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Confusing Problem
4

Confusing Problem

Confusing Problem

(OP)
Hello Everyone,

I am a product designer so my mechanical engineering is a bit rusty. I have been tasked with putting together a test program for a traffic sign. One of the tests is a simulated impact. The way this is done is by hitting the product with an (undetermined) mass using a pendulum. The arm has to be 1.25 meters long and the impact energy has to be 150 Nm.

My problem is that, that is the only information the standard gives to work with. I have estimated (as best I can) the period of the pendulum:

T=2π√L/G For which I have calculated T=2π√1.25/9.81 = 2.25

However After that I am not sure were to go, I need to prove that the energy that will impact the sign is 150 Nm.

If possible a break down of how the Pendulum period works and how to calculate pendulum energy would be excellent.

Many Thanks,

04jchatter

RE: Confusing Problem

Perhaps a description or title of the standard might be useful. What seems to be missing is the mass at the end of pendulum.

TTFN
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RE: Confusing Problem

First why care about the period? Provided the sign has to survive the impact the period is irrelevant as all of the energy will be transmitted to the sign and there is no oscillation.

Second the energy is a function of the height, Potential energy = mgh or weight*height. All you need is the center of mass of the pendulum and any added weight at the end. Use the change in height of the center of mass and the weight of your system and you have the energy.

RE: Confusing Problem

This is a second law problem. With the trajectory you can come up with the velocity at impact. Once you've determined the mass, the only thing missing is how long it takes for the velocity to reach zero after first impact. Once you can determine that then F=m*a.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

RE: Confusing Problem

(OP)
@Dought115, my mistake I should have stated that the sign is supposed to be impacted and bounce back up. @GregLocock & IRstuff, I am Trying to deturmin the Mass of the weight that is what I want to know. I have the energy in the system as 150Nm but I do not have the mas of the weight.

Thank you to everyone for you help.
@imcjoek - I am a student but I am currently undertaking a work placement so this Is a real Industry problem but thank you for the Link

Kind Regards,

04jchatter

RE: Confusing Problem

Am I missing something here or is there an error in describing energy as Nm. To me Nm is a torque value and energy is measured in Joules??

Hence the answer would seem to be simple - if 150Nm is actually 150Joules, then just work out for whatever mass you have what the height above the vertical position you need to generate 150J on the basis that all the potential energy is translated into kinetic energy and all is then transferred to the sign when it hits it. You might need to experiment with different masses to get the velocity right to have the best effect.

The practical difficulty will be that the pendulum will either bounce back after hitting the sign, or continue along its path (just much slower) or will get in the way of the sign bouncing back.

A diagram always helps!

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Confusing Problem

charpy impact test ?

RE: Confusing Problem

"To me Nm is a torque value and energy is measured in Joules"

The primary units of a joule is newton*meter, i.e., force*distance

TTFN
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RE: Confusing Problem

(OP)

Hello everyone,

Firstly many thanks for all your helpful advice, what you have all said has provoked me to think about the problem differently.

For those who were interested, This is the final result I got from the equations I found, the values I got seem reasonable to me to if what I have done makes sense please say.

(Also if it makes no sense please also say)

KE=1/2 x M x V²

Where: - M = Mass of bumper
V = Velocity
KE = Kinetic Energy

V= √2xGxL(1-Cos(∂max))

Where: - G = Acceleration due to Gravity (9.81m/s²)
L = Length of arm
∂ = Max Angle
So: - V= √2xGxL(1-Cos(∂max))

Therefore: V= √2x9.81x1.25(1-Cos(90max))
= 5.96m/s²

KE=1/2 x M x V²

Therefore: 150 =1/2 x M x 5.96²
Rearrange to find M: -

(1⁄2 x M x 〖5.96〗^2)/150 So: - 1/M=(1⁄2 x 〖5.96〗^2)/150=0.1184

M=1/0.1184=8.44 kg

Mass of Bumper = 8.44kg

Once again, Many thanks
04jchatter

RE: Confusing Problem

Two things.

Your velocity units should be m/s

The max angle is not 90° it is 180° (1-cos180=2). (But don't use a lift of 180° as there is a chance the weight may go over the other side and hit your sign from the back.)


RE: Confusing Problem

Makes no sense to me and I've looked at it a couple of times now. your v I calculate as 4.95. when you do that you get 12.23 - but see simpler way below.

[b]Always[/b] double check your calculations......

Why not use the energy balance.

Assuming a the bottom of the pendulum you have your sign and all the potential energy turns into kinetic energy, then

energy = 150J = mass x G x height at 90 degrees

So height - 1.25m at 90 degrees
Mass = 150 / 9.81 x 1.25) = 12.23kg.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Confusing Problem

I agree with LI here, except... I get 12.237 winky smile

that's just from the round off of using 9.81 instead of 9.80665 (that's just what's built-in with Mathcad)

The OP's calculator appears to be broken; the PE calculation should have resulted in 4.9514 m/s, which, when plugged into the KE equation would have resulted 12.237 kg. Note also, their first calculation has m/^2 for units of V.

