Impact Energy Formula Needed

lets see....hmmm, 32m/m²? no.... lets see here....100kg times D of 1m.

Nope.

Force on a steel plate huh....would it be more force than on a paper plate?

Plate thickness?

Material?

Support?

Not enough data to give any serious answer even if it is a rough estimation.

Within the elastic range of the plate material total load=P= kx. Where k is the stiffness and x is the deformation of the plate under total load P (weight+impact).

x=x1+x2, k=(A*E)/t

x₁=(t/A*E)*W (Deformation under static load W)

x₂=((2*h*t*W)/(A*E))^1/2 (Deformation under impact load)

(Potential energy of W: (PE=W*h=(1/2)*k*(x₂)²)

Total load P=k*x =(A*E/t)*(W*t/A*E+(2*W*t*h/A*E) ^1/2

Total load P= W+(W*(2*A*E*h/t))^1/2

Where;

A : Crssectional area of the plate

t : Thickness of the plate

E : Modulus of elasticity of the plate material

Does it seem reasonable?

Hi can you please email me the source of these formulaes

I am looking for the source of this formulaes

Regards,

Vismay

With impact loading, stresses produced are simply a multiple of the static stress that would occur if the weight was simply resting on the plate. Since your plate appears to be fully supported by the ground, the static stress of the mass resting on it would be a contact stress; equal to the weight (m*g), divided by the area of contact (.1 m^2). Since the plate is also enduring the momentum of the mass due to its fall, you would multiply this stress by a factor proportional to the fall height. this factor is equal to 1 + (1+2*(h/d))^1/2. where, h = fall height, and d = static strain (i.e. strain deformation that would occur if the mass were simply resting on the plate). d (assuming your material is linear) is equal to S.s divided by the modulus of elasticity of your material. Please note that this will change if your plate, beam or whatever is being impacted, becomes supported over a span instead of fully supported by the ground. this is because, when you have a span, the plate will be allowed to oscillate which will dissipate some of the energy of the moving mass. this is accounted for by reducing the energy (i.e. the fall height) of the falling mass by a factor that varies based on your boundary support conditions. these factors can be found in most structural books like Roark and Young's stress and strain, or mark's engineering handbook.

in summary:

S.i = impact stress

S.s = static stress

S.i = S.s *(impact factor)

S.s = (m*g)/(impact area)

impact factor = 1 + (1+2*(h/d))^1/2

h = fall height

d = S.s/(modulus of elasticity)

note: this impact factor will also give the impact strain when mulitplied by static strain.

thank you for the replies. if the area of the plate is 1m^2 and the area of contact is only 0.1 m^2. should we take 0.1 m^2 in the calculation? and secondly what can be the effective area of the plate that is stressed due to the impact? is it 0.1 m^2 or a bit more than that? if the plate is resting on the concrete floor, shall we have to consider the foundation modulus of concrete?