# Question 3.18: A stepped rod is made up of three uniform elements, or membe......

A stepped rod is made up of three uniform elements, or members, as shown in Fig. 1. The rod exactly fits between rigid walls when no external forces are applied, and the ends of the rod are welded to the rigid walls. (a) Determine the axial stress, $σ_i$ , in each of the three elements. Let $F_1$, the internal force in member AB, be the redundant force.
(b) Determine the joint displacements $u_B$ and $u_C$.

Step-by-Step
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Plan the Solution As always, we will first set up the three fundamental sets of equations—equilibrium equations, force-deformation equations, and compatibility equations. In solving these equations we are asked to let $F_1$ be taken as the redundant force. As illustrated in Fig. 2, by making a single (imaginary) “cut” anywhere along element (1) we make the resulting structure “statically determinate,” with $F_1$ treated as a known load rather than as an unknown internal force. There is one compatibility (constraint) equation, namely, that the total elongation is zero. By writing this equation in terms of the redundant force $F_1$, we will be able to solve this one equation for the one unknown.

(a) Determine the axial stresses $σ_i$.
Equilibrium: Having selected $F_1$ as the redundant internal force, we need to draw two free-body diagrams that can be used to express $F_2$ and $F_3$ in terms of the redundant force $F_1$. By cutting on both sides of the joint at B, we produce the free-body diagram of Fig. 3a, which relates $F_2$ to $F_1$. Similarly, by cutting through elements (1) and (3) we get the free-body diagram in Fig. 3b, which relates $F_3$ to the redundant force $F_1$. Note that we have applied the method of sections in drawing the free-body diagrams and writing the equilibrium equations. You will find that the method of sections is the best way to draw free-body diagrams for Force-Method solutions.

For Fig. 3a: $\underrightarrow{+} \sum{F_x} = 0: \rightarrow -F_1 + F_2 + P_B$ = 0

For Fig. 3B: $\underrightarrow{+} \sum{F_x} = 0: \rightarrow -F_1 + P_B + P_C + F_3$ = 0

As indicated in Step 1 of the Force-Method Procedure, we rearrange these equilibrium equations so that they express the determinate forces $F_2$ and $F_3$ in terms of the one redundant force $F_1$ and the external loads.

$F_2 = F_1 – P_B$          Equilibrium    (1)

$F_3 = F_1 – P_B – P_C$

Note that, once we have solved for the redundant force $F_1$, we will come back to Eqs. (1) to solve for the remaining unknown forces.
Element Force-Deformation Behavior: The force-deformation equations for the three elements can be written in the form given by Eq. 3.14, namely,

e = fF, where $f ≡ \frac{L}{AE}$    (3.14)

$e_1 = f_1F_1, f_1 = (L/AE)_1$Element Force-Deformation Behavior

$e_2 = f_2F_2, f_2 = (L/AE)_2$

$e_3 = f_3F_3, f_3 = (L/AE)_3$

Geometry of Deformation: At this point we have five equations and six unknowns. Since we have one redundant force, $F_1$, we need one equation of geometric compatibility. We can see (Fig. 1) that the total length of the three-element rod system must remain constant, that is, there is no elongation of the rod. The appropriate compatibility equation is

$e_{\text{total}} = e_1 + e_2 + e_3 = 0$    Compatibility (3)

Force-Method Solution of the Fundamental Equations: Note that the equilibrium equations have been written in a form that expresses the determinate forces, $F_2$ and $F_3$, in terms of the redundant force, $F_1$, and the two nodal loads. Following Step 4 in the Force-Method Procedure, we first combine Eqs. (1) and (2) so that all of the force-deformation equations will be expressed in terms of the known loads and the one redundant force, $F_1$.

$e_1 = f_1F_1, e_2 = f_2(F_1 – P_B), e_3 = f_3(F_1 – P_B – P_C)$    (4)

Next (Step 5), we substitute Eqs. (4) into the compatibility equation, Eq. (3), to get

$( f_1 + f_2 + f_3)F_1 = f_2P_B + f_3(P_B + P_C)$  Compatibility in Terms of the Redundant Force       (5)

We have reduced six equations in six unknowns to $\underline{\text{one equation in one unknown}}$ , which we can easily solve for the redundant force (Step 6).

$F_1 = \frac{( f_2 + f_3)P_B + f_3P_C}{f_1 + f_2 + f_3}$     (6a)

This is the $\underline{\text{key solution step}}$ .We have determined the redundant internal force in terms of the given external loads and properties of the structure. The approach we have used is called the Force Method because the major solution step gives us a force quantity. It is sometimes referred to as the Flexibility Method because it is convenient to use flexibility coefficients, $f_i$, in the solution; and it is sometimes referred to as the Method of Consistent Deformations because the key equation(s) in the solution is (are) the deformation-compatibility equation(s).
Knowing the redundant force $F_1$, we can now use the equilibrium equations, Eqs. (1), to obtain the determinate forces $F_2$ and $F_3$, thus completing Step 6.

$F_2 = F_1 – P_B = \frac{-f_1P_B + f_3P_C}{f_1 + f_2 + f_3}$

$F_3 = F_1 – P_B – P_C = \frac{-f_1P_B – ( f_1 + f_2)P_C}{f_1 + f_2 + f_3}$          (6b,c)

Since we were asked for the stresses in the three elements, we need to divide the forces in Eqs. (6) by their respective cross-sectional areas, that is

$σ_1 = \frac{F_1} {A_1}, σ_2 = \frac{F_2} {A_2}, σ_3 = \frac{F_3} {A_3}$       (a) (7)

(b) Obtain the displacements $u_B$ and $u_C$. We need to use geometry of deformation equations. We can note (Fig. 1) that

$u_B = e_1, u_C = -e_3$      Geometry of Deformation      (8)

The first and third of Eqs. (2) may be substituted into these to give

$u_B = f_1F_1, u_C = -f_3F_3$        (9)

Then, forces $F_1$ and $F_3$ from Eqs. (6) can be substituted into Eqs. (9) giving

$\left\{\begin{matrix}u_B = \frac{f_1[(f_2 + f_3)P_B + f_3P_C]} {f_1 + f_2 + f_3} \\ u_C = \frac{f_3[f_1 P_B + (f_1 + f_2)P_C]} {f_1 + f_2 + f_3} \end{matrix} \right\}$         (b) (10)

Review the Solution We can easily check the dimensions in each of the answers and see that the answers are dimensionally homogeneous. For example, dimensionally, the f ’s in the numerator and the f ’s in the denominator of Eqs. (6) cancel, leaving the dimensional equation F = F.As a second check, we can see (Fig. 1) that, if $P_B$ > 0 and $P_C$ = 0, element (1) will be in tension while elements (2) and (3) will be in compression.
From Eqs. (6) we can see that this is the case.

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