How do you Convert distributed load to point load?

Answer 1

To convert a distributed load to a point load, you need to calculate the total load acting over the distributed area. This is done by multiplying the intensity of the distributed load by the area over which it is acting. Once you have the total load, you can then consider it as a point load acting at the centroid of the distributed area. This simplification allows for easier analysis and calculations in structural engineering and mechanics.

Answer 2

For an evenly distributed load (example: F=10 N/m):Simply multiply the distributed load times the span of the load.

If you have distributed load of w = 150 kN/m2 on a span of L = 10 m and width of B= 5m then you will have

W = w x B = 150 x 5 = 750 kN/m

P = W x L = 750 x 10 = 7500 kN as a point load acting n the centre of your area L x B

For an uneven distributed load (example: F=.5x^2 N/m)Converting this kind of distributed load into a point load involves calculating two things: 1) The total load 2) The point at which the load acts

To calculate (1), integrate the load function over the length. So for a beam of length 12m with distributed load F = 5x^2 N/m, you would integrate 5x^2 from 0 to 12.

To calculate (2), you need to find the place along the beam where the sum of the moments on either side = 0. You do this by solving the following equation for a:

Int[F(x)(a-x),x,0,a] = Int[F(x)(x-a),x,a,L]

Where:

Int[a,b,c,d] = integration of function a, with respect to b, from c to d.

F(x) = the distributed load

a = the distance at which the concentrated load acts

L = the total length the distributed load acts over Solving this for F(x) = kx^n shows that:

Loads proportional to x act at 2/3 L

Loads proportional to x^2 act at 3/4 L

Loads proportional to x^3 act at 4/5 L

etc.

Answer 3

AnswerFor an evenly distributed load (example: F=10 N/m):

Simply multiply the distributed load times the span of the load.

If you have distributed load of w = 150 kN/m2 on a span of L = 10 m and width of B= 5m then you will have

W = w x B = 150 x 5 = 750 kN/m

P = W x L = 750 x 10 = 7500 kN as a point load acting n the centre of your area L x B

For an uneven distributed load (example: F=.5x^2 N/m)

Converting this kind of distributed load into a point load involves calculating two things:

1) The total load

2) The point at which the load acts

To calculate (1), integrate the load function over the length. So for a beam of length 12m with distributed load F = 5x^2 N/m, you would integrate 5x^2 from 0 to 12.

To calculate (2), you need to find the place along the beam where the sum of the moments on either side = 0. You do this by solving the following equation for a:

Int[F(x)(a-x),x,0,a] = Int[F(x)(x-a),x,a,L]

Where:

Int[a,b,c,d] = integration of function a, with respect to b, from c to d.

F(x) = the distributed load

a = the distance at which the concentrated load acts

L = the total length the distributed load acts over

Solving this for F(x) = kx^n shows that:

Loads proportional to x act at 2/3 L

Loads proportional to x^2 act at 3/4 L

Loads proportional to x^3 act at 4/5 L

etc.

Read more at link