21-Design of Simple Shear Connections (Steel Structural Design & Prof. Shehab Mourad)

21-Design of Simple Shear Connections (Steel Structural Design & Prof. Shehab Mourad)

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 . F v . Ab . Nb . Ns

Where;
F
v : Shear strength of bolts
(for A325 bolts = 400 MPa)
A
b : gross area of bolt
N
b : Number of bolts
N
s : 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
t
min : of group A bolts is less of tw or 2tangle
of group B t flange 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/t
w = 756 / 26.4 = 28.6 < 2.45 2 x 10
5
= 69

250 ∴ φVn = 0.9 x 0.6 Fy . Aw where Aw = (690 + 48 ) x 26.4 = 19483 mm
2


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 (πd
2
/4) x 400 x Nb x 2 = Vu = 1000 x 10
3


∴ d
2
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
F
u for steel = 400 MPa
F
y for steel = 250 MPa

For W 840 x 473
d= 893 mm
h = 756 mm
t
w = 26.4 mm
tf = 48 mm
For W 920 x 653
d = 972 mm
t
w = 34.5 mm
W 840 x 473

W 920 x 653

893 mm 690 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 10
3

∴ 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 > V u = 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
L
v = 6.5 s = 6.5 x 98.5 = 640.25 mm
A
v= 640.25 x 2 x 6.35 = 8131.2 mm
2

A
vn = 8131.2 – (6.5 x 21 x 6.35 x 2) = 6397.65 mm
2

Gage distance g = 35 mm , L
t = 65 - 35 = 30 mm
A
tg = 30 x 6.35 x 2 = 381 mm
2

A
tn = 381 – 0.5 x 21 x 6.35 x 2 = 247.65 mm
2

0.6 F
u Avn = 0.6 x 400 x 6397.65 x 10
-3
= 1535.44 kN
F
u Atn = 400 x 247.65 = 99.06 kN
0.6 F
u Avn > Fu Atn
φ Rn = 0.75 [0.6 F u Avn + Fy Atg]
φ Rn = 0.75 [ 1535.4 + 250 x 381 x 10
-3
] = 1223 kN > V u = 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