Page 209 - Shigley's Mechanical Engineering Design

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184 Mechanical Engineering Design
greater than (l/k) 1 . . Otherwise, use one of the methods in the sections that follow. See
Examples 4–17 and 4–18.
Most designers select point T such that P cr /A = S y /2. Using Eq. (4–43), we find
the corresponding value of (l/k) 1 to be
1/2
2
l 2π CE
= (4–45)
k
1 S y
4–13 Intermediate-Length Columns with Central Loading
Over the years there have been a number of column formulas proposed and used for the
range of l/k values for which the Euler formula is not suitable. Many of these are based
on the use of a single material; others, on a so-called safe unit load rather than the crit-
ical value. Most of these formulas are based on the use of a linear relationship between
the slenderness ratio and the unit load. The parabolic or J. B. Johnson formula now
seems to be the preferred one among designers in the machine, automotive, aircraft, and
structural-steel construction fields.
The general form of the parabolic formula is
2
P cr l
= a − b (a)
A k
where a and b are constants that are evaluated by fitting a parabola to the Euler curve
of Fig. 4–19 as shown by the dashed line ending at T . If the parabola is begun at S y ,
then a = S y . If point T is selected as previously noted, then Eq. (4–42) gives the value
of (l/k) 1 and the constant b is found to be

2
S y 1
b = (b)
2π CE
Upon substituting the known values of a and b into Eq. (a), we obtain, for the parabolic
equation,
2

P cr S y l 1 l l
= S y − ≤ (4–46)
A 2π k CE k k
1
4–14 Columns with Eccentric Loading
We have noted before that deviations from an ideal column, such as load eccentrici-
ties or crookedness, are likely to occur during manufacture and assembly. Though
these deviations are often quite small, it is still convenient to have a method of
dealing with them. Frequently, too, problems occur in which load eccentricities are
unavoidable.
Figure 4–20a shows a column in which the line of action of the column forces is
separated from the centroidal axis of the column by the eccentricity e. From Fig. 4–20b,
2
2
M =−P(e + y). Substituting this into Eq. (4–12), d y/dx = M/EI, results in the
differential equation
2
d y P Pe
+ y =− (a)
dx 2 EI EI