# Question 3.9: The structural assemblage in Fig. 1 is made up of three unif......

The structural assemblage in Fig. 1 is made up of three uniform elements, or members. Element (1) is a solid rod. Element (2) is a pipe that surrounds element (3), which is a solid rod that is identical to element (1) and collinear with it. The three elements are all attached at B to a rigid plate of negligible thickness. With no external force at B, the three-element assemblage exactly fits between the rigid walls at A and C; its ends are then attached to the two walls. Determine expressions for the axial forces in the three elements when an external force $P_B$ is applied at B.

Step-by-Step
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Plan the Solution We can think of this structure as being composed of three uniform elements and one connecting node (joint), and we can write an equilibrium equation for node B. Using Eq. 3.14, we can write a force-deformation equation for each of the three elements. We can relate the element elongations to each other through the displacement at B.
Finally, we can use the Basic Force Method to combine these three sets of fundamental equations to get expressions for the forces in the individual elements.

$e = fF, \text{where} f ≡ \frac{L}{AE}$        (3.14)
If the external force $P_B$ acts to the right (i.e., if it is positive), we should find that the left-hand element is in tension and the two righthand elements are in compression.

Equilibrium: From the free-body diagram of node B in Fig. 2,

$\underrightarrow{+} \sum{F_x}=0: -F_1 + F_2 + F_3 + P_B = 0$  Equilibrium  (1)

Equation (1) relates the three unknown internal element forces to the known external load. Since there are three unknown forces, but only one equilibrium equation, this system is statically indeterminate and there are two redundant forces.

Element Force-Deformation Behavior: We have three uniform, axial-deformation elements, and for each one we can write an element force-deformation equation like Eq. 3.14.We have called the element forces
$F_1, F_2,$ and $F_3$ (tension positive), so we have

$e_1 = f_1F_1$,  where  $f_1 = (L_1/A_1E_1)$

$e_2 = f_2F_2$,  where  $f_2 = (L_2/A_2E_2)$  Element Force-Deformation Behavior    (3a–c)

$e_3 = f_3F_3$,  where  $f_3 = (L_3/A_3E_3)$

In Eqs. (2) the e’s are the element elongations. A positive Fi (tension) produces a positive $e_i$ (element gets longer), since the $f_i$’s are, by definition, positive.

Geometry of Deformation: Referring to Fig. 1, we can easily relate the elongation of each of the three elements to the displacement $u_B$ by using the definition of elongation of an element, that is, e = u(L) – u(0). So,

$e_1 = -e_2 = -e_3 = u_B$

Here we have used the fact that the displacements at joints A and C are zero. Note that, since e is positive when an element gets longer, a displacement of joint B to the right by an amount $u_B$ implies a shortening of elements (2) and (3) by that amount; hence the minus sign for $e_2$ and $e_3$.We can eliminate $u_B$ and write the above as two compatibility equations:

$e_2 = -e_1$      Geometry of Deformation    (3a,b)

$e_3 = e_2$

The fact that there are two compatibility equations is consistent with the fact that there are two redundant forces in the equilibrium equation. There will always be as many compatibility equations as there are redundant forces!

Solution of the Equations: If we count equations and unknowns, we find that we have six equations and six unknowns. Rather than just combine Eqs. (1) through (3) in some arbitrary order, we will follow the Basic Force Method.
Substitute Eqs. (2) (element force-deformation) into Eqs. (3) (deformation compatibility) to obtain the compatibility equations written in terms of forces.

$f_2F_2 = -f_1F_1$    Compatibility in Terms of Element Forces    (4a,b)
$f_3F_3 = f_2F_2$

We now have three equations, Eqs. (1) and (4a,b), in three unknowns, the three element forces. We $\underline{\text{solve these equations simultaneously}}$ to get the following expressions for the three unknown element forces:

$F_1 = \left( \frac{f_2 f_3} {f_1 f_2 + f_2 f_3 + f_1 f_3}\right)P_B$

$F_2 = \left( \frac{-f_1 f_3} {f_1 f_2 + f_2 f_3 + f_1 f_3}\right)P_B$        Ans. (5a–c)

$F_3 = \left( \frac{-f_1 f_2} {f_1 f_2 + f_2 f_3 + f_1 f_3}\right)P_B$

Review the Solution As one check of our work, we can substitute Eqs.
(5) back into Eqs. (4a,b) to see if deformation compatibility is satisfied.

Are $f_2F_2 = -f_1F_1$      and  $f_3F_3 = f_2F_2$?    Yes.

The fact that the compatibility equations, Eqs. (4a,b), are satisfied by our answers means that we have probably not made errors in our solution. Also, from Eqs. (5) we see that, when $P_B$ is positive, element (1) is in tension and elements (2) and (3) are in compression. This is what we expected to find.

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