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

Question

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.

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

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}

result,

\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}

$$P \le 4411.765 N$$

Answer 1

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

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

where:

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

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)

$$P\le 4411.76$$