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Consider a beam that is simply supported at both ends by fixed supports and a point load is applied directly in the middle. The force (F), length of beam (L), Young's Modulus (E) and tensile and crushing strength (σ). The amount of deflection is not known.

Let's say 1,000,000kN was applied to the beam, how would I calculate how large the cross-section would need to be in order for it to not fail and break?

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  • $\begingroup$ This has the appearance of a homework question. Have you tried to determine a solution? Also, 1,000,000 kN is a very strange number, I appreciate that it is a number plucked from the air, but apart from being very large number, expressing it as 1 GN would be more in keeping with the SI system of units. $\endgroup$
    – Fred
    Commented Nov 9, 2016 at 4:11
  • $\begingroup$ The answer to this equation is also going to depend on the shape of the beam, instead of just some estimate of the total cross-sectional area. $\endgroup$
    – Marchi
    Commented Nov 9, 2016 at 14:44
  • $\begingroup$ You're doing something wrong. 10kN is one metric tonne, or approximately 1 long ton. 10,000kN is then 1,000 tonnes, and 1,000,000kN is about 100,000 long tons. How much is that? One fully-loaded Nimitz class aircraft carrier. As a point load. On a beam. At this point you're getting beyond the reasonable assumptions for the equation given as an answer below. You're more likely to simply cut through the beam than to fracture it by bending. Want to distribute the load to avoid cutting or plastic deformation? Add it to your question. $\endgroup$
    – Chuck
    Commented Nov 9, 2016 at 16:40
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    $\begingroup$ @Chuck - Amusing though your comment is, I get the impression you actually think the OP wants to design something for 1GN. Personally, I infer that he's not talking about an actual situation and just plucked a number out of the air. $\endgroup$
    – AndyT
    Commented Nov 9, 2016 at 17:20

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$$\begin{gather} M = \dfrac{FL}{4} \\ \sigma = \dfrac{My}{I} = \dfrac{FLy}{4I} \end{gather}$$

  • $F$ is your vertical force at midspan
  • $L$ is your span length
  • $y$ (distance from neutral axis to extreme fibre) and $I$ (second moment of area) depend on the shape and size of the section. For a rectangular section of width $b$ and depth $d$, $y = \dfrac{d}{2}$ and $I = \dfrac{bd^3}{12}$

This is assuming your material is linear elastic, and $\sigma$ is your elastic limit. It also ignores any safety factor(s).

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  • $\begingroup$ What is the "distance from neutral axis to extreme fibre"? $\endgroup$
    – Dale
    Commented May 27, 2017 at 16:39
  • $\begingroup$ @Dale - When a beam bends one face (a.k.a fibre) goes into tension and the opposite side goes into compression; the neutral axis is level at which there is no strain (tensile or compressive). The shape of a cross-section can be such that the neutral axis isn't half way up, and one face is further from it than the other. In this case, assuming linear elastic behaviour where the strain varies linearly through the depth, one face has greater strain than the other face; the face with greater strain is the "extreme fibre". $\endgroup$
    – AndyT
    Commented Jun 5, 2017 at 13:54

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