Find elongation of prismatic bar of length l, cross sectional area a, and elongation modulus e
Answers
Members in Uni – axial state of stress
Introduction: [For members subjected to uniaxial 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 continuosly 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
i.e. the axial force Pxand area of the cross – section Ax must be expressed as functions of x. If the expressions for Pxand 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.
Let
a = cross sectional area of the bar at a chosen section XX
then
Stress s = p / a
If E = Young's modulus of bar then the strain at the section XX can be calculated
Î = s / E
Then the extension of the short element d x. = Î .original length = s / E. dx
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
therefore, the diameter 'y' at the X-section is
or = d + 2k
Hence the cross –section area at section X- X will be
hence the total extension of the bar will be given by expression
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
The same results are obtained if the bar is turned upside down and loaded as a column as shown in the figure below:
IIIustrative Problem 1: Calculate the overall change in length of the tapered rod as shown in figure below. It carries a tensile load of 10kN at the free end and at the step change in section a compressive load of 2 MN/m evenly distributed around a circle of 30 mm diameter take the value of E = 208 GN / m2.
This problem may be solved using the procedure as discussed earlier in this section
IIIustrative Problem 2: 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
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:
consider the above figure let r1 be the radius at the smaller end. Then at a X crosssection XX located at a distance x from the smaller end, the value of radius is equal to