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Unformatted text preview: 11Ch11.qxd 9/27/08 870 3:39 PM Page 870 CHAPTER 11 Columns Solution 11.4-10
L 2.0 m Q a 72 GPa
200 kN n 1.0 m CRITICAL LOAD 3.0
d Pcr P#n 200 mm
Pcr FREE-BODY DIAGRAM OF THE BEAM I Pcr 1200 kN p2EI
4PcrL2 I p2E 27.019 * 106 mm4 MOMENT OF INERTIA
64 I 4 d
P 0P 2Q A (d
d4 2t)4 d
p 2 t min 10.0 mm ; 400 kN Problem 11.4-11 The frame ABC consists of two
members AB and BC that are rigidly connected at joint
B, as shown in part (a) of the figure. The frame has pin
supports at A and C. A concentrated load P acts at joint
B, thereby placing member AB in direct compression.
To assist in determining the buckling load for member
AB, we represents it as a pinned-end column, as shown in
part (b) of the figure. At the top of the column, a rotational
spring of stiffness b R represents the restraining action of
the horizontal beam BC on the column (note that the
horizontal beam provides resistance to rotation of joint
B when the column buckles). Also, consider only bending
effects in the analysis (i.e., disregard the effects of axial
(a) By solving the differential equation of the
deflection curve, derive the following buckling
equation for this column:
(kL cot kL
EI 1) k 2L2 0 in which L is the length of the column and EI is
its flexural rigidity.
(b) For the particular case when member BC is
identical to member AB, the rotation stiffness b R
equals 3EI/L (see Case 7, Table G-2, Appendix G).
For this special case, determine the critical load Pcr. x
B B L L EI y A (a) A (b) ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.
- Fall '11
- Moment Of Inertia