Posted: 12/26/2013 6:32:24 AM EDT
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Arfcom engineers....let's say that a 15 lb sledge hammer (or similar size/weight metal object) is accidentally dropped from a measured 50-foot height to the ground. Some unlucky individual, dressed in a hard hat and all appropriate safety gear is standing on the ground directly underneath it and is hit atop the hardhat by the falling hammer.
How would the force of the blow be measured? Any estimates of how hard it would hit? This is for an actual work related accident that I am currently investigating. Thanks! |
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It depends on the hit.
Does the hammer skip off the guys hard hat and continue a downward path? Does the hammer hit the guys hard hat and stop completely? Does the hammer hit the guys hard hat and bounce? You asked an engineer, so you get the engineer answer: "I need more information" 
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a is constant, so integrate for velocity and position. solve for velocity and KE=MV^2 Quoted:
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f=m*a a is constant, so integrate for velocity and position. solve for velocity and KE=MV^2 Velocity = (2*a*distance)^1/2 Velocity = 56 feet per second KE = (1/2) 50 lb (56^2) = 80,000 lb-ft^2/sec^2 |
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s=1/2 a t ^2+ Vo t+So to find velocity
as previously stated Ke=1/2 mv^2 so assuming zero initial velocity the fall took 1.76227 seconds this gives a velocity of 56.745 FPS at impact. 56.745*mass= energy mass should be in slugs and i dont feel like looking up the formula, but you get the idea |
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It depends on the hit. Does the hammer skip off the guys hard hat and continue a downward path? Does the hammer hit the guys hard hat and stop completely? Does the hammer hit the guys hard hat and bounce? You asked an engineer, so you get the engineer answer: "I need more information" The worker was facing down, concentrating on his task at ground level and the falling object apparently "skipped" on the back of his hardhat. Yes, he's still alive. I am not privileged to his medical info (HIPPA and all that stuff...) All I know is that the client is arranging for continuing care. Thanks! |
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How tall was the worker that got hit? Good point! My answer assumes he was standing in a hole, and the top of his hard hat was at ground level. I will now re-calculate, assuming that his hard hat is 6 foot off the ground, just for my convenience. ETA, my new and improved answer is 660 ft-lbs. |
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This should help...
Impact Force from a Falling Object The dynamic energy in a falling object at the impact moment can be expressed as E = m g h (4) where g = acceleration of gravitation (9.81 m/s2, 32.17405 ft/s2) h = falling height (m) The equation can be combined with the equationof work as F = m g h / s (5) Example - Falling Car The same car as above falls from a height of 14.2 m and crashes with the front down on a massive concrete tarmac. The front impacts 0.5 m as above. The impact force can be calculated as F = (2000 kg) (9.81 m/s2) (14.2 m) / (0.5 m) = 557 kN Note that a car driving in 90 km/h (25 m/s) compares to a fall from 32 m(!) Example - a Person falling from a Table A person with weight (gravitational force) of 200 lbs (lbf) falls from a table 4 feet high. The energy of the falling body when it hits the ground can be calculated as E = (200 lbf) (4 ft) = 800 ft lb The impact on a human body can be difficult to determine since it depends on how the body hits the ground - which part of the body, the angle of the body and/or if hands are used to protect the body and so on. For the case of this example we use an impact of 3/4 inch to calculate the impact force: F = (800 ft lb) / ((3/4 in) (1/12 ft/in)) = 12800 lbf I like 44*15 = 660 ft lbs Whoops - I need to go to Force...so 660 & lets say 1/2 in...so 3960 pounds of force.... |
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The worker was facing down, concentrating on his task at ground level and the falling object apparently "skipped" on the back of his hardhat. Yes, he's still alive. I am not privileged to his medical info (HIPPA and all that stuff...) All I know is that the client is arranging for continuing care. Thanks! Quoted:
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It depends on the hit. Does the hammer skip off the guys hard hat and continue a downward path? Does the hammer hit the guys hard hat and stop completely? Does the hammer hit the guys hard hat and bounce? You asked an engineer, so you get the engineer answer: "I need more information" The worker was facing down, concentrating on his task at ground level and the falling object apparently "skipped" on the back of his hardhat. Yes, he's still alive. I am not privileged to his medical info (HIPPA and all that stuff...) All I know is that the client is arranging for continuing care. Thanks! Well then we need high speed video at an angle where we can tell the change of direction and speed of the hammer after the impact. Then you know what acceleration the hammer experienced when it struck the helmet, and you know the direction vector. You also know the mass, and can solve for the force at that point. Seriously. 
