A 0.75-in.-thick polymer [E = 470,000 psi; ν = 0.37] casting is subjected to biaxial stresses of $\sigma_{x}$ = 2,500 psi and $\sigma_{y}$ = 8,300 psi, acting in the directions shown in Figure P13.50. The dimensions of the casting are b = 12.0 in. and h = 8.0 in. Determine
(a) the change in length of edges AB and AD.
(b) the change in length of diagonal AC.
(c) the change in thickness of the plate.

Question

A 0.75-in.-thick polymer [E = 470,000 psi; ν = 0.37] casting is subjected to biaxial stresses of [latex]\sigma_{x}[/latex] = 2,500 psi and [latex]\sigma_{y}[/latex] = 8,300 psi, acting in the directions shown in Figure P13.50. The dimensions of the casting are b = 12.0 in. and h = 8.0 in. Determine
(a) the change in length of edges AB and AD.
(b) the change in length of diagonal AC.
(c) the change in thickness of the plate.

Accepted Answer

(a) From the generalized Hooke’s Law equations for plane stress, the normal strains produced in the plate can be computed from Eqs. (13.21):

[latex]\begin{aligned}&\varepsilon_{x}=\frac{1}{E}\left(\sigma_{x}-v \sigma_{y} ight)=\frac{1}{470,000 psi }[2,500 psi -(0.37)(-8,300 psi )]=11,853.19 imes 10^{-6} ext { in./in. } \&\varepsilon_{y}=\frac{1}{E}\left(\sigma_{y}-v \sigma_{x} ight)=\frac{1}{470,000 psi }[-8,300 psi -(0.37)(2,500 psi )]=-19,627.66 imes 10^{-6} ext { in. } / in . \&\varepsilon_{z}=-\frac{v}{E}\left(\sigma_{x}+\sigma_{y} ight)=-\frac{0.37}{470,000 psi }[2,500 psi +(-8,300 psi )]=4,565.96 imes 10^{-6} ext { in. } / in .\end{aligned} [/latex]

Plate edge AB is aligned with the x direction; therefore, the change in length of edge AB can be computed from the product of [latex]\varepsilon_{x}[/latex] and plate dimension b:

[latex]\delta_{A B}=\varepsilon_{x} b=\left(11,853.19 imes 10^{-6} ext { in./in.)(12 in. } ight)=0.1422 in .[/latex]

Plate edge AD is aligned with the y direction; therefore, the change in length of edge AD can be computed from the product of [latex]\varepsilon_{y}[/latex] and plate dimension h:

[latex]\delta_{A D}=\varepsilon_{y} h=\left(-19,627.66 imes 10^{-6} in . / in . ight)(8.00 in .)=-0.1570 in.[/latex]

(b) A strain transformation equation [Eq. (13.3)]

[latex]\varepsilon_{n}=\varepsilon_{x} \cos ^{2} heta+\varepsilon_{y} \sin ^{2} heta+\gamma_{x y} \sin heta \cos heta[/latex]

can be written to determine the normal strain in the direction of diagonal AC. Since there is no shear stress acting on the plate, we know that [latex]\gamma_{x y}=0[/latex] (since Hooke’s Law relating shear stress and shear strain is [latex] au_{x y}=G \gamma_{x y}[/latex]).

The angle θ between edge AB and diagonal AC is:

[latex] an heta=\frac{8.00 in .}{12.00 in .}=0.6667 \quad herefore heta=33.690^{\circ}[/latex]

Therefore, the normal strain in the direction of diagonal AC is:

[latex]\begin{aligned}\varepsilon_{A C} &=\left(11,853.19 imes 10^{-6} in . / in . ight) \cos ^{2}\left(33.690^{\circ} ight)+\left(-19,627.66 imes 10^{-6} in . / in . ight) \sin ^{2}\left(33.690^{\circ} ight) \&=2,166.77 imes 10^{-6} in . / in .\end{aligned}[/latex]

The initial length of diagonal AC is:

[latex]L_{A C}=\sqrt{(12.00 in .)^{2}+(8.00 in .)^{2}}=14.42221 in.[/latex]

The change in length of diagonal AC is computed from the product of [latex]\varepsilon_{A C}[/latex] and the initial diagonal length:

[latex]\delta_{A C}=\varepsilon_{A C} L_{A C}=\left(2,166.77 imes 10^{-6} in . / in. ight)(14.42221 in .)=0.0328 in .[/latex]

Alternate Method: The change in length of diagonal AC can also be computed using the Pythagorean theorem. After deformation, the final length of side AB of the plate is 12.14224 in. The final length of side AD is 7.84298 in. From the Pythagorean theorem, the deformed length of diagonal AC is:

[latex]L_{A C}^{\prime}=\sqrt{(12.14224 in .)^{2}+(7.84298 in .)^{2}}=14.45497 in.[/latex]

and therefore, the elongation of diagonal AC is

[latex]\delta_{A C}=14.45497 ext { in. }-14.42221 in. =0.00328 in .[/latex]

(c) The change in plate thickness is computed from the product of [latex]\varepsilon_{z}[/latex] and the plate thickness:

[latex]\delta_{ ext {thick }}=\varepsilon_{z}( ext { thickness })=\left(4,565.96 imes 10^{-6} ext { in. } / in . ight)(0.750 in .)=0.00342 in .[/latex]

Question 13.50: A 0.75-in.-thick polymer [E = 470,000 psi; ν = 0.37] casting...

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