Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

column is as shown in Figure 16.8b. From this free-body diagram, the bending moment at
any section can be expressed as
Σ MA
= M + Pv + Pe = 0
∴ M = −Pv − Pe
691
THE SECANT FORMuLA
If the stress does not exceed the proportional limit and deflections are small, the differential
equation of the elastic curve becomes
EI d 2
v + Pv = −Pe
2
dx
or
2
d v P
+ = −
dx EI v P
EI e
2
As in the Euler derivation, the term P/EI will be denoted by k 2 [Equation (16.2)] so that the
differential equation can be rewritten as
The solution of this equation has the form
2
d v
2 2
+ k v = −ke
2
dx
B
x
e
P
v = C sinkx + C cos kx − e
(a)
1 2
Two boundary conditions exist for the column. At pin support A, the boundary condition
v(0) = 0 gives
v(0) = 0 = C sin k(0) + C cos k(0)
− e
1 2
∴ C = e
At pin support B, the boundary condition v(L) = 0 gives
2
vL ( ) = 0 = C sin kL + C coskL − e = C sin kL − e(1 − cos kL)
Using the trigonometric identities
1 2 1
⎡1−
cos kL ⎤
∴ C1
= e⎢
⎥
⎣ sin kL ⎦
2
θ
θ θ
1− cosθ
= 2sin and sinθ
= 2sin cos 2 2 2
allows Equation (a) to be rewritten as
C
1
2
2sin ( kL/2)
kL
= e
⎡ e tan
⎣ ⎢ ⎤
⎥ =
2sin( kL/2)cos( kL/2)
⎦ 2
With this expression for C 1 , the solution of the differential equation preceding Equation (a)
becomes
kL
v = etan sinkx + ecos
kx − e
2
kL
= e
⎡
tan sinkx
+ cos kx − 1
⎤
⎣⎢ 2 ⎦⎥
(16.18)
L
y, v
x
y, v
A
A
x
e
v max
v
P
(a) Pinned–pinned column
M
P
P
(b) Free-body diagram
FIGURE 16.8 Pinned–pinned
column with eccentric load.