Can somebody help me find the equation for the deformation of a simply supported beam under trapezoidal load (with intensity Q1 and Q2, at starting point X1 and end point X2)?
I am trying to find it myself using double integration, but without luck. I already have a numerical solution - but it's too slow (i'm doing thousands of calculations).
(edit 3/31/2018. Completed all equations)
I would split the trapezoidal load
into the sum of a uniform load
and triangular load
.
The reactions $R_1$, $R_5$, $R_4$, and $R_8$ can be calculated from statics. $$R_4=Q_1L_{23}\frac{L_{12}+\frac 12L_{23}}{L_{14}}$$ $$R_1=Q_1L_{23}-R_4$$ $$R_5=\frac 12Q_2L_{67}(\frac 13L_{67}+L_{78})\frac{1}{L_{58}}$$ $$R_8=\frac 12Q_2L_{67}-R_5$$
The uniform and triangular loads can be integrated section by section to get the shear, moment, slope, and displacement. In each of these equations, x is measured from the previous point, not the beginning of the beam. Using the following equations, it is easy to create a spreadsheet to calculate the value at each point $V_1, V_2, \ldots, \delta_7, \delta_8$ or at any points in between using the equations that are functions of (x).
Uniform load, shear: $V(x)=V_i+\int load(x)\,dx$ where i is at the start of the segment and load(x)=$Q_1$ between points 2 and 3. $$V_{1}=R_1$$ $$V_{12}(x)=V_1$$ $$V_{2}=V_1$$ $$V_{23}(x)=V_2-Q_1x$$ $$V_3=V_2-Q_1L_{23}$$ $$V_{34}(x)=V_3$$ $$V_4=V_3$$
Uniform load, moment: $M(x)=M_i+\int V(x)\,dx$ $$M_{1}=0$$ $$M_{12}(x)=M_1+V_1x$$ $$M_2=M_1+V_1L_{12}$$ $$M_{23}(x)=M_2+V_2x-\frac 12Q_1x^2$$ $$M_{3}=M_2+V_2L_{23}-\frac 12Q_1L_{23}^2$$ $$M_{34}(x)=M_3+V_3x$$ $$M_{4}=M_3+V_3L_{34}$$ $M_4$ should equal 0.
Uniform load, slope: $\theta(x)=\theta_i+\int \frac{M(x)}{EI}\,dx$
$$\theta_{1}=unknown$$ $$\theta_{12}(x)=\theta_1+\frac{1}{EI}(M_1x+\frac 12V_1x^2)$$ $$\theta_{2}=\theta_1+\frac{1}{EI}(M_1L_{12}+\frac 12V_1L_{12}^2)=\theta_1+c_{s2}$$ $$\theta_{23}(x)=\theta_2+\frac{1}{EI}(M_2x+\frac 12V_2x^2-\frac 16Q_1x^3)$$ $$\theta_{3}=\theta_2+\frac{1}{EI}(M_2L_{23}+\frac 12V_2L_{23}^2-\frac 16Q_1L_{23}^3)=\theta_2+c_{s3}$$ $$\theta_{34}(x)=\theta_3+\frac{1}{EI}(M_3x+\frac 12V_3x^2)$$ $$\theta_{4}=\theta_3+\frac{1}{EI}(M_3L_{34}+\frac 12V_3L_{34}^2)$$
Uniform load, displacement: $\delta (x)=\delta_i+\int \theta(x)\,dx$ $$\delta_1=0$$ $$\delta_{12}(x)=\delta_1+\theta_1x+\frac{1}{EI}(\frac 12M_1x^2+\frac 16V_1x^3)$$ $$\delta_2=\delta_1+\theta_1L_{12}+\frac{1}{EI}(\frac 12M_1L_{12}^2+\frac 16V_1L_{12}^3)=\delta_1+\theta_1L_{12}+c_{d2}$$ $$\delta_{23}(x)=\delta_2+\theta_2x+\frac{1}{EI}(\frac 12M_2x^2+\frac 16V_2x^3-\frac{1}{24}Q_1x^4)$$ $$\delta_3=\delta_2+\theta_2L_{23}+\frac{1}{EI}(\frac 12M_2L_{23}^2+\frac 16V_2L_{23}^3-\frac{1}{24}Q_1L_{23}^4)=\delta_2+\theta_2L_{23}+c_{d3}$$ $$\delta_{34}(x)=\delta_3+\theta_3x+\frac{1}{EI}(\frac 12M_3x^2+\frac 16V_3x^3)$$ $$\delta_4=\delta_3+\theta_3L_{34}+\frac{1}{EI}(\frac 12M_3L_{34}^2+\frac 16V_3L_{34}^3)=\delta_3+\theta_3L_{34}+c_{d4}$$
The last equation gives $\delta_4=0=\theta_1+known\; values$ after you make all of the substitutions, so this gives the result for $\theta_1$ as follows (where $c_s$ and $c_d$ are simplification of constant terms defined above):
$$\theta_1=-\frac{\delta_1+c_{d2}+c_{d3}+c_{d4}+c_{s2}L_{23}+(c_{s2}+c_{s3})L_{34}}{L_{14}}$$
All values are now known for the beam with a uniform load.
