Equivalent Bending Moment due to Impact Formula?

[Jamie Gough]

Hi guys and girls,

I've been working on this problem for a little while now but I'm not really getting the result that I want. This is what I have:

I have what is essentially a big pair of vice grips controlled by a hydraulic cylinder. The cylinder opens and closes both arms of the vice grips. The cylinder has a pressure, P, cross sectional area, A, and flow rate Q.

The arm rotates on a pin connection. The force from the cylinder on the arm is a distance, L away from the pin connection. The moment on the arm the cylinder produces is, M=P*A*L. The arm has a rectangular cross sectional area, a*b.

The rotational energy given to the arm by the cylinder is, U=(1/2)*Im*w^2, where I am is the mass moment of inertia of the arm about the pin, and w is the angular speed. w=(Q/A)*(1/L).

I know the elastic energy imparted to the arm is (1/2)*Me*theta, where theta is the angular deflection of the arm and Me is the equivalent moment produced by the impact.

So,

(1/2)*Im*w^2=(1/2)*Me*theta, where theta=Me*(theta_static/M), when theta_static is the deflection caused by M=P*A*L as a static load. We also know that M=(EI/L)*theta_static, where EI/L is the bending stiffness, and E is the Modulus of Elasticity, and I is the area moment of inertia.

Some rearranging I get:

Me= M*sqrt((Im*w^2*E*I)/(M^2*L)),

Where the impact factor, G=sqrt((Im*w^2*E*I)/(M^2*L))

I have:
Q=21 gpm,
P=2500 psi,
A=4.91 in^2,
L=10.4 in,
Im = 19.1 slug*ft*in,
a=5 in,
b=2 in,
E=30 Mpsi
I=(1/12)*a*b^3=3.3 in^4

Unfortunately, this leaves my impact factor, G=0.17. Of course the impact factor needs to be greater than or at least equal to unity. If anyone would kindly explain where I have gone wrong it would be very much appreciated.

Thank you kindly,

Jamie

 

[Nidum]

Sorry but I cannot make any sense of this at all .

Please describe the construction and working of this device more completely and provide a drawing .

 

[Jamie Gough]

Thanks Nidum,

It's alright, I just managed to get it. I used U=PQt instead of the rotational kinetic energy and that worked much, much better.