Anything else is a guesstimate based on assumptions. Realistically the best your going to get is an accurate calculation of the hammers kinetic energy at time of impact. Without knowing exactly what happened to the hammer after impact you can't really definitively say how much of that energy was transferred to the hard hat. Thus, the kinetic energy doesn't really mean much. |
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If he isn't dead, and the story is true, I wanna know who made that hardhat.  Quoted:
If he isn't dead, and the story is true, I wanna know who made that hardhat. All I know is that the client is arranging for continuing care. There is living, and there is 'living'. Just because your skull is still in one piece does not mean your brain is. 
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I assume the worker survived?
I could see that might be a fatal blow. I work for an Eng. company and our largest client requires us to go thru a yearly mandatory 10hr (OSHA and MSHA) training. You better have your head on a swivel at these plants, dangerous place. There are all kinds of hazards just in the daily plant operation, then you throw in any construction work. Safety has to be your top priority, no exceptions. |
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There is living, and there is 'living'. Just because your skull is still in one piece does not mean your brain is. Quoted:
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If he isn't dead, and the story is true, I wanna know who made that hardhat. All I know is that the client is arranging for continuing care. There is living, and there is 'living'. Just because your skull is still in one piece does not mean your brain is. Indeed, but at even at 660ft/lbs (the lowest estimate posted here) the fact that his head isn't evenly distributed across the jobsite is pretty impressive. Damn good hat. |
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Oh snap...
I completely forgot to mention that the injured worker was already on his knees, facing downward and disconnecting 2 widgets with his hands at ground level, so I'm guessing his head was 2-3 feet above ground. Thanks again! ETA: I also believe that emergency surgery involved removing bone fragments from his brain. |
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The sledge has negligible aerodynamic drag, so it is reasonable to calculate its motion from basic point particle motion, F=ma, V=at, momentum p=mV, and the kinetic energy, E = mv^2/2 at impact. But all that is just a start.
The dynamics at the helmet are hugely more complicated by the helmet suspension and the boundary conditions presented by the human underneath. This starts as an impact problem with damping and it is not amenable to a back of the envelope calculation. If I were you, I would track down the certification specs for the helmet, and I would start with OSHA requirements for performance of the helmet. Then I would find out how the helmet manufacturer proves it meets the requirements, and that means there will be test results. I would expect to find a requirement that the helmet has to protect the wearer against a smooth edged weight dropped from a certain height without inducing a certain level of acceleration into the head and without piercing the helmet, and most likely another requirement for protection from a certain weight and height of a sharp edged object. That research will lead to other specs from organizations such as the ASTM, ASME, AISI, and probably others that I don't deal with on a regular basis. |
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Velocity = (2*a*distance)^1/2 Velocity = 56 feet per second KE = (1/2) 50 lb (56^2) = 80,000 lb-ft^2/sec^2 Quoted:
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f=m*a a is constant, so integrate for velocity and position. solve for velocity and KE=MV^2 Velocity = (2*a*distance)^1/2 Velocity = 56 feet per second KE = (1/2) 50 lb (56^2) = 80,000 lb-ft^2/sec^2 50lbs is not the mass, nor is 15lbs. 15lbs is the weight (Force due to gravity). The mass is 15/32.1 = 0.47 slugs KE = 1/2mv^2 = 1/2*0.47*(56^2) = 2930.8 foot pound-force |
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Velocity = (2*a*distance)^1/2 Velocity = 56 feet per second KE = (1/2) 50 lb (56^2) = 80,000 lb-ft^2/sec^2 Quoted:
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f=m*a a is constant, so integrate for velocity and position. solve for velocity and KE=MV^2 Velocity = (2*a*distance)^1/2 Velocity = 56 feet per second KE = (1/2) 50 lb (56^2) = 80,000 lb-ft^2/sec^2 This is what I calculated. 56fps~38MPH=OUCH! |
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Well, I'm not about to do all that math.
