How to find the location on the 2-support beam where the deflection will be biggest?

Question

Consider the uniform indeterminate beam as shown above. This beam is to be subjected to a single, 80N point load (acting down ) somewhere along its length.

What is the procedure (i.e steps and logic) to calculate the location where application of this load will cause the largest deflection (in mm), i.e. where the "worst case scenario" occurs? My initial intuition tells to assume the maximum sagging will occur if we apply 80N force in the middle, but checking with the online beam calculators proved that the point is closer to the roller rather than equally distanced from both supports.

Answer 1

Conceptually, you need to establish function for the value of the maximum deflection $w_{max}$ depending on the load location $x_F$ (using zero slopes $\theta$, $L$ is the length of the beam):

$$w_{max}(x_F) = \max(w_i(x_i)\in [0.. L] | \theta_i(x_i) = 0)$$

Then you can try putting a partial derivative of $w_{max}(x_F)$ with respect to $x_F$ to zero: $$\frac{\partial{w_{max}(x_F)}}{\partial{x_F}} = 0$$

The $x_F$ from this condition should be the load location causing the largest deflection (in some cases, you may also need to check endpoints of the beam or of each of its segments, as they can have extreme deflections even without the derivative being 0).

Answer 2

1) - Use table

2) - Intergrating the second order differential equation $EI\dfrac{d^2y}{dx^2} = M$

$\theta(x) = \dfrac{1}{EI} \int M(x)dx$

$\delta(x) = \int \theta(x)dx$

Note, that for a beam supported on both ends, the maximum deflection occurs at a location where the slope of rotation angle ($\theta$) is zero.