Table 2 Results for vertical point load... | Emerald Publishing

Table 2

Results for vertical point load at the surface

(a) Displacements
T2.1	Note	$λ_{1}^{'} = λ_{2}$ and $λ_{2}^{'} = λ_{1}$
T2.2	Radial	$U_{r} = \frac{P_{z}}{4 π G_{vh}} \sum_{i = 1}^{2} (- λ_{i}^{'} \frac{R_{i}^{*}}{r R_{i}})$
T2.3	Circumferential	U_θ = 0
T2.4	Vertical	$U_{z} = \frac{P_{z}}{4 π G_{vh}} \sum_{i = 1}^{2} (\frac{- m_{i} λ_{i}^{'}}{R_{i}})$
(b) Strains
T2.5	$ϵ_{r r} = \frac{P_{z}}{4 π G_{vh}} \sum_{i = 1}^{2} [λ_{i}^{'} (\frac{z_{i}}{R_{i}^{3}} - \frac{R_{i}^{*}}{r^{2} R_{i}})]$
T2.6	$ϵ_{θ θ} = \frac{P_{z}}{4 π G_{vh}} \sum_{i = 1}^{2} [λ_{i}^{'} \frac{R_{i}^{*}}{r^{2} R_{i}}]$
T2.7	$ϵ_{z z} = \frac{P_{z}}{4 π G_{vh}} \sum_{i = 1}^{2} [- λ_{i}^{'} m_{i} u_{i} \frac{z_{i}}{R_{i}^{3}}]$
T2.8	$γ_{r θ} = γ_{θ z} = 0$
T2.9	$γ_{r z} = \frac{P_{z}}{4 π G_{vh}} \sum_{i = 1}^{2} [- \frac{λ_{i}^{'} (u_{i} + m_{i})}{R_{i}^{3}}]$
(c) Stresses
T2.10	$σ_{r r} = \frac{P_{z}}{4 π} {\sum_{i = 1}^{2} [λ_{i}^{'} u_{i} (u_{i} + m_{i}) \frac{z_{i}}{R_{i}^{3}}] + 2 \frac{G_{hh}}{G_{vh}} \sum_{i = 1}^{2} (- λ_{i}^{'} \frac{R_{i}^{*}}{r^{2} R_{i}})}$
T2.11	$σ_{θ θ} = \frac{P_{z}}{4 π} {\sum_{i = 1}^{2} [λ_{i}^{'} u_{i} (u_{i} + m_{i}) \frac{z_{i}}{R_{i}^{3}}] - 2 \frac{G_{hh}}{G_{vh}} \sum_{i = 1}^{2} [λ_{i}^{'} (\frac{z_{i}}{R_{i}^{3}} - \frac{R_{i}^{*}}{r^{2} R_{i}})]}$
T2.12	$σ_{z z} = \frac{P_{z}}{4 π} \sum_{i = 1}^{2} [- \frac{λ_{i}^{'}}{u_{i}} (u_{i} + m_{i}) \frac{z_{i}}{R_{i}^{3}}]$
T2.13	τ_rθ = τ_θr = 0
T2.14	$τ_{r z} = \frac{P_{z}}{4 π} \sum_{i = 1}^{2} [- λ_{i}^{'} (u_{i} + m_{i}) \frac{r}{R_{i}^{3}}]$

Based on the general solutions by Liao and Wang (1998)