Stress and Strain in Bars in various conditions20 min read

Stress and Strain in a bar due to its self-weight

A bar of length l metre and area  A m^{2} is fixed at one end. Let the density of material be ρ kg/m^{3}. Let us consider a very small shaded strip of bar ABCD of thickness dy at a distance y from the bottom end as shown in fig.strain self weight

Now the force is acting down at CD

= weight of bar CDEF

=    Ayρ

The stress at the section CD is given by

 \Rightarrow  \sigma = \frac{\mbox{ Force at CD}}{\mbox{ area of cross-section of the bar}}

 \Rightarrow  \sigma = \frac{A \times y \times \rho \times 9.81 }{ A}

 \Rightarrow  \sigma = 9.81 \rho \times y      N/ m^2

Which means stress at any section due to self weight of the bar is directly propotional to y.

∴      stress at bottom end = 0

and stress at top end,  \sigma_{max} = \mbox{9.81 \rho  l}

If we assume dy is very small, the thickness in section AB and CD are practically equal.


⇒ strain in length,   dy = \frac{ \sigma}{E} = \frac{ \mbox{9.81  \rho  y}}{E}

Extension in length , dy = \frac{9.81 \rho y }{E} dy

∴ Total elongation of the bar,

⇒      \delta l =\int_{0}^{l} \frac{9.81 \rho y}{E}dy

⇒      = \frac{9.81 \rho }{E}\int_{0}^{l} y.dy

⇒      =  \frac{9.81 \rho }{E}\left [ \frac{y^2}{2} \right ]_{0}^{l}

⇒     \boxed{\delta l = \frac{9.81 \rho l^2}{2E} }

Stress & Elongation in a bar due to rotation

A bar of length  l is rotating about y-axis at an angular speed of ω rad/sec.rotation bar strain

Tensile force on the element AB:

Centrifugal force P on the part BC = \omega^2 rM

where M is the mass of the part BC

Thus, if

A = Area of cross-section of the bar (m^2)

and  ρ = Density of the material ( kg/m^3)


 P = \rho A \left [ \frac{1}{2}- x \right ] \omega^2 \left [ x + \frac{1}{2}\left ( \frac{1}{2}-x \right )\right ]

or  P = 1/2 \rho A \omega^2\left ( l^2 /4 -x^2 \right ) Newton

i.e. Stress,  \sigma_c = 1/2 \rho \omega^2 \left( l^2/4 - x^2\right) N/m^2       . . . . . (1)

Maximum stress will occur at

x = 0     i.e.

 \sigma_{max} = \frac{1}{8} \rho \omega^2 l^2                       . . . . .(2)

Extension of the element AB

 = \frac{\sigma_c}{E}dx

Thus substituting the value  \sigma_c from equation 1 and integration, the total extension in the bar is given by:

 \delta l = 2 \int_{0}^{\frac{1}{2}} \frac{1}{2E} \rho \omega^2 \left ( l^2/4 - x^2 \right )dx

 \delta l = \frac{\rho \omega^2}{E}\left [ \frac{l^2}{4}x - \frac{x^3}{3} \right ]_0^\frac{1}{2}

i.e.  \boxed{\delta l = \frac {\rho \omega^2 l^3}{12E}} Metre

Elongation in case of a taper rod

A rod of length l tapers uniformly from diameter d_1 to d_2. Its wider end is fixed and the lower end is subjected to an axial tensile load P. AC and BD are produced to meet at N.elongation in taper bar

Consider a small length “dy” at a distance “y” from the lower end as shown in fig.

Let A_1 to A_2 and A be the cross-sectional areas of the tapered rod at the top, bottom and at UV respectively and let stresses due to the load P on these areas be  \sigma_1,  \sigma_2 and  \sigma respectively.

Then,  A_1 = \frac{ \pi}{4}d_1^2 ;  A_2 = \frac{ \pi}{4}d_2^2 ;  A = \frac{ \pi}{4}d^2

and  P = \sigma_1.A_1 =  \sigma_2.A_2 = \sigma.A

or  \sigma = \frac{\sigma_2 . A_2}{A}

Elongation of the small strip, STUV =  \frac{\sigma}{E}.dy

 \frac{\sigma_2 . A_2}{A.E}.dy = \frac{\sigma_2}{E}.\frac{d_2^2}{d^2}.dy       ….(1)

from similar triangles CDN and UVN

 \frac{d_2}{d} = \frac{l' - l}{l' - l + y}

Putting the value of  \frac{d_2}{d} in eqn (1) , We get

elongation in small strip STVU

 = \frac{\sigma_2}{E}. \frac{(l' - l)^2}{(l' - l +y)^2}

∴ Total elongation in the taper Rod,

 \delta l = \int_{0}^{1}\frac{\sigma_2}{E}.\frac{(l' - l)^2}{(l' - l + y)^2}.dy

 \frac{\sigma_2(l' - l)^2}{E} \int_{0}^{1} \frac{1}{(l' - l +y)^2}.dy

 \frac{\sigma_2(l' - l)^2}{E} \left | \frac{-1}{l' - l + y} \right |_0^l

 \frac{\sigma_2(l' - l)^2}{E}\left | \frac{-1}{l' - l + l} - \frac{-1}{l' -l + 0} \right |

 \frac{\sigma_2(l' - l)^2}{E} \times \frac{l}{l'(l' - l)}

 \frac{\sigma_2(l' - l)}{l'.E} .l            ….(2)

Also from similar triangles CDN and ABN

 \frac{l' - l}{l'} = \frac{d_2}{d_1}

Subtituting this value in eqn (2), we get

 \delta l = \frac{\sigma_2. l}{E}. \frac{d_2}{d_1}

or       \delta l = \frac{P. l }{A_2. E}. \frac{d_2}{d_1} = \frac{P.l}{ \frac{\pi}{4}.d_2^2.E}. \frac{d_2}{d_1}

i.e.     \boxed{ \delta l = \frac{4.P.l}{ \pi. E.d_1.d_2}}

Stresses induced in compound ties or struts

Generally ties consists of two materials, fastened together to prevent uneven straining of the two materials. It will be assumed that the two materials also very symmetrically distributed about the axis of the bar, as with the cylindrical rod in cased in the tube.compound bar

If an axial load P is applied to the bar,

 P = \sigma_1.A_1 + \sigma_2. A_2

where  \sigma_1 and  \sigma_2 are the stresses induced and  A_1 and  A_2 are cross-sectional areas of the materials.

The strains produced  e_1 and  e_2 are equal.

∴     e_1 = e_2

∴       \boxed{ \frac{\sigma_1}{E_1} = \frac{\sigma_2}{E_2}   \Rightarrow \frac{\sigma_1}{\sigma_2} =\frac{E_1}{E_2} }



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