# Deformation of a fixed-fixed beam with perfectly plastic yield stresses

Here's the setup:

The problem asks to find the relationship between the force $P$ and the displacement of B ($u_B$). I have done this correctly:

$$P = \left(\dfrac{E_1A_1}{L_1}+\dfrac{E_2A_2}{L_2}\right)u_B$$

Now imagine that members AB and BC both have yield stresses ($\sigma_{Y_1}$ and $\sigma_{Y_2}$) and are perfectly plastic after yield.

I need to find the relationship between $u_B$ and $P$ in the following cases:

1. No member yields - ($0 < u_B < \sigma_{Y_1}$) & ($0 < P < P_{y_1}$)
2. Member AB yields, but not member BC - ($u_{Y_1} < u_B < u_{Y_2}$) & ($P_{y,1} < P < P_{y,2}$)
3. Both members yield - ($u_B > u_{Y_2}$) & ($P > P_{y,2}$)

I'm having trouble finding explicit expressions for $P_{y_{1/2}}$ and $u_{Y_{1/2}}$. I'm not quite sure how to relate yield stress to elongation and displacement of the members.

To be honest, this problem has tied my brain in a knot and I believe my conceptual understanding (or misunderstanding) of the problem is limiting my ability to solve it mathematically. Any help at all would be greatly appreciated!

Edit: Here is the exact problem

From Hooke's Law, we know that

$$\epsilon = \dfrac{\sigma}{E}$$

This is valid up until the yield point. Considering a perfectly plastic material, the deformation $\epsilon$ is constant after the yield point, which gives us the following graph:

So, to answer your question, what is the value of $u_{Y_1}$? Well, it is equal to the displacement under the yield stress:

$$u_{Y_1} = \dfrac{\epsilon}{L} = \dfrac{\sigma_{Y_1}}{E_1L_1}$$

Likewise for $P_{Y_1}$:

$$P_{Y_1} = \sigma_{Y_1}A_1$$

Now, for the rest of the question, I think it is worth expanding your equation:

$$P = \dfrac{E_1A_1u_B}{L_1} + \dfrac{E_2A_2u_B}{L_2}$$

This is equivalent to

$$P = \sigma_1A_1 + \sigma_2A_2 = P_1 + P_2$$

(where $\sigma_1$ and $\sigma_2$ are variables and proportional to $u_B$) which means that the force applied is equal to the sum of the forces applied by each member, which each are equal to the product of the stress and the area of the member. Keep that in mind.

Now, the question states that we can assume that AB will become plastic first. So we know that the situation will change when $u_B = u_1$. In this case, the equation above will change to

$$P = \sigma_{Y_1}A_1 + \sigma_2A_2 = P_{Y_1} + \dfrac{E_2A_2u_B}{L_2}$$

So, as of that point, member AB will no longer increase how much it resists deformation; it is no longer proportional to $u_B$, but is instead a constant equal to $P_{Y_1} = \sigma_{Y_1}$. Member BC, however, has yet to yield, and therefore remains proportional to $u_B$.

As the force increases, $u_B$ will eventually reach $u_2$, at which point BC's collaboration will also become a constant and equal to $P_{Y_2}$, so we have

$$P = \sigma_{Y_1}A_1 + \sigma_{Y_2}A_2 = P_{Y_1} + P_{Y_2}$$

This tells us that once both members cross into the plastic region, the deformation will tend to infinity.

P = \begin{cases} \begin{aligned} &\left(\dfrac{E_1A_1}{L_1} + \dfrac{E_2A_2}{L_2}\right)u_B &&\text{ for }u_B \in [0, u_1] \\ &\sigma_{Y_1}A_1 + \dfrac{E_2A_2}{L_2}u_B &&\text{ for }u_B \in [u_1, u_2] \\ &\sigma_{Y_1}A_1 + \sigma_{Y_2}A_2 &&\text{ for }u_B \geq u_2 \end{aligned} \end{cases}
• @Vitruvius: My graph has $E_1 > E_2$ and $P_{Y_1} > P_{Y_2}$, but that doesn't have to be the case. You could make $E_1 < E_2$, which would also imply that $P_{Y_1} < P_{Y_2}$ (since $u_{Y_1} < u_{Y_2}$). And of course you could make $E_1=E_2$, which would also imply that $P_{Y_1} < P_{Y_2}$ for the same reason. – Wasabi Feb 1 '17 at 19:47