Similar equations are written for the triangular load.
triangular load, shear: $V(x)=V_i+\int load(x)\,dx$ where load(x)=$Q_2\frac xL$ between points 6 and 7. $$V_{5}=R_5$$ $$V_{56}(x)=V_5$$ $$V_{6}=V_5$$ $$V_{67}(x)=V_6-\frac 12Q_2\frac{x^2}{L_{67}}$$ $$V_7=V_6-\frac 12Q_2L_{67}$$ $$V_{78}(x)=V_7$$ $$V_8=V_7$$
triangular load, moment: $M(x)=M_i+\int V(x)\,dx$ $$M_{5}=0$$ $$M_{56}(x)=M_5+V_5x$$ $$M_6=M_5+V_5L_{56}$$ $$M_{67}(x)=M_6+V_6x-\frac 16Q_2\frac{x^3}{L_{67}}$$ $$M_{7}=M_6+V_6L_{67}-\frac 16Q_2L_{67}^2$$ $$M_{78}(x)=M_7+V_7x$$ $$M_{8}=M_7+V_7L_{78}$$ $M_8$ should equal 0.
triangular load, slope: $\theta(x)=\theta_i+\int \frac{M(x)}{EI}\,dx$
$$\theta_{5}=unknown$$ $$\theta_{56}(x)=\theta_5+\frac{1}{EI}(M_5x+\frac 12V_5x^2)$$ $$\theta_{6}=\theta_5+\frac{1}{EI}(M_5L_{56}+\frac 12V_5L_{56}^2)=\theta_5+c_{s6}$$ $$\theta_{67}(x)=\theta_6+\frac{1}{EI}(M_6x+\frac 12V_6x^2-\frac{1}{24}Q_2\frac{x^4}{L_{67}})$$ $$\theta_{7}=\theta_6+\frac{1}{EI}(M_6L_{67}+\frac 12V_6L_{67}^2-\frac{1}{24}Q_2L_{67}^3)=\theta_6+c_{s7}$$ $$\theta_{78}(x)=\theta_7+\frac{1}{EI}(M_7x+\frac 12V_7x^2)$$ $$\theta_{8}=\theta_7+\frac{1}{EI}(M_7L_{78}+\frac 12V_7L_{78}^2)$$
triangular load, displacement: $\delta (x)=\delta_i+\int \theta(x)\,dx$ $$\delta_5=0$$ $$\delta_{56}(x)=\delta_5+\theta_5x+\frac{1}{EI}(\frac 12M_5x^2+\frac 16V_5x^3)$$ $$\delta_6=\delta_5+\theta_5L_{56}+\frac{1}{EI}(\frac 12M_5L_{56}^2+\frac 16V_5L_{56}^3)=\delta_5+\theta_5L_{56}+c_{d6}$$ $$\delta_{67}(x)=\delta_6+\theta_6x+\frac{1}{EI}(\frac 12M_6x^2+\frac 16V_6x^3-\frac{1}{120}Q_2\frac{x^5}{L_{67}})$$ $$\delta_7=\delta_6+\theta_6L_{67}+\frac{1}{EI}(\frac 12M_6L_{67}^2+\frac 16V_6L_{67}^3-\frac{1}{120}Q_2L_{67}^4)=\delta_6+\theta_6L_{67}+c_{d7}$$ $$\delta_{78}(x)=\delta_7+\theta_7x+\frac{1}{EI}(\frac 12M_7x^2+\frac 16V_7x^3)$$ $$\delta_8=\delta_7+\theta_7L_{78}+\frac{1}{EI}(\frac 12M_7L_{78}^2+\frac 16V_7L_{78}^3)=\delta_7+\theta_7L_{78}+c_{d8}$$
The last equation gives $\delta_8=0=\theta_5+known\; values$ after you make all of the substitutions, so this gives the result for $\theta_5$ as follows (where $c_s$ and $c_d$ are terms defined above):
$$\theta_5=-\frac{\delta_5+c_{d6}+c_{d7}+c_{d8}+c_{s6}L_{67}+(c_{s6}+c_{s7})L_{78}}{L_{58}}$$
All values are now known for the beam with a triangular load.
