How to calculate axial force
Calculation of truss axial force with node method:
two unknown forces can be obtained from a node equation, generally starting from the support node and proceeding in turn. For a node, if the member is removed and replaced by force along the member direction, it can be assumed that it is tensile force (if the force is negative, it means pressure). The equilibrium equations in X and Y directions are listed respectively (the forces are projected into the equilibrium equations in X and Y directions respectively): ∑ x = 0 ∑ y = 0. The specific form may be as follows: f1cosa + f2cosb + acosc = 0, f1sina + f2sinb + asinc = 0, where a represents the known force, F1 F2 is the unknown force. The unknown forces F1 and F2 can be obtained by solving the equations. The positive value is the tensile force and the negative value is the pressure
The axial force of AC section is - 20KN, not - 10kN. Because - 10kN acts on point C, the AC is disconnected, the left part is taken as the isolator, and only the left end bears the axial force of - 20KN, so the axial force is - 20KN. Similarly, the axial force of CD segment is - 10kN, and that of de segment is + 10kN
For columns with large slenderness ratio, the initial eccentricity caused by various accidental factors cannot be ignored. With the increase of the load, the lateral deflection also increases. The compression deformation and bending deformation of the member occur at the same time. Finally, the member is destroyed under the combined action of axial pressure and additional bending momentfirstly, the concavity compressive concrete is crushed, the longitudinal reinforcement is bent and bulged out, and the concrete cover is peeled off; At the same time, when the convex surface is under tension, the concrete will proce horizontal cracks, the lateral deflection will increase sharply, and the column will be damaged
Extended data:
for short columns with longitudinal bars and stirrups, the strain of the whole section is basically uniform under axial load. When the load is small, the concrete and steel are in the elastic stage. As the load continues to increase, the lateral deformation of the concrete increases, the fiber stress at the edge of the section first reaches the tensile strength of the concrete, and micro cracks begin to appear in the column
Afterbecause the elastic molus of steel bar is greater than that of concrete, the stress of steel bar increases rapidly, the stress of column longitudinal bar first reaches the tensile strength of steel bar and is crushed, and micro cracks begin to appear in the column
in the second picture, the projection forces of two triangles are added, but the two forces have opposite directions, one positive and one negative, and a negative sign is added in front of the bracket, so the symbols of the two forces are different.
the downward moment of point P to point B, f100knx3m = 300KN. M
the upward force of point a to point B, 300KN. M / 4.5m = 66.667kn
Balance first, then draw left / up and right / down
Of course, we should the results on the graph:N3 = 2F (pull), N2 = - f (pressure), N1 = f (pull)
using node method to calculate truss axial force:
one node equation can solve two unknown forces, generally starting from the support node, in turn. For a node, if the member is removed and replaced by force along the member direction, it can be assumed that it is tensile force (if the force is negative, it means pressure). The equilibrium equations in X and Y directions are listed respectively (the forces are projected into the equilibrium equations in X and Y directions respectively): ∑ x = 0 ∑ y = 0. The specific form may be as follows: f1cosa + f2cosb + acosc = 0, f1sina + f2sinb + asinc = 0, where a represents the known force, F1 F2 is the unknown force. The unknown forces F1 and F2 can be obtained by solving the equations. The positive value is the tensile force and the negative value is the pressure