## Maximum deflection distance for a simply supported beam Under triangular load-Numerically.

We want to get the maximum defection distance for a triangular load acting in a simply supported beam of length=6.0m. The maximum intensity of the load is 20 kNn/m. Our task is to find the x max distance using the Numerical method. We will solve this by using the Newton -Raphson method for roof findings.

### The video I used for the illustration.

In the second part of the video, there is a given example where we will have a triangular load with a maximum value=2KN/m.

The support length would be 6 meters, and the EI value 8*10^4 Newton*m2.

The reason we are using the EI value is that we have to substitute for being able to use numerical methods for evaluation; we recall that y”=1/EI*PL x/6- PX^3/6L) as we add y’ and y, as shown in the slides based on the calculation of the double integration method, the x max lies apart from the y-axis by distance 0.519L in our case, it will be 3.114m. For the content of this post, please start to watch from time 7:14 till the end for the discussion of the triangular load.

You can click on any picture to enlarge, then press the small arrow at the right to review all the other images as a slide show.

### The three equations for the change of slope & slope and deflection values.

This is a list of three items; the first item is the change of curvature equation and the diagram shape for a simply supported beam under triangular load, where P is the maximum value of the triangular load, x is the distance for which we want to calculate the change of curvature.

The second item is the slope of curvature equation and the diagram shape for a simply supported beam under triangular load, where P is the maximum value of the triangular load, and x is the distance for which we want to calculate the slope of curvature.

The third item is the slope of curvature equation and the diagram shape, for a simply supported beam under triangular load, where P is the maximum value of the triangular load, x is the distance for which we want to calculate the deflection value.

The next slide image shows the slope and deflection equations for a simply supported beam under a uniformly Triangular distributed load.

It is required to estimate the maximum deflection distance from the left support **x max**. The point of the maximum deflection is the point for which the slope of curvature is zero.

Normally, we use the Newton – Raphson method we want to get the point where Y value =0. the function value f(x) was on the numerator side, rerefer to the relation, where _{X0} is the starting point, f(x_{0}) is the function value at x0, f'(x_{0}) is the slope value of the function at point X_{0}.

## The revised form of the Newton-Raphson equation is based on the slope equation.

To use the Newton-Raphson equation, we are going to deal with y’ curve, since it is changing from+ve value to negative value, then a root point for zero slopes will give us the point of maximum deflection, instead of dealing with y curve as in the case of the original form of Newton-Raphson equation, we are now dealing with y’ function.

Please refer to the next slide image for more details.

## The steps for estimating x max by the analytical solution.

The modified equation for the root point will be as, for the first point after setting an initial point of, which will be given a value of 2.00m, less than the L/2.

### Our first trial value of x_{0} as =2.00m.

We need f'(2) and f”(2) to apply it in the adjusted equation. We have the expression for F'(x) for the slope value of the curve due to the triangular load also y” the change of curvature of the beam due to the triangular load. We plug in with x_{0}=2.0m and get the two values of f'(2) and f”(2), as shown in the next slide. We need to use 1/EI=1/8*10^4.

Our first x_{0} with a starting value of 2.0m will lead us to a modified value of 3.30m, as the maximum deflection distance. along with y’ function.

### Our second trial value of x_{0}.

Then we plug in with x_{0} as a second trial=3.30m, we get the second value for the maximum deflection distance of 3.1169 m. Our third x_{0} with a starting value of 3.1169 m will lead us to a modified value of 3.1159778m, as the maximum deflection distance.

Please refer to the next slide image for more details.

Then we plug in with x_{0} as a second trial=3.1159m, we get the third value for the maximum deflection distance of 3.115977 m.

The following table shows the result of x_{0} when we start with an initial point of 2.0m for four iterations.

Please refer to the next slide image for more details.

### In case our first trial value of x_{0} is =3.00m.

If somebody wishes to choose a starting point as=3.00m. x_{0} with a starting value of 3.0m will lead us to a modified value of 3.1166m, as the maximum deflection distance. Then we plug in with x_{0} as a second trial=3.1166m, we get the second value for the maximum deflection distance of 3.115977 m.

The following table shows the result of x_{0} when we start with an initial point of 3.0mfor four iterations.

If somebody wishes to choose a starting point as=4.0. x_{0} with a starting value of 4.00m will lead us to a modified value of 3.09m, as the maximum deflection distance. Then we plug in with x_{0} as a second trial=3.09m, we get the second value for the maximum deflection distance of 3.1160 m.

The following table shows the result of x_{0} when we start with an initial point of 4.0m for four iterations.

Our x distance is 3.114m is our exact value for the maximum deflection distance. From the numerical analysis, we can get a closer value of 3.1159m as previously shown.

This is the pdf used in the illustration of this post.

This is a good external reference. Holistic numerical method.

This is a useful link for a numerical analysis calculator.