w(l^2)/8 w = 38N l = 5m
It is parabolic, or second order:M = q x squared/2An excellent software to view the profiles of Shear force & Bending moment diagrams.http://www.mdsolids.com/
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
no you can not
When a cantilever beam is loaded with a Uniformly Distributed Load (UDL), the maximum bending moment occurs at the fixed support or the point of fixation. In other words, the point where the cantilever is attached to the wall or the ground experiences the highest bending moment. A cantilever beam is a structural element that is fixed at one end and free at the other end. When a UDL is applied to the free end of the cantilever, the load is distributed uniformly along the length of the beam. As a result, the bending moment gradually increases from zero at the free end to its maximum value at the fixed support. The bending moment at any section along the cantilever can be calculated using the following formula for a UDL: Bending Moment (M) = (UDL × distance from support) × (length of the cantilever - distance from support) At the fixed support, the distance from the support is zero, which means that the bending moment at that point is: Maximum Bending Moment (Mmax) = UDL × length of the cantilever Therefore, the maximum bending moment in a cantilever beam loaded with a UDL occurs at the fixed support. This information is essential for designing and analyzing cantilever structures to ensure they can withstand the applied loads without failure.
in order to distribute the load uniformly from top to bottom and to increase the thickness at the region where maximum bending can occur.
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
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
You can calculate the combined effect of bending and torsional stress on a rotating pipe using the outside diameter. The angle of rotation and the shearing stress should also be considered.
erm i don't know liak sorry.
Cry man, cry!
Parabolic, max moment at midspan of value wL^2/8 where w is the distributed load and L the length of the beam.
"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 :)