.606`

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

`V_(dpb)=(2.5**0.606**20**12**410)/1.25`

`=119260.8 N`

`=119.260\ KN`

Thus, the design strength of a single bolt is `119.260\ KN`

The design strength of the connection of 6 bolts is `6**119.260=715.564\ KN`

Check for plates:

`T_(dg)=(f_yA_g)/gamma_(m0)=818.182>715.566\ KN`

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

`=906.854>715.566\ KN`

Again, for the design strength due to block shear,

`A_(vg)=2**(40+2**100)**12=5760\ mm^2`

`A_(vn)=2**(40+2**100-2.5**22)**12=4440\ mm^2`

`A_(tg)=2**50**12=1200\ mm^2`

`A_(tn)=(50-22/2)**2**12=936\ mm^2`

Now,

`T_(db1)=[(A_(vg)*f_y)/(root()(3)**gamma_(m0))+(0.9**A_(tn)**f_u)/gamma_(m1)]` 

`=1032.111\ KN`

Also, `T_(db2)=[(0.9**A_(vn)**f_u)/(root()(3)**gamma_(m1))+(A_(tg)**f_y)/gamma_(m0)]`

`=1029.453\ KN`

Thus, `T_(db)=1029.453\ KN > 715.566\ KN`