Eccentric loading of columns, Secant formula (Strength of Materials ...

Problem :
Eccentric loading of columns, Secant formula
(Strength of Materials - II, Final Exam-43-2)
A brass pipe having the cross section shown has an axial load P applied 5 mm from its geometric
axis. Using E = 120 GPa, determine (a) the load P for which the horizontal deflection at the midpoint
C is 5 mm, (b) the corresponding maximum stress in the column.
Solution :
The moment of inertia of the beam section is
I = π ¡ 4
r0 − r
4
4¢ π ¡ 4 4
i = 60 − 54
4
¢ =3. 500 5 × 10 6 mm 4
The effective length of the column is
The critical load is
Pcr = π2 EI
l 2
le = l =2.8 m
= π2 × 120 × 10 3 × ¡ 3. 500 5 × 10 6¢
2800 2
Pcr = 528.8 × 10 3 N
The maximum deflection occurs at the midpoint C
" Ã r
π P
ymax = e sec
2
! #
− 1
à r !
π P
sec
2 Pcr
à r !
π P
cos
2
Pcr
P
Pcr
Pcr
= ymax + e
e
=
=
e
ymax + e
∙
2
π arccos
µ ¸2 5
5+5
Dr. M. Kemal Apalak 1

The maximum bending moment is
P = 0.444 44Pcr
P = 0.444 44 × 528.8
P = 235. 02 kN
Mmax = P (ymax + e) = 235. 02 × 10 3 × (5 + 5) × 10 −3
Mmax = 2350.2 N.m
The cross-sectional area is
A = π ¡ r 2 0 − r 2¢ ¡ 2 2
i = π 60 − 54 ¢
A = 2148. 8 mm 2
The maxium normal stress is
σmax = P
A + Mmax × c
I
σmax =
235. 02 × 103
2148. 8
σmax = 149. 66 MPa
+ 2350.2 × 103 × 60
3. 500 5 × 10 6
Dr. M. Kemal Apalak 2