A constant strain triangular element has corners 1(0,0), 2(4,0) and 3(2,2) and is 1 unit thick. If the elasticity matrix [D] has elements D11 = D22 = a, D12 = D1 = b, D13 = D23 = D31 = D32 = 0 and D33 = c derive the stiffness matrix for the element.

Question

A constant strain triangular element has corners 1(0,0), 2(4,0) and 3(2,2) and is 1 unit thick. If the elasticity matrix [D] has elements [latex]D_{11}=D_{22}=a, D_{12}=D_1=b[/latex], [latex]\mathrm{D}_{13}=D_{23}=D_{31}=D_{32}=0 ext { and } \mathrm{D}_{33}=c[/latex] derive the stiffness matrix for the element.

Accepted Answer

Assume displacement functions

[latex]\begin{aligned}&u(x, y)=\alpha_1+\alpha_2 x+\alpha_3 y \&v(x, y)=\alpha_4+\alpha_5 x+\alpha_6 y\end{aligned}[/latex]

Then

[latex]\begin{aligned}&u_1=\alpha_1 \&u_2=\alpha_1+4 \alpha_2 \&u_3=\alpha_1+2 \alpha_2+2 \alpha_3\end{aligned}[/latex]

Solving

[latex]\alpha_2=\frac{u_2-u_1}{4} \quad \alpha_3=\frac{2 u_3-u_1-u_2}{4}[/latex]

Therefore

[latex]u=u_1+\left(\frac{u_2-u_1}{4} ight) x+\left(\frac{2 u_3-u_1-u_2}{4} ight) y[/latex]

or

[latex]u=\left(1-\frac{x}{4}-\frac{y}{4} ight) u_1+\left(\frac{x}{4}-\frac{y}{4} ight) u_2+\frac{y}{2} u_3[/latex]

Similarly

[latex]v=\left(1-\frac{x}{4}-\frac{y}{4} ight) v_1+\left(\frac{x}{4}-\frac{y}{4} ight) v_2+\frac{y}{2} v_3[/latex]

Then, from Eqs (1.18) and (1.20)

[latex]\left.\begin{array}{rl}\varepsilon_x & =\frac{\partial u}{\partial x} \\varepsilon_y & =\frac{\partial v}{\partial y} \\varepsilon_z & =\frac{\partial w}{\partial z}\end{array} ight\}[/latex] (1.18)

[latex]\left.\begin{array}{l}\gamma_{x z}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z} \\gamma_{x y}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} \\gamma_{y z}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}\end{array} ight\}[/latex] (1.20)

[latex]\begin{aligned}\varepsilon_x &=\frac{\partial u}{\partial x}=-\frac{u_1}{4}+\frac{u_2}{4} \\varepsilon_y &=\frac{\partial v}{\partial y}=-\frac{v_1}{4}-\frac{v_2}{4}+\frac{v_3}{2} \\gamma_{x y} &=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=-\frac{u_1}{4}-\frac{u_2}{4}-\frac{v_1}{4}+\frac{v_2}{4}\end{aligned}[/latex]

Hence

[latex][B]\left\{\delta^e ight\}=\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]=\frac{1}{4}\left[\begin{array}{rrrrrc}-1 & 0 & 1 & 0 & 0 & 0 \0 & -1 & 0 & -1 & 0 & 2 \-1 & -1 & -1 & 1 & 2 & 0\end{array} ight]\left\{\begin{array}{l}u_1 \v_1 \u_2 \v_2 \u_3 \v_3\end{array} ight\}[/latex]

But

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

so that

[latex][D][B]=\frac{1}{4}\left[\begin{array}{rrrrrr}-a & -b & a & -b & 0 & 2 b \-b & -a & b & -a & 0 & 2 a \-c & -c & -c & c & 2 c & 0\end{array} ight][/latex]

and

[latex][B]^{\mathrm{T}}[D][B]=\frac{1}{16}\left[\begin{array}{cccccc}a+c & b+c & -a+c & b-c & -2 c & -2 b \b+c & a+c & -b+c & a-c & -2 c & -2 a \-a+c & -b+c & a+c & -b-c & -2 c & 2 b \b-c & a-c & -b-c & a+c & 2 c & -2 a \-2 c & -2 c & -2 c & 2 c & 4 c & 0 \-2 b & -2 a & 2 b & -2 a & 0 & 4 a\end{array} ight][/latex]

[latex] ext { Since }\left[K^e ight]=[B]^{\mathrm{T}}[D][B] imes 4 imes 1[/latex]

[latex]\left[K^e ight]=\frac{1}{4}\left[\begin{array}{cccccc}a+c & & & & & ext { SYM } \b+c & a+c & & & & \-a+c & -b+c & a+c & & & \b-c & a-c & -b-c & a+c & & \-2 c & -2 c & -2 c & 2 c & 4 c & \-2 b & -2 a & 2 b & -2 a & 0 & 4 a\end{array} ight][/latex]

Question p.6.13: A constant strain triangular element has corners 1(0,0), 2(4...

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