Question 5.63: The pin-connected structure shown in Figure P5.63 consists o...

Equilibrium
Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin C gives the best information for this situation:

Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles:

\frac{v_{A}}{950  mm }=\frac{v_{B}}{600  mm }=\frac{v_{D}}{720  mm }                                   (b)

Since there are no gaps or clearances at either pin A or pin B, the deformations of members (1) and (2) will equal the deflections of the rigid bar at A and B, respectively.

\frac{\delta_{1}}{950  mm }=\frac{\delta_{2}}{600  mm }                                        (c)

Force-Temperature-Deformation Relationships
The relationship between the internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2)

\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}                                        (d)

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

\frac{1}{950  mm }\left[\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1}\right]=\frac{1}{600  mm }\left[\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}\right]                                   (e)

Solve the Equations
Solve Eq. (e) for F_{1}:

F_{1}=\left\{\frac{950}{600}\left[\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}\right]-\alpha_{1} \Delta T_{1} L_{1}\right\} \frac{A_{1} E_{1}}{L_{1}}                              (f)

and substitute this expression into Eq. (a):

(950  mm )\left\{\frac{950}{600}\left[\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}\right]-\alpha_{1} \Delta T_{1} L_{1}\right\} \frac{A_{1} E_{1}}{L_{1}}+(600  mm ) F_{2}=(720 mm )(36  kN )                                   (g)

For this structure, the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below:

Substitute these values along with \Delta T_{1}=\Delta T_{2}=+25^{\circ} C into Eq. (g) and calculate F_{2} = –3,989.18 N.
Backsubstitute into Eq. (f) to calculate F_{1} = 29,803.69 N.

(a) Normal stresses: 

(b) Deflections of the rigid bar
Calculate the deformation of member (1):

Since the pin at A is assumed to have a perfect connection, v_{A}=\delta_{1}=0.5985  mm From Eq. (b),