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

3/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

Early experiments

R.Hooke (1635-1703),„De Potentia Restitutiva” ,1678

E.Mariotte (1620-1684)

4/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

Modern testing machine

5/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

6/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

7/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

8/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

9/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

10/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

11/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

12/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

13/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

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

14/13M.Chrzanowski: Strength of Materials

SM2-02: Tension/compression

stop