The standard dead load of an elevator is about 450 pounds. This translates to a kN value of around 0.75.
bearing capacity is the capacity of soil or strata that can be able to sustain the load of superstructure in the unit of load per m2 either ton/m2 or KN/m2 bearing pressure is nothing but bearing capacity example when you apply 100 KN on a unit area, equal opposite pressure will rise from the soil. load / area = bearing capacity
600mm^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 haveW = w x B = 150 x 5 = 750 kN/mP = W x L = 750 x 10 = 7500 kN as a point load acting n the centre of your area L x BFor 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 actsTo 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 loada = the distance at which the concentrated load actsL = 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 LLoads proportional to x^2 act at 3/4 LLoads proportional to x^3 act at 4/5 Letc.
ST 4 concrete is the stronger out of the two. ST4 is equivalent of 16 kn at 28 day strength where as ST2 is only 8 kn at 28 day strength.
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.
1 kN = 225 pounds
When it's not breathing...
Vehicular loads are typically based on the AASHTO H-25 or HS-25configuration, Figure 2-2, which represents a 25 ton (222 kN) semi-truck.Some specifiers use an H-20 or HS-20 load; the load distribution is thesame as an H-25 or HS-25, but the resulting load is about 20% lower.Similarly in railroad applications, the standard load is represented by theCooper E-80 configuration at 80,000 lbs/ft (1167 kN/m) of trackAASHTO H-25 Highway Load-The intensity of the vehicular load decreases as the depth increases,however, the area over which the force acts increases.Live LoadsVehicular loads are typically based on the AASHTO H-25 or HS-25 configuration, Figure 2-2, which represents a 25 ton (222 kN) semi-truck. Some specifiers use an H-20 or HS-20 load; the load distribution is the same as an H-25 or HS-25, but the resulting load is about 20% lower. Similarly in railroad applications, the standard load is represented by the Cooper E-80 configuration at 80,000 lbs/ft (1167 kN/m) of track. Figure 2-2 - AASHTO H-25 Highway Load
"kN.m is a unit of bending moment. kN/m is a unit of udl (uniformly distributed load) as far as i know, there isn't kN.m2 but there is kN/m2 kN/m2 is a unit of pressure acting on an area. Please check your question again." I think you have misunderstood the question. The asker can correct me if i'm wrong but I think they mean, for example, that if you have a uniformly distributed load over an floor area in kN/m2 and you have say a beam running across this floor that you would like to run an analysis on, what would be the value of the load in kN/m on the beam? would it simply be the same value in kN/m or would the conversion affect the value? I say this because I'd also like to know the answer :)
Based on E-M Strength (120 kn) 45 KN 70 KN 90 KN 120 KN 160 KN 210 KN 320 KN etc
bearing capacity is the capacity of soil or strata that can be able to sustain the load of superstructure in the unit of load per m2 either ton/m2 or KN/m2 bearing pressure is nothing but bearing capacity example when you apply 100 KN on a unit area, equal opposite pressure will rise from the soil. load / area = bearing capacity
11.52 kN
why is kn reading as n
4 kN4 kN4 kN4 kN
367.7 kN
V=kn
kN x 101.97 = kg