The following interpolation formula is suggested as a displacement function for deriving the stiffness of a plane stress rectangular element of uniform thickness t shown in Fig. P.6.14.

Question

[latex]u=\frac{1}{4 a b}\left[(a-x)(b-y) u_1+(a+x)(b-y) u_2+(a+x)(b+y) u_3+(a-x)(b+y) u_1 ight][/latex]

Form the strain matrix and obtain the stiffness coefficients [latex]K_{11}[/latex] and [latex]K_{12}[/latex] in terms of the material constants c, d and e defined below.

In the elasticity matrix [D]

[latex]D_{11}=D_{22}=c \quad D_{12}=d \quad D_{33}=e \quad ext { and } \quad D_{13}=D_{23}=0[/latex]

Accepted Answer

For a = 1, b = 2

[latex]u=\frac{1}{8}\left[(1-x)(2-y) u_1+(1+x)(2-y) u_2+(1+x)(2+y) u_3+(1-x)(2+y) u_4 ight][/latex]

Similarly for v
Then

[latex]\begin{gathered}\frac{\partial u}{\partial x}=\frac{1}{8}\left[-(2-y) u_1+(2-y) u_2+(2+y) u_3-(2+y) u_4 ight] \\frac{\partial v}{\partial y}=\frac{1}{8}\left[-(1-x) v_1-(1+x) v_2+(1+x) v_3-(1-x) v_4 ight] \\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=\frac{1}{8}\left[-(1-x) u_1-(2-y) v_1-(1+x) u_2+(2-y) v_2+(1+x) u_3 ight. \\left.+(2+y) v_3+(1-x) u_4-(2+y) v_4 ight]\end{gathered}[/latex]

In matrix form

[latex]\left[\begin{array}{c}\frac{\partial u}{\partial x} \\frac{\partial v}{\partial y} \\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\end{array} ight][/latex]

[latex]=\frac{1}{8}\left[\begin{array}{cccccccc}-(2-y) & 0 & (2-y) & 0 & (2+y) & 0 & -(2+y) & 0 \0 & -(1-x) & 0 & -(1+x) & 0 & (1+x) & 0 & (1-x) \-(1-x) & -(2-y) & -(1+x) & (2-y) & (1+x) & (2+y) & (1-x) & -(2+y)\end{array} ight][/latex] [latex]\left\{\begin{array}{l}u_1 \v_1 \u_2 \v_2 \u_3 \v_3 \u_4 \v_4\end{array} ight\}[/latex]

Also

[latex]D=\left[\begin{array}{lll}c & d & 0 \d & c & 0 \0 & 0 & e\end{array} ight][/latex]

Then
[D][B]

[latex]=\frac{1}{8}\left[\begin{array}{ccccc}-c(2-y) & -d(1-x) & e(2-y) & -d(1+x) & e(2+y) &d(1 + x)&−c(2 + y)& d(1 − x)\-d(2-y) & -c(1-x) & d(2-y) & -c(1+x) & d(2+y) &e(1 + x)&−d(2 + y)&c(1 − x)\-e(1-x) & -e(2-y) & -e(1+x) & e(2-y) & e(1+x) &e(2 + y)&e(1 − x)&−e(2 + y)\end{array} ight][/latex]

Then

[latex][B]^{\mathrm{T}}[D][B]=\frac{1}{64}[/latex] [latex]\left[\begin{array}{ccc}-(2-y) & 0 & -(1-x) \0 & -(1-x) & -(2-y) \& \vdots & \& \vdots & \& \vdots & \& \vdots & \& \vdots & \& \vdots &\end{array} ight][/latex] [latex]\left[\begin{array}{lllll}-c(2-y) & -d(1-x) & \ldots & \ldots & \ldots \-d(2-y) & -c(1-x) & \ldots & \ldots & \ldots \-e(1-x) & -e(2-y) & \ldots & \ldots&\ldots\end{array} ight][/latex]

Therefore

[latex]K_{11}=\frac{t}{64} \int_{-2}^2 \int_{-1}^1\left[c(2-y)^2+e(1-x)^2 ight] \mathrm{d} x \mathrm{~d} y[/latex]

which gives [latex]K_{11}=\frac{t}{6}(4 c+e)[/latex]

[latex]K_{12}=\frac{t}{64} \int_{-2}^2 \int_{-1}^1[d(2-y)(1-x)+e(1-x)(2-y)] \mathrm{d} x \mathrm{~d} y[/latex]

which gives [latex]K_{12}=\frac{t}{4}(d+e)[/latex].

Question p.6.14: The following interpolation formula is suggested as a displa...

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