assuming the point load acts in the centre, take the value under it as P*L / 4
where P=point load (kN)
L=length between supports
if its not in the middle, take it as P*a*b / 8
a=dist from left hand support to load
b=dist from right hand support to load
thanks,
Abdul wahab
The " in not in the middle formula" is incorrect.
Your Welcome
Paul
A uniformly distributed load will have the same effect on bending moment as a point load of the total weight of the distributed load applied at the center of the distributed load.
If you reduce the uniformly distributed load to a point load, it can be calculated in the same manner as, and along with, the other point loads.
A bending moment is the distance times the point load, remembering to take direction into account. They can be added algebraically.
Example: assume we have a 10 meter beam with a point load of 10 kg at 1 m and a uniformly distributed load of 10 kg/m from 2 m to 4 m. Reduce the distributed load to (10 kg/m * 2 m) or 20 kg applied 3 m out. You now have two point loads: 10 kg at 1 m, and 20 kg at 3 m. The problem then becomes (10 kg * 1m) + (20 kg * 3 m) or 10 kg-m + 60 kg-m = 70 kg-m.
If the load is not uniformly distributed, this same principle can still be applied by finding the total weight applied and the center of gravity of the distributed load and applying that full load as a point load at the point of the center of gravity.
Example: assume we have a distributed load of the form 2 kg/m from 0 to 6 m. We know that the general shape of this load distribution will be triangular. We know that the center of gravity of a triangle is 1/3 the width of the triangle from the highest point of the triangle. The highest point will occur at 6 m. 1/3 in from that is at 4 m. That's our point of application. Our total weight is the integral from 0 to 6 of 2 kg/m dx (or 1/2 6 * 12) = 36 kg. So we assume a point load of 36 kg applied at 4 m for a bending moment of 144 kg-m.
simply supported beam formula when load is at L/4 lenth
bending moment will be zero(0)..
Bending moment is wl/4,
where w is load, l is effective length
Parabolic, max moment at midspan of value wL^2/8 where w is the distributed load and L the length of the beam.
a slab of a house suupported with RCC frame column & beam, What will be the bending moment in different spans. ER. J.S.DEORI
It is a beam that is attached at aboundary that is free to rotate, like a hinge. It cannot develop a bending moment. It is often used to idealize a simply supported beam
The strength, S, of the beam is Mc/I where M = max moment to fail = PL/4 for load concentrated in the middle of the beam or WL/8 for uniformly distributed load. Here P is the concentrated load, W = distributed load, c = distance to outer fiber from neutral axis and I the area moment of inertia of the beam. L = length Solving for load maximum, P = 4IS/Lc for concentrated center load W = 8IS/Lc for distributed load
It actually depends on the type of beam it is. If it is a cantilever, the formula would be PL/2 and for a simply supported beam it would be PL/4
It depends on the loading conditions of the beam, it will generally occur close to the middle of the span.
zero
Parabolic, max moment at midspan of value wL^2/8 where w is the distributed load and L the length of the beam.
a slab of a house suupported with RCC frame column & beam, What will be the bending moment in different spans. ER. J.S.DEORI
It is a beam that is attached at aboundary that is free to rotate, like a hinge. It cannot develop a bending moment. It is often used to idealize a simply supported beam
The slabs that are supported only at end are called simply supported slabs i.e. there is no intermediate supports in the slab and there will be no support moment acting on the slab.
b'coz in smply supported beam it has two fixed ends
Yes, as long as your beam is relatively slender (i.e. L/d greater than about 2)
The strength, S, of the beam is Mc/I where M = max moment to fail = PL/4 for load concentrated in the middle of the beam or WL/8 for uniformly distributed load. Here P is the concentrated load, W = distributed load, c = distance to outer fiber from neutral axis and I the area moment of inertia of the beam. L = length Solving for load maximum, P = 4IS/Lc for concentrated center load W = 8IS/Lc for distributed load
It actually depends on the type of beam it is. If it is a cantilever, the formula would be PL/2 and for a simply supported beam it would be PL/4
Reinforcements is provided to resist moment and shear force, in a simply supported beam maximum moment at centre and its reduces towards (zero)support. so no 100% reinforcments at support required, so curtailment is possible (max 50%) at ends.
Sand Bending is a variation of Earth Bending, as sand is simply "little chunks of earth".