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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.
Wiki User
∙ 15y ago
Wiki User
∙ 13y agoSimply 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.
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What size laminated timber beam is needed for a 23 foot clear span under a 15 foot roof in between load bearing walls?
38 feet longIt is not possible to give a definite answer without more detail. The construction and type of roof will determine the load (weight) bearing on the beam, and factors such as wind load and earthquake load, and the way the roof bears on the beam (truss/point load/distributed load) will all affect the sizing.
What is uniformly distributed loads?
Uniformly distributed loads, also known as uniformly distributed loads (UDL), refer to loads that are evenly distributed over a given length or area of a structural element. They exert a constant magnitude per unit length or unit area along the specified region. In the case of one-dimensional structural elements like beams or slabs, a uniformly distributed load applies a constant force or weight per unit length. For example, a beam with a UDL of 10 kN/m means that there is a load of 10 kilonewtons acting on every meter of the beam's length. In two-dimensional elements like plates or surfaces, uniformly distributed loads apply a constant pressure or weight per unit area. For instance, a floor slab with a UDL of 5 kN/m² means that there is a load of 5 kilonewtons per square meter acting on the entire surface area of the slab. Uniformly distributed loads are commonly encountered in various structural applications, such as floor loads in buildings, self-weight of structural elements, dead loads, or evenly distributed loads from equipment or storage. They allow for simplified analysis and design calculations since the load intensity remains constant over the specified area or length. When analyzing or designing structures subjected to uniformly distributed loads, engineers consider the load magnitude, the span or length of the element, and the support conditions. By applying principles of structural mechanics and equilibrium, they can determine the internal forces, moments, deflections, and overall behavior of the structure under the UDL. It's important to note that UDLs are an idealization of real-life loading conditions. In practice, actual loads may vary or have different distributions, requiring engineers to consider more complex load patterns and combinations to accurately analyze and design structures.
Calculation uniform distributed load UDL?
material = PVC w = 384/5 X ymax XE X l/4 X l where L = 1 meter. deflection = 6 mm
How do you calculate the maximum bending moment for a simply supported beam of 6m long carrying uniformly distributed load only till 4.5 m long?
I assume this is a cantilever beam with one end fixed and the other free, the load starts at the free end and continues for 4.5 m if w is the load distribution then it has a force at centroid of 4.5 w acting at distance of (6.5 - 4.5/2 )from the end, or 4.25 m The max moment is 4.5 w x 4.25 = 19.125
What is tensile load?
Think of a tensile load as a "pulling" force. A tensile load is the only type of load that can be taken by a rope, for instance.
What are point loads in engineering?
A point load is a load which is localized to a specific location on a structure. (Even though it is usually really not applied at a sharp point) The alternate kind of a load is a distributed load, which is pread accross a large area. For example, a pedestrian standing on a footbridge is considered a point load. Snow on the same footbridge is considered distributed load.
Which formula used to convert uvl to point load?
Load * Distance ., will act on the CG
Shall you please tell me hw the loads are taken by a beam?
loads are carried out as point load uniformly distributed load and uniformly varying load
What is the Bending moment of a simply supported beam with point load at the middle?
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
What is a distributed load?
A uniformly distributed load (UDL) is a load which is spread over a beam in such a way that each unit length is loaded to the same extent.
How do you calculate reactions for a simply supported beam loaded with a uniformly distributed load?
For finding reactions for simply supported beam with uniformly distributed load, first we have to convert the u.d.l into a single point load. And then we have to consider it to be a simply supported beam with a point load and solve it. I think you know how to calculate the reactions for beam with point load.
What are uniformly varying loads?
udl is converted into point load by multiplying the value of udl with the length of the section of the beam over which the udl is acting.these converted point load is acted at the middle of the section.
What is couple load?
couple load is the combination of both concentrated and distributed loads.
An uniformly distributed load is one which?
A uniformly distributed load is one which the load is spread evenly across the full length of the beam (i.e. there is equal loading per unit length of the beam).
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 does the number of arches in an arch bridge affect the load it can carry?
An arch is a curved structure with supports at each end to take up the load which is evenly distributed across the arch plane with the stress at the Linc pin block but distributed to the supporting pillars . The greater the length of the bridge the greater is the number of arches to have a distributed load.
How thick does a 2X2 foot piece of glass need to be to hold 50 lbs supported only at the corners?
If the load is uniformly distributed over the area it needs to be 0.21 inch thick using a safety factor of 5 on the glass strength to account for flaws. The support at the corners must not be a point load but rather distributed by a soft or plastic pad
