Question 5.61 : A load P will be supported by a structure consisting of a ri...

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

\Sigma M_{A}=(30 \text { in. }) F_{1}+(84 \text { in. }) F_{2}-(66 \text { in. }) P=0                                            (a)

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_{B}}{30 \text { in. }}=\frac{v_{D}}{84  in. }             (b)

Since there are no gaps, clearances, or other misfits at pins B and D, the deformation of member (1) will equal the deflection of the rigid bar at B and the deformation of member (2) will equal the deflection of the rigid bar at D. Therefore, Eq. (b) can be rewritten in terms of the member deformations as:

\frac{\delta_{1}}{30 \text { in. }}=\frac{\delta_{2}}{84  in. }                                (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}{30 \text { in. }}\left[\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1}\right]=\frac{1}{84 \text { in. }}\left[\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}\right]                                     (e)

Solve the Equations
Note that the temperature change for both axial members is the same \text { (i.e., } \left.\Delta T_{1}=\Delta T_{2}=\Delta T\right) . Solve Eq. (e) for F_{1}:

Substitute this expression into equilibrium equation (a):

and solve for F_{2}:

F_{2}=\frac{(66 \text { in. }) P-\frac{(30  in .)^{2}}{84 \text { in. }} \alpha_{2} \Delta T L_{2} \frac{A_{1} E_{1}}{L_{1}}+(30 \text { in. }) \alpha_{1} \Delta T A_{1} E_{1}}{\frac{(30 in .)^{2}}{84 \text { in. }} \frac{L_{2}}{L_{1}} \frac{A_{1}}{A_{2}} \frac{E_{1}}{E_{2}}+84  in .}                                (g)

For this structure, P = 26 kips, and the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below:

Substitute these values along with ΔT = 70°F into Eq. (g) and calculate F_{2} = 19.3218 kips. Backsubstitute into Eq. (f) to calculate F_{1} = 3.0991 kips.

Normal Stresses
The normal stresses in each axial member can now be calculated:

Deflections of the rigid bar
Calculate the deformation of member (2):

\delta_{2}=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}=\frac{(19.3218  kips )(96  in .)}{\left(2.00  in .^{2}\right)(10,000  ksi )}+\left(12.5 \times 10^{-6} /{ }^{\circ} F \right)\left(70^{\circ} F \right)(96  in .)=0.1767 \text { in. }                                 (h)

Since there are no gaps at pin D, the rigid bar deflection at D is equal to the deformation of member (2);
therefore: