I have trouble with following problem (ii):

A uniform beam $AOB$, $O$ being the mid point of $AB$, mass M, rests on three identical vertical springs with stiffness constants $k_1$, $k_2$ and $k_3$ at $A$, $O$ and $B$ respectively. The bases of the springs are fixed to a horizontal platform. Determine the compression of the springs and their compressional forces in the case:

(i) $k_1 = k_3 = k$ and $k_2 = 2k$

(ii) $k_1 = k$, $k_2 = 2$k and $k_3=3k$

In (i) the beam is not inclined because the two springs at $A$ and $B$ have the same $k$. Then the weight of the beam is compensated by the compressional forces of the springs and every spring experiences the same compression $x$:

$$ Mg=kx+2kx+kx $$ which can be rearranged for $x$ and inserted into $F=kx$ to get the force exerted by each spring.

The problem in (ii) is that the beam is inclined because there isn't the symmetry of $k$'s as in (i).

How would one solve (ii)? You would have to find an expression for the component of the weight of the beam which compensates the forces of the springs, right?

  • 1
    $\begingroup$ Since the k’s in part 2 are linearly related, the answer is readily determined. $\endgroup$
    – Jon Custer
    Commented Oct 28, 2017 at 0:17