1. Calculate the strength of a 20mm diameter bolt of grade 4.6 connecting main plates of 12mm
thickness if connection is :
a. Lap joint
b. Single cover butt joint with cover plate thickness 10mm
c. Double cover butt joint with cover plate thickness 8mm
Solution : Nominal diameter = 20mm
Area of bolt at shank = Asb= (3.14/4)x20x20 = 314.16mm2
Area at threaded portion = Anb = 0.78 x Asb = 0.78 x 314.16 = 245.04 mm2
Grade = 4.6  fub = 400N/ mm2 and fy = 240N/ mm2
fup = 410MPa
For lap joint,
nn = 1, ns=0
Design strength of bolt in shear = Vdsb = 𝛾
𝑓𝑢𝑏
𝑚𝑏 √3
400
1.25 𝑥 √3
(𝑛𝑛. 𝐴𝑛𝑏 + 𝑛𝑠. 𝐴𝑠𝑏) =
(1 ∗ 245.04 + 0) = 45.26𝑘𝑁
Design strength of bolt in bearing 
end distance = e = 1.5do = 1.5 x 22 = 33mm round off to 35mm
pitch distance = p = 2.5d = 2.5 x 20 = 50mm
kb => minimum of all :




e/(3do) = 35/(3x22) = 0.53
[p/(3do)]-0.25 = [50/(3x22)]-0.25 = 0.51
fub/fup = 400/410 = 0.97
1
kb=0.51
Vdpb = 2.5kbdtfup/𝛾𝑚𝑏 = 2.5x0.51x20x12x410/1.25 = 100.4kN
Bolt value = minimum of (Vdsb,Vdpb) = 45.26kN
b)Single cover butt joint
Design strength in shear = Vdsb = 𝛾
=
400
1.25 𝑥 √3
𝑓𝑢𝑏
𝑚𝑏 √3
(𝑛𝑛. 𝐴𝑛𝑏 + 𝑛𝑠. 𝐴𝑠𝑏)
(1 ∗ 245.04 + 0)
= 45.26𝑘𝑁
Design strength of bolt in bearing 
end distance = e = 1.5do = 1.5 x 22 = 33mm round off to 35mm
pitch distance = p = 2.5d = 2.5 x 20 = 50mm
kb => minimum of all :




e/(3do) = 35/(3x22) = 0.53
[p/(3do)]-0.25 = [50/(3x22)]-0.25 = 0.51
fub/fup = 400/410 = 0.97
1
kb=0.51
Vdpb = 2.5kbdtfup/𝛾𝑚𝑏 = 2.5x0.51x20x10x410/1.25 = 83.6kN
Bolt value = minimum of (Vdsb,Vdpb) = 45.26kN
c) Double cover butt joint
nn = 2, ns=0
Design strength in shear = Vdsb =
=
400
1.25 𝑥 √3
𝑓𝑢𝑏
(𝑛𝑛. 𝐴𝑛𝑏
𝛾𝑚𝑏 √3
+ 𝑛𝑠. 𝐴𝑠𝑏)
(2 ∗ 245.04 + 0)
= 90.52𝑘𝑁
Design strength of bolt in bearing 
end distance = e = 1.5do = 1.5 x 22 = 33mm round off to 35mm
pitch distance = p = 2.5d = 2.5 x 20 = 50mm
kb => minimum of all :




e/(3do) = 35/(3x22) = 0.53
[p/(3do)]-0.25 = [50/(3x22)]-0.25 = 0.51
fub/fup = 400/410 = 0.97
1
kb=0.51
Vdpb = 2.5kbdtfup/𝛾𝑚𝑏 = 2.5x0.51x20x12x410/1.25 = 100.4kN
Bolt value = minimum of (Vdsb,Vdpb) = 90.52kN