I can tell you that I had a patient once that was hit by a 5# sledge that was dropped off the tower of a drilling rig and struck a guy standing on the platform. There was a hammer shaped dent in his aluminum hardhat, a corresponding dent in his skull, and a corresponding dent in his brain. He did not do well. Is that pretty much what you were asking??? |
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50lbs is not the mass, nor is 15lbs. 15lbs is the weight (Force due to gravity). The mass is 15/32.1 = 0.47 slugs KE = 1/2mv^2 = 1/2*0.47*(56^2) = 2930.8 foot pound-force Quoted:
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f=m*a a is constant, so integrate for velocity and position. solve for velocity and KE=MV^2 Velocity = (2*a*distance)^1/2 Velocity = 56 feet per second KE = (1/2) 50 lb (56^2) = 80,000 lb-ft^2/sec^2 50lbs is not the mass, nor is 15lbs. 15lbs is the weight (Force due to gravity). The mass is 15/32.1 = 0.47 slugs KE = 1/2mv^2 = 1/2*0.47*(56^2) = 2930.8 foot pound-force I hate it when that happens. Slug = 1/32 lbf*sec^2/ft makes the units work out to lbf-ft of energy too. Good work bro. I think.
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Well, I'm not about to do all that math. I can tell you that I had a patient once that was hit by a 5# sledge that was dropped off the tower of a drilling rig and struck a guy standing on the platform. There was a hammer shaped dent in his aluminum hardhat, a corresponding dent in his skull, and a corresponding dent in his brain. He did not do well. Is that pretty much what you were asking??? yep, pretty much... My prayers go out for the worker and his family. Needless to say, it makes for a sucky Christmas. |
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It's impossible to give a meaningful answer without data on the performance of the hard hat, since that will change the way to momentum is transferred to head, and indeed will change the total energy imparted to the head.
As an example, if you stand in street and let yourself fall forward onto the cement, hitting your head, you will likely suffer serious brain injury. If you do the same thing wearing a motorcycle helmet, you will be fine. Same scenario, same energy, same total momentum, but the helmet liner changes the way the momentum is transferred. Incidentally, this is why I wear a helmet while riding. While many accidents are unsurvivable with or without, the probability of having a "stupid" accident where you hit your head on the cement from a height of about 6 feet is statistically the most likely. I'd rather not wear diapers for the rest of my life because I low-sided from some sand on a corner. |
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It depends on the hit. Does the hammer skip off the guys hard hat and continue a downward path? Does the hammer hit the guys hard hat and stop completely? Does the hammer hit the guys hard hat and bounce? You asked an engineer, so you get the engineer answer: "I need more information" The answer either begins with that or "it depends..."
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I hate it when that happens. Slug = 1/32 lbf*sec^2/ft makes the units work out to lbf-ft of energy too. Good work bro. I think. Quoted:
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f=m*a a is constant, so integrate for velocity and position. solve for velocity and KE=MV^2 Velocity = (2*a*distance)^1/2 Velocity = 56 feet per second KE = (1/2) 50 lb (56^2) = 80,000 lb-ft^2/sec^2 50lbs is not the mass, nor is 15lbs. 15lbs is the weight (Force due to gravity). The mass is 15/32.1 = 0.47 slugs KE = 1/2mv^2 = 1/2*0.47*(56^2) = 2930.8 foot pound-force I hate it when that happens. Slug = 1/32 lbf*sec^2/ft makes the units work out to lbf-ft of energy too. Good work bro. I think. Either way, it's a friggin skull crusher 
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The sledge has negligible aerodynamic drag, so it is reasonable to calculate its motion from basic point particle motion, F=ma, V=at, momentum p=mV, and the kinetic energy, E = mv^2/2 at impact. But all that is just a start. The dynamics at the helmet are hugely more complicated by the helmet suspension and the boundary conditions presented by the human underneath. This starts as an impact problem with damping and it is not amenable to a back of the envelope calculation. If I were you, I would track down the certification specs for the helmet, and I would start with OSHA requirements for performance of the helmet. Then I would find out how the helmet manufacturer proves it meets the requirements, and that means there will be test results. I would expect to find a requirement that the helmet has to protect the wearer against a smooth edged weight dropped from a certain height without inducing a certain level of acceleration into the head and without piercing the helmet, and most likely another requirement for protection from a certain weight and height of a sharp edged object. That research will lead to other specs from organizations such as the ASTM, ASME, AISI, and probably others that I don't deal with on a regular basis. OSHA uses ANSI z89.1 for hard hat requirements and testing. |
