(a) A steel column is 4m long with pin jointed ends but restrained from movement normal to the column's length. The column has the rectangular cross section shown in Q1(a). Calculate the force for Euler buckling about the axis shown. Esteel = 205GPa. P n=1 Pe=n² 1² El (² n'El L² Le=LPc = (1²) (205x109) (7.29x106) = 4m 4² P=9218518936N 921 85 KN 1 = least second moment of area for cross section. Therefore, in this case 1xx < lyy Ixx in mm = 7.29 x 106 m4

Elements Of Electromagnetics
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(a) A steel column is 4m long with pin jointed ends but restrained from movement
normal to the column's length. The column has the rectangular cross section
shown in Q1(a). Calculate the force for Euler buckling about the axis shown. Esteel
= 205GPa.
Р
n=1
Pe=n² 1² EI
(²
- n²EI
L²
=L
Le = Pc- (nº) (205 × 10°) (729x10²6)
= 4m
2
4²
Pc = 921851-8936N
= 921-85 KN
Answer: Euler Buckling Force =
Answer:
1 = least second moment
of area for cross
section.
Therefore, in this
case 1xx <lyy
1xx in mmy
= 7.29 x 106 m4
921852
(b) Will the column buckle about this axis, i.e. is the force calculated in Q2(a) the
minimum buckling force? Circle or delete as appropriate.
N
Yes/No
Transcribed Image Text:(a) A steel column is 4m long with pin jointed ends but restrained from movement normal to the column's length. The column has the rectangular cross section shown in Q1(a). Calculate the force for Euler buckling about the axis shown. Esteel = 205GPa. Р n=1 Pe=n² 1² EI (² - n²EI L² =L Le = Pc- (nº) (205 × 10°) (729x10²6) = 4m 2 4² Pc = 921851-8936N = 921-85 KN Answer: Euler Buckling Force = Answer: 1 = least second moment of area for cross section. Therefore, in this case 1xx <lyy 1xx in mmy = 7.29 x 106 m4 921852 (b) Will the column buckle about this axis, i.e. is the force calculated in Q2(a) the minimum buckling force? Circle or delete as appropriate. N Yes/No
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