Question 5.64: The pin-connected structure shown in Figure P5.64 consists o...

Equilibrium
Consider a FBD of joint D. Assume tension forces in bars (1) and (2).

Geometry-of-deformation relationship
Draw a deformation diagram. We imagine temporarily removing the pin at D and allowing the members to elongate independently. Then, after deformation, bars (1) are rotated so that they intersect the lower end of bar (2), and the pin at D is reinserted. If we assume that the deformations will be small, then the angle after deformation is approximately the same as the angle before deformation. From the deformation diagram geometry, we can assert that:

(b)

Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for bars (1) and (2) as:

$\delta_{1}=\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1} \quad \delta_{2}=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}$                              (c)

Compatibility Equation

Substitute the force-deformation relationships (c) into the geometry-of-deformation relationship (b) to derive the compatibility equation:

$\frac{1}{\sin 60.9454^{\circ}}\left[\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1}\right]=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}$                                       (d)

Solve the Equations
From the problem statement, $A_{1}=A_{2}$. The bar lengths are:

Additionally, the temperature change is the same for all members. Use these facts and solve Eq. (d) for $F_{2}$:

Substitute this result in Eq. (a) and solve for $F_{1}:$

The force in bar (2) is therefore:

(a) Normal stresses in bars (1) and (2). The normal stresses in bars (1) and (2) can now be calculated:

(b) Downward displacement of pin D. The downward displacement of pin D is simply equal to the deformation of bar (2).