1. The document describes the design of a simple shear connection between a beam and column using bolts. It provides equations to check the shear strength of the bolts and bearing strength of the plate. 2. An example is presented to determine the number and size of bolts needed to resist an ultimate shear force of 1000 kN between two beams. It is determined that 7 bolts with 18 mm diameter and 98.5 mm spacing will suffice. 3. The document also checks the strength of double angles used in the connection to transfer the force and confirms the chosen angles are adequate.

1. 1 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Design of Simple Shear Connections
(Bearing type connection)
Strength of bolts is governed by;
1- Shear failure of bolts
φ Rn = 0.75 . Fv . Ab . Nb . Ns
Where;
Fv : Shear strength of bolts
(for A325 bolts = 400 MPa)
Ab : gross area of bolt
Nb : Number of bolts
Ns : Number of shear planes,
For group A, Ns = 2, for group B Ns= 1
2- Bearing failure of plate
φ Rn = 0.75 * 2.4 Fu (d . tmin) . Nb
Where:
Fu : ultimate tensile strength of plates
d : diameter of bolts
tmin : of group A bolts is less of tw or 2tangle
of group B tflange of column or tangle
Group A bolts
Group B Bolts
Sec A- A
A
A
Simple Shear Connection between a
beam and column
Secondary
beam
Secondary
beam
Main beam
Simple Shear Connection between
Secondary beams and main beam

2. 2 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example
Solution:
Check the shear strength at critical section
For W 480 x 473, h/tw = 756 / 26.4 = 28.6 < 2.45 2 x 105
= 69
250
∴ φVn = 0.9 x 0.6 Fy . Aw where Aw = (690 + 48 ) x 26.4 = 19483 mm2
φVn = 0.9 x 0.6 x 250 x 19483 x 10-3
= 2630 > Vu = 1000 kN ∴ OK
Strength of Bolts
a) Group (A) bolts (Double Shear)
1- Shear Failure, φ Rn = 0.75 x (πd2
/4) x 400 x Nb x 2 = Vu = 1000 x 103
∴ d2
Nb = 2122.07
Try , if
Nb d (mm) Spacing = 690/Nb
3 26.5 230mm = 8.67 d > 6d not OK
4 23 172 mm = 7.5 d > 6d not OK
5 20.6 138 mm = 6.7 d > 6d not OK
6 18.8 115 mm =6.1 d > 6d not OK
7 17.4 98.5mm = 5.6 d < 6d O.K
Determine the number of bolts
needed to support an ultimate
shear force = 1000 kN,
transformed from beam
W 840 x 473 to a main beam
W 920 x 653 with double
angles 65 x 65 with a length of
690 mm
F v for bolts = 400 MPa
Fu for steel = 400 MPa
Fy for steel = 250 MPa
For W 840 x 473
d= 893 mm
h = 756 mm
tw = 26.4 mm
tf = 48 mm
For W 920 x 653
d = 972 mm
tw = 34.5 mm
W 840 x 473
W 920 x 653
893 mm690 mm
155 mm
48 mm
Critical section
for shear
Group A bolts
Group B bolts

3. 3 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
2- Check bearing strength for 7 M18 bolts
φ Rn = 0.75 x 2.4 Fu x d x tmin x Nb
= 0.75 x 2.4 x 400 x 18 x t min x 7 = 1000 x 103
∴ t min = 11.02 = 2 t angle
∴ tmin of angle = 5.5 mm
∴ Choose double angles 65 x 65 x 6.35 mm
b) Group (B) bolts (Single Shear)
1- Check Shear Failure,
φ Rn = 0.75 x (π 18 2
/4) x 400 x 14 x 1 x 10-3
= 1068.8 kN > Vu = 1000 kN
2- Check bearing strength
φ Rn = = 0.75 x 2.4 x 400 x 18 x 6.35 x 14 x 10-3
= 1152 kN > 1000 kN
Check shear block of double angles
D hole = 18 + 3 = 21 mm
Lv = 6.5 s = 6.5 x 98.5 = 640.25 mm
Av= 640.25 x 2 x 6.35 = 8131.2 mm2
Avn = 8131.2 – (6.5 x 21 x 6.35 x 2) = 6397.65 mm2
Gage distance g = 35 mm , Lt = 65 - 35 = 30 mm
Atg = 30 x 6.35 x 2 = 381 mm2
Atn = 381 – 0.5 x 21 x 6.35 x 2 = 247.65 mm2
0.6 Fu Avn = 0.6 x 400 x 6397.65 x 10-3
= 1535.44 kN
Fu Atn = 400 x 247.65 = 99.06 kN
0.6 Fu Avn > Fu Atn
φ Rn = 0.75 [0.6 Fu Avn + Fy Atg]
φ Rn = 0.75 [ 1535.4 + 250 x 381 x 10-3
] = 1223 kN > Vu = 1000 kN
∴ Choose 7 bolts with
diameter 18 mm with
spacing = 98.5 mm for
group A
∴ Use 14 M18 for group B bolts
∴ double angles 65 x 65 x 6.35 mm with 7 M 18 A325 bolts with spacing 98.5 mm
and gage distance g = 35 mm are sufficient to resist Vu = 1000 kN
s/2
690 mm
s
s
s
s
s
s
s/2
s = 98.5 mm
g
s/2
s
s
s
s
s
s
s/2
g Lt
Lv