弹性地基梁验证
原理
《弹性地基梁的三解级数解法》（杨维加）
$$ \begin{aligned} & p(\xi)=\sum_{n=0}^{\infty} a_{\mathrm{n}} \cos n \pi \xi+\sum_{n=1}^{\infty} b_{\mathrm{n}} \sin (n-0.5) \pi \xi \qquad \text{(1-14)}\ \end{aligned} $$
$$ \begin{aligned} & t=\frac{\pi E_0 b l^3}{4 E I} \qquad \text{(2-17)}\ \end{aligned} $$ $$ \begin
\end{aligned} $$
$$ \begin{aligned} & S(\xi, n)=\frac{1}{n \pi}\left[\cos n \pi \xi f_{\mathrm{s}}(\xi, n)-\sin n \pi \xi f_{\mathrm{c}}(\xi, n)\right] \qquad \text{(2-41)}\ & S(\xi, n)=-\frac{1}{(n \pi)^2}\left[s(\xi, n)+\frac{2 \cos n \pi}{1-\xi^2}\right] \quad(2-50) \ \end{aligned} $$
$$ \begin{aligned} & z(\xi, n) =-m \pi\left[\sin m \pi \xi f_{\mathrm{s}}(\xi, m)+\cos m \pi \xi f_{\mathrm{c}}(\xi, m)\right]-\frac{2 \xi \sin m \pi}{1-\xi^2} \qquad \text{(2-113)}\ & s(\xi, n)=n \pi\left[\sin n \pi \xi f_{\mathrm{c}}(\xi, n)-\cos n \pi \xi f_{\mathrm{s}}(\xi, n)\right]-\frac{2 \cos n \pi}{1-\xi^2} \qquad \text{(2-49)}\ \end{aligned} $$
$$ \begin{aligned} \end{aligned} $$ $$ \begin
\end{aligned} $$
$$ \begin
\end{aligned} $$
$$ \begin
\end{aligned} $$ $$ \begin{aligned} & Z(\xi, n)=-\frac{1}{(m \pi)^2}\left[z(\xi, n)+\frac{2 \xi \sin m \pi}{1-\xi^2}\right] \qquad \text{(2-114)}\ \end{aligned} $$
$$ \begin{aligned} & S_k(\xi, n)=S_{\infty}(\xi, n)-2 \cos n \pi \int_0^{10 / k} \frac{1-f(k \rho)}{\rho^2-(n \pi)^2} \sin \rho \cos \rho \xi d \rho \qquad \text{(4-29)}\ \end{aligned} $$ $$ \begin{aligned} & f_{\mathrm{s}}(\xi, n)=\int_0^{1+\xi} \frac{\sin n \pi \rho}{\rho} \mathrm{~d} \rho+\int_0^{1-\xi} \frac{\sin n \pi \rho}{\rho} \mathrm{~d} \rho \qquad \text{(2-40)}\ & f_{\mathrm{c}}(\xi, n)=\int_{1-\xi}^{1+\xi} \frac{\cos n \pi \rho}{\rho} \mathrm{~d} \rho \qquad \text{(2-40)}\ & r(\xi, n)=\frac{2}{(n \pi)^2}(\cos n \pi-\cos n \pi \xi) \qquad \text{(2-54)}\ & r(\xi, 0)=-(1 -\xi^2) \qquad \text{(2-56)}\ & M_{\mathrm{p}}(\xi)=\frac{l^2}{2} \sum_{\mathrm{n}=0}^{\infty} a_{\mathrm{n}} r(\xi, n) \qquad \text{(2-55)}\ & m=n-0.5 \ & w(\xi, n)=\frac{2}{(m \pi)^2}(\sin m \pi-\sin m \pi \xi) \qquad \text{(2-117)}\ & M_{\mathrm{p}}(\xi)=\frac{l^2}{2} \sum_{\mathrm{n}=1}^{\infty} b_{\mathrm{n}} w(\xi, n) \qquad \text{(2-118)}\ & f(\alpha)=\frac{\operatorname{sh}^2 \alpha}{\alpha+\operatorname{sh} \alpha \operatorname{ch} \alpha} \qquad \text{(4-2)}\ & k=\frac{H}{l} \qquad \text{(4-6)} \ & s_k(\xi, n)=s_{\infty}(\xi, n)+2 \cos n \pi \int_0^{10 / k} \rho^2 \frac{1-f(k \rho)}{\rho^2-(n \pi)^2} \sin \rho \cos \rho \xi d\rho \qquad \text{(4-30)}\ & s_{\infty}(\xi, n)=n \pi\left[\sin n \pi \xi f_{\mathrm{c}}(\xi, n)-\cos n \pi \xi f_{\mathrm{s}}(\xi, n)\right]-\frac{2 \cos n \pi}{1-\xi^2} \qquad \text{(2-49)}\ & z_k(\xi, n)=z_{\infty}(\xi, n)-2 \sin m \pi \int_0^{10 / k} \rho^2 \frac{1-f(k \rho)}{\rho^2-(m \pi)^2} \cos \rho \sin \rho \xi d\rho \qquad \text{(4-38)(注：积分前改为负号，积分内改n为m)}\ & z_{\infty}(\xi, n) =-m \pi\left[\sin m \pi \xi f_{\mathrm{s}}(\xi, m)+\cos m \pi \xi f_{\mathrm{c}}(\xi, m)\right]-\frac{2 \xi \sin m \pi}{1-\xi^2} \qquad \text{(2-113)}\ &\mathbf{F}{\mathrm{a}}=\mathbf{s}+t \mathbf{r}\ & \mathbf{F}{\mathrm{b}}=\mathbf{z}+t \mathbf{w} \ &\mathbf{F}{\mathrm{a}} \mathbf{X}{\mathrm{a}}=\mathbf{\Delta}{\mathrm{a}}\ &\mathbf{F}{\mathrm{b}} \mathbf{X}{\mathrm{b}}=\mathbf{\Delta}{\mathrm{b}}\ \end{aligned} $$
$$ \begin{aligned} \Delta_{q^{\prime}}(\xi)= & 2 \xi \frac{q_1 \rho_2-q_2 \rho_1}{\rho_2-\rho_1}\left(\frac{1}{\left(\rho_1^2-\xi^2\right)}-\frac{1}{\left(\rho_2^2-\xi^2\right)}\right. \ & +\frac{q_2-q_1}{\rho_2-\rho_1}\left[\ln \frac{\left(\rho_1+\xi\right)\left(\rho_2-\xi\right)}{\left(\rho_1-\xi\right)\left(\rho_2+\xi\right)}\right. \ & \left.+2 \xi\left(\frac{\rho_1}{\rho_1^2-\xi^2}-\frac{\rho_2}{\rho_2^2-\xi^2}\right)\right]\qquad \text{(2-114)}\ &\Delta_{\mathrm{q}^{\prime}}(\xi)=\frac{4 P \alpha \xi}{l\left(\alpha^2-\xi^2\right)^2}\qquad \text{(2-125)}\ \end{aligned} $$
