# Question 3.15: 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. (a) Use the Displacement Method to determine an expression that relates the displacement $u_B$ at node B to the axial force $P_B$ applied there. (b) Determine expressions for the axial forces in the three elements in terms of the external force $P_B$.

Step-by-Step
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Plan the Solution We can follow the steps outlined in the Displacement-Method Procedure. Step 1: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. Step 2: Using Eq. 3.15, we can write a force-deformation equation for each of the three elements. Step 3:We can relate the element elongations to the displacement at B. Steps 4–6: Finally, we can use the Displacement Method to combine these three sets of fundamental equations to get expressions for the displacement for node B. Step 7: The forces in the individual elements can then be determined by substituting this displacement into the element force-deformation equations.
Step 8: If the external force $P_B$ acts to the right (i.e., if it is positive), we should find that node B moves to the right, and we should find that the left-hand element is in tension and that the two right-hand elements are in compression.

$F = ke, \text{where} k = \frac{AE} {L}$        (3.15)

(a) Determine the displacement $u_B$ of node B.
Equilibrium: For node B, shown in the free-body diagram 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 problem is statically indeterminate. That is, the internal forces cannot be determined by statics alone. Thus, to find additional equations we must look to element force-deformation behavior and to the geometry of deformation.

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. 3.15.We have called the element forces $F_1, F_2$, and $F_3$ (tension positive), so we have

$F_1 = k_1e_1, \text{where} k_1 = (A_1E_1/L_1)$

$F_2 = k_2e_2, \text{where} k_2 = (A_2E_2/L_2)$ Element Force-Deformation  (2)

$F_3 = k_3e_3, \text{where} k_3 = (A_3E_3/L_3)$

In Eqs. (2) the e’s are the element elongations. A positive $F_i$ (tension) produces a positive $e_i$ (element gets longer), since the $k_i$’s are, by definition, positive.
So far, we have four equations in six unknowns—three F’s and three e’s. We must still enforce the deformation compatibility of the three elements with each other and with the walls at A and C.

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,

$e_1 = u_B$      Geometry of Deformation    (3)

$e_2 = e_3 = -u_B$

Here we have used the fact that the displacements at nodes A and C are zero. Equations (3) can also be called compatibility equations since they express mathematically the fact that A and C are fixed ends and the fact that the three elements are joined together at B. (Note that, since e is positive when the element gets longer, a displacement of node B to the right by an amount $u_B$ implies a shortening of elements (2) and (3) by that amount; hence the minus sign in the equations for $e_2$ and $e_3$.)

Displacement-Method Solution: Now, 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 can note that, since there is only one node, there is only $\underline{\text{one equilibrium equation}}$; correspondingly, there is only $\underline{\text{one nodal displacement}}$ , $u_B$. By substituting Eqs. (3) into Eqs. (2), we are able to write the three F’s in terms of the nodal displacement, $u_B$. If we then substitute these equations, call them Eqs. (4), into Eq. (1), we will have one equilibrium equation expressed in terms of the one unknown nodal displacement.
Substitute Eqs. (3) (deformation geometry) into Eqs. (2) (element force-deformation) to obtain

$F_1 = k_1u_B, F_2 = -k_2u_B, F_3 = -k_3u_B$        (4)

Now substitute Eqs. (4) into Eq. (1).

$(k_1 + k_2 + k_3)u_B = P_B$              Node Equilibrium in Terms of Node Displacement      (5)

The solution of this equation is

$u_B = \frac{P_B} {k_1 + k_2 + k_3}$          (a) (6)

The strategy of substituting deformation geometry equations (3) into element force-deformation equations (2) into equilibrium equations (1) is called the Displacement Method because the major solution step, Eq. (6), gives an answer that is a displacement. It is also sometimes referred to as the Stiffness Method since stiffness coefficients appear in the final solution.

(b) Determine the element forces. This is simple to do, because we only need to substitute the nodal displacement $u_B$ into Eqs. (4). Thus, we get

$\left\{\begin{matrix}F_1 = k_1u_B = \frac{k_1P_B} {k_1 + k_2 + k_3} \\ F_2 = -k_2u_B = \frac{-k_2P_B} {k_1 + k_2 + k_3} \\ F_3 = -k_3u_B = \frac{-k_3P_B} {k_1 + k_2 + k_3} \end{matrix} \right\}$     (b) (7)

Review the Solution As one check of our work, we can substitute Eqs. (7) back into Eq. (1) to see if equilibrium is satisfied.
Is $F_1 – F_2 – F_3 = P_B$? Yes.
The fact that equilibrium is satisfied by our answers means that we have probably not made errors in our solution. Also, from Eqs. (7) 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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