# How do you calculate the bending moments for a uniformly distributed loads with point loads?

# How do you Convert distributed load to point load?

Answer .
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/m 2 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..

# What is a Uniformly 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 the maximum bending moment for a light 5m beam carrying a central point load 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

# What is uniformly distributed load?

UDL = Uniformly Distributed Load UDSWL = Uniformly Distributed Safe Working Load UDL describes the way in which a load or weight is spread across a shelf area. Imagine a… fish tank exactly the same size as the shelf; as you fill it with water, it finds its' own level so the load transmitted to the shelf is uniformly distributed.

# How do you calculate the maximum bending moment for a light 5m beam carrying a uniformly distributed load of 38kN?

w(l^2)/8.
w = 38N.
l = 5m

# Bending moment diagram fixed point load?

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

# 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).

# Bending moment equation for simply supported with point load?

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

# What is an example of a Uniformly distributed load?

Zero amps on the neutral at the distribution panel. In a real life example - a fish tank (full of water) the same size as the panel. The water evenly distributes the load o…ver the entire surface area of the panel / shelf. follow this link to find out more www.blog.rapidracking.com

# 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 t…o 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 is a non-uniformly distributed load?

A distributed load that varies across a surface and thus can only be represented bij first or second degree functions, and not a single number

# What is the Shape of bending moment diagram for cantilever beam carrying uniformly distributed load?

It is parabolic, or second order: .
M = q x squared/2\n .
An excellent software to view the profiles of Shear force & Bending moment diagrams. .
http://www.mdsolids.com/ .…

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# What is the Bending moment of a simply supported beam with point load at the middle?

bending moment will be zero(0).. Bending moment is wl/4, where w is load, l is effective length

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# How you can convert uniformly distributed load to point load and apply on beam?

Answer 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/m… 2 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.

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# How do you calculate fixed end moment for cantilever beam with uniformly distributed load?

I'm not sure if you're asking for just the equation for the reactions, or how to calculate the FEM of a cantilever via the Moment Distribution method (aka Hardy Cross method),… so I'll anser both. If you assume clockwise moments are negative... If your cantilevered span has a length " L ", and a uniform load " W " acting downward along the entire length of the span, then the fixed end moment is ( W * L 2 )/8 If the fixed end is to the left, the equation is positive (the load is acting clockwise, so the reacting FEM is counter-clockwise, therefore positive.) If the fixed end is on the right, the equation becomes -( W * L 2 )/8 If your uniform load W is acting downward, and is acting along length " d " (which is NOT the entire span), its easiest to convert the uniform load W to a point load " P ", where P = W * d Then your FEM = [ P * a * b *(2* L - a )]/(2* L 2 ) where: L = entire length of span a = distance P is acting from the fixed end, b = distance P is acting from the cantilevered end (also, b = L - a) OR, if W is centered along the span (therefore P is also centered), then the equation becomes (3* P * L )/16 When going through the distributive process, the cantilevered end will ALWAYS have a final moment of zero. At the fixed point, the cantilevered side has a distribution factor of zero, while the other side has a D.F. of one. So, if you have length ABCD, where A is a cantelever, and B, C and D are assumed fixed, FEM AB = 0, D.F. BA = 0, D.F. BC = 1, and D.F. CB = D.F. CD = D.F. DC = 0.5 So what ends up happening is that the final FEM for the cantilever (" AB ") is zero, and the opposite end of the span (" BA ") remains as its initial value. The reason for this is when you sum up the reactions at the point (in this case B ) supporting the cantilever and flip the sign, the distribution factors tell you to distribute everything to the side of the support opposite the cantilever ( -(FEM BA + FEM BC ) is added to FEM BC ). So you can go through the Cross Method as normal, but you don't touch AB or BA , and BC is adjusted only the once that I just mentioned (you don't carry over the 1/2 reactions from CB ).

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# 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

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# What is uniformly distributed loads?

Uniformly distributed loads are loads which have loading distributed evenly across a span of length "L". Written as kips/ft in U.S. customary units. Uniformly distributed load…s can be considered a point load acting at the center of a simply supported span when you multiply load per foot by the length of the span. EX) A uniform 200 kips/ft load is placed on a simply supported beam 10 feet in length. (200 kips/ft)*(10ft)=2000 kips concentrated load acting at the mid span, (L=5ft). This information can be used to determine shear and moment diagram for design considerations.