Thursday, October 13, 2022

Using AISC column specifications, determine the safe axial loads on W 310x107 section used under following conditions (a) hinged ends and a length of 10 m (b) built-in ends and unsupported length of 10 m (c) built-in ends and a length of 10 m braced at mid point. Use proportional limit for steel as 380 MPa and Modulus of elasticity equal to 200 GPa. Take properties of section from internet or recommended books. Furthermore discuss in detail about the variation of column capacity with the variation of conditions.

 

Q# Using AISC column specifications, determine the safe axial loads on W 310x107 section used under following conditions (a) hinged ends and a length of 10 m  (b) built-in ends and unsupported length of 10 m (c) built-in ends and a length of 10 m braced at mid point.  Use proportional limit for steel as 380 MPa and Modulus of elasticity equal to 200 GPa.  Take properties of section from internet or recommended books.  Furthermore  discuss in detail about the variation of column capacity with the variation of conditions.                                                                                                                           

 

SOLUTION

Given data

Properties of steel section

 Column section                                 W 310×107

Gross area of cross sectipn                 Ag = 13600 mm²

Radii of gyration                                     rx= 135mm

                                                                 ry = 77.2mm

 

Restrain

Along x direction                                                                       Along Y direction

Unsupported length of column        Lx = 10m                      unsupported length of column    Ly = 10m

Bottom end                                          : pinned                        Bottom end                                          : pinned

Top end                                                 : pinned                        top end                                                 : pined

 

Calculations

Yield stress of steel                          Fy = 380 Mpa

Modulus of elasticity of steel          E = 200000 Mpa

Effective length factors                    Kx =  1

                                                              Ky = 1

 

                                                  

                                                                           = 74.07


for;

λe    , 1.5,  Fa = (0.658 λe²)Fy 

 

λe  › 1.5,    Fa = (0.877/ λe²) Fy

                                                                              = 129.53

 Since Kx Lx / rx   Ky Ly / ry ,  KyLy/ ry controls

λe = [i] π² E

                      = 1.80  > 1.50

 

 for;

λe    , 1.5,  Fa = (0.658 λe²)Fy 

λe  › 1.5,    Fa = (0.877/ λe²) Fy

 

therefor,c ritical stress,                            Fa =  103.2 Mpa

hence,

 

design axial strength of column,               ɸͤe P n = ɸe Fa Ag

 

                                                                              =0.9 ×103.2×13600

                                                                              = 1263.0 kN

 

 

 

 

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