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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 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 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

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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 . 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
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