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## Q. 12.2

A constant time section of lo00 h taken through a series of strain-time creep curves obtained for a particular polymer at various stress levels yields the following isochronous stress-strain data.

$σ(kN/m^{2})$        1.0   2.25   3.75   5.25   6.54    7.85   9.0

ε(%)                       0.23    0.52  0.85  1.24   1.68    2.17     2.7

The polymer is now used to manufacture:

(a) a disc of thickness 6 mm, which is to rotate at 500 rev/min continuously,
(b) a diaphragm of the same thickness which is to be subjected to a uniform lateral pressure of $16 N/m^{2}$ when clamped around its edge.
Determine the radius required for each component in order that a limiting stress of $6 k N/m^{2}$ is not exceeded after l000 hours of service. Hence find the maximum deflection of the diaphragm after this lo00 hours of service.

The lateral strain ratio for the polymer may be taken as 0.45 and its density as $1075 kN/m^{3}$.

## Verified Solution

(a) From eqn. (4.11) the maximum stress at the centre of a solid rotating disc is given by:

$\sigma _{r_{max} }=\sigma _{\theta _{max} }=(3+\nu )\frac{\rho \omega ^{2}R^{2}}{8}$    (4.11)

For the limiting stress condition, therefore, with Poissons ratio ν replaced by the lateral strain ratio:

$6\times 10^{3}=3.45\times 1075\times \frac{(500\times 2\pi )^{2}}{60} \times \frac{R^{2}}{8}$

From which          $R^{2} = 0.00472$

and                R = 0.0687 m = 68.7 mm.

(b) For the diaphragm with clamped edges the maximum stress is given by eqn. (22.24) as:

$\sigma _{r_{max} }=\frac{3qR^{2}}{4t^{2}}$    (22.24)
$6\times 10^{3}=\frac{3\times 16\times R^{2}}{4\times (6\times 10^{-3})^{2}}$

From which R = 0.134 m = 134 mm.

The maximum deflection of the diaphragm is then given in Table 7.1 as:

 Loading  condition maximum deflection $y _{{max}}$ Maximum stresses $\sigma _{r_{max} }$ $\sigma _{z_{max} }$ Uniformly loaded, edges clamped $\frac{3qR^{4}}{16Et^{3}}(1-\nu^{2})$ $\frac{3qR^{2}}{4t^{2}}$ $\frac{3qR^{2}}{8t^{2}(1+\nu)}$ Uniformly loaded, edges freely supported $\frac{3qR^{4}}{16Et^{3}}(5+\nu)(1-\nu)$ $\frac{3qR^{2}}{8t^{2}(3+\nu)}$ $\frac{3qR^{2}}{8t^{2}(3+\nu)}$ Central load F , edges clamped $\frac{3FR^{2}}{4\pi Et^{3}}(1-\nu^{2})$ $\frac{3F}{2\pi t^{2}}$ $\frac{3ν F}{2\pi t^{2}}$ Central load F, edges freely supported $\frac{3FR^{2}}{4\pi Et^{3}}(3+\nu)(1-\nu)$ From $\frac{3F}{2\pi t^{2}}(1+\nu)\log_{e}\frac{R}{r}$ From $\frac{3F}{2\pi t^{2}}[(1+\nu)\log_{e}\frac{R}{r}+(1-\nu)]$
$\delta _{max} =\frac{3qR^{4}}{16Et^{3}} (1-\upsilon ^{2})$

Here it is necessary to replace Young’s modulus E by the secant modulus obtained from the isochronous curve data and Poisson’s ratio by the lateral strain ratio. The lo00 hour isochronous curve has been plotted from the given data .in Fig. 12.16 producing a secant modulus of $405 kN/m^{2}$ at the stated limiting stress of $6 kN/m^{2}$; this being the slope of the line from the origin to the $6 kN/m^{2}$ point on the isochronous curve.

$\delta _{max} =\frac{3\times 16\times (134\times 10^{-3})^{4}\times (1-0.45^{2})}{16\times 405\times 10^{3}\times (6\times 10^{-3})^{3}}$
$=0.0088 m=8.8 mm$