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

ExAmpLE 10.5
A beam is loaded and supported as shown. Assume that EI is
constant for the beam. Determine
v
w 0 πx
w(x) = w 0 cos
2L
(a) the equation of the elastic curve in terms of w 0 , L, x, E, and I.
(b) the deflection of the right end of the beam.
(c) the support reactions A y and M A at the left end of the beam.
Plan the Solution
Since the equation for the load distribution is given and the
moment equation is not easy to derive, Equation (10.3) will be
used to determine the deflections.
A
L
B
x
SolutioN
The upward direction is considered positive for a distributed load w; therefore,
Equation (10.3) is written as
π
EI d 4
v = = −
⎛ x
wx ( ) w cos
⎞
4 0 (a)
dx
⎝ 2L ⎠
integration
Equation (a) will be integrated four times to obtain the elastic curve equation:
3
dv
EI
3
dx
π
= = − ⎛ ⎝ ⎜ 2wL
0 ⎞ ⎛ x
V( x)
⎞
π ⎠
⎟ sin
⎝ L ⎠ + C 2
1
(b)
EI d 2
v
2
dx
π
= = ⎛ ⎝ ⎜ 4wL
0 2 ⎞ ⎛ x
Mx ( )
⎞
π ⎠
⎟ cos
⎝ L ⎠ + Cx + C
2
2 1 2 (c)
dv
EI
dx
π
= θ = ⎛ ⎝ ⎜ 8wL
0 3 ⎞ ⎛ x
EI
⎞
π ⎠
⎟ ⎝ L ⎠ + C x 2
sin + Cx + C
3 1 2 3 (d)
2 2
EIv π
=− ⎛ ⎝ ⎜ 16wL 0 4 ⎞ ⎛ x ⎞
π ⎠
⎟ ⎝ L ⎠ + C x 3
+ C x 2
cos + Cx + C
4 1 2 3 4 (e)
2 6 2
Boundary Conditions and Constants
The four constants of integration are determined by applying the boundary conditions:
16wL
At x = 0, v = 0; therefore, C4
=
4
π
dv
At x = 0, = 0; therefore, C3
= 0
dx
2wL
0
At x = L, V = 0; therefore, C1
=
π
2wL
0 2
At x = L, M = 0; therefore, C2
=
π
0 4
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