Calculate the residual stresses in the bars and the residual displacement of the $ C $ point when the $ P $ force is removed

Question

Consider the rigid bar $ ABC $ linked to two bars, $ AD $ and $ BF $, as shown in the figure. All bars are made of mild steel which is admitted to be an elastoplastic material ($ E $ = 210 GPa, $ \sigma_Y $ = 250 MPa) but the bar $ AD $ has a uniform and rectangular cross section of 20 mm $ \times $ 6 mm while the $ BF $ bar consists of two sections, $ BE $ and $ EF $, with section area respectively equal to $ A_ {BE} = 1200 $ mm $ ^ 2 $ and $ A_ {EF} = 2400 $ mm $ ^ 2 $. The strength of the $ P $ force applied in $ C $ is gradually increased until the displacement of the $ C $ point reaches 2.5 mm.

Calculate the residual stresses in the bars and the residual displacement of the $ C $ point when the $ P $ force is removed.

The resolution says:

Now consider the extreme case where the $ C $ point offset is increased to 2.5 mm. What will happen to the displacements and forces at the other points of interest on the rigid bar $ ABC $?

Since the maximum force $ P $ does not change during the path $ \delta_C in [1.206 \ , ~ 2.5] $ mm the forces on the bars $ AD $ and $ BF $ do not change:

\begin{equation} \begin{cases} F_{BF} &= P + F_{AD} = \left(\displaystyle\frac{L_{BC}}{L_{AB}} + 1\right) P \\ F_{AD} &= \left(\displaystyle\frac{L_{BC}}{L_{AB}}\right) P \end{cases} \end{equation}

In this case, the displacement of the point $ B $ remains fixed ($ F_ {BF} $ constant) and the geometric rotation of the mechanism occurs around $ B $. At constant force, there will be a continuous increment of variation in the length of the bar $ AD $ being the total displacement value in $ \delta_A ^ t $ (for $ \delta_ {C_p} = $ 2.5 mm) given by:

\begin{equation} \begin{aligned} \frac{\delta_A^t+\delta_B}{L_{AB}} = \frac{\delta_A^t+\delta_{C_p}}{L_{AC}} \end{aligned} \quad\Leftrightarrow\quad \begin{aligned} \delta_A^t = \frac{\delta_{C_p}L_{AB} - \delta_BL_{AC} }{L_{AC}-L_{AB}} \end{aligned} \end{equation}

deltaA tot = 1.6983 mm

I can't understand how they calculated deltaA tot ? I also tried to use the relationship $\frac{\delta}{L}$ but i'm not getting it.

Could someone explain it to me?

Answer 1

Delta A total is the sum of Delta A Plastic and Delta A Elastic. Below and up to the yeild strenght it will deform elastically given the modulus of elasticity. At and after the yeild strenght it will "fail" and plasticaly deform. (F/A)*EL for elastic deformation before failure.

"Thus, a plastic deformation will occur at point A. This value can be determined considering that the total deformation is the sum of the elastic deformation with the plastic deformation:

δtA=δeA+δpA⇔δpA=δtA−δeA

deltaA plast = 0.9244 mm "