I am working on calculating the stiffness of a cantilever beam by applying a distributed load. I wanted to ask if the distribution of the load on the beam effects its stiffness.
For eg., if I apply a point load at the free end vs a constant distributed load throughout the beam. Will the stiffness I get be different? How do I calculate the stiffness in later case?


2 Answers 2


Stiffness is a murky term frequently used ambiguously in engineering.

However, the most common definition of stiffness is the product of a beam's Young's Modulus $E$ (which is a function of its material) and its moment of inertia $I$ (which is a function of its cross-section). So $\text{Stiffness} = EI$.

Loading has nothing to do with stiffness according to this definition, which you could say describes an isolated beam's stiffness. This value allows you to say whether one beam would be stiffer than another in identical circumstances (whatever those may be).

Now, another possible definition is stiffness as the deflection a beam or structure suffers under load. This would be an analogy with a spring's stiffness (which is literally measured in force needed to move the spring a unit distance).

This could be described as the whole structure's stiffness (even if that structure is a single cantilever beam). Obviously, a structure with pinned supports will be less stiff than one with fixed supports. Likewise, a cantilever beam with a concentrated load at midspan will deflect less (be stiffer) than one where the load is at the free end.

Now, as mentioned in @kamran's answer, different loading patterns will lead to different deflections even if the total applied load is equal:

  • A cantilever beam with a uniformly distributed load will have a deflection at the free end of $$\delta = \dfrac{qL^4}{8EI}$$
  • The same beam with a concentrated force (equal to $P=qL$ to keep the same total force) at the free end will have $$\delta = \dfrac{PL^3}{3EI} = \dfrac{qL^4}{3EI}$$
  • The same beam where the concentrated force is applied at the midspan will have $$\delta = \dfrac{5PL^3}{48EI} = \dfrac{5qL^4}{48EI}$$

Note that all these solutions have the constant $EI$ given as the more common definition of stiffness above. So a stiffer beam (one with greater $EI$) will deflect less than a more flexible one in any and all cases. However, how much any given beam will deflect depends on the loading and support conditions.

Given this, the answer to your question depends on what you're looking for.

If you're looking for the isolated beam's stiffness, then the result will be the same no matter your loading pattern (since you'll use different equations depending on the loading to get that beam's stiffness).

If, however, you're looking to get the stiffness as "force per deflection", then the loading will change your result. Any loading is valid, but your result will only be valid for that type of loading (i.e. this beam needs 10 kN/mm at midspan, but only 3.2 kN/mm at the free end).


The stiffness of a beam does not change with the loading if the equivalent loads and their points of action on the beam are equal.

First lets do the stiffness of the beam under q uniform load. $$\delta = \frac{qL^4}{8EI} $$

Now let's load a cantilever beam with a point load equivalent to uniform load. in the distribuited load we have total load $ \ P=qL \ $ acting at the center witch is L/2.

$$\delta= \frac{P(L/2)^2}{6EI}\cdot(3L-a_{distance\ from\ end}) \\ \therefore \delta=\frac{qL^3}{24EI}(3L-0)= \frac {qL^4}{8EI}$$

So as we see the stiffness doesn't change under different loading.

Just to remind us a simple point load at the end of a cantilever beam causes deflection $ \delta=PL^3/3EI.$ If we compare this to the above result we see placing the same load on the beam but uniformly distributed causes 3/8 of point load deflection, even though the stiffness remains the same.


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