# 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).

The material properties are:
Properties
Value
Friction Coefficient
$f=0.15$
Modulus of Elasticity
2.1e+11 Pa
Poisson's Ratio
0.3
Five load cases are considered. In all load cases the lower beam bottom is fixed, the upper beam is loaded on its top with total vertical load of 2 N uniformly distributed over the surface. The right end of the upper beam is loaded either with vertical load $Fy$ or with horizontal load $Fx$ uniformly distributed over the surface.
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
Beams were simulated as two solids (Figure 2). Contact condition at the connection was set to "Separating" with friction coefficient 0.15.

## 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.

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:

$F-Fx=0$

Where, $F$ is total friction force.

Maximum total friction force is:

$F=f\ast P$

Total slippage (or total tangential separation) starts when . When , only partial slippage is possible. Maximum total friction force is .

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.