Calculation of 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 unit of axial force is n or kn< 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
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however, in the case of fixed rotating shaft (such as "hinge"), because the direction of force on the rotating shaft is complex, the rotating shaft must be taken as the "rotating shaft" to formulate the equation.
when estimating the area of foundation bottom, the factor of foundation buried depth should also be considered
for reference only. I wish you all the best.
1、 The necessary condition for cutting tool to cut into rock is p < sub > y < / sub > ≥ s < sub > 0 < / sub > 0· σ Where: P < sub > y < / sub > is the axial pressure on a cutting tool; S < sub > 0 < / sub > is the contact area between cutting tool and rock; σ Is the critical compressive strength of rock
Fig. 1-3-8 force system balance diagram of cutting tool when cutting into rock γ It moves downward in the direction of; γ The angle depends on the friction coefficient of rock to metal and the cutting edge angle of the cutting tool β Therefore, on the front ob, positive pressure n < sub > 2 < / sub > and frictional resistance n < sub > 2 < / sub > Tan are generated ring the cutting process φ tan φ Equal to the friction coefficient f). Similarly, positive pressure n < sub > 1 < / sub > and frictional resistance n < sub > 1 < / sub > Tan are proced on the back slope φ， See figure 1-3-8
The balance relationship of each force is as follows:
rock breaking engineering
after simplification:
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∑ f < sub > y < / sub > = 0
rock breaking engineering
after simplification:
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substitute formula (1-3-2) into formula (1-3-3), According to the conditions of cutting tool cutting into rock:
rock breaking engineering
where: B is the width of cutting tool; B is the width of cutting tool; C is the width of cutting tool; C is the width of cutting tool; σ< Sub > n < / sub > is the normal pressure (or stress) on the surface; σ Is the pressure perpendicular to ab plane, equal to the compressive strength of rock
by substituting equation (1-3-5) into equation (1-3-4), the calculation formula of axial force is obtained as follows:
rock breaking engineering
after mathematical arrangement of equation (1-3-6), the cut in depth h < sub > 0 < / sub > should be:
rock breaking engineering
cos < sup > 2 < / sup > in the right bracket of equation (1-3-7) φ/ sin β+ two φ= Z. There are:
rock breaking engineering
where Z is the cutting edge angle β And the friction angle between cutting tool and rock φ In general, z = 0.88-0.97
Theformula (1-3-8) is basically confirmed for plastic rocks. That is to say, the cutting depth is basically proportional to the axial pressure P < sub > y < / sub >, but not to the cutting tool width B and the cutting edge angle β And the compressive strength of rock. For brittle rocks, the breaking depth is greater than the cutting depth
(2) calculation of horizontal force when large shear occurs in rock e to horizontal force, the cutting tool must approximately overcome the shear resistance of rock mass with area CC ′ B ′ B and side area ABC and a ′ B ′ C ′ in Fig. 1-3-9 and the friction between the cutting tool and groove bottom Figure 1-3-9 the resistance of the cutting tool in large shear is as follows: the area of CC ′ B ′ B is equal to, and the side area of ABC and a ′ B ′ C is equal to Whenshear aa'bb'cc ', the shear resistance is equal to:
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where: σ< Sub > 0 < / sub > is the shear strength of rock
When shearing AA ′ BB ′ CC ′ rock mass, the total resistance to be overcome is equal to:
rock breaking engineering
where f < sub > 1 < / sub > is the internal friction coefficient of rock
If equation (1-3-9) and equation (1-3-10) are equal, the relationship between P < sub > x < / sub > and P < sub > y < / sub > can be obtained:
rock breaking engineering
according to the formula (1-3-11), P < sub > x < / sub > force is related to B, h,
x < / sub > force σ< Sub > 0 < / sub >, P < sub > y < / sub >, F, and COS β In inverse proportion