Civil Engineering Reference
In-Depth Information
Solution
89000
0.01
10
9
Pa
s =
0.225
0.2225 GPa
0.04
=
*
=
*
0.1
100
=
e =
0.001 mm/mm
s
e
=
0.2225
0.001
E
=
=
222.5 GPa
Sample Problem 3.3
A steel specimen is tested in tension. The specimen is 1 in. wide by 0.5 in. thick in the
test region. By monitoring the load dial of the testing machine, it was found that the
specimen yielded at a load of 36 kips and fractured at 48 kips.
a. Determine the tensile stresses at yield and at fracture.
b. If the original gauge length was 4 in., estimate the gauge length when the speci-
men is stressed to 1/2 the yield stress.
Solution
a. Yield stress
1
s
y
2
=
36/
1
1
*
0.5
2
=
72 ksi
Fracture stress
1
s
f
2
=
48/
1
1
*
0.5
2
=
96 ksi
10
6
psi
b. Assume
E
=
30
*
A
1
2
B
10
3
/
10
6
e =
s
y
/E
=
1
1/2
2
*
72
*
1
30
*
2
=
0.0012 in./in.
ยข
L
=
L
e =
4
*
0.0012
=
0.0048 in.
Final gauge length
=
4
+
0.0048
=
4.0048 in.
Steel is generally assumed to be a homogenous and isotropic material.
However, in the production of structural members, the final shape may be
obtained by cold rolling. This essentially causes the steel to undergo plas-
tic deformations, with the degree of deformation varying throughout the
member. As discussed in Chapter 1, plastic deformation causes an increase
in yield strength and a reduction in ductility. Figure 3.15 demonstrates
that the measured properties vary, depending on the orientation of the
sample relative to the axis of rolling (Hassett, 2003). Thus, it is necessary
to specify how the sample is collected when evaluating the mechanical
properties of steel.
Search WWH ::
Custom Search