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Unformatted text preview: Columns: 2. A ﬂag pole is subjected to compressive load P. The geometry of the pole is as shown. p * E = 29 x 106 Ib/in2 Column is of circular cross section PM Find the maximum possible P if column diameter is 6 inches. t = 0.25 in
30 ft Solution: A=%(D§ —Di2)=%62in2 —11.5 in2)=9.23 in2 _£ 4_ 4 _£ .4_ .4 _ .4
1—64(DO D,)64(12m [1.5m )—59.3m 7:251 _ n2 x29x10" 1b: ml x1593 in“
L3 12 in 2 Pm = 87,863 lb Pm = = 87,863 lb [2.0x30fx I, Columns: 3. What should be the minimum diameter of the cane, (1, such
that the cane (length L and modulus E) will not buckle while supporting half
of Charlie’s weight, 129. W/2 ? Assume pinpin boundary conditions for the
cane. Solution:
For pinpin boundary conditions
12 anI W WL’
Pa = 2 = — —) I = 2
L 2 27: E
For a circular cross section
m1“
1 = —
64
Solving the above two equations for d yields
d. = 32sz
2
7: E Columns: 5. Beam ACE is simply supported by a roller at A, a pin at B, and by column CE which is
pinned to the beam at C and pinned externally at E. Find the load F that will buckle column CE. Beam
ACB has stiffness EI and the column is of circular cross section and has stiffness 0.25EI. Note that for a
simply supported beam of length L and stiﬁness EI loaded at midspan by load P, the deﬂection at P13 Solution: Let EB, 13 denote the modulus of elasticity and moment of inertia of beam ACB, respectively, and EC, I C the modulus of elasticity and moment of inertia of column CE. The beam is subjected to a force
FR, where R denotes the force in column CE. F R L3
Deﬂection of beam at C: 56 = M
485,1,
’ RLCE
Displacement of column CE at C: 6C =
ECA 3
Compatibility implies the 2 above displacements are equal. This yields m = h
4813,], ECA 2
7! E I
At critical load R = P“, =%
LCE The above 2 equations yield
wild/11543, +48EBIBLCE) _ 7:110 (3,415C + 215,13)
ALZBLiE 27A
_ Ir’ECICOA +81C)
27A F = and since ERIE = 4ECIC F Columns: 6 Consider the truss shown in the ﬁgure subjected to a force P. Each rod of the truss is made of
a material of elastic modulus E; has a circular cross section (radius of cross section R). All joints are pin
joints. What is the critical force P at which one or both rods buckle (in terms of E, R, H)? For a circle of ItR‘ radius R, I =
4 _____________)._____ Solution Let F denote the force in each bar. Equilibrium of the joint, and since symmetry implies the two forces
are equal: P= 2F~00360° =F .
From geometry of the truss: H = L '005 600 —) L = 2H 7:2E1 anI 7:2 E] 7:12“ 7:3 ER‘ Then
3
P” = 7r El:4
16H Columns: 12 Column CD of length L and rectangular crosssection bxh, where h>b, is ﬁxed at D and at
C it is braced (motion restricted) in the CrosHecﬁom
‘Weak” ydirection but not braced in the bxh Y
“strong” zdirection. For what ratio of h/b T x; H :[b 
will the critical load for buckling in the p
“weak” direction be equal to the critical ,—.
load for buckling in the “strong” direction? L h Solution, an bh3
bh3 — 2E bh3
For buckling in the “strong” direction, I = — , and Le =2L, thus P = —122 = u—z—
12 " (2L) L 48
. . . . . hb3
For buckling in the “weak” direction, With the brace, I = E , and Le=0.7L, thus
hb3
2
P TC “ZE hb3 3 3 2
The two above critical loads are equal when hb = bh > h = 48 > h = 2.857 .
5.88 48 b 5.88 b Thus, for h/b=2.85 7, bracing will make the buckling load in the “weak” direction equal to the
buckling load in the “strong” direction. ...
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This note was uploaded on 08/25/2011 for the course CE 215 taught by Professor Frantziskonis during the Summer '08 term at University of Arizona Tucson.
 Summer '08
 FRANTZISKONIS

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