PROBLEM 11.2
KNOWN:
A steel rod (circular crosssection) having:
10
0.3 m
5 mm
20 10 Pa
50 kg
m
L
D
E
m
=
=
=×
=
FIND:
The change in length of the rod,
L
δ
.
ASSUMPTIONS:
Rod is elastically deformed, such that
m
E
σ
ε
=
⋅
SOLUTION:
The force resulting from 50 kg in standard gravity is
()
( )
2
2
50 kg
9.8 m/s
kg m
1.0
s N
N
c
ma
F
g
==
⎛⎞
⎜⎟
⎝⎠
490 N
N
F
=
The resulting uniaxial stress is
N
a
c
F
A
=
where
2
35
2
6
5
2
51
0
1
.
9
61
0 m
4
490 N
25 10 Pa
1.96 10 m
c
a
A
π
−−
×
×
and
6
6
10
125 10
a
a
m
E
−
×
×
The change in length is then
66
0.3 125 10
37.5 10 m
a
LL
δε
=⋅ =
×
=
×
0.3 m
D=5 mm
F
N
Cross
Section, A
c
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View Full DocumentPROBLEM 11.10
KNOWN:
A steel member (
p
ν
= 0.3) subject to simple axial tension.
Strain gauges
are mounted on top center, and bottom center.
2,
All
'
120
GF
R s
=
=Ω
10
V and
10 V,
120
and
=2
oi
EE
R
G
F
δ
μ
==
=
Ω
FIND:
Bridge constant, for gauge locations 1 and 4.
Is the system temperature
compensated?
Determine the axial and transverse strains.
SOLUTION:
The configuration is
Since for any 4 gauges
()
0
1243
4
i
E
GF
E
ε
εεε
=−
+
−
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 Spring '11
 Dr.ChrisMechefske
 Strain, Trigraph, σa, δ Eo

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