Question 17.CS.2: Steel Plate in Tension A cantilever plate of depth h, length......

The discretized plate is depicted in Figure 17.15(b). The origin of coordinates is placed at node 1, for convenience; however, it may be located at any point in the x, y plane. The area of each element is

A=\frac{1}{2} h L=\frac{1}{2}(10)(20)=100 \text { in. }^2

The statically equivalent forces at nodes 2 and 3,  [ 4(10 × 1 / 2) / 2=10  kips] , are shown in the figure. For plane stress, elasticity matrix [D] is given by Equation (17.29a).

[D]=\frac{E}{1-\nu^2}\left[\begin{array}{ccc} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & (1-\nu) / 2 \end{array}\right]      (17.29a)

Stiffness matrix. For element a , on assigning i=1, j=3 \text {, and } m=4 \text {, } Equation (17.34) gives

\begin{array}{ccc} a_i=x_j y_m-y_j x_m & a_j=y_i x_m-x_i y_m & a_m=x_i y_j-y_i x_j \\ b_i=y_j-y_m & b_j=y_m-y_i & b_m=y_i-y_j \\ c_i=x_m-x_j & c_j=x_i-x_m & c_m=x_j-x_i \end{array}     (17.34)

\begin{aligned} & b_1=y_3-y_4=10-10=0 \\ & b_3=y_4-y_1=10-0=10 \\ & b_4=y_1-y_3=0-10=-10 \\ & c_1=x_3-x_4=0-20=-20 \\ & c_3=x_1-x_4=0-0=0 \\ & c_4=x_3-x_1=20-0=20 \end{aligned}         (a)

Substitution of these and the given data into Equation (17.40), after performing the matrix multiplications, results in stiffness matrix [k]_a . Similarly, for element b , assignment of i =1, j =2, and m =3 into Equation (17.34) leads to

[K]_e=[B]^T[D][B] t A      (17.40)

\begin{aligned} & b_1=y_2-y_3=0-10=-10 \\ & b_2=y_3-y_1=10-0=10 \\ & b_3=y_1-y_2=0-0=0 \\ & c_1=x_3-x_2=20-20=0 \\ & c_2=x_1-x_3=0-20=-20 \\ & c_3=x_2-x_1=20-0=20 \end{aligned}       (b)

and [k]_b is determined. The displacements u_2, \upsilon _2 and u_4, \upsilon _4 are not involved in elements a and b , respectively. So, before summing [k]_a and [k]_b to form the system matrix, rows and columns of zeros must be added to each element matrix to account for the absence of these displacements, as mentioned in Section 17.3. Finally, superimposition of the resulting matrices gives the system matrix [K] .

Nodal displacements. The boundary conditions are u_1=\upsilon_1=u_4=\upsilon_4=0 . The force displacement relationship of the system is

\left\{\begin{array}{c} R_{1 x} \\ R_{1 y} \\ 10 \\ 0 \\ 10 \\ 0 \\ R_{4 x} \\ R_{4 y} \end{array}\right\}=[K]\left\{\begin{array}{c} 0 \\ 0 \\ u_2 \\ \upsilon_2 \\ u_3 \\ \upsilon_3 \\ 0 \\ 0 \end{array}\right\}

Next, to compare the quantities involved, we introduce the results without going through the computation of the [K] . It can be verified [2] that the preceding derivations yield

\left\{\begin{array}{c} 10 \\ 0 \\ 10 \\ 0 \end{array}\right\}=\frac{187.5}{0.91}\left[\begin{array}{cccc} 48 & 0 & -28 & 14 \\ 0 & 87 & 12 & -80 \\ -28 & 12 & 48 & -26 \\ 14 & -80 & -26 & 87 \end{array}\right]\left\{\begin{array}{l} u_2 \\ \upsilon_2 \\ u_3 \\ \upsilon_3 \end{array}\right\}

Solving,

\left\{\begin{array}{l} u_2 \\ \upsilon_2 \\ u_3 \\ \upsilon_3 \end{array}\right\}=\left\{\begin{array}{l} 2.4383 \\ 0.0163 \\ 2.6548 \\ 0.4261 \end{array}\right\}\left(10^{-3}\right) in .       (b)

Stresses. For element a , carrying Equations (a) and (b) into (17.37), we obtain the strain matrix \{\varepsilon\}_a . Equation (17.28), [D] \{\varepsilon\}_a , then results in

\left\{\begin{array}{l} \varepsilon_4 \\ \varepsilon_5 \\ \gamma_{x y} \end{array}\right\}=\frac{1}{2 A}\left[\begin{array}{cccccc} b_i & 0 & b_i & 0 & b_m & 0 \\ 0 & c_i & 0 & c_j & 0 & c_m \\ c_i & b_i & c_j & b_j & c_m & b_m \end{array}\right]\{\delta\}_e       (17.37)

\{\sigma\}_e=\frac{E}{1-\nu^2}\left[\begin{array}{ccc} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & (1-\nu) / 2 \end{array}\right]\left\{\begin{array}{c} \\ε_x \\ \\ε_y \\ \gamma_{x y} \end{array}\right\}_e        (17.28a)

\{\sigma\}_e=[D]\{\varepsilon\}_e       (17.28b)

\left\{\begin{array}{c} \sigma_x \\ \sigma_y \\ \tau_{x y} \end{array}\right\}_a=\left\{\begin{array}{c} 4020 \\ 1204 \\ 9.6 \end{array}\right\} psi

Element b is treated in a like manner.

Comments: Due to constant x -directed stress of 4000 psi applied on the edge of the plate, the normal stress is expected to be about 4000 psi in the element a (or b ). The foregoing result for \sigma _x is therefore quite good. Interestingly, the support of the element a at nodes 1 and 4 causes a relatively high stress of \sigma _y =1204 psi. Also note that the value of shear stress \tau _{xy} is negligibly small, as anticipated.