# Question 3.23: Let us take a special case of the statically indeterminate s......

Let us take a special case of the statically indeterminate system in Fig. 3.28. This special case is shown in Fig. 1a. If the load P is increased until $ε_2 = -1.5 ε_Y$ and is then removed, what residual stresses will be left in the two elements? What is the permanent deformation? Let $A_1 = A_2 = A, L_1 = 2L_2$ = 2L, and let both elements be made of the elastic-plastic material depicted in Fig. 1b.

Step-by-Step
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Plan the Solution We can divide the analysis into a loading phase and an unloading phase. The analysis of the loading phase can be taken directly from the discussion of elastic-plastic analysis of statically indeterminate members at the beginning of Sect. 3.11. For the unloading phase we must set P = 0 in the equilibrium equation and use linearly elastic material behavior to represent the unloading σ − ε paths of the two members.

Geometry of Deformation: Let us consider the geometry of deformation first, since this is the same for both loading and unloading. The member elongations are related to the displacement of node B by

$e_1 = u, e_2 = -u$     (1)

Since the strain is uniform along each member, the geometric compatibility equations can be written in terms of strains as

$ε_1 = \frac{e_1} {2L} = \frac{u} {2L}$    Geometric Compatibility    (2)

$ε_2 = \frac{e_2} {L} = \frac{-u} {L}$

These strain equations will prove to be more convenient than Eqs. (1), since we have to carefully examine the loading and unloading paths on the stress-strain diagram.

Equilibrium: Equilibrium of node B (Fig. 2) gives

$F_1 – F_2 = P$      Equilibrium (3)

Loading Phase: You should follow the loading and unloading strain paths on Fig. 1b.We are given that the load is increased until load P produces a strain $ε^′_2 = -1.5ε_Y$ .For member 2 this corresponds to a path O → C → 2′ in Fig. 1b. Combining this information with Eq. (2b) we get

$ε^′_2 = -1.5ε_Y = \frac{-u^′} {L}$ (4)

where the primes denote quantities when the load has been increased until it produces the given strain in member 2. Point 2′ on Fig. 1b characterizes the state for member 2. From Eq. (4),

u′ = 1.5 $ε_Y$L    (5)

Combining Eqs. (2a) and (5) gives

$ε^′_1 = \frac{1.5 ε_YL} {2L} = 0.75ε_Y$  (6)

Therefore, member 1 is still elastic at the point designated 1′ on Fig. 1b.
Material Behavior—Loading Phase: Since |$ε^′_2| > ε_Y$ and |$ε^′_1| < ε_Y$ the appropriate material-behavior equations are given by Case 2 of the earlier elastic-plastic analysis, namely

$F_1 = Aσ_1 = AEε_1$   Material Behavior—Loading      (7)

$F_2 = Aσ_2 = -σ_YA$

To determine the external load P′ that corresponds to the given state (i.e., that makes $ε^′_2 = -1.5 ε_Y)$, we can combine Eqs. (7), (2a), and (5) to obtain the member forces

$F^′_1 = \frac{AE}{2L} (1.5ε_YL) = 0.75σ_YA$  (8)

$F^′_2 = -σ_YA$

These may be substituted into the equilibrium equation, Eq. (3), giving the external load

P′ = 1.75$σ_Y$A     (9)

From Eq. (5) the corresponding displacement of node B is

$u^′ = 1.5 \frac{σ_YL} {E}$ (10)

Unloading Phase: When the load is removed, members 1 and 2 unload to points 1″ and 2″, respectively, on Fig. 1b. Since P″ = 0, the equilibrium equation, Eq. (1), becomes

$F^″_1 – F^″_2 = 0$    (11)

The geometry of deformation is still described by Eqs. (1) and (2), but the material-behavior equations must now correspond to the unloading paths indicated in Fig. 1b. Point 1″ lies on the original linear portion, AC, of the σ – ε  diagram, but point 2″ lies along the unloading path 2′ –  2″ that intersects the ε axis at $ε^′_2 + ε_Y = -0.5 ε_Y$. Hence, along these two respective paths,

$F_1 = Aσ_1 = AEε_1$    Material Behavior—Unloading  (12)

$F_2 = Aσ_2 = AE(ε_2 + 0.5ε_Y)$

Substituting Eqs. (2) into Eqs. (12), we get the following expressions for the internal forces:

$F_1 = AE \left( \frac{u} {2L}\right)$

$F_2 = AE \left( -\frac{u} {L} + \frac{σ_Y}{2E}\right)$    (13)

The permanent displacement of point B, u″, at which these forces satisfy the equilibrium equation, Eq. (11), is

$u^″ = \frac{1} {3} \frac{σ_YL}{E}$    Ans. (14)

with corresponding element forces

$F^″_{1} = F^″_{2} = \frac{1} {6} σ_YA$    (15)

Therefore, the two members are left with a residual stress of

$σ^″_{1} = σ^″_{2} = \frac{1}{6} σ_Y$  Ans. (16)

Review the Solution A key equation is the geometric-compatibility equation, Eq. (2). It tells us that for any increment of displacement Δu, $Δε_1$ = Δu/2L and $Δε_2$ = -Δu/L. Therefore, we can easily check points 2′ and 1′ on Fig. 1b. From the corresponding stresses, $σ_1 = 3/4σ_Y$ and $σ_2 = -σ_Y$, and the nodal equilibrium equation we can verify that the force P′ given in Eq. (9) is correct.
Next, to determine the residual stresses we know that $F^″_1 = F^″_2$, so $σ^″_1 = σ^″_2$.We also know that $Δε_1 = -\frac{1}{2}Δε_2$ .We see in Fig. 1b that points 1″ and 2″ have equal stress and that the strain increments, $Δε_1 ≡ (ε_{1^″} – ε_{1^′})$ and $Δε_2 ≡ (ε_{2^″} – ε_{2^′})$ satisfy this strain-increment equation. Therefore, our residual stress answers in Eqs. (16) appear to be correct.

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