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

F y

or with horizontal load

F x

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:

Figure 3.

Consider the equilibrium of the upper beam. If $F y$ =0, then force N is equilibrated by the reaction force in contact. As $F y$ 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 $F y$ which results in full separation can be found from moments equilibrium equation.(1)
$\sum M (A) = P * (\frac{L}{2}) - F y * L = 0$

This ultimate value is $F y$ =1 N.

SimSolid result for this value of $F y$ 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.

Figure 4.
Case 2:
In this load case, force $F y$ =0.9 N is not sufficient to cause full separation of the contact (Figure 5).

Figure 5.
Case 3:

Figure 6.

In this load case, normal separation does occur because there is no detaching force - $F y$ =0. Shear force $F x$ 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)
$F - F x = 0$

Where, $F$ is total friction force.

Maximum total friction force is:(3)
$F = f * P$

Total slippage (or total tangential separation) starts when $F x \geq f * P$ . When $F x < f * P$ , only partial slippage is possible. Maximum total friction force is $F m a x = 0.15 * 2 = 0.3 N$ .

The result for $F x$ =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.

Figure 7.
Case 4:
In this load case the shifting force $F x$ 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.

Figure 8.
Case 5:
In this load case the shifting force $F x$ 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.

Figure 9.