What is the maximum load-bearing capacity of channel 12M span 100?
When flat, the span of one meter is concentrated, and the solution is as follows: M=W[,]=15610*215=3356150N-mm=3.356kN-mM=L/4*N+q*L*L/8N= (M-0.1*1*1/8), *4= (3.356-0.1*1*1/8), *4=13.374kN (1.3 tons)When standing, the span of one meter is concentrated, and the solution is as follows: M=W[,]=39660*215=8526900N-mm=8.527kN-mM=L/4*N+q*L*L/8N= (M-0.1*1*1/8), *4= (8.527-0.1*1*1/8), *4=34.058kN (3.4 tons)
Suppose that the ends are hinged (other forms may be alternative), according to the mechanics of materialsBending moment M = 1/4PL = 18917.5Nm; P=56752N=56.8kNThe safety factor is taken as 1, and if the other is taken, the carrying capacity P is correspondingly reduced.
Maximum load-bearing capacity of channel steel with span 12M 100 can be obtained by calculation.Suppose the channel material is Q235, model 10, vertical and 12M, center force, and the maximum load is P. First check the manual to get yield stress 235MPa, then check the channel manual to get the bending coefficient of 80.5, multiply the two, and get the bending moment of 18917.5Nm.
