Simple Stress (Strength of Materials - I, Midterm Exam-40-1 ...

Problem :
Simple Stress
(Strength of Materials - I, Midterm Exam-40-1)
1. A steel structure
In the steel structure shown, a 6-mm-diameter pin is used at C and 10-mm-diameter pins are used
at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress
is 400 MPa in link BD. Knowing that a factor of safety of 3 is desired, determine the largest load P
which may be applied at A. Note that link BD is not reinforced around the pin holes.
Solution :
Theallowableshearstress
Theallowablenormalstress
Equilibrium equation
i) BD link
ii) Pin B
τ al = τ ul 150
= =50MPa (1)
FS 3
σal = σul
FS
= 400
3
X MC = 0, −120 × FBD + 280 × P =0
P = 3
7 FBD
= 133.3 MPa (2)
σal =
µ
F
,
A BD
FBD = σal × ABD
FBD = 133.3 × (18 − 10) × 6
FBD = 6398. 4 N (4)
P = 3
× 6398. 4, P = 2742. 2 N (5)
7
Dr. M. Kemal Apalak 1
(3)

iii) Pin C
Equilibrium equation
TheallowableloadP
, FBD = τ al × AB
FBD =
³
π
50× × 102´
4
FBD = 3927.0 N (6)
τ al = FBD
AB
P = 3
× 3927.0, P = 1683.0 N (7)
7
τ al
C
=
=
C
, C = τ al × 2AC
2AC
³
π
50× 2 × 62´
4
C = 2827. 4 N (8)
X Fx = 0, Cx =0
Cy = C = 2827. 4 N (9)
X MB = 0, −120 × Cy + 160 × P =0
P = 3
4 Cy
(10)
P = 3
× 2827. 4, P = 2120. 6 N (11)
4
P = 1683.0 N
Dr. M. Kemal Apalak 2