The pin-connected structure shown in Figure P5.64 consists of two cold-rolled steel [E = 30,000 ksi; α = 6.5 × 10^−6/°F] bars (1) and a bronze [E = 15,000 ksi; α = 12.2 × 10^−6/°F] bar (2) that are connected at pin D. All three bars have cross-sectional areas of 1.250 in.^2 . Assume an initial

Question

The pin-connected structure shown in Figure P5.64 consists of two cold-rolled steel [E = 30,000 ksi; α = 6.5 × [latex]10^{-6}[/latex]/°F] bars (1) and a bronze [E = 15,000 ksi; α = 12.2 × [latex]10^{-6}[/latex]/°F] bar (2) that are connected at pin D. All three bars have cross-sectional areas of 1.250 [latex] ext { in. }^{2}[/latex] . Assume an initial geometry of a = 10 ft and b = 18 ft. A load of P = 34 kips is applied to the structure at pin D, and the temperature increases by 60°F. Calculate:
(a) the normal stresses in bars (1) and (2).
(b) the downward displacement of pin D.

Accepted Answer

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

[latex]\begin{aligned} an heta &=\frac{b}{a}=\frac{18 ft }{10 ft }=1.8 \quad heta=60.9454^{\circ} \\Sigma F_{x} &=-F_{1} \cos heta+F_{1} \cos heta=0 \\Sigma F_{y} &=F_{1} \sin heta+F_{2}+F_{1} \sin heta-P=0 \& herefore 2 F_{1} \sin heta+F_{2}=P (a)\end{aligned}[/latex]

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:

[latex]\frac{\delta_{1}}{\sin 60.9454^{\circ}}=\delta_{2}[/latex]

(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:

[latex]\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}[/latex] (c)

Compatibility Equation

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

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

Solve the Equations
From the problem statement, [latex]A_{1}=A_{2}[/latex]. The bar lengths are:

[latex]L_{1}=\sqrt{(10 ft )^{2}+(18 ft )^{2}}=20.5913 ft =247.0951 in . \quad L_{2}=18 ft =216 in.[/latex]

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

[latex]\begin{aligned}\frac{F_{2} L_{2}}{A_{2} E_{2}} &=\frac{1}{\sin 60.9454^{\circ}}\left[\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1} ight]-\alpha_{2} \Delta T_{2} L_{2} \F_{2}=& \frac{A_{2} E_{2}}{L_{2} \sin 60.9454^{\circ}}\left[\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1} ight]-\alpha_{2} \Delta T_{2} L_{2} \frac{A_{2} E_{2}}{L_{2}} \=& F_{1}\left(\frac{L_{1}}{L_{2} \sin 60.9454^{\circ}} ight)\left(\frac{E_{2}}{E_{1}} ight)+\left[\frac{\alpha_{1} L_{1}}{L_{2} \sin 60.9454^{\circ}}-\alpha_{2} ight] A_{2} E_{2} \Delta T \=& F_{1}\left(\frac{20.5913 ft }{(18 ft ) \sin 60.9454^{\circ}} ight)\left(\frac{15,000 ksi }{30,000 ksi } ight) \&+\left[\frac{\left(6.5 imes 10^{-6} /{ }^{\circ} F ight)(20.5913 ft )}{(18 ft ) \sin 60.9454^{\circ}}-\left(12.2 imes 10^{-6} /{ }^{\circ} F ight) ight]\left(1.25 in .{ }^{2} ight)(15,000 ksi )\left(60^{\circ} F ight) \=& F_{1}(0.654321)-4.155555 kips\end{aligned}[/latex]

Substitute this result in Eq. (a) and solve for [latex]F_{1}:[/latex]

[latex]\begin{aligned}2 F_{1} \sin 60.9454^{\circ}+F_{1}(0.654321)-4.155555 ext { kips } &=34 ext { kips } \& herefore F_{1}=\frac{38.155555 ext { kips }}{2.402636}=15.8807 ext { kips }\end{aligned}[/latex]

The force in bar (2) is therefore:

[latex]F_{2}=(15.8807 kips )(0.654321)-4.155555 ext { kips }=6.2355 ext { kips }[/latex]

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

[latex]\begin{aligned}&\sigma_{1}=\frac{F_{1}}{A_{1}}=\frac{15.8807 ext { kips }}{1.25 in .^{2}}=12.70 ksi \&\sigma_{2}=\frac{F_{2}}{A_{2}}=\frac{6.2355 kips }{1.25 in .^{2}}=4.99 ksi\end{aligned}[/latex]

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

[latex]\begin{aligned}\delta_{2} &=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}=\frac{(6.2355 kips )(216 in .)}{\left(1.25 in .^{2} ight)(15,000 ksi )}+\left(12.2 imes 10^{-6} /{ }^{\circ} F ight)\left(60^{\circ} F ight)(216 in .)=0.2299 in . \& herefore v_{D}=0.230 in \downarrow\end{aligned}[/latex]

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

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