Question 7.4: Determine the maximum deflection of a circular plate on an e...

Determine the maximum deflection of a circular plate on an elastic foundation subjected to a concentrated load F in the center of the plate.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

From Table 7.1, it is seen that as r approaches infinity, Z_{1} and Z_{2} also approach infinity. Therefore, C_{1} and C_{2} must be set to zero. Thus,

w=C_{3} Z_{3}(\alpha r)+C_{4} Z_{4}(\alpha r)

and

\theta=\frac{ d w}{ d r}=C_{3} \alpha Z_{3}^{\prime}(\alpha r)+C_{4} \alpha Z_{4}^{\prime}(\alpha r)

As r approaches zero, 𝜃 must be zero due to symmetry. But from Table 7.1, Z_{4}^{\prime} approaches infinity as r approaches zero. Hence, C_{4} must be set to zero.

Thus,

w=C_{3} Z_{3}(\alpha r)

 

\frac{ d w}{ d r}=C_{3} \alpha Z_{3}^{\prime}(\alpha r)

 

\frac{ d ^{2} w}{ d r^{2}}=C_{3} \alpha^{2}\left[Z_{4}(\alpha r)-\frac{1}{\alpha r} Z_{3}^{\prime}(\alpha r)\right]

 

\frac{ d ^{3} w}{ d r^{3}}=C_{3} \alpha^{2}\left[\alpha Z_{4}^{\prime}(\alpha r)-\frac{1}{r} Z_{3}^{\prime \prime}(\alpha r)\right]

Substituting these derivatives into Eq. (7.10) and equating this to F gives

Q=D\left(\frac{ d ^{3} w}{ d r^{3}}+\frac{1}{r} \frac{ d ^{2} w}{ d r^{2}}-\frac{1}{r^{2}} \frac{ d w}{ d r}\right)                      (7.10)

 

C_{3}=\frac{F}{4 \alpha^{2} D}

and

w=\frac{F}{4 \alpha^{2} D} Z_{3}(\alpha r)

and

w_{\max }=\frac{F}{8 \alpha^{2} D}

 

Table 7.1 Limits of Z functions

Function Limit as x → 0 Limit as x → ∞
Z_{1}(x) 1.0 \zeta \cos \kappa
Z_{2}( x ) \frac{-x^{2}}{4}   -\zeta \sin \kappa
Z_{3}(x) 0.5 v \sin \psi
Z_{4}(x) \frac{2}{\pi} \ln \frac{\gamma x}{2} -v \cos \psi
\frac{ d Z_{1}(x)}{ d x} \frac{-x^{3}}{16} \frac{\zeta}{\sqrt{2}}(\cos \kappa-\sin \kappa)
\frac{ d Z_{2}(x)}{ d x} \frac{-x}{2} \frac{-\zeta}{\sqrt{2}}(\cos \kappa+\sin \kappa)
\frac{ d Z_{3}(x)}{ d x} \frac{x}{\pi} \ln \frac{\gamma x}{2} \frac{v}{\sqrt{2}}(\cos \psi-\sin \psi)
\frac{ d Z_{4}(x)}{ d x} \frac{2}{\pi x} \frac{v}{\sqrt{2}}(\cos \psi+\sin \psi)
\zeta=\frac{1}{\sqrt{2 \pi x}} \exp \frac{x}{\sqrt{2}} v=\sqrt{\frac{2}{\pi x}} \exp \frac{-x}{\sqrt{2}}
\kappa=\frac{x}{\sqrt{8}}-\frac{\pi}{8} \psi=\frac{x}{\sqrt{2}} +\frac{\pi}{8}
\zeta=\frac{1}{\sqrt{2 \pi x}} \exp \frac{x}{\sqrt{2}} v=\sqrt{\frac{2}{\pi x}} \exp \frac{-x}{\sqrt{2}}
\kappa=\frac{x}{\sqrt{8}}-\frac{\pi}{8} \psi=\frac{x}{\sqrt{2}}+\frac{\pi}{8}

Related Answered Questions