Ouch.. this last response really ...stings.
Well, Engineers and brain surgeons are incomparable. That is why a 10-year experienced engineer is paid only a fraction of what a new graduate brain surgeon is paid.
Four thousand years ago Man built pyramids, towers, bridges, and all kinds of...
I’d like to hear especially from those who have experience in designing high-rise buildings. We are going to design a 26-story high building. It is our first of this size. What are the most significant considerations that we have to keep in mind? We have not decided yet the type of lateral...
EddyC,
Given E=200000N/mm^2, I=37e6mm^4, L=3000mm, h=5000, g=9.81, and m=30kg,
k=13155 N/mm
x=14.9 mm, and
P=196766N or 197kN
Means the force of falling 30kg object is equavalent to 197kN standing still on the baem? or I am having units probem?
Desertfox,
your last line of calc. yielded 49.52...
EddyC,
Sounds too complicated especially that I don't know the size of the beam. A trial and error is required I guess. Any practical conservative solution?
This is a temporary overhead canopy for the enterance of a building where some constuction is being done. The structure is wood.
Thanks
Well, I rememer that F=ma is the force needed to keep (m) moves at a rate of acceleration (a). This force obviously is independant of the height.
My problem is that I am required to account for the load applied on a beam due to the fall of the object,( 30kg from 5m). what is that load in Newton...
My physics information are not that fresh so I need some help please.
I’d like to know how to calculate the force of a falling 30 kg object from 5m high.
Is it 1/2*m*v^2 where v^2=g.h
Thanks
According to the Canadian standard CSA A23.3-1994:
1) Yes
2) both the longitudinal rebars and stirrups are shear Friction reinforcement,
3) stirrups crossing the failure plane are assumed to be at yield, shear forces are limited so that will not cause compression failure in the concrete, and the...
