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Assume the horizontal member can be treated as a beam and assuming the diagonal member may be treated as a rod, what is its axial force as a function of the applied loading $P$ and length parameters $L_i$. Assume that the connection between the beam and the rod may be treated as as socket joint.

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Firstly, I have defined the vertical, horizontal and moment reactions at the top and bottom built-in ends as $V_1, H_1, M_1, V_2, H_2,$ and $M_2$ respectively. I have also defined the axial force (assumed tensile) in the diagonal member as $T$.

Vertical equilibrium of the beam:

$P + V_1 - Tsin(\phi) = 0$

Horizontal equilibrium of the beam:

$H_1 - T cos(\phi) = 0$

Moment equilibrium of the beam:

$M_1 + P(L_1 + L_2) - T sin(\phi) L_1 = 0$

Vertical equilibrium of the rod:

$T \sin{\phi} + V_2 = 0$

Horizontal equilibrium of the rod:

$T cos(\phi) + H_2 = 0$

Moment equilibrium of the rod:

$M_2 = 0$

Global vertical equilibrium:

$P + V_1 + V_2 =0$

Global horizontal equilibrium:

$H_1 + H_2 = 0$

When I attempt to solve the simultaneous equation, I get an indeterminate problem and can not solve for the value of $T$.

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    $\begingroup$ Welcome to Engineering! This looks like a homework question. In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$
    – Wasabi ♦
    Jan 31, 2017 at 0:49
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    $\begingroup$ I have added my working attempts to the solution $\endgroup$
    – aerodokuu
    Jan 31, 2017 at 8:01

1 Answer 1

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You are correct that the problem is indeterminate. The relative axial stiffness of the rod and bending stiffness of the beam are required. In order to solve this problem you need to find the stiffnesses and use the stiffness method.

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  • $\begingroup$ Hi. Could you elaborate on the stiffness method? $\endgroup$
    – aerodokuu
    Jan 31, 2017 at 10:48
  • $\begingroup$ I don't think you can assume the beam is a rod. That would require a full hinge between the bars, which would make the small cantilever ($L_2 + L_3$) hypostatic. $\endgroup$
    – Wasabi ♦
    Jan 31, 2017 at 14:05
  • $\begingroup$ @Wasabi - I'm not sure what the term "hypostatic" means, and googling leads to either something religious or something to do with liquids. But, looking at it, I agree that L<sub>2</sub> and L<sub>3</sub> must be acting as beams, with one of the other members also acting as a beam to provide moment support to the cantilever. Which throws out any possibility of members acting axially only. I've now removed that section of my answer. $\endgroup$
    – AndyT
    Feb 1, 2017 at 9:51
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    $\begingroup$ @AndyT: Huh, indeed, hypostatic doesn't seem to exist in english. In portuguese, it's the opposite of hyperstatic (statically indeterminate): a system with insufficient constraints and which is therefore a mechanism, not a structure. Apparently the preferred terms are "unstable" or "partially constrained". $\endgroup$
    – Wasabi ♦
    Feb 1, 2017 at 12:14

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