Project 1 - Design a simple double cleat angle shear connection
Q-Design a simple double cleat angle shear connection to connect (ISMB 500) secondary beam with a primary beam (ISMB 600) in a mezzanine deck of a building to transfer factored shear force of 200 kN. Consider bolt to connect angle cleat with primary as well as secondary beam.
Do manual design and also prepare a spread sheet to design this shear beam-beam connection
Solution :
AIM & INTRODUCTION : Simple double cleat angle shear connection to connect secondary beam with primary beam in mezzanine deck of a building is required to be designed with following inputs-
Secondary Beam : ISMB 500 ,
`t_w = 10.2 mm` '
`t_f= 17.2 mm`
D = 500 mm
`b_f`= 180 mm
Primary Beam : ISMB 600
`t_w = 12 mm
t_f = 20.8 mm`
D = 600 mm
`b_f`= 210 mm
Factored Shear Force : 200 kN
Yield strength of members `f_y`= 250 `N/(mm^2)`
Ultimate strength of plate `f_u = 410 N//mm^2`
Grade of bolt : Design with 4.6 ordinary grade as well as 8.8 HSFG grade bolt
: 400 MPa and 800 MPa
Shear plane : 2 as bolt is in double shear
Assuming diameter of bolt 'd' : 16 mm
Stress Area of bolt : `0.78xx(pi//4)xxd^2` = 156.75 `mm^2`
Minimum pitch distance , p : 2.5x dia of bolt = 40 mm , provide p = 50 mm
Edge distance, e : 30 mm
Diameter of bolt hole '`d_0` : 18 mm
Partial safety factor for bolt material '`gamma_(mb)` = 1.2
DESIGN USING 4.6 GRADE ORDINARY GRADE BOLT
Shear capacity of bolt (in double shear)
Nominal Shear strength of bolt '`V_(nsb) = (A_(nb)xxn_n + n_sxxA_(sb))f_(ub)//sqrt3`
Design Shear strength of bolt ' `V_(dsb) = V_(nsb)//gamma_(mo)`
= `f_u(n_nxxA_(nb)+n_sxxA_(sb))/(sqrt3xxgamma_(mb)`
`n_n=2` ( number of shear plane passing through the thread of the bolt)
`n_s=0` ( number of shear plane passing through the shank of the bolt)
= 156.75x2x 400/1.73x 1.25 = 57988.439 N = 57.988
Cnnection between web cleat and web of the secondary beam
Bearing strength of bolt: `V_(dpb)=V_(npb)/gamma_(mb)= 2.5k_bdtf_u//gamma_(mb)`
where `k_b` is least of the following-
`k_b= e/(3d_o)` = 30 /3x30 = 0.556
`k_b= p/(3d_0)-2.5`= (50/3x18)-0.25 =0.676
`k_b= f_(ub)/f_u` = 400/410 = 0.976
`k_b =1`
therefore `k_b`=0.556
Assuming cleat angle as IAS 110X110X10 where thickness of leg of angle cleat = 10 mm
`V_(dpb)`= 2.5x0.556x 16x10x 410/1.25 = 72947.2 N = 72.947 kN
Strength of bolt is lesser of shear strength and bearing strength.
So strength of bolt = 57.988 kN
Number of bolts required = 200/57.988 = 3.45
Provide 4 bolts of grade 4.6
Cnnection between web cleat and web of the main beam
Main beam is in single shear:
So Shear capacity of bolt in single shear `V_(dsp)`= 57.988/2 = 28.994 kN
Bearing capacity of bolt `V_(dpb)` = 72.947 kN
Bolt value = 28.994kN
Number of bolt required = 200/28.994 = 6.89
Providing 8 Bolts in two column i.e. 4 in each column
Length of cleat angle= 3p+2e
p= 50 mm & e = 30 mm
Length of cleat angle = 3x50+ 2x30 = 210 mm
provide cleat angle of 110x110x10 mm, 225 mm long on each side of web
DESIGN USING 8.8 GRADE HSFG BOLT
Nominal Shear strength of bolt '`V_(nsb) = A_(nb)xxn_sxxf_(ub)//sqrt3`
Design Shear strength of bolt ' `V_(dsb) = V_(nsb)//gamma_(mo)`
= `A_(nb)xxn_sxxf_(ub)//sqrt3xxgamma_(mo)`
= 156.75x2x 800/1.73x 1.25 = 115842.956 N = 115.843 kN
Bearing strength of bolt: `V_(dpb)=V_(npb)/gamma_(mb)= 2.5k_bdtf_u//gamma_(mb)`
where `k_b` is least of the following-
`k_b= e/(3d_o)` = 30 /3x30 = 0.556
`k_b= p/(3d_0)-2.5`= (40/3x18)-0.25 =0.678
`k_b= f_(ub)/f_u` = 400/410 = 0.976
`k_b =1`
therefore `k_b`=0.556
`V_(dpb)`= 2.5x0.556x 16x10x 410/1.25 = 72947.2 N = 72.947 kN
Strength of bolt is lesser of shear strength and bearing strength.
So strength of bolt = 72.947 kN
Number of bolts required = 200/72.947 = 2.74
Provide 3 bolts of grade 8.8
Fronr and Side view of Beam to Beam bolted shear connection using double cleat angle are shown below-
FRONT VIEW
SIDE VIEW
` SIDE VIEW
Shear strength of HSFG bolt
`V_(nGf)=mu_f.n_e.k_h.F_o`
providing HSFG bolts of grade 8.8
`f_(ub)= 800 MPa`
Net tensile stress area of bolt `A_(nb) = 0.78xx pi/4xxd^2`
= 0.78x 3.14/4x 16x16 = 156.75 sqmm
For `Fe 410` grade steel `f_u= 410 MPa`
Slip critical condition:
`F_0= 0.7 f_uxxA_(nb)`= 0.7x800`xx10^-3`x156.75 = 87.78 kN
Slip resistance of the bolts `V_(ds f) =mu_fxxn_exxk_hxxF_0//gamma_(mf)`
= 0.5x1x1x87.78/1.25 = 35.112 kN
Number of bolts = 200/35.112 = 5.7 say 6 bolts
If slip is permitted :
Bearing type connection
`V_(dsb)= A_(nb)xxf_(ub)//sqrt3xxgamma_(mb)`
= 156.75x800/1.732x1.25x1000 = 57.92kN
`V_(dpb)= 2.5.k_b.d.t.f_u//gamma_(mb)`
= 2.5x 0.556x16x10x410/1.25 X1000 = 72.95 kN
Thus, Shear strength of bolt =57.92 kN
Number of bolts = 200/52.92 = 3.78 say 4 bolts
RESULT : Thus beam to beam bolted shear connection using double cleat angle is designed with-
- 4.6 grade bolt
- 8.8 grade bolt
- considering slip resistance of HSFG bolts
- Spread sheet has also been attached as sheet-1 &2
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