Question

In: Civil Engineering

A strut W 200 x 300 x 19.3 is used as a 10m long fixed connected...

A strut W 200 x 300 x 19.3 is used as a 10m long fixed connected column. Using the Euler's Formula determine the column's critical load and the critical stress.

Solutions

Expert Solution

Solution:- the values given in the question are as follows:

length of column(L)=10 m

size of wide flange column(strut)=W 200*300*19.3

bf=width of flange=300 mm

dw=depth of web=161.4 mm

Calculating Euler's load(critical load)(Pe):-

according to euler's formula critical load is given by-

, [Eq-1]

where, Pe=critical load(euler's load)

Le=effective length

Le=L/2=10/2 , for both end fixed

Le=5 m , or 5000 mm

E=young's modulus of elaticity

E=2*10^5 N/mm^2 , for steel

Imin= minimum moment of inertia about centroid

Imin=minimum of {Ixx , Iyy}

Ixx=moment of inertia about centroid parallel X-axis

Iyy=moment of inertia about centroid parallel Y-axis

Calculating Ixx and Iyy:-

the cross-section of column is symmetrical about both X-axis and Y-axis, so the centroid of section is occur middle of the web shown in above figure.

Ixx=2*moment of inertia of flange about centroid parallel to X-axis+moment of inertia of web about centroid parallel to X-axis.

Ixx=2[(bf*t^3)/12+{bf*t*(h/2-t/2)^2}]+t*dw^3/12

Ixx=2*[(300*19.3^3)/12+{300*19.3*(100-9.65)^2}]+19.3*161.4^3/12

Ixx=94888411.4+6762178.867

Ixx=101650590.3 mm^4

Iyy=2*moment of inertia of flange about centroid parallel to Y-axis+moment of inertia of web about centroid parallel toY-axis.

Iyy=2*[t*bf^3/12]+bw*t^3/12

Iyy=2*[19.3*300^3/12]+161.4*19.3^3/12

Iyy=86850000+96692.81665

Iyy=86946692.82 mm^4

Ixx is greater than Iyy, so minimum moment of inertia is Iyy

Imin=Iyy

Imin=86946692.82 mm^4

values put in above equation-(1) and calculate the value of pe

Pe=(3.14^2*2*10^5*86946692.82)/(5000^2)

Pe=6865035.697 N

pe=6865.0356 kN

Column's critical load(Pe)=6865.0356 kN

Calculating direct stress(d) of column:-

direct critical stress(d)=force(Pe)/area of column

where, force=Pe=6865035.697 N

area of column(A)=2*bf*t+dw*t

area of column(A)=2*300*19.3+161.4*19.3

area of column(A)=14695.02 mm^2

direct stress(d)=6865035.697/14695.02

direct critical stress(d)=467.1674 N/mm^2

critical stress(d)=467.1674 N/mm^2

[Ans]


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