The figure represents a rigid bar $ BCE $ connected to two steel ties $ AB $ and $ CD $ ($ E $ = 200 GPa). The rods are connected by joints at the ends and have a uniform cross section of 6 $ \times $ 25 mm $ ^ 2 $. Determine the value of the largest $ P $ load that can be suspended at the $ E $ point without the vertical displacement at that point exceed 0.25 mm.

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I calculate $\delta_B$, $\delta_C$ using FL/AE and with the equilibrium equations and equilibrium moments i got $F_{AB}=1.5P$ and $F_{CD}=2.5P$ And finally using virtual work i got $P*\delta_C - F_{AB}*\delta_B - F_{CD}*\delta_C$ =0 i got P=4405Kn is my resolution also correct? Since theres a small diference between my result and the resultions result...

In the solutions they have done

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From the analysis of the deformed diagram of the $ BCE $ bar it is concluded, by the equality of triangles:

\begin{equation} \tan\alpha = \displaystyle\frac{\delta_\mathrm{B}}{x} = \displaystyle\frac{\delta_\mathrm{C}}{l-x} \quad\Leftrightarrow\quad \displaystyle\frac{\delta_\mathrm{B}}{\delta_\mathrm{C}} = \displaystyle\frac{x}{l-x} \end{equation}

Also taking into account that the ratio of length variations $\delta_\mathrm{B}/\delta_\mathrm{C}$ can be expressed by:

\begin{equation} \displaystyle\frac{\delta_\mathrm{B}}{\delta_\mathrm{C}} = \displaystyle\frac{F_\mathrm{AB}L_\mathrm{AB}/A_\mathrm{AB} E} {F_\mathrm{CD}L_\mathrm{CD}/A_\mathrm{CD} E} \end{equation}

since $L_\mathrm{AB}=L_\mathrm{CD}$ and $A_\mathrm{AB}=A_\mathrm{CD}$,, the previous expression is simplified to:

\begin{equation} \displaystyle\frac{\delta_\mathrm{B}}{\delta_\mathrm{C}} = \displaystyle\frac{F_\mathrm{AB}}{F_\mathrm{CD}} = \displaystyle\frac{1.5 P}{2.5 P} = 0.6 \end{equation}


\begin{equation} \frac{\delta_\mathrm{B}}{\delta_\mathrm{C}} = \displaystyle\frac{x}{l-x} = 0.6 \quad\Leftrightarrow\quad x = \frac{0.6}{1.6}\,l = 0.374~l \quad\mbox{(mm)} \end{equation}

$$x = 93.750 mm$$

Due to the similarity of the triangles comes:

  • $\delta_B,~x,~\delta_E ~\longrightarrow~P$ :

\begin{equation} \tan\alpha = \displaystyle\frac{\delta_\mathrm{B}}{x} = \displaystyle\frac{\delta_\mathrm{E}}{L-x}~, \quad \delta_\mathrm{E} \leq 0.25~\mbox{mm}~, \quad x = 93.75~\mbox{mm}~, \quad \delta_\mathrm{B} = 10^{-5} P \end{equation}

whence it results:

\begin{equation} P \leq 25000 \frac{x}{L-x} \end{equation}

  • $\delta_C,~x,~\delta_E ~\longrightarrow~P$ :

\begin{equation} \tan\alpha = \displaystyle\frac{\delta_\mathrm{C}}{l-x} = \displaystyle\frac{\delta_\mathrm{E}}{L-x}~,\quad \delta_\mathrm{E} \leq 0.25~\mbox{mm}~, \quad x = 93.75~\mbox{mm}~, \quad \delta_\mathrm{C} = 1.667e-05 P \end{equation}

whence it results:

\begin{equation} P \leq 15000 \frac{l-x}{L-x} \end{equation}

$$P \le 4411.765 N$$

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  • $\begingroup$ where does 0.675 come from? The length of the ties is 200[mm] $\endgroup$
    – NMech
    Jan 6 '21 at 15:23
  • $\begingroup$ its a lever. when you push down E, then the BCE will rotate and move downwards. To counteract the rotation the center of rotation needs to be between the points (so that the moment ther reactions generate is opposite to the moment P generates). $\endgroup$
    – NMech
    Jan 6 '21 at 19:56

There is nothing wrong with the approach of virtual Work, however what you got wrong, was the displacement at E and its contributtion to the work. i.e. it should be

$$P*\delta_E - F_{AB}*\delta_B - F_{CD}*\delta_C=0$$


  • $\delta_B = \frac{P_{AB} L}{E A}=\frac{1.5 P L}{E A}$
  • $\delta_C = \frac{P_{CD} L}{E A}=\frac{2.5 P L}{E A}$
  • $\delta_E = 0.00025[m]$

if you plug all those in and substitute you should come up with the following equation:

$$0.00025P - 8.5 \frac{P^2 L}{E A}=0$$

$$P(0.00025 - 8.5 \frac{P L}{E A})=0$$

$$\begin{cases} P=0\\ 0.00025 - 8.5 \frac{P L}{E A}=0 \end{cases}$$

From the second equation:

$$0.00025 - 8.5 \frac{P L}{E A}=0\Rightarrow P = \frac{0.00025 E A}{8.5 L}$$

which if you substitute:

  • $E=200*10^9 [N/m^2]$
  • $A=6*25*10^-6[m^2]$
  • $L=0.2[m]$ the length of the ties

you will get (see online calculation)

$$P\le 4411.76$$

  • $\begingroup$ I corrected a $P^2$ that I had forgotten. However, I double checked and I got the same result 4411.76 for P. $\endgroup$
    – NMech
    Jan 6 '21 at 14:33

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