A concentrated load P is supported by two bars as shown in Figure P3.23. Bar (1) is made of cold-rolled red brass [E = 16,700 ksi; α = 10.4 × 10^−6 /°F] and has a cross-sectional area of 0.225 in.^2 . Bar (2) is made of 6061-T6 aluminum [E = 10,000 ksi; α = 13.1 × 10^−6 /°F] and has a

Question

A concentrated load P is supported by two bars as shown in Figure P3.23. Bar (1) is made of cold-rolled red brass [E = 16,700 ksi; α = 10.4 × [latex]10^{-6}[/latex] /°F] and has a cross-sectional area of 0.225 [latex]in. ^{2}[/latex]. Bar (2) is made of 6061-T6 aluminum [E = 10,000 ksi; α = 13.1 × [latex]10^{-6}[/latex] /°F] and has a crosssectional area of 0.375 [latex]in. ^{2}[/latex] . After load P has been applied and the temperature of the entire assembly has increased by 50°F, the total strain in bar (1) is measured as 1,400 με (elongation). Determine:
(a) the magnitude of load P.
(b) the total strain in bar (2).

Accepted Answer

Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore, the equilibrium equations can be written as:

[latex]\begin{aligned}&\Sigma F_{x}=F_{2} \cos 55^{\circ}-F_{1} \cos 40^{\circ}=0 (a) \&\Sigma F_{y}=F_{2} \sin 55^{\circ}+F_{1} \sin 40^{\circ}-P=0 (b)\end{aligned}[/latex]

From Eq. (a):

[latex]F_{2}=F_{1} \frac{\cos 40^{\circ}}{\cos 55^{\circ}}[/latex] (c)

Substitute this expression into Eq. (b) to obtain:

[latex]\begin{aligned}\left[F_{1} \frac{\cos 40^{\circ}}{\cos 55^{\circ}} ight] \sin 55^{\circ}+F_{1} \sin 40^{\circ} &=P \F_{1}\left[\cos 40^{\circ} an 55^{\circ}+\sin 40^{\circ} ight] &=P \& herefore P=F_{1}[1.736812] (d)\end{aligned}[/latex]

(a) The total strain in bar (1) is measured as 1,400 με (elongation). Part of this strain is due to the stress acting in the bar and part of this strain is due to the temperature increase. The strain caused by the temperature change is

[latex]\varepsilon_{T}=\alpha \Delta T=\left(10.4 imes 10^{-6} /{ }^{\circ} F ight)\left(50^{\circ} F ight)=520 imes 10^{-6} ext { in. } / ext { in. }[/latex]

Since the total strain is [latex]\varepsilon=1,400 \mu \varepsilon=0.001400 ext { in./in. }[/latex] , the strain caused by the stress in bar (1) must be:

[latex]\varepsilon_{\sigma}=\varepsilon-\varepsilon_{T}=0.001400 ext { in./in. }-0.000520 ext { in./in. }=0.000880 ext { in./in. }[/latex]

From Hooke’s law, the stress in bar (1) is therefore:

[latex]\sigma_{1}=E \varepsilon_{1}=(16,700 ksi )(0.000880 ext { in./in. })=14.696 ksi[/latex]

and thus, the force in bar (1) is:

[latex]F_{1}=\sigma_{1} A_{1}=(14.696 ksi )\left(0.225 in. ^{2} ight)=3.3066 kips[/latex]

From Eq. (d), the applied load P is:

[latex]P=F_{1}[1.736812]=(3.3066 kips )(1.736812)=5.7429 kips =5.74 kips[/latex]

(b) From Eq. (c), the force in bar (2) is:

[latex]F_{2}=F_{1} \frac{\cos 40^{\circ}}{\cos 55^{\circ}}=(3.3066 kips )(1.335558)=4.4162 kips[/latex]

The normal stress in bar (2) is therefore:

[latex]\sigma_{2}=\frac{F_{2}}{A_{2}}=\frac{4.4162 kips }{0.375 in .{ }^{2}}=11.7764 ksi[/latex]

The normal strain due to this stress is:

[latex]\varepsilon_{2}=\frac{\sigma_{2}}{E_{2}}=\frac{11.7764 ksi }{10,000 ksi }=1,177.64 imes 10^{-6} ext { in./in. }[/latex]

The strain caused in bar (2) by the temperature change is:

[latex]\varepsilon_{T}=\alpha \Delta T=\left(13.1 imes 10^{-6} /{ }^{\circ} F ight)\left(50^{\circ} F ight)=655 imes 10^{-6} ext { in. } / ext { in. }[/latex]

and thus, the total strain in bar (2) is:

[latex]\begin{aligned}\varepsilon&=\varepsilon_{\sigma}+\varepsilon_{T}=1,177.64 imes 10^{-6} ext { in./in. }+655 imes 10^{-6} ext { in./in. } \&=1,832.64 imes 10^{-6} ext { in. } / ext { in. }=1,833 \mu \varepsilon\end{aligned}[/latex]

Question 3.23: A concentrated load P is supported by two bars as shown in F...

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