# SS-V: 5060 Separate Beams

Test No. VNL07 Find total or partial beam separation/slippage depending on the applied loads.

## Definition

Two similar beams with dimensions 10 x 10 x 100 mm are in contact (Figure 1).

**Properties****Value**- Friction Coefficient
- $f=0.15$
- Modulus of Elasticity
- 2.1e+11 Pa
- Poisson's Ratio
- 0.3

Load Case | Fx, [N] | Fy, [N] |
---|---|---|

1 | 0 | 1 |

2 | 0 | 0.9 |

3 | 0.3 | 0 |

4 | 0.31 | 0 |

5 | 0.29 | 0 |

## Results

- Case 1:Consider the equilibrium of the upper beam. If $Fy$ =0, then force N is equilibrated by the reaction force in contact. As $Fy$ grows, the beam deforms, and the contact separation starts at its right end and expands to the left. Ultimately, the contact fully separates and contact area degenerates into a line (point A in Figure 3). The value of force $Fy$ which results in full separation can be found from moments equilibrium equation.
(1) $$\sum M\left(A\right)=P*\left(\frac{L}{2}\right)-Fy*L=0$$This ultimate value is $Fy$ =1 N.

SimSolid result for this value of $Fy$ is shown in Figure 4. Full separation occurred and the contact occurs only along the single edge which causes stress concentration at the beam corner. - Case 2:In this load case, force $Fy$ =0.9 N is not sufficient to cause full separation of the contact (Figure 5).
- Case 3:In this load case, normal separation does occur because there is no detaching force - $Fy$ =0. Shear force $Fx$ tends to cause beam slippage which is resisted by friction forces distributed over the contact area. In equilibrium the force projectes onto the horizontal axis:
(2) $$F-Fx=0$$Where, $F$ is total friction force.

Maximum total friction force is:(3) $$F=f\ast P$$Total slippage (or total tangential separation) starts when $Fx\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{}f*P$ . When $Fx\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}<\text{}f*P$ , only partial slippage is possible. Maximum total friction force is $Fmax=0.15*2=0.3\text{}N$ .

The result for $Fx$ =0.3 N is shown in Figure 7. Full slippage develops in contact area, but left lower edge of the upper beam is still almost coincides with the edge of the lower beam. - Case 4:In this load case the shifting force $Fx$ is increased to 0.31 N. The force exceeds the maximum friction force and the upper beam not only deforms but also starts moving as a rigid body.
- Case 5:In this load case the shifting force $Fx$ is decreased to 0.29 N in order to not exceed the maximum friction force. Only partial slippage occurs in this case. There is a clear "sticking" contact area at the left end of the beams.