TTFN
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RE: Confusing Problem

The equations were kind of sloppy above. The velocity equation should be:

v=(2*g*L*(1-cos(α))^0.5

(the way the square root symbol was written above said "square root of 2 times g times L times the quantity 1-cos(α) when the real velocity is the square root of everything. It doesn't take much to do a simple unit analysis and see that the way it written gives you a "velocity" of m2/s2

If L = 1.25 m then v = 4.95 m/s

So
m = KE/v^2= 150 N*m/(4.95 m/s)^2 = 6.118 N*s^2/m = 6.118 kg

If the pendulum is raised to 179.5 degrees then the velocity becomes 7.11 m/s and the required mass drops to 3.059 kg.

I can't figure out why so many engineers have had so much trouble with this bit of high school math.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

RE: Confusing Problem

Dang I'm glad that I wasn't too mean about checking work. Yes I did lose the 1/2. So at 90° it should be 12.2 kg, just like LitteInch said. At 180° (minus) it is 6.118 kg.

I really can't understand why I joined the engineers that can't do this high school math problem.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

RE: Confusing Problem

12.2 kg on a lever-arm only 1.25 meters?

28 some-odd lbs swinging naturally from rest on a pivot arm only 49 inches long?

That's not even the energy of a bicycle at near-walking speed! For a road sign specification? (I am assuming it hits somewhere near the ground - not at waist-high. ) You'd put a dent in a 4x4 fence post. Maybe. )

RE: Confusing Problem

(OP)
Hello everyone,

Thank you all for your responses, maybe I am being a bit dense and maybey as IRstuff says my calculator may be broken, but I cant seem to see where you have calculated my Velocity? I seem to always get a value of 5.942m/s ? If there is anyway someone could explain where they are getting (4.95m/s) I would be very grateful.

Racookpe1978 - The impact will be 250mm of the ground, but the bumper is quite big (500mm/250mm plywood with weights on the back) Its a test to ensure there is no determinant deflection on the base of the product after it rebounds.

Once again thank you to everyone for your help,

04jchatter

RE: Confusing Problem

(OP)
GregLocock - The sign is designed to rebound so hopefully 12.2Kg should be sufficient, I just need to prove its the right mass

RE: Confusing Problem

Hi 04jchatter

A diagram of your setup would help however you need to find the center of percussion of your pendulum because what you will design is a compound pendulum not a simple one.
A simple pendulum is where the mass is connected to a piece of string and the string is effectively massless.
In your case the rod compared to the pendulum bob will not be negligible so search for the mechanics of a compound pendulum.

RE: Confusing Problem

04,

As zdas04 says, the equation where you get velocity is v=(2*g*L*(1-cos(α))^0.5

or V squared = 2*g*L*(1-cos(α)

thus V^2 = 2 x 9.81 x 1.25 x (1-0)

= 24.525

Thus V = Sqrt 24.525 = 4.952 m/sec

I've tried a few ways of incorrectly calculating this and can't get to your figure.

Anyway, checking it against the simple energy equation shows it is incorrect.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Confusing Problem

Also, bear in mind that the calculations performed provide only the MINIMUM required mass. Any losses, like friction, will lower the KE, and to regain the desired KE, additional mass or velocity is required.

btw, you stated 5.96 in your first result, and now, you're stating 5.942, so something changed. The calculation is pretty much bulletproof, so it's unclear where your math is coming from:

TTFN
FAQ731-376: Eng-Tips.com Forum Policies

Need help writing a question or understanding a reply? forum1529: Translation Assistance for Engineers


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com: http://www.engineering.com/AskForum/aff/32.aspx

RE: Confusing Problem

Note also that MIL-S-901D, which is the US Navy specification for shipboard shock testing uses a similar kind of pendulum, and there is a graduated scale that's used to determine the angle of the pendulum to achieve a desired impact. In the case of 901D, the UUT is mounted on an "anvil" which is struck by the pendulum hammer.

TTFN
FAQ731-376: Eng-Tips.com Forum Policies

Need help writing a question or understanding a reply? forum1529: Translation Assistance for Engineers


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com: http://www.engineering.com/AskForum/aff/32.aspx

RE: Confusing Problem

Change your calculator from radians to degrees or use pi/2 radians.

Also was the 150Nm the amount of energy absorbed by the sign or how much energy the pendulum has to have when it hits? The equations we have been using are to calculate the energy of the pendulum at impact but if "bounce" is required then some of that energy will be reflected back to the pendulum decreasing the energy absorbed by the sign. There is also the mechanics of the impact, sharp point, blade edge, rubber bumper.... Each of these will result in different test results.

Z and LI you keep dropping a ")" from your equations.smile

RE: Confusing Problem

That's because I copy and pasted his!

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Confusing Problem

What is the shape of the pendulum where it hits the plywood?

"Weighted plywood" => but NOT anchored sign post!! => you've suddenly changed the impact problem to that of a moving (pointed ?) object hitting a stationary object and forcing it sideways (some), up (a little bit because the plywood will rotate up and away from its pivot point on the BACK side of the weighted bottom flat plywood, and in deformation (crushing) of the plywood at point of impact.

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