# Pin joint system displacement question

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,

$$cos\theta=\frac{\delta_{AB}}{\delta_{AC}}$$

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.

• It is a common approximation to assume that deflections are small compared to the scale of the structure. – ingenørd Sep 16 at 16:12
• $\delta_{AB}<<AB$ - 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. – Jonathan R Swift Sep 16 at 17:55
• Is such an assumption the only way to solve statically indeterminate systems like the above? – Wb16 Sep 18 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. $$F_{AC} = 0.435\,P$$.

• HI! Could you explain what do you mean by to first order in x using binomial? – Wb16 Sep 18 at 1:43
• 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 – alephzero Sep 18 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 $$l_c= Pl(\sqrt{3}/2) + \frac{Pl\sqrt{3}/2}{2.73205EA}$$

• What happened to $T_1$ after the first equality? The relation between $T$ and $T_1$ is not at all "obvious" IMO. – alephzero Sep 16 at 22:12
• 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. – alephzero Sep 17 at 0:51