A three-element truss has the configuration shown in Fig. 1. Each member has a cross-sectional area of 1.2 in², and all of them are made of aluminum, with E = 10 × 10³ ksi. When the truss members were fabricated, members (1) and (3) were manufactured correctly (i.e., correct lengths of L_1 = 40 in. and L_3 = 30 in.). However, member (2) has a distance between hole centers of L_2, = 49.90 in., rather than the correct value of 50.00 in. If member (2) is stretched so that a pin can be inserted to connect all three members together at B, and then the system is released,

(a) what displacements,u ≡ u_B and ν ≡ ν_B will occur at node B?

(b) What forces will be induced in the three members?

Step-by-Step

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**Plan the Solution** As always, we need to formulate equations for equilibrium, element force-deformation behavior, and geometry of deformation. We must incorporate the “misfit” condition in the deformation-compatibility equations.

Because member (2) will tend to return to its original length but will be restrained by members (1) and (3), we should expect member (2) to be in tension and the other two members to be in compression.

(a) Use the Displacement Method to solve for the displacements u and ν.

Equilibrium: The node at B (Fig. 2) is acted on only by the element forces, since there is no external load. As usual, the axial forces in the elements are taken positive in tension.

\underrightarrow{+}\sum{F_x} = 0: -F_1-\frac{4}{5} F_2 = 0,+\uparrow \sum{F_y} = 0: -\frac{3}{5}F_2 – F_3 = 0

F_1+\frac{4}{5} F_2 = 0, Equilibrium (1)

\frac{3}{5}F_2 + F_3 = 0

Element Force-Deformation Behavior: We have three elements, so we need to write Eq. 3.28 for each of these elements. It will be helpful to compile a table of element properties first.

F = k(e – αLΔT), k = \frac{AE}{L} (3.28)

k_1 = \left(\frac{ AE}{L}\right)_1 = \frac{(1.2 in^2)(10 × 10^3 ksi)}{(40 in.)} = 300 kips/in.

k_2 = \left(\frac{ AE}{L}\right)_2 = \frac{(1.2 in^2)(10 × 10^3 ksi)}{(50 in.)} = 240 kips/in.

k_3 = \left(\frac{ AE}{L}\right)_3 = \frac{(1.2 in^2)(10 × 10^3 ksi)}{(30 in.)} = 400 kips/in.

Then, from Eq. 3.28,

F_1 = k_1e_1 = 300e_1 Element Force-Deformation Behavior (2)

F_2 = k_2e_2 = 240e_2F_3 = k_3e_3 = 400e_3

Geometry of Deformation: We sketch a deformation diagram (Fig. 3) so that we can see how to relate the element elongations to the system displacements u and ν. We greatly exaggerate the gap between the end of element (2) and node B, where it is supposed to connect to elements (1) and (3).

First, we can use Eq. 3.29 to express the elongation of each element in terms of the displacement at its “free” end, that is, end B.

e = e_u + e_ν = u cos θ + ν sin θ (3.29)

e_i = u_i \cos θ_i + ν_i \sin θ_i i = 1, 2, 3

However, we will have to modify the equation for element (2) to account for the “misfit.” Since all elements are pinned together at joint B,

u_1 = u_2 = u_3 ≡ u, ν_1 = ν_2 = ν_3 ≡ ν

Combining these equations, and referring to the Table of Element Properties, we get

e_1 = u \cos θ_1 + ν \sin θ_1 = ue_2 = u \cos θ_2 + ν \sin θ_2 + \bar{δ} Deformation Compatibility (3)

= 0.8u + 0.6ν + \bar{δ}

e_3 = u \cos θ_3 + ν \sin θ_3 = νDisplacement-Method Solution: All we need to do is combine Eqs. (1) through (3) in the Displacement-Method order: (3) → (2) → (1).

F_1 = 300u, F_2 = 192u + 144ν + 24.00, F_3 = 400ν (4)

Then, from Eqs. (1) and (4), we have

300u + 0.8(192u + 144ν + 24.00) = 0

0.6(192u + 144ν + 24.00) + 400ν = 0

or

453.6u + 115.2ν = -19.20 Nodal Equilibrium in Terms of Nodal Displacement (5)

115.2u + 486.4ν = -14.40

Solving these simultaneous algebraic equations, we get

u = 3.70(10^{-2}) in. Ans. (a) (6)

ν = -2.08(10^{-2}) in.

(b) Solve for the element forces. The displacements u and ν can be substituted into Eqs. (4) to give

F_1 = 300u = -11.11 kips

F_2 = 192u + 144ν + 24.00 = 13.89 kips

F_3 = 400ν = -8.33 kips

\left\{\begin{matrix} F_1 = 11.11 kips (C) \\F_2 = 13.89 kips (T) \\F_3 = 8.33 kips (C) \end{matrix} \right\} Ans. (b) (7)

**Review the Solution** One way to verify our results is to check equilibrium (using full calculator precision, not the reported rounded values).

Is F_1 +0.8F_2 = 0? Yes.

Is 0.6F_2 + F_3 = 0? Yes.

Also, we note that all of the element forces have the signs that we expected them to have, that is, member (2) is in tension, while members (1) and (3) are in compression.

**Table of Element Properties**

Element, i | L_i (in.) | k_i (kips/in.) | \cos θ_i | \sin θ_i |

1 | 40 | 300 | 1.0 | 0.0 |

2 | 50 | 240 | 0.8 | 0.6 |

3 | 30 | 400 | 0.0 | 1.0 |

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