Three rods of different materials are connected and placed between rigid supports at A and D, as shown in Figure P5.66/67. Properties for each of the three rods are given below. The bars are initially unstressed when the structure is assembled at 20°C. After the temperature has been increased to 100°C, determine:
(a) the normal stresses in the three rods.
(b) the force exerted on the rigid supports.
(c) the deflections of joints B and C relative to rigid support A.

Question

[latex]\begin{array}{|l|l|l|l|}\hline ext { Aluminum (1) } & ext { Cast Iron (2) } & ext { Bronze (3) } \L_{1}=440 mm & L_{2}=200 mm & L_{3}=320 mm \A_{1}=1,200 mm ^{2} & A_{2}=2,800 mm ^{2} & A_{3}=800 mm ^{2} \E_{1}=70 GPa & E_{2}=155 GPa & E_{3}=100 GPa \\alpha_{1}=22.5 imes 10^{-6} /{ }^{\circ} C & \alpha_{2}=13.5 imes 10^{-6} /{ }^{\circ} C & \alpha_{3}=17.0 imes 10^{-6} /{ }^{\circ} C \\hline\end{array}[/latex]

Accepted Answer

Equilibrium
Consider a FBD at joint B. Assume that both internal axial forces will be tension.

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

Similarly, consider a FBD at joint C. Assume that both internal axial forces will be tension.

[latex]\Sigma F_{x}=-F_{2}+F_{3}=0 \quad herefore F_{3}=F_{2}[/latex] (b)

Geometry of Deformations Relationship

[latex]\delta_{1}+\delta_{2}+\delta_{3}=0[/latex] (c)

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} \quad \delta_{3}=\frac{F_{3} L_{3}}{A_{3} E_{3}}+\alpha_{3} \Delta T_{3} L_{3}[/latex] (d)

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}+\frac{F_{3} L_{3}}{A_{3} E_{3}}+\alpha_{3} \Delta T_{3} L_{3}=0[/latex] (e)

Solve the Equations
From Eq. (a), [latex]F_{1}=F_{2}[/latex], and from Eq. (b), [latex]F_{3}=F_{2}[/latex]. The temperature change is the same for all members;
therefore, [latex]\Delta T_{1}=\Delta T_{2}=\Delta T_{3}=\Delta T[/latex]. Eq. (e) then can be written as:

[latex]\frac{\left(F_{2} ight) L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T L_{1}+\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T L_{2}+\frac{\left(F_{2} ight) L_{3}}{A_{3} E_{3}}+\alpha_{3} \Delta T L_{3}=0[/latex]

Solving for [latex]F_{2}[/latex] :

[latex]\begin{gathered}\frac{F_{2} L_{1}}{A_{1} E_{1}}+\frac{F_{2} L_{2}}{A_{2} E_{2}}+\frac{F_{3} L_{3}}{A_{3} E_{3}}=-\alpha_{1} \Delta T L_{1}-\alpha_{2} \Delta T L_{2}-\alpha_{3} \Delta T L_{3} \F_{2}\left[\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}}+\frac{L_{3}}{A_{3} E_{3}} ight]=-\Delta T\left[\alpha_{1} L_{1}+\alpha_{2} L_{2}+\alpha_{3} L_{3} ight] \F_{2}=-\frac{\Delta T\left[\alpha_{1} L_{1}+\alpha_{2} L_{2}+\alpha_{3} L_{3} ight]}{\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}}+\frac{L_{3}}{A_{3} E_{3}}} (f)\end{gathered}[/latex]

Substitute the problem data along with ΔT = +80°C into Eq. (f) and calculate [latex]F_{1}[/latex] = −148.80 kN. From Eq. (a), [latex]F_{1}[/latex] = −148.80 kN and from Eq. (b), [latex]F_{3}[/latex] = −148.80 kN.

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

[latex]\begin{aligned}&\sigma_{1}=\frac{F_{1}}{A_{1}}=\frac{-148,800 N }{1,200 mm ^{2}}=-124.00 MPa =124.0 MPa ( C ) \&\sigma_{2}=\frac{F_{2}}{A_{2}}=\frac{-148,800 N }{2,800 mm ^{2}}=-53.143 MPa =53.1 MPa ( C ) \&\sigma_{3}=\frac{F_{3}}{A_{3}}=\frac{-148,800 N }{800 mm ^{2}}=-186.0 MPa =186.0 MPa ( C )\end{aligned}[/latex]

(b) Force on Rigid Supports
The force exerted on the rigid supports is equal to the internal axial force:

[latex]R_{A}=R_{D}=148.8 kN[/latex]

(c) Deflection of Joints B and C
The deflection of joint B is equal to the deformation (i.e., contraction in this instance) of rod (1). The deformation of rod (1) is given by:

[latex]\begin{aligned}\delta_{1} &=\frac{F_{1} L_{1}}{A_{1} E_{1}}+\alpha_{1} \Delta T_{1} L_{1} \&=\frac{(-148,800 N )(440 mm )}{\left(1,200 mm ^{2} ight)\left(70,000 N / mm ^{2} ight)}+\left(22.5 imes 10^{-6} /{ }^{\circ} C ight)\left(80^{\circ} C ight)(440 mm )=0.01257 mm\end{aligned}[/latex]

The deflection of joint B is thus:

[latex]u_{B}=\delta_{1}=0.01257 mm ightarrow[/latex]

The deformation of rod (2) is given by:

[latex]\begin{aligned}\delta_{2} &=\frac{F_{2} L_{2}}{A_{2} E_{2}}+\alpha_{2} \Delta T_{2} L_{2} \&=\frac{(-148,800 N )(200 mm )}{\left(2,800 mm ^{2} ight)\left(155,000 N / mm ^{2} ight)}+\left(13.5 imes 10^{-6} /{ }^{\circ} C ight)\left(80^{\circ} C ight)(200 mm )=0.14743 mm\end{aligned}[/latex]

The deflection of joint C is:

[latex]u_{C}=u_{B}+\delta_{2}=0.01257 mm +0.14743 mm =0.1600 mm ightarrow[/latex]

Question 5.67: Three rods of different materials are connected and placed b...

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