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How to calculate the standard value of axial force

Publish: 2021-05-24 12:26:37
1.

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

2. It doesn't seem to make sense.. Bending moment diagram, shear force diagram and stress diagram all have the meaning of area representation and axial diagram.. I think it's Muyou..
3. It's very simple. Analyze the force at point B, and calculate the sum of longitudinal force and transverse force as 0.
4. The axial force at the bottom of the column is the axial force at the top of the column plus the self weight of the column itself, which is only considered when calculating the vertical dead load. When calculating the live load, the axial force at the top and bottom of the column take the same value.
5. This has something to do with the type of Jack you are using. You can check (or calculate) the cross-sectional area of the jack. In square meters, the prestressing force can not be applied according to the design value of the steel support axial force, but should be applied according to the design value of the prestressing force (generally in KN)
suppose:
the designed applied prestress value is 300KN
the sectional area of the jack is 0.017671 M2
the applied prestress = the designed applied prestress value / sectional area of the jack = 300 / 0.017671/1000 = 1.70mpa
6. In the basic man-machine input of jccad, the load input can be seen, which has the force under various working conditions
7. I usually send integrated package
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