solid mechanics practice final

For part A I have $\Delta$A + $\Delta$B = $\delta$. For part B I've tried summing forces at the joints but keep getting lost. Any help is appreciated my final is on Friday.

  • $\begingroup$ are the horizontal bars fixed at the ends or are they allowed to rotate? $\endgroup$
    – NMech
    May 11 at 4:47

First of all some geometry, and some simplifications:

enter image description here

The horizontal bars (from the wall to point A) is going to rotate and extend.

  • The extension of each horizontal will be equal to $\Delta L$. From the detail of AKD (right angle at K) we see that $\Delta L = \delta_A \sin\theta$ (so $\Delta L \le \delta_A $)

  • angle with the horizontal will be equal to $\sin \theta = \frac{\delta_A}{L+\delta L}$. But because from the above we have $\Delta L \le \delta_A$ and $\delta\ge\delta_A\ge \delta L$ and also $\delta << L$

$\sin \theta = \frac{\delta_A}{L+\delta L}\approx \frac{\delta_A}{L}$

Force and extension of the vertical bar (B).

The vertical bar will be extended by : $$\delta_B =\frac{F_B L}{EA}$$

therefore the force on it will be equal to $F_B =\frac{EA}{L}\delta_B $

Force and extension of the vertical bar (A).

The horizontal bar will be extended by : $$\delta_L =\frac{F_A L}{EA}$$ However, because $\delta_L = \delta_A \sin\theta= \delta_A \frac{\delta_A}{L}=\frac{\delta_A^2}{L} $

$$\frac{\delta_A^2}{L} =\frac{F_A L}{EA}$$

therefore the force on it will be equal to $F_A =\frac{EA}{L^2}\delta_A^2 $

putting it all together

As you said the displacement should be $$\delta_A + \delta_B =\delta$$

Also, with respect to forces, when points A, and B are merged you will have the following shape

enter image description here

So the equilibrium in the y-axis will be :

$$2 (F_A \sin\theta) = F_B$$

Substituting $F_A, F_B, \sin\theta$:

$$2 (\frac{EA}{L^2}\delta_A^2 \frac{\delta_A}{L}) = \frac{EA}{L}\delta_B$$ $$2 \frac{\delta_A^3}{L^2} = \delta_B$$

So you end up with the following two equations to solve:

$$\begin{cases} \delta_A + \delta_B =\delta\\ 2 \frac{\delta_A^3}{L^2} = \delta_B \end{cases}$$

After, you solve for $\delta_A,\delta_B$ you can substitute in the forces and obtain the Forces on each bar.

I am pretty certain that you can do more substantial assumptions, to reduce the problem (the assumptions will be related to the geometry of the problem again). That will probably reduce the order of the problem and make it easier to obtain an analytical solution. For me this is a fair compromise.


enter image description here

a) From Fig 1, $P = \delta EA/L$ ----- (1)

From Fig 2, $\sum F_Y = 0, ---> P = 2P_A + P_B$ -----(2)

From Fig 3, $\Delta = (L + L*)sin\theta = (L + TL/EA)sin\theta = (L +P_AL/sin\theta EA)sin\theta; P_A = Tsin\theta$

$\Delta = Lsin\theta + P_AL/EA = P_BL/EA$ -----(3)

Solving $P_B$ from (3), $P_B = (Lsin\theta + P_AL/EA)EA/L = sin\theta EA +P_A$

Plug $P_B$ into (2), $2P_A + sin\theta EA + P_A = P ---> P_A =(P - sin\theta EA)/3$

$P_B = P - 2P_A ---> P_B = (P + 2sin\theta EA)/3$

b) PLug (1) into $P_A$ & $P_B$,

$P_A =(P - sin\theta EA)/3 = (\delta EA/L - sin\theta EA)/3 = (\delta/L - sin\theta)EA/3$

$P_B = (P + 2sin\theta EA)/3 = (\delta EA/L + 2sin\theta EA)/3 = (\delta/L + 2sin\theta)EA/3$

c) from (3), $\Delta = P_BL/EA = (\delta/L + 2sin\theta)EA/3*(L/EA) = (\delta + 2Lsin\theta)/3$

d) The relationship above indicates that for small deflection $(\theta$ approaches $0)$, $\Delta \approx \delta/3$

Please check and advise mistakes.


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