A steel [E = 30,000 ksi, α = 6.6 ×10^−6/°F] pipe column (1) with a crosssectional area of A1 = 5.60 in.^2 is connected at flange B to an aluminum alloy [E = 10,000 ksi, α = 12.5 × 10^−6/°F] pipe (2) with a crosssectional area of A2 = 4.40 in.^2 . The assembly (shown in Figure P5.60) is connected to

Question

A steel [E = 30,000 ksi, α = 6.6 ×[latex]10^{-6}[/latex]/°F] pipe column (1) with a crosssectional area of [latex]A_{1}=5.60 in. ^{2}[/latex] is connected at flange B to an aluminum alloy [E = 10,000 ksi, α = 12.5 × [latex]10^{-6}[/latex]/°F] pipe (2) with a crosssectional area of [latex]A_{2}=4.40 in. ^{2}[/latex] . The assembly (shown in Figure P5.60) is connected to rigid supports at A and C. It is initially unstressed at a temperature of 90°F.
(a) At what temperature will the normal stress in steel pipe (1) be reduced to zero?
(b) Determine the normal stresses in steel pipe (1) and aluminum pipe (2) when the temperature reaches –10°F.

Accepted Answer

Equilibrium: Consider a FBD of flange B. Sum forces in the horizontal direction to obtain:

[latex]\Sigma F_{x}=-F_{1}+F_{2}-60 kips =0[/latex] (a)

Geometry of Deformations:

[latex]\delta_{1}+\delta_{2}=0[/latex] (b)

Force-Temperature-Deformation Relationships:

[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 Eqs. (c) into Eq. (b) to derive the compatibility equation:

[latex]\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1}+\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2}=0[/latex] (d)

Solve the Equations: Set [latex]F_{1}[/latex] = 0 and solve Eq. (a) to find [latex]F_{2}[/latex] = 60 kips. Substitute these values for[latex]F_{1}[/latex] and [latex]F_{2}[/latex] into Eq. (d) along with the observation that the temperature change for both axial members is the same (i.e., [latex]\Delta T_{1}=\Delta T_{2}=\Delta T[/latex]) and solve for ΔT:

[latex]\Delta T=\frac{-\frac{F_{1} L_{1}}{A_{1} E_{1}}-\frac{F_{2} L_{2}}{A_{2} E_{2}}}{\alpha_{1} L_{1}+\alpha_{2} L_{2}}=\frac{0-\frac{(60 kips )(144 in .)}{\left(4.40 in .^{2} ight)(10,000 ksi )}}{\left(6.6 imes 10^{-6} /{ }^{\circ} F ight)(120 in .)+\left(12.5 imes 10^{-6} /{ }^{\circ} F ight)(144 in .)}=-75.758^{\circ} F[/latex]

Since the pipes are initially at a temperature 90°F, the temperature at which the normal stress in steel pipe (1) is reduced to zero is

[latex]T=90^{\circ} F -75.748^{\circ} F =14.24^{\circ} F[/latex]

(b) Solve Eq. (a) for [latex]F_{2}[/latex] to obtain

[latex]F_{2}=F_{1}+60 kips[/latex] (e)

When the temperature reaches −10°F, the total change in temperature is ΔT = −100°F. Substitute this value along with Eq. (e) into the compatibility equation [Eq. (d)] and derive an expression for [latex]F_{1}[/latex]:

[latex]\begin{gathered}\frac{F_{1} L_{1}}{A_{1} E_{1}}+\frac{\left(F_{1}+60 kips ight) L_{2}}{A_{2} E_{2}}=-\alpha_{1} \Delta T L_{1}-\alpha_{2} \Delta T L_{2} \F_{1}\left[\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}} ight]+\frac{(60 kips ) L_{2}}{A_{2} E_{2}}=-\Delta T\left[\alpha_{1} L_{1}+\alpha_{2} L_{2} ight] \F_{1}\left[\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}} ight]=-\Delta T\left[\alpha_{1} L_{1}+\alpha_{2} L_{2} ight]-\frac{(60 kips ) L_{2}}{A_{2} E_{2}} \F_{1}=\frac{-\Delta T\left[\alpha_{1} L_{1}+\alpha_{2} L_{2} ight]-\frac{(60 kips ) L_{2}}{A_{2} E_{2}}}{\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}}}\end{gathered}[/latex]

and compute [latex]F_{1}[/latex]:

[latex]F_{1}=\frac{-\left(-100^{\circ} F ight)\left[\left(6.6 imes 10^{-6} ight)(120 in .)+\left(12.5 imes 10^{-6} ight)(144 in .) ight]-\frac{(60 kips )(144 in .)}{\left(4.40 in.^{2} ight)(10,000 ksi )}}{\frac{120 in .}{\left(5.60 in .^{2} ight)(30,000 ksi )}+\frac{144 in .}{\left(4.40 in .^{2} ight)(10,000 ksi )}}[/latex]

[latex]\begin{aligned}&=\frac{0.259200 in. -0.196364 in .}{714.2857 imes 10^{-6} in . / kip +3,272.7273 imes 10^{-6} in . / kip }=\frac{0.062836 in .}{3,987.0130 imes 10^{-6} in . / kip } \&=15.7602 kips\end{aligned}[/latex]

From Eq. (a), [latex]F_{2}[/latex] has a value of

[latex]F_{2}=F_{1}+60 kips =15.7602 kips +60 kips =75.7602 kips[/latex]

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

[latex]\begin{aligned}&\sigma_{1}=\frac{F_{1}}{A_{1}}=\frac{15.7602 kips }{5.60 in .^{2}}=2.8143 ksi =2.81 ksi ( T ) \&\sigma_{2}=\frac{F_{2}}{A_{2}}=\frac{75.7602 kips }{4.40 in .^{2}}=17.2182 ksi =17.22 ksi ( T )\end{aligned}[/latex]

Question 5.60: A steel [E = 30,000 ksi, α = 6.6 ×10^−6/°F] pipe column (1) ...

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