Skip to main content

Thin plate: nodal displacements and stresses

Short answer

Model as one CST in plane stress with node 1 pinned and node 2 restrained horizontally. Free displacements are v2=-3.556e-4 in, u3=1.022e-3 in, v3=-2.667e-3 in. Element stresses are sigma_x=1333.3 psi, sigma_y=-666.7 psi, tau_xy=-1333.3 psi.

A thin plate is subjected to the loads as shown in the figure below. The plate thickness is 0.3 in and the other dimensions are shown in the figure. Given that the Poisson's ratio =0.3= 0.3 and the modulus of elasticity E=30×106E = 30\times10^6 psi, determine the nodal load displacements and the elemental stresses.

[Figure: triangular plate with node 1 at the bottom-left (pinned), node 2 directly above node 1 at 10 in (supported), node 3 at 20 in to the right of node 1 and 10 in above it; at node 3 a 2000 lb horizontal load (to the right) and a 2000 lb downward load]

The plate is a single CST in plane stress. Units: lb and in.

Assumptions

  • Node coordinates (in): 1 (0, 0), 2 (0, 10), 3 (20, 10). Nodes are taken in anticlockwise order 1-3-2 (i=1, j=3, m=2i=1,\ j=3,\ m=2) so that the area is positive.
  • Node 1 is pinned (u1=v1=0u_1=v_1=0); node 2 is on a roller that stops horizontal movement (u2=0u_2=0). Free dofs: v2, u3, v3v_2,\ u_3,\ v_3.
  • Loads at node 3: F3x=+2000F_{3x}=+2000 lb, F3y=−2000F_{3y}=-2000 lb.
  • E=30×106E=30\times10^6 psi, ν=0.3\nu=0.3, t=0.3t=0.3 in.

Step 1: Geometry

A=12(20)(10)=100 in2A=\tfrac12(20)(10)=100\ \text{in}^2

bi=yj−ym=0, bj=ym−yi=10, bm=yi−yj=−10b_i=y_j-y_m=0,\ b_j=y_m-y_i=10,\ b_m=y_i-y_j=-10; ci=xm−xj=−20, cj=xi−xm=0, cm=xj−xi=20c_i=x_m-x_j=-20,\ c_j=x_i-x_m=0,\ c_m=x_j-x_i=20.

[B]=1200[00100−1000−2000020−20001020−10][B]=\frac{1}{200}\begin{bmatrix}0&0&10&0&-10&0\\0&-20&0&0&0&20\\-20&0&0&10&20&-10\end{bmatrix}

(columns in the order u1,v1,u3,v3,u2,v2u_1,v_1,u_3,v_3,u_2,v_2).

Step 2: Elasticity matrix

[D]=E1−ν2[1ν0ν10001−ν2]=[3.2967×1079890110098901103.2967×1070001.1538×107] psi[D]=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\ \nu&1&0\\0&0&\frac{1-\nu}{2}\end{bmatrix}=\begin{bmatrix} 3.2967\times10^{7} & 9890110 & 0 \\ 9890110 & 3.2967\times10^{7} & 0 \\ 0 & 0 & 1.1538\times10^{7} \end{bmatrix}\ \text{psi}

Step 3: Stiffness and reduced equations

[k]=tA[B]T[D][B][k]=tA[B]^T[D][B] is assembled in global dof order (u1,v1,u2,v2,u3,v3)(u_1,v_1,u_2,v_2,u_3,v_3). Removing the rows and columns of u1,v1,u2u_1,v_1,u_2:

[1.0755×1071483516−865385148351624725270−8653850865385][v2u3v3]=[02000−2000]\begin{bmatrix} 1.0755\times10^{7} & 1483516 & -865385 \\ 1483516 & 2472527 & 0 \\ -865385 & 0 & 865385 \end{bmatrix}\begin{bmatrix}v_2\\u_3\\v_3\end{bmatrix}=\begin{bmatrix}0\\2000\\-2000\end{bmatrix}

Step 4: Nodal displacements

v2=−3.556×10−4 in,u3=1.022×10−3 in,v3=−2.667×10−3 inv_2=-3.556\times10^{-4}\ \text{in},\quad u_3=1.022\times10^{-3}\ \text{in},\quad v_3=-2.667\times10^{-3}\ \text{in}

Reactions (check): R1x=4000R_{1x}=4000, R1y=2000R_{1y}=2000, R2x=−6000R_{2x}=-6000 lb. Then ΣFx=4000−6000+2000=0\Sigma F_x = 4000-6000+2000=0 and ΣFy=2000−2000=0\Sigma F_y = 2000-2000=0.

Step 5: Element stresses, {σ}=[D][B]{u}\{\sigma\}=[D][B]\{u\}

σx=1333.3 psi,σy=−666.7 psi,τxy=−1333.3 psi\sigma_x=1333.3\ \text{psi},\quad \sigma_y=-666.7\ \text{psi},\quad \tau_{xy}=-1333.3\ \text{psi}

Principal stresses: σ1=2000.0\sigma_1=2000.0 psi, σ2=−1333.3\sigma_2=-1333.3 psi, at θp=−26.57∘\theta_p=-26.57^\circ from the xx-axis.

Answer: v2=−3.556×10−4v_2=-3.556\times10^{-4} in, u3=1.022×10−3u_3=1.022\times10^{-3} in, v3=−2.667×10−3v_3=-2.667\times10^{-3} in; σx=1333.3\sigma_x=1333.3, σy=−666.7\sigma_y=-666.7, τxy=−1333.3\tau_{xy}=-1333.3 psi.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Computational Techniques in Civil Engineering chapter 4 questions →