Uniaxial Stresses

Members In Uni – Axial State Of Stress:

For a prismatic bar loaded in tension by an axial force P, the elongation of the bar can be determined as

Suppose the bar is loaded at one or more intermediate positions, then equation (1) can be readily adapted to handle this situation, i.e. we can determine the axial force in each part of the bar i.e. parts AB, BC, CD, and calculate the elongation or shortening of each part separately, finally, these changes in lengths can be added algebraically to obtain the total charge in length of the entire bar.

When either the axial force or the cross – sectional area varies continuously along the axis of the bar, then equation (1) is no longer suitable. Instead, the elongation can be found by considering a deferential element of a bar and then the equation (1) becomes

d δ=Px d x

E Ax

δ=∫0

l Px d x

E Ax

i.e. the axial force Px and area of the cross – section Ax must be expressed as functions of x. If the expressions for Px

and Ax are not too complicated, the integral can be evaluated analytically, otherwise Numerical methods or techniques can be used to evaluate these integrals.

Stresses In Non – Uniform Bars:

Consider a bar of varying cross section subjected to a tensile force P as shown below.

Leta = cross sectional area of the bar at a chosen section XXthenStress = p / aσIf E = Young's modulus of bar then the strain at the section XX can be calculatedε = / EσThen the extension of the short element  x. = . original length =δ ε  σ / E. δx

¿ PEδxa

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Uniaxial Stresses

Thus, the extension for the entire bar is

δ=∫0

l Px

Eδ xa

Total Extension= PE∫0

lδ xa

Now let us for example take a case when the bar tapers uniformly from d at x = 0 to D at x = l

In order to compute the value of diameter of a bar at a chosen location let us determine the value of dimension k, from similar triangles

(D−d ) /2l

= kx

k=(D−d ) x2 l

therefore, the diameter 'y' at the X-section isor y = d + 2k

y=d+(D−d ) x

lHence the cross –section area at section X- X will be

Ax∨a=π4y2=π

4 [d+(D−d ) xl ]2

hence the total extension of the bar will be given by expression

Total Extension= PE∫0

lδ xa

=π4PE∫0

lδ x

π4 [d+(D−d ) x

l ]2=4 PlπE [ 1D−1

d ]= 4 PlπEDd

An interesting problem is to determine the shape of a bar which would have a uniform stress in it under the action of its own weight and a load P.let us consider such a bar as shown in the figure below: The weight of the bar being supported under section XX is

¿∫0

x

ρga d x

where is density of the bar. Thus the stress at XX is ρ

σ=P+∫

0

x

ρgad x

a

σ .a=P+∫0

x

ρga d x

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Uniaxial Stresses

Differentiating the above equation with respect to x we get

σ .d ad x

=ρ .g .a

integrating

∫ d aa

=∫ ρ .gσ

d x

log ea=ρ . g . xσ

+constant

applying boundary conditions

at x = 0; a = a0 thus constant = log eab

log ea=ρ . g . xσ

+loge a0

log e( aa0 )= ρ .g . xσ

eρ . g . x

σ = aa0

also at x = 0

σ= Pa0

Thus

aa0

=eρ. g . x a0

P

Illustrative Problem: A round bar, of length L, tapers uniformly from radius r1 at one end to radius r2at the other.

Show that the extension produced by a tensile axial load P is PL

2πE r12 . If r2 = 2r1 , compare this extension with that

of a uniform cylindrical bar having a radius equal to the mean radius of the tapered bar.Solution:

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Uniaxial Stresses

Consider the above figure let r1 be the radius at the smaller end. Then at a X cross-section XX located at a distance x from the smaller end, the value of radius is equal to

¿ r1+xL

(r2−r 1)=r1 (1+kx )

where k=( r2−r1L ). 1r1

Stress at section XX= loadarea

= P

π r12 (1+kx )2

Strainat this section= stressE

= P

E .π r12 (1+kx )2

Thus for a small length dx of the bar at this section the extension is

P .d x

Eπ r12 (1+kx )2

integrating

Extension=∫0

LP .dx

Eπ r12 (1+kx )2

= PEπ r1

2∫0

L

(1+kx )−2d x= PEπ r1

2 [ (1+kx )−1

−k ]0

L

= PEπ r 1

2 [ (1+kx )−1

−k− 1

−k ]¿ PEπ r1

2 [1− 11+kL ]= PL

Eπ r12 (1+k L )

since k=( r2−r1r1 . L )

Thus 1+kL=r 2r 1

Thus , Extension= PLEπ r1r 2

Comparing of Extensions:

For the case when r2 = 2.r1, the value of computed extension as above becomes equal to P

2Eπ r12  

The mean radius of taper bar = 1 / 2(r1 + r2) = 1 / 2(r1 +2 r2) = 3 / 2 .r1

Therefore, the extension of uniform bar = Original length. Strain

¿ L σE

= LE.

P

π ( 32 r1)2= 4 PL

gπEr12

Extensionof uniformExtension of tapered

=

4 PL

gπEr 12

P2 Eπ r1

2

=8g

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