On May 21, the bridge pier of the G65 Baomao Expressway was hit by stones and was completely damaged. The article assumes that the kinetic energy of the stone is completely converted into impact energy, and the weight of the stone is about 110kg, the speed is about 40km/h, and the size is about 40cm, which arouses heated discussions among readers. Typical points of view are:
1. After the stone hits the pier, the speed is not necessarily 0;
2. If the kinetic energy of the stone is completely converted into the impact energy, then for an object with a small mass and high velocity (such as a bullet), if its kinetic energy is the same as that of the stone, can it also break the pier?
For the above questions, the owner also wanted to know the answer, so he further analyzed and proposed a second model.
1. What happens if the speed is not 0 after the stone hits the pier
Assuming that the velocity v1 before the impact and the velocity v2 after the impact, and then assuming that the displacement of the stone and the displacement of the pier during the impact are the same, the following equations can be listed:
Solving the above equations yields:
That is, knowing the mass m of the stone and the velocity v1 before the impact and the equivalent mass M of the pier, the velocity v2 after the impact and the velocity V corresponding to the equivalent mass of the pier can be obtained. For the Charpy model, the equivalent mass M of the pier is 1/2 of the mass of the column.
2. The minimum mass of the stone that breaks the pier is about 400kg, and the size is about 60cm
For the G65 Baomao high-speed stone impact pier problem, the impact energy is still 7400J, that is, 1/2*M*V²=7400J, and the equivalent mass of the pier is 4574kg, so the pier velocity V=1.8m/s after impact can be obtained.
If the velocity v1 before the impact of the stone is still 11.8m/s, then according to the relation:
It can be found that the mass m of the stone is 378kg, the side length of the cube is 54cm, and the velocity of the stone after impact v2=-10m/s, that is, the stone rebounds back.
3. Can a bullet break a bridge pier?
The mass of the stone is m=378kg, the velocity v1=11.8m/s, and the kinetic energy is 26316J. It is roughly equivalent to the sum of the kinetic energy of 3 rifle bullets. Taking the kinetic energy of 26500J unchanged, adjusting different masses m and muzzle velocity v1, the calculation results are as follows:
As can be seen from the table:
1. When the kinetic energy of the stone before impact remains unchanged, the smaller the mass of the stone, the higher the speed, and the smaller the pier velocity after impact, the smaller the kinetic energy (impact energy).
2. When the mass of the stone is less than 350kg, the impact energy is less than 7000J, and the pier will not be broken; If the mass of the stone is greater than 400kg, the impact energy is greater than 7800J, and the pier will be broken.
3. Three rifle bullets hit the bridge pier at the same time, and the bullets rebounded after impact, and the kinetic energy (impact energy) was only 1J, which could be omitted.
4. What determines the severity of the impact is the momentum before the impact of the stone, not the kinetic energy.
IV. Conclusions
(1) Assuming that the displacement of the stone during the impact process is equal to the displacement of the bridge pier, a new model can be constructed.
(2) The minimum number of stones that break the pier is about 400kg, and the size is about 60cm.
(3) The severity of the impact is determined by the momentum before the impact of the stone, not the kinetic energy. The impact of the bullet on the pier was negligible.
(4) Further analysis shows that the impact time t is only related to the equivalent mass M and stiffness k of the pier, and for this example, the impact time is 0.0165 seconds.