Page 211 - Shigley's Mechanical Engineering Design

bud29281_ch04_147-211.qxd 11/27/2009 7:54 pm Page 186 pinnacle s-171:Desktop Folder:Temp Work:Don't Delete (Jobs):MHDQ196/Budynas:

186 Mechanical Engineering Design
Figure 4–21

Comparison of secant and
S
Euler equations for steel with 2 y
ec/k = 0.1
S y = 40 kpsi.
Unit load P/A 1.0 0.3 Euler's curve
0.6

0 50 100 150 200 250
Slenderness ratio l/k

yield strength of 40 kpsi. Note how the P/A contours asymptotically approach the Euler
curve as l/k increases.
Equation (4–50) cannot be solved explicitly for the load P. Design charts, in the
fashion of Fig. 4–21, can be prepared for a single material if much column design
is to be done. Otherwise, a root-finding technique using numerical methods must
be used.

EXAMPLE 4–16 Develop specific Euler equations for the sizes of columns having
(a) Round cross sections
(b) Rectangular cross sections
Solution (a) Using A = πd /4 and k = √ I/A = [(πd /64)/(πd /4)] 1/2 = d/4 with Eq. (4–44)
4
2
2
gives
64P cr l 2 1/4
Answer d = (4–51)
3
π CE
(b) For the rectangular column, we specify a cross section h × b with the
restriction that h ≤ b. If the end conditions are the same for buckling in both directions,
then buckling will occur in the direction of the least thickness. Therefore

bh 3 2 h 2
I = A = bh k = I/A =
12 12
Substituting these in Eq. (4–44) gives
12P cr l 2
Answer b = h ≤ b (4–52)
2
π CEh 3
Note, however, that rectangular columns do not generally have the same end conditions
in both directions.