We are tasked with calculating the resultant force and its location for a distributed load, represented by the given function $y = a \cdot f(x)$, where:

$f(x) = 1 + A\left(\frac{x}{L}\right) + B\left(\frac{x^2}{L^2}\right) + C\left(\frac{x^3}{L^3}\right) + D\left(\frac{x^4}{L^4}\right)$

The problem provides the following parameters:

• $a = 2.0 \, \text{kN/m}$
• $L = 2 \, \text{m}$
• $A = -1$
• $B = 3$
• $C = 1$
• $D = 1$

1. Resultant Force $R$

The resultant force $R$ is calculated by integrating the load function over the domain from $x = 0$ to $x = 2L$.

$R = \int_0^{2L} a \cdot f(x) \, dx$

First, substitute the expression for $f(x)$ into the equation:

$R = \int_0^{2L} 2.0 \cdot \left[ 1 + \left(\frac{-1 \cdot x}{L}\right) + \left(\frac{3 \cdot x^2}{L^2}\right) + \left(\frac{x^3}{L^3}\right) + \left(\frac{x^4}{L^4}\right) \right] dx$

Now, let's integrate term by term. Each term will involve powers of $x$, and we can compute this explicitly.

2. Location of the Resultant Force $x_R$

The location $x_R$ of the resultant force is found using the moment about the origin:

$x_R = \frac{\int_0^{2L} x \cdot a \cdot f(x) \, dx}{R}$

Again, substitute the load function into the integral and calculate both the numerator and denominator.

I will now proceed with the integration and the final calculations.The calculations yield the following results:

1. Resultant Force $R$: $73.6 \, \text{kN}$
2. Location of Resultant Force $x_R$: $3.09 \, \text{m}$

This means the total load is 73.6 kN, and it acts at a point approximately 3.09 meters from the origin along the x-axis.

Would you like a detailed breakdown of the integration steps or have any other questions? Here are five related questions you might find useful:

1. How does varying the parameters $A$, $B$, $C$, and $D$ affect the resultant force?
2. Can you explain how to derive the moment about the origin for irregular loads?
3. What is the significance of the resultant location $x_R$ in practical applications?
4. How does the length $L$ influence the load distribution and resultant?
5. How would this result change if the load function was discontinuous?

Tip: When calculating the location of a resultant force, ensure the integration for both force and moment are over the same limits for consistent results.