Image one.

I solved it using the logic that the vertical link of the beam is basically there to make the beam act like a cantilever. So I think it is a cantilever problem.

Therefore, $$EI\frac{d^2y}{dx^2}=-px=M$$

and further integrating it we get $$y_{max}=\frac{PL^3}{3EI}$$

But the book people say that the answer is wrong and the answer it 4 times a this answer. Why?

  • 3
    $\begingroup$ The load P is applying a moment to the vertical beam, so the joint will rotate. You assumed the rotation at the joint was zero. $\endgroup$
    – alephzero
    Commented Jan 29, 2017 at 21:30
  • $\begingroup$ The reason I had to make an assumption @alephzero because it is a structure problem and the links are capable of bending but not of rotating. So I did the derivation but what I was not able to understand is how is it 4 times the normal derivation? $\endgroup$ Commented Jan 30, 2017 at 14:33
  • $\begingroup$ On what basis do you maintain that the links will not rotate? There's nothing in the question saying that there is no rotation at the corner, and while the support conditions are poorly defined, I think the intent is that the corner is a position but not a rotation restraint. $\endgroup$
    – achrn
    Commented Jan 30, 2017 at 22:02

1 Answer 1


I think your 'book people' are wrong, but so are you. The support conditions are not well defined, but if I've interpreted them correctly, I think the tip vertical displacement is (b) $-\dfrac{2PL^3}{3EI}$.

I am assuming that the support points are intended to be fixed in position, but not providing rotational restraint. If the restraint at the corner was supposed to be a moment restraint, it would just be represented as a built-in support and there would be no need of the vertical leg.

The answers all only have $EI$ terms, so it is evident that the author of the question only wants to consider flexural effects (not shear or axial deformations - we'd need terms with shear or axial area in them if those effects were included).

The question asks about logic, so this is my thought sequence:

I start at the loaded tip. From there all the way to the corner it's just a cantilever, so we'll get cantilever behaviour. If it were a rigid support, this would give the classic result tip $\delta=\dfrac{PL^3}{3EI}$. This is the result the questioner derives.

However, at the corner we have a moment ($PL$) that must be resisted. There's no restraint at the corner support, but restraint is provided by the vertical leg. In isolation, this member is just a pin-ended beam with a moment applied at one end. That's another standard result - in that situation the beam end rotation (at the loaded end) is $\dfrac{ML}{3EI}$.

However, we know $M = PL$, so our corner rotation is $\dfrac{PL^2}{3EI}$. As far as the horizontal leg is concerned, that will a rigid body rotation superimposed on the bending deformation of the horizontal leg. A rigid rotation of $\dfrac{PL^2}{3EI}$ about the corner moves the tip downwards $\dfrac{PL^3}{3EI}$.

So we have $\dfrac{PL^3}{3EI}$ bending deformation in the horizontal leg, plus $\dfrac{PL^3}{3EI}$ due to rotation about the corner, hence a total tip deflection of $\dfrac{2PL^3}{3EI}$, which is answer (b).

  • $\begingroup$ I made a simple model and can confirm that $\delta = \dfrac{2PL^3}{3EI}$. $\endgroup$
    – Wasabi
    Commented Jan 31, 2017 at 14:15

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