The results for the trapezoidal load are thus $$R_A=R_1+R_5$$ $$R_D=R_4+R_8$$ $$V_A=V_1+V5$$ $$V_B=V_2+V6$$ $$\ldots$$ $$\delta_C=\delta_3+\delta_7$$ $$\delta_D=\delta_4+\delta_8$$
An example which I verified with FEA is as follows: $$L_{AB}=20$$ $$L_{BC}=30$$ $$L_{CD}=40$$ $$Q_1=10$$ $$Q_2=15$$ $$E=1E6$$ $$I=0.75$$ gives results of $$\theta_1=-0.1976\;radians$$ $$\delta_3=-5.2370$$ $$\theta_5=-0.1513\;radians$$ $$\delta_7=-4.2267$$ The point of maximum displacement occurs at $\theta_{23}(x)+\theta_{67}(x)=0$ which is a solution to a 5th order equation which I did not attempt to solve. An approximate solution is at x=23.283 from B to C which gives a maximum displacement of (-5.412)+(-4.333)=(-9.745).
You should get the same results if I did not make any typos.
I found a better way to solve this problem by using singularity functions. Not only is the solution simple and elegant, one equation can be written for the entire beam! Also, the format for each equation is similar, so it is easy to copy the formula in a spreadsheet or program for one result (such as for the shear) and revise the equation for other results (such as the displacement) -- just change the multiplier and exponent. Furthermore, you do not need to calculate all of the variables (shear, moment, slope) to calculate any of the the other equations (such as displacement).
Not to go into too much detail, but singularity functions are like a switch to turn a load on. Furthermore, they can be integrated just like a normal equation. The singularity functions work as follow:
$$ \begin{align} \langle x-a \rangle ^{-1} &= \begin{cases} 0, & x \neq a \\ 1, & x=a \end{cases} \\ \langle x-a \rangle ^0 &= \begin{cases} 0, & x \neq a \\ 1, & x=a \end{cases} \\ \text{if n>0, } \langle x-a \rangle ^n &= \begin{cases} 0, & \text{x<a} \\ (x-a)^n, & \text{x>a} \end{cases} \\ \end{align} $$
$$ \begin{align} & \int \langle x-a \rangle ^{-1} dx = \langle x-a \rangle ^0 \\ & \int \langle x-a \rangle ^0 dx = \langle x-a \rangle ^1 \\ & \int \langle x-a \rangle ^n dx = \frac{1}{n+1} \langle x-a \rangle ^{(n+1)} \end{align} $$
(https://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect12.d/IAST.Lect12.pdf is a good example of using singularity functions to solve beam deflections.)
Using the singularity functions, the beam to solve
is represented as follows:
where $slope=(Q_C-Q_B)/(x_C-x_B)$, the load beginning with $Q_B$ continues for the remaining length, and the load beginning with $Q_C$ turns the load "off". (The singularity function turns a load on but not off, so any load that ends before the end of the beam requires a second load and singularity function to remove the load.)
Letting $x_A=0$ and $x_D=L$, the equation for the load $q(x)$ is $$ \bbox[yellow] { \begin{align} q(x)=R_A\langle x \rangle ^{-1} + R_D\langle x-L \rangle ^{-1}&-Q_B\langle x-x_B \rangle ^0-slope\langle x-x_B \rangle ^1 \\ &+Q_C\langle x-x_C \rangle ^0+slope\langle x-x_C \rangle ^1 \end{align} \tag 1 } $$
The equation for the shear is $V(x)=\int q(x)dx$, $$ \bbox[yellow] { \begin{align} V(x)=R_A\langle x \rangle ^{0} + R_D\langle x-L \rangle ^{0}&-Q_B\langle x-x_B \rangle ^1-\frac{1}{2}slope\langle x-x_B \rangle ^2 \\ &+Q_C\langle x-x_C \rangle ^1+\frac{1}{2}slope\langle x-x_C \rangle ^2 \end{align} \tag 2 } $$
The equation for the moment is $M(x)=\int V(x)dx$, $$ \bbox[yellow] { \begin{align} M(x)=R_A\langle x \rangle ^{1} + R_D\langle x-L \rangle ^{1}&-\frac {1}{2}Q_B\langle x-x_B \rangle ^2-\frac{1}{6}slope\langle x-x_B \rangle ^3 \\ &+\frac {1}{2}Q_C\langle x-x_C \rangle ^2+\frac{1}{6}slope\langle x-x_C \rangle ^3 \end{align} \tag 3 } $$
From here on, singularity functions that begin at x=0 can be converted to regular functions, and singularity functions that begin at X=L can be dropped.
