下面求解式(3-43), 令(12N₀/mL³)(T-T₂)=θ,且tanθ=K, 则：

w(T)=[w₁+w₂+(M_0U+M_0C)/N₀]·cosθ·[√(12N₀/mL³)]3R²v₁/L²·sinθ-M_0U+M_0C/N₀

=[w₁+w₂+(M_0U+M_0C)/N₀]·cosθ·{1+[√(mL³/12N₀)]sinθ/[w₁+w₂+(M_0U+M_0C)/N₀]cosθ}

=[w₁+w₂+(M_0U+M_0C)/N₀]·cosθ·(1+tan²θ)-(M_0U+M_0C)/N₀

=[w₁+w₂+(M_0U+M_0C)/N₀]·(1/cosθ)-(M_0U+M_0C)/N₀

=[w₁+w₂+(M_0U+M_0C)/N₀]·[1+tan²(θ/2)]/[1-tan²(θ/2)]-(M_0U+M_0C)/N₀                     (3-44)

由tanθ=[√(mL³/12N₀)]·3R²v₁/{L²[w₁+w₂+(M_0U+M_0C)/N₀]}=K,可求得：

tan(θ/2)={[√(1+K²)]-1}/K                                                           (3-45)

将式(3-45)代入式(3-44),得：

w(T)=[w₁+w₂+(M_0U+M_0C)/N₀]·〔1+{[√(1+K²)]-1}/K〕/〔1-{[√(1+K²)]-1}/K〕-(M_0U+M_0C)/N₀

=[w₁+w₂+(M_0U+M_0C)/N₀]·〔K²+{1+K²-2[√(1+K²)]+1}/K²-{1+K²-2[√(1+K²)]+1}〕-(M_0U+M_0C)/N₀

=[w₁+w₂+(M_0U+M_0C)/N₀]·[√(1+K²)]-(M_0U+M_0C)/N₀

=〔√{[√(mL³/12N₀)]·(3R²v₁/L²)}²+[w₁+w₂+(M_0U+M_0C)/N₀]²〕-(M_0U+M_0C)/N₀               (3-46)

上式即为板中心点的最终挠度。

因此，夹芯板中心点的最终响应时间和最大挠度分别为:

T=T₂+[√(mL³/12N₀)]·tan^-1·{3R²v₁/L²[√(12N₀/mL³)][w₁+w₂+(M_0U+M_0C)/N₀]}               (3-47)

w(T)=〔√{[√(mL³/12N₀)]·(3R²v₁/L²)}²+[w₁+w₂+(M_0U+M_0C)/N₀]²-(M_0U+M_0C)/N₀              (3-48)