Exercise in tensile testing
Exercise:
A round test specimen with d0=10 mm initial diameter was tensile tested. After the test process the next data were measured:
• offset yield point at 0,2% strain: Fp0,2=22 kN;
• ultimate force: Fm=29 kN;
• fracture force: Fu=23 kN;
• necking rate: Z=60%;
• diameter at the ultimate force: dm=9,2 mm.
Let us determine 3 points of the true stress-strain diagram and the approximate value of the specific work of rupture!
Solution:
The 1st point is the offset yield point. 0,2 means that the force belonging to it was measured at 002
, 0
% 2 ,
0 =
ε = engineering elongation. Knowing it we can calculate a ϕp0,2 true elongation and then a dp0,2 diameter a σp0,2 true stress.
On the one hand:
001998 ,
0 ) 002 , 0 1 ln(
) 1
2 ln(
, 0
p = +ε = + =
ϕ .
On the other hand:
mm 99 , 9 e
d e
d d d
ln d 2
2 001998 , 0
0 2
0 2 , 0 p 2 , 0 p
0 2
, 0
p = ⇒ = p0,2 = =
ϕ ϕ .
Now we can calculate a true stress:
( )
280,67MPa4 mm 99 , 9
N 22000 4
d F S
F
2 2
2 , 0 p
2 , 0 p 2 , 0 p
2 , 0 p 2 , 0
p =
= π
= π
=
σ .
It is approximately equal to the engineering stress:
( )
280,11MPa4 mm 10
N 22000 4
d
R F2 2
0 2 , 0 p 2 , 0
p =
= π
= π .
The 2nd point is the point of ultimate strenght, where we can use the next equations:
1168 , mm 0 2 , 9
mm ln 10
d 2 ln d 2
m 0
m ⎟=
⎠
⎜ ⎞
⎝
= ⎛
=
ϕ ,
and
( )
436,25MPa4 mm 2 , 9
N 29000 4
d F S F
2 2
m m m m
m =
= π
= π
=
σ .
The 3rd point is the point of rupture, but firstly we can calculate the cross sectional are of the specimen.
From the equation of necking:
( )
22 2 0 0
u 0
u
0 (1 0,6) 31,4159mm
4 mm ) 10
Z 1 4 ( ) d Z 1 ( S S S
S
Z S π − =
= π −
=
−
=
− ⇒
= .
Now we can calculate a real elongation:
9163 , mm 0 4159 , 31
mm 5398 , ln 78 S
ln S 2
2
u 0
u ⎟⎟⎠=
⎜⎜ ⎞
⎝
= ⎛
⎟⎟⎠
⎜⎜ ⎞
⎝
= ⎛
ϕ ,
and the true stress before the rupture:
MPa 11 , mm 732
4159 , 31
N 23000 S
F
2 u
u
u = = =
σ .
Now the 3 point of the true stress-strain curve has been calculated, what can be plotted in a σ–
φ diagram. In this diagram the area below the curve is equal to the specific work of rupture.
The approximate value of specific work of rupture can be calculated by the next equation below:
u 3 u m
c cm
1855 J 9163
, 2 0
MPa ) 11 , 732 11 , 280 ( 2
W ≈ R +σ ϕ = + ⋅ = .
Because:
3 3
3
2 cm
J cm 001 , 0
J 001 , 0 mm
1Nmm mm
1 N = = = .
