Question 3.13: The rigid beam BD in Fig. 1 is supported by a wire AC of len......
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 beam, the beam is horizontal and the wire and rod are stress-free. Assume that the tip displacement is very small, that is δ_D \ll L.
Determine expressions for the axial forces in wire AC and rod DE when a load of intensity w per unit length of beam is uniformly distributed over CD, and wire AC is \underline{\text{simultaneously cooled}} by an amount ΔT (i.e., ΔT_1=-ΔT, ΔT_2 = 0).
Plan the Solution This is a statically indeterminate structure that is somewhat similar to the determinate structure in Example 3.3, but we will need to add rod DE and take into account the temperature change of wire AC. The Basic Force Method will provide a systematic way of combining the three sets of equations—equilibrium, element force-temperature-deformation behavior, and compatibility.
Cooling wire AC should cause tension in both the wire and the rod; applying a downward load over CD should cause tension in the wire, but it should cause compression of the rod. In the expressions for the force in the wire and the force in the rod, the temperature change should lead to terms having the form AEαΔT.
Equilibrium: The free-body diagram in Fig. 2 will enable us to write an equilibrium equation that relates F_1 (the force in wire AC) and F_2 (the force in rod DE) to the external load. We do not need equations for the reaction components at B.
+\circlearrowleft \left(\sum{M} \right) _B = 0: \left(\frac{\sqrt{3}}{2}F_1\right)\left(\frac{L}{2}\right) – \left(\frac{wL}{2}\right)\left(\frac{3L}{4}\right) – F_2L = 0
or
2\sqrt{3}F_1 – 8F_2 = 3wL Equilibrium (1)
Element Force-Temperature-Deformation Behavior. From Eq. 3.27, the total elongation of the wire and rod, respectively, are
e = fF = αLΔT, f = \frac{L}{AE} (3.27)
e_1 = f_1F_1 + α_1L_1ΔT_1 Element Force-Temperature-Deformation Behavior (2a,b)
e_2 = f_2F_2where it has been noted that ΔT_2 = 0, and where
f_1 = \left(\frac{ L} {AE}\right)_1 = \frac{L}{AE}, f_2 = \left(\frac{ L} {AE}\right)_2 = \frac{(L/2)}{(2AE)} =\frac{1}{4} \left(\frac{ L} {AE}\right)Geometry of Deformation; Compatibility Equation: Figure 3 is a deformation diagram that enables us to determine an equation that relates the \underline{\text{total elongation}} , e_1, of wire AC to the vertical displacement of point D, and, hence, to the shortening of rod DE. From the deformation diagram in Fig. 3, we can write the following deformation equations:
e_1 = \frac{\sqrt{3}}{2}\left(\frac{δ_D}{2}\right) = \frac{\sqrt{3}}{2} δ_D Geometry of Deformation
e_2 = -δ_D
from which we can eliminate δ_D to obtain the compatibility equation
e_1 = -\frac{\sqrt{3}}{4}e_2 Compatibility (3)
Since it enforces the condition that the end of the wire AC and the top of the rod DE remain attached to a rigid beam that rotates about end A, Eq. (3) is called a compatibility equation.
Solution by the Basic Force Method: Following the procedure of the Basic Force Method, we can eliminate the e’s by substituting Eqs. (2) into Eq. (3), giving the following compatibility equation in terms of forces:
f_1F_1 + α_1L_1ΔT_1 = -\left(\frac{\sqrt{3}}{4}\right)f_2F_2 Compatibility in Terms of Element Forces (4)
Noting that ΔT_1 = -ΔT, and substituting in the values of the flexibility coefficients f_1 and f_2, we can now solve Eqs. (1) and (4) simultaneously to get the forces
F_1 = \frac{64} {67} (AEαΔT) + \frac{3\sqrt{3}} {134} (wL) (5)
F_2 = \frac{16\sqrt{3}} {67} (AEαΔT) – \frac{24} {67} (wL)
Review the Solution The expressions in Eqs. (5) for the unknown forces F_1 and F_2 have the proper forms that were anticipated in the “Plan the Solution” section. Therefore, the answers we have obtained seem reasonable.
