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

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.

• Does this answer your question? Deriving the deflection force equation for a beam that is fixed on both ends Sep 5, 2022 at 12:02
• @SolarMike Thank you for the link. However, that does not answer my question, as the answer in the link (as many other similar answers) assumes the location of the applied force is known. In my question, it isnt - I have to choose (and prove) the position on the beam where application of 80N force will cause the biggest deflection. Sep 5, 2022 at 12:08
• Then iirc, you need to do some differentiation... But you should show some effort towards a solution. This is not a free homework completion site. Sep 5, 2022 at 12:10
• @SolarMike Yes, I agree - there will be differentiation. However, I do not know how to approach it and where to start... Sep 5, 2022 at 12:11
• @TomášLétal Imagine that the beam is the bridge, and a pedestrian crosses over. If we model the pedestrian as a point load, at what location on the bridge does the pedestrian have to stand to cause the most deflection? Sep 5, 2022 at 12:45

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).

• I ended up solving the problem numerically: 1. Derived the function f(x) for roller reaction using superposition and compatibility equation. 2. Again, using superposition, derived max deflection function for the entire beam as a function of x, where x is position of applied force. 3. Used matlab to repeatedly solve these two equations while incrementing x by 0.0001 each iteration. Sep 20, 2022 at 1:34

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.