# Question 3.21: A three-element truss has the configuration shown in Fig. 1.......

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
The 'Blue Check Mark' means that this solution was answered by an expert.

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)

$F_2 = k_2e_2 = 240e_2$

$F_3 = k_3e_3 = 400e_3$

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

$e_1 = u \cos θ_1 + ν \sin θ_1 = u$

$e_2 = u \cos θ_2 + ν \sin θ_2 + \bar{δ}$    Deformation Compatibility    (3)

= 0.8u + 0.6ν + $\bar{δ}$

$e_3 = u \cos θ_3 + ν \sin θ_3 = ν$

Displacement-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

Question: 3.6

## The rigid, weightless beam AC in Fig. 1 is supported at end A by a vertical column, and at end C it is supported by a vertical rod CD that is attached at D to a leveling jack, that is, a jack that can support the required load in rod CD but can also move the vertical position of D upward or ...

Plan the Solution A free-body diagram of beam AC m...
Question: 3.15

## 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 ...

Plan the Solution We can follow the steps outlined...
Question: 3.14

## A 14-mm-diameter steel bolt, and a steel pipe with 19-mm ID and 25-mm OD are arranged as shown in Fig. 1. For both the steel bolt and steel pipe, E = 200 GPa. The pitch of the (single) thread is 2 mm. What stresses will be produced in the steel bolt and sleeve if the nut is tightened by 1/8 turn? ...

Plan the Solution As we discussed earlier in this ...
Question: 3.13

## The rigid beam BD in Fig. 1 is supported by a wire AC of length L, cross-sectional area A, modulus of elasticity E, and coefficient of thermal expansion α; and by a rod DE of length L/2, cross-sectional area 2A, and modulus E. Neglect the weight of the beam. When there is no load acting on the ...

Plan the Solution This is a statically indetermina...
Question: 3.8

## A load P is hung from the end of a rigid beam AD, which is supported by a pin at end A and by two uniform hanger rods (see Fig. 1). How much load, to the nearest 100 pounds, can be applied without exceeding an allowable tensile stress of σallow = 30 ksi in either of the rods? Assume small-angle ...

Plan the Solution This problem is not quite as str...
Question: 3.18

## 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, σi , in each of the three elements. Let ...

Plan the Solution As always, we will first set up ...
Question: 3.1

## A bimetallic bar is made of two linearly elastic materials, material 1 and material 2, that are bonded together at their interface, as shown in Fig. 1. Assume that E2 > E1. Determine the distribution of normal stress that must be applied at each end if the bar is to undergo axial deformation, and ...

Plan the Solution The bar is said to undergo axial...
Question: 3.23

## 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 ε2 = -1.5 εY and is then removed, what residual stresses will be left in the two elements? What is the permanent deformation? Let A1 = A2 = A, ...

Plan the Solution We can divide the analysis into ...
Question: 3.22

## The two support rods in Fig. 1 are made of structural steel that may be assumed to have a stress-strain diagram like Fig. 3.28b, with E = 30(10³) ksi and σY = 36 ksi. (a) Construct a load-displacement diagram that relates the load P to the vertical displacement at D. Make the usual small-angle ...

(a) Determine the element stresses σ_1[/lat...