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

λ_e = [i] √ π² E

                      = 1.80  > 1.50

 

 for;

λe    ‹_, 1.5,  F_a = (0.658 ^λe²)F_y 

λe  › 1.5,    F_a = (0.877/ λe²) F_y

 

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

 

 

 

 

An aluminium strut 10 ft long has a rectangular cross section 1.5in by 3 in. A bolt through each end secure the strut in such a way that is acts as hinged column about an axis perpendicular to 3 in dimension; and as fixed column about an axis perpendicular to 1.5 in dimension. Determine the safe central load with a factor of safety of 2 and E= 10300000 psi. Also discuss in detail about the governing column capacity and reasons behind it.

Q#.  An aluminium strut 10 ft long has a rectangular cross section 1.5in by 3 in. A bolt through each end secure the strut in such a way that is acts as hinged column about an axis perpendicular to 3 in dimension; and as fixed column about an axis perpendicular to 1.5 in dimension. Determine the safe central load with a factor of safety of 2 and E= 10300000 psi.

Also discuss in detail about the governing column capacity and reasons behind it.          

SOLUTION

L = 10 ft

   P safe =  ?   Fos = 2

 

  Moment of inertia ;

 Ix and Iy

Ix = bh³/12   ⇉  1.5(3)³/12 

    Ix = 3.37in⁴

rx = √Ix/A ⇉   √3.37/ 1.5 × 3  ⇉  0.86´´ 

    rx = 0.86in

 

Iy = bh³/12  ⇉   3 (1.5)³ / 12  = 0.843

    Iy= 0.843 in⁴

ry = √Iy/A = √ 0.843 / 1.5× 3 ⇉    0.483in

   ry = 0.483in

 

when moment is about  X- axis                            (    E= 10.3×10⁶;  and Le =120)

 hinsed = Le=L

 

 P = E Ix π ² / Le ²  ⇉    (10.3×10⁶ ) ×3.37× π ²  /(120)²                 

            P= 23790.5 lbs

                                     Psafe = 23790.5/2

                                  = 11895.25 lbs                                                   

 

 When moment is about to y-axis

P = E Iy π ² / Le ²  ⇉    (10.3×10⁶ ) ×0.843× π ²  /(0.5×120)²

 P=23780.53    ⇉   psafe= 23780.53/2

 

                                       psafe= 7934.88lbs  is governing value for colum capacity

Draw the Mohr,s cilcle for the stress element shown in figure. Determine the principal stress and maximum shear stress and also draw respective stress elements

 Q# Draw the Mohr,s cilcle for the stress element shown in figure. Determine the principal stress and maximum shear stress and also draw respective stress elements.                                             

 

solution

 

Here

σx = -80Mpa

σy = 50 Mpa

τxy = -25Mpa

 

Principle stress

In mohr's circle  OD = σ1

                             OC= σ2

 

`From measurent           σ1 =        55Mpa

                                      σ2 = -85Mpa

To verify it

σ1, σ2 =  σx + σy /2 ± √(σx - σy / 2)² + τxy²

 

            =  -80+50/2   ± √(-80-50/2)² +  25²

σ1= 54̥◦641941 Mpa

σ2= 84◦641941 Mpa

 

Max shear stress

 In mohr's circle τmax = O'G

from measurement;         τmax= 70Mpa 

From equation

τmax =  √(σx–σy/2)² + τxy²

√(-80-50 / 2)² + 25²

τmax = 69◦641941Mpa

 

σavg = σx + σy/2

           = -80+50/2       = -15 Mpa

 

Principal stress element

Maximum shear  stress element

The strength of the longitudinal joint is 30 kips/ft, whereas for the girth joint it is 15 kips/ft. Calculate the maximum diameter of the cylindrical tank if the internal pressure is 120 psi. Also discuss the joint strength and diameter of tank requirement for the given conditions

 Q# The strength of the longitudinal joint is 30 kips/ft, whereas for the girth joint it is 15 kips/ft. Calculate the maximum diameter of the cylindrical tank if the internal pressure is 120 psi. Also discuss the joint strength and diameter of tank  requirement for the given conditions                                                                                                                                                 (     

solution

D= ?

 p=  120psi / 17280lbs/ft² 

 ( for lognitudinal joints)   (consider  1ft length)  

 

F= 2T

P×D×1 = 2(t×1) σ t

lets simplify it;

tangential stress

σ t = pD/2t                                                                                      δt = PD/2t                                

 

30,000lb/ft²/ t  =  17280 ×D/ 2(t)

 

After simplification;

 

D =3.472ft   or 41.667in 

 

For girth joints                                                   

 F=P

P(π/4 ×D²)  = σ l(πDt) 

σ l = pD/ 4t 

15000lb/ft /t   ⇉    17280 ×D /4(t)         (t and t are cut to each other )

 15000×4/ 17280   

 

D= 3.472ft or 41.667in  Governoring diameter ids  D=41.667in 

so in the above case  the smaller diameter is the governing diameter  because  we know that the smaller diameter will be the governring diameter .