$$ \begin{aligned} \Delta_q(\xi)=2 & {\left[\frac{q_1 \rho_2-q_2 \rho_1}{\rho_2-\rho_1}\left(\frac{\rho_1}{\rho_1^2-\xi^2}-\frac{\rho_2}{\rho_2^2-\xi^2}\right)\right.} \ & \left.+\frac{q_2-q_1}{\rho_2-\rho_1}\left(\frac{1}{2} \ln \frac{\rho_2^2-\xi^2}{\rho_1^2-\xi^2}+\frac{\rho_1^2}{\rho_1^2-\xi^2}-\frac{\rho_2^2}{\rho_2^2-\xi^2}\right)\right] \qquad \text{(2-62)}\ & \Delta_{\mathrm{q}^{\prime}}(\xi)=\frac{2 P}{l} \frac{\alpha^2+\xi^2}{\left(\alpha^2-\xi^2\right)^2} \qquad \text{(2-63)}\ \end{aligned} $$ $$ \begin{aligned} \left.\begin{array}{l} & a_0=\bar{q} \ \sum_{n=0}^{\infty} a_n[s(\xi, n)+\operatorname{tr}(\xi, n)]=-\frac{2 t}{l^2} M_q(\xi)-\Delta_{q^{\prime}}(\xi) \end{array}\right}\ &\Delta_{\mathrm{bi}}= \begin{cases}\frac{\pi^2}{8 l^2} M_0 & i=1 \ -\frac{2 t}{l^2} M_{\mathrm{q}}\left(\xi_{\mathrm{i}-1}\right)-\Delta_{\mathrm{q}^{\prime}}\left(\xi_{\mathrm{i}-1}\right) & i>1\end{cases}\ \end{aligned} $$
$$ \begin
\end{aligned} $$ $$ \begin{aligned} \left.\begin{array}{l} \sum_{\mathrm{n}=1}^{\infty} \frac{(-1)^{\mathrm{n}-1}}{(2 n-1)^2} b_{\mathrm{n}}=\frac{\pi^2}{8 l^2} M_0 \ \sum_{\mathrm{n}=1}^{\infty} b_{\mathrm{n}}[z(\xi, n)+t w(\xi, n)]=-\frac{2 t}{ l^2} M_{\mathrm{q}}(\xi)-\Delta_{\mathrm{q}^{\prime}}(\xi) \end{array}\right} \end{aligned} $$ $$ \begin{aligned} \left.\begin{array}{l} a_0=\bar{q} \ \sum_{n=0}^{\infty} a_n[s(\xi, n)+\operatorname{tr}(\xi, n)]=-\frac{2 t}{l^2} M_q(\xi)-\Delta_{q^{\prime}}(\xi) \end{array}\right} \end{aligned} $$
$$ \begin
\end{aligned} $$
P0、M0为梁上的荷载全力及其对0点之矩
2-19
验证
1.1.1.1 《弹性地基梁的计算》（龙驭球）
P126 例题 4-1 弹性地基梁处于平面应力状态，已知梁的长度为 4.5m，截面尺寸 b×h=1.0m×0.75m，在梁的中点承受集中荷载 P=1000kN。梁的抗弯刚度为 EI=73820kN·m³(E=2.1E6 kN/m²)，地基的弹性模量为 E₀=3×10⁶kN/m²。求地基梁的弯矩。结果：梁分为 9 段，M₁=-5.5 kN·m，M₂=22.0kN·m，M₃=111 kN·m，M₄=298 kN·m。
P138 例题 4-3 水闸底板计算，水闸底板厚 1.6m，3 孔，每孔 10m，边墙 0.9m 厚，中墩 1.2m 厚，全长 34.2m，受均布力 38.7 kN/m，集中力 P1~P4 分别为 18.7、83、83、18.7kN（力位置为 0+11.5+11.2+11.5m）。设 E₀=5000kN/m²，E=1.44×10⁷kN/m²。求底板弯矩。结果：以中心为 0 点，与中心距 0、0.1、0.3、0.5、0.7、0.9、1.0 倍一半长度的弯矩分别为 1127、1120、1070、724、342、44、0 kN·m，两边对称。
取水输水建筑物丛书水闸（第3版）
P291例6-1
上游段完建期工况仅考虑板带上荷载时，M最大值为287.8，位于0.5*(L/2)处。
计算精度受分段数影响，分段数为20时（教材），M最大值为312.67；当分段数为80时，M为289.41，与教材相当。
《水闸设计规范》（SL265-2016）第7.5.2条 开敞式水美国味道加室底板应力分析可采用下列方法：
1
2 当采用弹性地基梁法