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force due to falling objects 2
- Thread starter FrustEng
- Start date
I find the equation can be made simpler by equating the energies.
The energy of the falling object as it strikes the beam (at the centre to be conservative)is 'mgh'.
Neglecting the losses due to impact (again to be conservative), this energy is stored as the strain energy in the beam while the beam deflects. The strain energy of the beam is given by (1/2EI) 0L?(Mx)2 dx
If P is the dynamic force causing the bending of the beam, the strain energy can be written in terms of P and the same can be equated to the potential energy of the falling mass. Solving for P the force can be obtained.
As Desertfox has stated in his post, we assume no bouncing during the clollision. Further, this impact will also make the beam vibrate and dissipate the additional energy before coming to rest in static equillibrium.
Trilinga
The energy of the falling object as it strikes the beam (at the centre to be conservative)is 'mgh'.
Neglecting the losses due to impact (again to be conservative), this energy is stored as the strain energy in the beam while the beam deflects. The strain energy of the beam is given by (1/2EI) 0L?(Mx)2 dx
If P is the dynamic force causing the bending of the beam, the strain energy can be written in terms of P and the same can be equated to the potential energy of the falling mass. Solving for P the force can be obtained.
As Desertfox has stated in his post, we assume no bouncing during the clollision. Further, this impact will also make the beam vibrate and dissipate the additional energy before coming to rest in static equillibrium.
Trilinga
FrustEng,
I dug a little further on this subject. The book Design Of Welded Structures by Omer Blodgett covers exactly the problem that you descibed. He uses the following formula:
U (Allowable Energy Storage Of Beam) = [(L)(I)((Fy)^2)]/[(6E)(c^2)]
You can set U=mgh (Potential Energy)
I dug a little further on this subject. The book Design Of Welded Structures by Omer Blodgett covers exactly the problem that you descibed. He uses the following formula:
U (Allowable Energy Storage Of Beam) = [(L)(I)((Fy)^2)]/[(6E)(c^2)]
You can set U=mgh (Potential Energy)
Frusteng,
going back to your post of the 23rd ... g = 9.81 m/sec2 ... but your other dimensions are in mm ... did you remember the conversion ?
btw, i don't like the expression "impulse force" ... there's something called "impulse" and there's something called "force", and forces have units of "N".
going back over some earlier posts, i agree with DBDavis, back on the 22nd; isn't the critical question how the cover (between the studs) is going to react ? this'll deflect a lot more than the stud and absorb the energy this way. you've dealing with wood ... maybe structures/civil guys are more comfortable with calculating stresses, but i don't know ... maybe if you've got good material properties. i'd prefer to answer the question along judgement lines.
going back over the analysis responses, we're assuming that the mass hits on one stud which reacts the load in isolation. how much load is distributed through the canopy to other beams ? if not much, 'cause they're too far apart or the canopy is too flexible, then you've got a whole different problem ! if the studs are closely pitched or the canopy is stiff (again judgement ... "close" or "stiff") then the load will be distributed through more than one stud, which isn't a problem if you are comfortable being conservative. ... but then if the (one stud) structure is insufficient, are you going to say it's no good, when it might be ?
as an analysis question, what you're doing is transferring KE from the mass to strain energy in the structure (be it a stud or a membrane/plate cover.
btw, could you test the set-up (off-shift?) ? and why 30kgs and 5m ? regulation requirements ?
going back to your post of the 23rd ... g = 9.81 m/sec2 ... but your other dimensions are in mm ... did you remember the conversion ?
btw, i don't like the expression "impulse force" ... there's something called "impulse" and there's something called "force", and forces have units of "N".
going back over some earlier posts, i agree with DBDavis, back on the 22nd; isn't the critical question how the cover (between the studs) is going to react ? this'll deflect a lot more than the stud and absorb the energy this way. you've dealing with wood ... maybe structures/civil guys are more comfortable with calculating stresses, but i don't know ... maybe if you've got good material properties. i'd prefer to answer the question along judgement lines.
going back over the analysis responses, we're assuming that the mass hits on one stud which reacts the load in isolation. how much load is distributed through the canopy to other beams ? if not much, 'cause they're too far apart or the canopy is too flexible, then you've got a whole different problem ! if the studs are closely pitched or the canopy is stiff (again judgement ... "close" or "stiff") then the load will be distributed through more than one stud, which isn't a problem if you are comfortable being conservative. ... but then if the (one stud) structure is insufficient, are you going to say it's no good, when it might be ?
as an analysis question, what you're doing is transferring KE from the mass to strain energy in the structure (be it a stud or a membrane/plate cover.
btw, could you test the set-up (off-shift?) ? and why 30kgs and 5m ? regulation requirements ?
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