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enter image description here

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.

Noideas
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  • Does this answer your question? [Deriving the deflection force equation for a beam that is fixed on both ends](https://engineering.stackexchange.com/questions/8203/deriving-the-deflection-force-equation-for-a-beam-that-is-fixed-on-both-ends) – Solar Mike Sep 05 '22 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. – Noideas Sep 05 '22 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. – Solar Mike Sep 05 '22 at 12:10
  • @SolarMike Yes, I agree - there will be differentiation. However, I do not know how to approach it and where to start... – Noideas Sep 05 '22 at 12:11
  • Largest deflection $\ne$ worst case scenario. Maximum bending moment can be at a different point than maximum deflection. – Tomáš Létal Sep 05 '22 at 12:36
  • Maximum defection will be at point where the slope $\theta$ is 0. – Tomáš Létal Sep 05 '22 at 12:38
  • @TomášLétal The beam (its EI) is being optimized for the least deflection. No matter where the load is applied on the beam, the deflection must not exceed 5mm down. So there must exist a point x where applying 80n load (or load of any magnitude) will deflect the beam the most. How to find this point? – Noideas Sep 05 '22 at 12:41
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    @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? – Noideas Sep 05 '22 at 12:45
  • @SolarMike I'm asking for conceptual help - how to approach these kinds of problems and what background knowledge is required. This isnt about any particular homework question. – Noideas Sep 05 '22 at 12:52

2 Answers2

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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).

Tomáš Létal
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  • 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. – Noideas Sep 20 '22 at 01:34
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1) - Use table

enter image description here

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.

r13
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