Simple Stress
(Strength of Materials - I, Midterm Exam-40-1)
Problem :
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 :
The allowable shear stress
τ ul
150
=
= 50 MPa
FS
3
(1)
σ ul
400
=
= 133.3 MPa
FS
3
(2)
τ al =
The allowable normal stress
σ al =
Equilibrium equation
X
MC
P
= 0,
=
−120 × FBD + 280 × P = 0
3
FBD
7
(3)
i) BD link
σ al
FBD
FBD
P =
ii) Pin B
Dr. M. Kemal Apalak
µ
¶
F
, FBD = σal × ABD
A BD
= 133.3 × (18 − 10) × 6
= 6398. 4 N
=
3
× 6398. 4,
7
P = 2742. 2 N
(4)
(5)
1
FBD
, FBD = τ al × AB
AB
³π
´
= 50 ×
× 102
4
= 3927.0 N
τ al
=
FBD
FBD
P =
3
× 3927.0,
7
τ al
=
iii) Pin C
C
C
(6)
P = 1683.0 N
(7)
C = τ al × 2AC
³π
´
= 50 × 2
× 62
4
= 2827. 4 N
(8)
C
,
2AC
Equilibrium equation
X
X
The allowable load P
MB
Fx
= 0,
Cy
= C = 2827. 4 N
= 0,
3
Cy
4
Cx = 0
(9)
−120 × Cy + 160 × P = 0
P
=
P =
3
× 2827. 4,
4
(10)
P = 2120. 6 N
(11)
P = 1683.0 N
Dr. M. Kemal Apalak
2
STRENGTH OF MATERIALS - I - 36 – 1
Simple stress
Two 5-kN vertical loads are applied to pin B of the assembly shown.
Knowing that a 12-mm-diameter pin is used at each connection, determine
the maximum value of the average normal stress in (a) link AB, (b) link
BC, (c) the average shearing stress in the pin at C, (d) the bearing stress at
C in member BC, (e) the bearing stress at B in member BC.
Equilibrium equations
∑F
= 0,
− FBA cos 30 − FBC cos 45 = 0,
∑F
= 0,
FBA sin 30 − FBC sin 45 − 10 = 0
x
y
FBA = − FBC
cos 45 ⎞
⎛
⎜ − FBC
⎟ sin 30 − FBC sin 45 − 10 = 0,
cos 30 ⎠
⎝
cos 45
FBA = − ( −8.966 )
, FBA = 7.32 kN
cos 30
(a) The maximum average normal stress in the link AB
Anet = ( 36 − 12 ) × 10 = 240 mm
(σ AB )max =
cos 45
cos 30
FBC = −8.966 kN
(d) The bearing stress at C in the member BC
2
FBA 7.32 kN
=
Anet 240 mm 2
(σ AB )max = 30.5 MPa (in tension)
(b) The maximum average normal
stress in the link BC: The section
of the link BC at the pin C does
not carry normal load; therefore,
A = 36 × 10 = 360 mm 2
σ BC =
FBC
8.966 kN
=−
A
360 mm 2
σ CD = −24.91MPa
( in compression )
(σ b )C =
FBC 8.966 kN
=
Ab 12 × 10 mm 2
(σ b )C = 74.7 MPa
(d) The bearing stress at B in the member BC
(c) The average shearing stress in the pin C
τC =
FBC
=
2A
8.966
kN
2
π × 12 mm 2
2×
4
τ C = 39.64 MPa
(σ b ) B
(
)
⎛ FBC ⎞ 8.966
⎜
2 ⎟⎠
2 kN
⎝
=
=
Ab
12 ×10 mm 2
(σ b ) B = 37.36 MPa
Dr. M. Kemal Apalak