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A pinned system comprising bars of the same material and cross-sectional area is subjected to a vertical load P.

I am required to form a relationship between FAB and P.

The first part of the solution is as such,

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But how is the angle $\theta$ the same after deflection? Shouldn't it be smaller? And I can't seem to form a relationship without assuming the theta as the same.

  • $\begingroup$ It is a common approximation to assume that deflections are small compared to the scale of the structure. $\endgroup$ – ingenørd Sep 16 '19 at 16:12
  • $\begingroup$ $\delta_{AB}<<AB$ - basically, The new length $\delta_{AB}+AB$ is almost exactly the same as the old length $AB$, so the angle is almost exactly the same. $\endgroup$ – Jonathan R Swift Sep 16 '19 at 17:55
  • $\begingroup$ Is such an assumption the only way to solve statically indeterminate systems like the above? $\endgroup$ – Wb16 Sep 18 '19 at 1:51

This is not a statically determinate system. To find how the load splits between the members, start by finding their stiffness.

Since the materials and cross-section areas are the same, the stiffness is inversely proportional to the length.

Length of AC = $L\cos\theta$. If the stiffness of AB is $K$, the stiffness of AC is $K/\cos\theta$.

Now consider a small downward displacement $x$ at point A. The length of AB changes from $L$ to $$\sqrt{L^2\sin^2\theta+ (L\cos\theta + x)^2} \approx \sqrt{L^2 + 2Lx\cos\theta}\\ \approx L + x\cos\theta$$ to first order in $x$, using the Binomial theorem .

So the tension in AB is $Kx\cos\theta$ and the downward component of the tension is $Kx \cos^2\theta$.

The tension in AC is $Kx/\cos\theta$.

So we have $P = 2Kx\cos^2\theta + Kx/\cos\theta$ and $F_{\text{AC}} = Kx/\cos\theta$.

$P = F_{\text{AC}}(2cos^3\theta + 1)$.

From the diagram $\cos\theta = \sqrt{3}/2$, so $P = F_{\text{AC}}(3\sqrt{3}/4 + 1)$.

i.e. $F_{AC} = 0.435\,P$.

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  • $\begingroup$ HI! Could you explain what do you mean by to first order in x using binomial? $\endgroup$ – Wb16 Sep 18 '19 at 1:43
  • $\begingroup$ First order means we ignore terms in $x^2$ or higher powers of $x$. I suppose I really meant the binomial series not the binomial theorem. See en.wikipedia.org/wiki/Binomial_series $\endgroup$ – alephzero Sep 18 '19 at 14:16

let's call the tension at the bars at B and D, T1 and at C, T. We see that $T_1=T\frac{ \sqrt3}{2}$

$$P = 2*T_1 +T= T +2 T\frac{\sqrt{3}}{2}=2.73205\cdot T$$

The share of C bar from P load will be $$T=P\frac{1}{2.73205} $$

$$ \delta_c= \frac{Pl_c}{2.73205EA}= \frac{Pl\sqrt{3}/2}{2.73205EA} $$

The extended length of the C bar will be $l_c= Pl(\sqrt{3}/2) + \frac{Pl\sqrt{3}/2}{2.73205EA} $

From here you can find your extended lengths and angles of the frame after the loading.

The loadings and strains on B and D bar will fall in place obligingly.

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  • $\begingroup$ What happened to $T_1$ after the first equality? The relation between $T$ and $T_1$ is not at all "obvious" IMO. $\endgroup$ – alephzero Sep 16 '19 at 22:12
  • $\begingroup$ @alephzero, remember the relation ship between the altitude and side of an equilateral triangle. Because this truss members are equal size bats the forces resolve as per geometry of the frame. So the frame is determinant. Or I am looking forward to see how you'd handle it. $\endgroup$ – kamran Sep 16 '19 at 23:57
  • $\begingroup$ my apologies for using my phone with 3 different language spell checkers mixing together. I meant: equal sized bars, not bats; also, determinate frame! $\endgroup$ – kamran Sep 17 '19 at 0:27
  • $\begingroup$ I don't know what "the forces resolve as per geometry of the frame" means, but I would have used the relative stiffness of the members to find the ratio of forces for a virtual displacement at A. If there is a general result about how this works out, fine, use it, but in the first line of your math it looks like $T_1$ just disappears by magic. $\endgroup$ – alephzero Sep 17 '19 at 0:51
  • $\begingroup$ I have plugged in T1 value.in terms of T which is 0.866T. $\endgroup$ – kamran Sep 17 '19 at 1:48

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