Exercise in tensile testing
Exercise:After tensile testing a round test specimen with d0=10 mm initial diameter (L0=10d0) the next data were measured:
• offset yield point at 0,2% strain: Fp0,2=6,3 kN;
• rate of necking: Z=75%;
• ultimate force: Fm=11,5 kN.
Marks on the specimen with d0 lenght were elongated as follows: 11, 11, 11, 12, 13, 16, 13, 12, 11, 11 mm.
Let us determine the value of offset yield strenght (Rp0,2), the true stress belonging to the ultimate force (σm), the engineering elongation (εm), and the maximum elongation in the true (ϕu) and engineering (A) system!
Solution:
The offset yield strenght in the engineering system:
( )
80,21MPa4 mm 10
N 6300 4
d F S
R F 2 2
0 2 , 0 p 0
2 , 0 p 2 , 0
p =
= π
= π
= .
For the true ultimate tensile strenght we should calculate the cross sectional area belonging to the ultimate force (Sm). Calculating it we should use the volumetric constancy:
m.
m 0
0l S l
S =
In the equation above S0 is the initial cross sectional area, Sm is the cross sectional area at the maximum force, l0 is the initial lenght of marks on the specimen (equal to 10 mm), lm is the lenght of marks after rupture, but not on the necking part (in this case lm=11 mm).
So:
( )
22
m 0 2 0 m 0 0
m 71,40mm
mm 11
mm 10 4
mm 10 l
l 4 d l S l
S = = π = π = ,
and:
MPa 06 , mm 161
40 , 71
N 11500 S
σ F 2
m
m = m = = .
The engineering elongation in this part:
% 10 mm 100
10
10mm - 100 11mm
l l ε l
0 0
m m− ⋅ = ⋅ =
= .
For calculating of the true elongation belonging to the last point we should calculate the cross sectional area of the specimen before rupture. For it we use the equation of necking rate:
( )
2 22 0 0
u 0
u
0 (1 0,75) 19,6349mm
4 mm ) 10
Z 1 4 ( ) d Z 1 ( S S S
S
Z=S − ⇒ = − = π − = π − = .
and
,3863 19,6349mm 1
4 (10mm) S ln
4 d S ln
lnS 2
2
u 2 0
u 0
u ⎟⎟⎠ =
⎜⎜ ⎞
⎝
⎛ π
⎟⎟ =
⎠
⎜⎜ ⎞
⎝
⎛ π
=
=
ϕ .
The value of maximum elongation in the engineering system:
% 21 mm 100
100
100mm -
121mm L 100
L A L
0 0
u− ⋅ = ⋅ =
= .
Where Lu is the sum of the length of marks on the specimen after rupture (in this case:
121 mm – implicitly: L0=100mm).
