Mar APR May 5 2013 2014 2015
1 capture 5 Apr 14 - 5 Apr 14

Close Help

  
Transition elements, BEAM71S1

Problem description: The cantilever beam shown is subjected to uniformly distributed loading l_y, l_z per unit length and the end force F_x, F_y, F_z. Calculate deflection and stresses for two load cases: i) the in-plane problem using l_y, F_x, F_y only and ii) the complete three-dimensional loading l_y, l_z, F_x, F_y, F_z.

Mesh: Two hexahedra, two semi-loofs--see appendix B.4.

Material properties: $E=2\times 10^5$ MPa, $\nu=0.3$.

Support: Clamped at x=0. Statically determinate.

u=v=w=0 node:	 1 
u=w=0 nodes: 	 2 25       
u=v=0 nodes: 	 4 28       
u=0 nodes: 	 3 26 27      

Loading: Given as

ly=-100 lz=200 unit: N/m 
Fx=20000 Fy=-20 Fz=10 unit: N      

Solution: The end force acts on edge L3 of the semi-loof element number 4, therefore it is replaced with the equivalent edge traction as in example II.5

$$l_x=\frac{20000}{0.02}=1000\times 10^3\,\mbox{[N/m]}~,~~~ l_y=-10^3\,\mbox{[N/m]} ~,~~~ l_z=0.5\times 10^3\,\mbox{[N/m]}$$

The distributed loading in y direction acts on faces S3 of the hexahedral elements number 1, 3 and on edges L2 of semi-loofs 2, 4. Following examples II.4, II.5 we replace this loading with the surface traction

$$q_n(\verb*\vert S3\vert)=\frac{l_y}{0.01}=-0.01\times 10^6\,\mbox{[Pa]}$$

for elements 1, 3 and the edge traction

$$l_{xh}(\verb*\vert L2\vert)=l_y=-100\,\mbox{[N/m]}$$

for elements 2 and 4.

Finally, the distributed loading in z direction is replaced with the surface traction

$$q_n(\verb*\vert S2\vert)=q_n(\verb*\vert S1\vert)= -\frac{l_z}{0.02}=-0.01\times 10^6\,\mbox{[Pa]}$$

acting on faces S2, S1 of elements 1, 3 and 2, 4, respectively.
 

Execute from prompt:
>rmd3 beam71s1.I1
>rpd3 beam71s1.I2
>srh3 beam71s1.I3
>fefs beam71s1.I4
>str3 beam71s1.I5