TENSION / COMPRESSION
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Transcript of TENSION / COMPRESSION
1/13M.Chrzanowski: Strength of Materials
SM2-02: Tension/compression
TENSION / COMPRESSION
2/13M.Chrzanowski: Strength of Materials
SM2-02: Tension/compression
N ≠ 0, Qy=0, Qz=0Formal definition: the case when set of internal forces reduces to the sum which is tangent to the bar axis
Example: a straight bar loaded by concentrated forces at its ends (truss strut or tie)
y
z
N= P
x
PP
P P
Mx=0, My=0, Mz=0
M=P·
N(x)=P
N
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Early experiments
R.Hooke (1635-1703),„De Potentia Restitutiva” ,1678
E.Mariotte (1620-1684)
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Modern testing machine
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PK B C D
K’ B’ C’
E
D’ E’
KuKK ' CuCC ' )(xuu Axial displacement is linear function of x !
x
),,( wvuuu
)(xx
u Normal strain is
constant along x axis!
Experimental approach
For constant P
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x
Bernoulli hypothesis
Axial displacement does not depend on
y i z variables
Normal strain does not depend on
y i z variables
Normal stress is constant over the
whole cross section
xx E
x
PK B C D
K’ B’ C’
E
D’ E’
KuKK ' CuCC ' )(xuu Axial displacement is linear function of x !
x
),,( wvuuu
)(xx
u Normal strain is
constant function of x !
Experimental approach
For constant P
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y
z
N
x
x x
The condition of equivalence of internal cross-sectional forces with its sum yields:
)()( xNdAxA
x
A
xx xNAxdAx )()()(
A
xNxx
)()(
Distribution of
x
xDistribution of
Experimental approach
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NconstxNx)(
A
Nx
For
EA
Nx
lEA
Nllxuu maxmax
xEA
Ndx
EA
Nu
?)?,,( wvuu
This is NOT a full solution
v
wz
y
zxz
yxy
Experimental approach
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x
000
000
00x
T
As stress tensor is given in principal axes with x axis which coincides with the cross-section centre of gravity thus axes y and z – are arbitrary orthogonal axes with origin at the centre of gravity
Z
De Saint-Venant hypothesis:Solution validity
range
xz
xy
x
T
00
00
00ANx /EANx /
EANxy /
Poisson coefficient
BVP approach
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BVP approach makes it easy to notice that at any other than principal axes the angular stresses arise . The maximum value of these appears – as we know from 3D strain analysis – on planes inclined by 45o to principal axes. These stresses are equal to half of the difference between two consecutive principal normal stresses, whereas the normal stress is equal to the half of sum of normal stresses there.
BVP approach
2max kj
i
2kj
vi
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1
2
3
P
P
1
2
3
02
max 321
22max 131
2
22max 121
3
02
321
v
22131
2
v
22121
3
v
·
·
0,0 321
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BVP solution demonstrates also that for the case of non-prismatic bar the stress tensor as found for prismatic bar does not satisfy Static Boundary Conditions if the side surface of the bar is free of traction. Therefore, the normal stresses in x,y,z, axes are not the principal stresses.
BVP approach
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000
000
00x
T
PPy
z
x
)1,0,0(
jijvj
)cos,0,sin(
SBC
0cos000)sin(01 xv
01000001 xv
tanxxz ?
xz
zx
0cos00)sin(03 zzxv
2tantan xzxz
zxx
z
z
)()(
xA
Pxx ?
xz
BVP approach
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stop