The equation for the slope is $EI\cdot\theta(x)=\int M(x)dx$, $$ \bbox[yellow] { \begin{align} EI\cdot\theta(x)=\frac {1}{2}R_Ax^2+C_1 &-\frac {1}{6}Q_B\langle x-x_B \rangle ^3-\frac{1}{24}slope\langle x-x_B \rangle ^4 \\ &+\frac {1}{6}Q_C\langle x-x_C \rangle ^3+\frac{1}{24}slope\langle x-x_C \rangle ^4 \end{align} \tag 4 } $$
The equation for the deflection is $EI\cdot\delta(x)=\int \theta(x)dx$, $$ \bbox[yellow] { \begin{align} EI\cdot\delta(x)=\frac {1}{6}R_Ax^3+C_1x+C_2 &-\frac {1}{24}Q_B\langle x-x_B \rangle ^4-\frac{1}{120}slope\langle x-x_B \rangle ^5 \\ &+\frac {1}{24}Q_C\langle x-x_C \rangle ^4+\frac{1}{120}slope\langle x-x_C \rangle ^5 \end{align} \tag 5 } $$
The reaction force $R_A$ can be determined from the moment equation. At x=L, M(x)=0,
\begin{align} 0=R_AL + R_D(0) &-\frac {1}{2}Q_B(L-x_B)^2-\frac{1}{6}slope(L-x_B)^3 \\ &+\frac {1}{2}Q_C(L-x_C)^2+\frac{1}{6}slope(L-x_C)^3\\ \tag 6 \end{align} \begin{align} R_AL =\frac {1}{6}[&+3Q_B(L-x_B)^2+slope(L-x_B)^3 \\ &-3Q_C(L-x_C)^2-slope(L-x_C)^3] \\ \tag 7 \end{align}
$$ \bbox[yellow] { \begin{align} \text{thus }R_A =\frac {1}{6L}[&+3Q_B(L-x_B)^2+slope(L-x_B)^3 \\ &-3Q_C(L-x_C)^2-slope(L-x_C)^3] \end{align} \tag 8 } $$
\begin{align} \text{and }R_D &= \text{total load }-R_A \\ &= \frac 12(Q_B+Q_C)(x_C-x_B)-R_A \tag 9 \end{align}
At x=0, $\delta=0$ from which $C_2=0$. At x=L, $\delta=0$ from which $C_1$ is calculated as follows: \begin{align} EI\cdot(0)=\frac 16R_AL^3+C_1L &-\frac{1}{24}Q_B(L-x_B)^4-\frac{1}{120}slope(L-x_B)^5\\ &+\frac{1}{24}Q_C(L-x_C)^4+\frac{1}{120}slope(L-x_C)^5\\ \tag{10} \end{align} \begin{align} -C_1L =\frac {1}{120}[20R_AL^3 &-5Q_B(L-x_B)^4- slope(L-x_B)^5\\ &+5Q_C(L-x_C)^4+slope(L-x_C)^5]\\ \tag{11} \end{align}
$$ \bbox[yellow] { \begin{align} \text{thus }C_1 =\frac {1}{120L}[-20R_AL^3 &+5Q_B(L-x_B)^4+ slope(L-x_B)^5\\ &-5Q_C(L-x_C)^4-slope(L-x_C)^5]\\ \end{align} \tag{12} } $$
All values are now known, so any of the first 5 equations can be used to calculate the quantity at any position $x$ long the length of the beam $0<x<L$.
Some example results for $x_B=20,\, x_C=50,\, x_D=L=90,\, E=1\times10^6,\, I=8.5$: \begin{array}{cc|cccc } Q_B &Q_C &\theta_A &\theta_D &\delta_B &\delta_C \\ \hline 20 &35 &-0.0482 &+0.0429 &-0.8872 &-1.2971 \\ 35 &10 &-0.0388 &+0.0325 &-0.7075 &-0.9958 \\ 100 &-100 &+0.0036 &-0.0140 &+0.0987 &+0.3516 \end{array}
