Drop Test Results: 24.75lb Weight at 1ft Height

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Discussion Overview

The discussion revolves around calculating the drop force of a 24.75lb weight dropped from various heights, specifically focusing on the accuracy of the force measurements and the underlying physics involved in impact testing. Participants explore concepts related to force, energy, and material properties in the context of drop tests used in engineering applications.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the validity of the current measurement of 24.75 ft/lb force at 1 foot, suggesting a calculated value of around 74.86 ft/lb instead.
  • Another participant proposes that the 24.75 figure might refer to kinetic energy per foot drop rather than force.
  • Concerns are raised about the impact force being dependent on the deceleration of the mass and the characteristics of the materials involved.
  • It is noted that the unit ft/lb is not a force, and a clarification is made that ft-lbf refers to energy.
  • A participant mentions that the impact force could be significantly higher than the weight of the object, depending on the material properties and deformation during impact.
  • Another participant introduces an equation for calculating impact load, emphasizing that the impact force will be more than twice the weight of the object.
  • Details about the specimen being tested are provided, including its dimensions and material properties, which may influence the calculations.
  • Questions arise regarding whether the specimen is being dropped onto a rigid surface or if a rigid object is being dropped onto the specimen.

Areas of Agreement / Disagreement

Participants express differing views on the correct interpretation of the measurements and the calculations involved. There is no consensus on the accurate calculation of drop force or the definitions of the terms used.

Contextual Notes

Participants highlight the importance of knowing the material properties and dimensions of the specimen to accurately predict impact forces. The discussion also touches on the limitations of the current understanding of the terms and measurements being used.

Who May Find This Useful

This discussion may be of interest to engineers, physicists, and students involved in material testing, impact analysis, and those seeking to understand the dynamics of drop tests in engineering contexts.

rampage11
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im trying to calculate the drop force of falling object. we do drop test at my work and believe the figures are wrong. we drop a 24.75lb weight from different heigths to check parts . they have it now set that dropping the given weight from 1 ft only produces 24.75 ft/lb force and they add 24.75 for each foot after that. I came up with around 74.86ft/lb at 1 foot. 24.75 seems low at 1 foot. any help is appreciated. brandon
 
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rampage11 said:
im trying to calculate the drop force of falling object. we do drop test at my work and believe the figures are wrong. we drop a 24.75lb weight from different heigths to check parts . they have it now set that dropping the given weight from 1 ft only produces 24.75 ft/lb force and they add 24.75 for each foot after that.
Are you sure it's not 24.75 ft-lbs of kinetic energy per foot drop, not force? That would make sense.
 
the parts has to hold 130lb of sheer force
 
The numbers in the OP don't make any seise to me. The force you apply during the impact depends very much on the deceleration of the mass, as well as the height you drop it from.

For an impact between two stiff and "hard" objects, the maximum impact force could be hundreds or thousands of times the weight of the dropped object.

Aside from that, the unit ft/lb isn't a "force" - maybe you meant ft-lbf, which is the energy of the dropped object, as Doc Al said.
 
ft-lbf is what I meant..the part moves about 1/2 inch at impact
 
130 psi of shear force from drop test is what our prints say
 
It is impossible to predict the force (average or peak) during the impact without knowing what material you are dropping the weight onto. I think you said that your specimen deformed a maximum of 1/2 inch during the test? That could be useful information, if we knew the area and thickness of the specimen, and if we knew how elastic the collision is.
 
  • #10
Hey,

Since the object is having sudden impact, the impact force will be more twice the load. The general equation of the impact load is : W+\sqrt{(W^{2} + 2AWhE/L)}

where,
W - weight of the object
A - surface area of impact
L- Length of the body perpendicular to surface of impact
h- height from which body is dropped
E- Modulus of elasticity of the object

Just put in the material and geometrical properties..
So the value comes out to be greater than twice the weight of the body. So 74.86 seems kinda ok. But I don't know the dimensions or the material dropped. So I can't tell.. If you want, just give me the details.. I'll try to work out the values :smile:
 
  • #11
ill have to remeasure it..but I believe its 4'' round stock steel..12'' long dropped from 1 ft increments..weighs 24.73lb
 
  • #12
Is the specimen being dropped onto something "rigid?" Or are you dropping something "rigid" onto the specimen? Your specimen is 4" diameter steel?
 
  • #13
rigid..dropping specimen onto a 5/8 steel rod in 1ft increments
 

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