Simple stress problem
(Strength of Materials - I, Mid-term Exam-44-1)
Problem :
Each of the two vertical links CF connecting the two horizontal members AD and EG
has a 10×40-mm uniform rectangular cross section and is made of a steel with an ultimate
strength in tension of 400 MPa, while each of the pins at C and F has a 20-mm diameter
and is made of a steel with an ultimate strength in shear of 150 MPa. Determine the
overall factor of safety for the links CF and the pins connecting them to the horizontal
members.
Solution :
The equilibrium equation
X
0.4FCF − 0.65 × 24 = 0
ME = 0,
FCF = 39 kN
FBE = FCF − 24 = 39 − 24
FBE = 15 kN
The links CF are in tension (critical)
A = (b − d) × t = (40 − 20) × 10
A = 200 mm2
σu =
Fu
2A
Fu = 2A × σu
Fu = 2 × 200 × 400
Fu = 160 × 103 N
Prof. Dr. M. Kemal APALAK
1
The pins at F are in double shear
π
202 = 314. 16 mm2
4
Fu
=
,
Fu = 2A × τu
2A
= 2 × 314. 16 × 150
= 94248 N
A =
τu
Fu
Fu
The critical force in link CF
Fcr = 94248 N
Factor of safety
Fcr
94248
=
FCF
39 × 103
F.S = 2. 416 6
F.S =
Prof. Dr. M. Kemal APALAK
2