answersLogoWhite

0


Best Answer

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

User Avatar

Wiki User

12y ago
This answer is:
User Avatar
More answers
User Avatar

Wiki User

15y ago

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.

This answer is:
User Avatar

User Avatar

Wiki User

12y ago

simply supported beam formula when load is at L/4 lenth

This answer is:
User Avatar

User Avatar

Wiki User

13y ago

bending moment will be zero(0)..

Bending moment is wl/4,

where w is load, l is effective length

This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: What is the Bending moment of a simply supported beam with point load at the middle?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

Where does the maximum bending moment occur in a simply supported beam?

It depends on the loading conditions of the beam, it will generally occur close to the middle of the span.


What are the Bending moment diagram for simply supported beam loaded in its center?

zero


What are the Bending moment diagram for simply supported beam with udl?

Parabolic, max moment at midspan of value wL^2/8 where w is the distributed load and L the length of the beam.


How will you know if it is a simply supported or a continuous slab?

a slab of a house suupported with RCC frame column & beam, What will be the bending moment in different spans. ER. J.S.DEORI


What is a hinged beam?

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


What is a simply supported slab?

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.


In simply supported beam why load is kept in the middle?

b'coz in smply supported beam it has two fixed ends


Is theory of simple bending applicable to simply supported beam?

Yes, as long as your beam is relatively slender (i.e. L/d greater than about 2)


What is formula for maximum load due to bending of 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


How do you work out the max bending moment for an i cross sectional beam supported at either side with a central force of 20kN?

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


Explain Why curtailment of main steel is possible close to ends of a simply supported element?

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.


Is their sand benders?

Sand Bending is a variation of Earth Bending, as sand is simply "little chunks of earth".