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

1 ⎛ 2wx
0 ⎞
Σ M =
⎛ ⎞
⎝
⎜
⎠
⎟
⎝ ⎠ − ⎛ ⎞
−
x x w 0 L
a a
x + M = 0
2 L 3 ⎝ 4 ⎠
Hence, the bending-moment equation for this beam is
0 0 3 (a)
M = wLx − w x L (valid for 0 ≤ x ≤ L/2)
4 3
Substitute this expression for M into Equation (10.1) to obtain
EI d 2
v
2
dx
wLx w x
= −
4 3L
0 0 3 (b)
integration
To obtain the elastic curve equation, Equation (b) will be integrated twice. The first
integration gives
EI dv
dx
2
wLx 0 w x
= − + C
8 12L
where C 1 is a constant of integration. Integrating again gives
3
wLx 0 w x
EIv = − + Cx+
C
24 60L
where C 2 is a second constant of integration.
0 4 1 (c)
0 5 1 2 (d)
Boundary Conditions
Moment equation (a) is valid only in the interval 0 ≤ x ≤ L/2; therefore, the boundary
conditions must be found in this same interval. At x = 0, the beam is supported by a pin
connection; consequently, v = 0 at x = 0.
A common mistake in trying to solve this type of problem is to attempt to use the
roller support at C as the second boundary condition. Although it is certainly true that the
beam’s deflection at C will be zero, we cannot use v = 0 at x = L as a boundary condition
for this problem. Why? We must choose a boundary condition that is within the bounds
of the moment equation—that is, within the interval 0 ≤ x ≤ L/2.
The second boundary condition required for evaluation of the constants of integration
can be found from symmetry. The beam is symmetrically supported, and the loading
is symmetrically placed on the span. Therefore, the slope of the beam at x = L/2 must be
dv/dx = 0.
Evaluate Constants
Substitute the boundary condition v = 0 at x = 0 into Equation (d) to find that C 2 = 0.
Next, substitute the value of C 2 and the boundary condition dv/dx = 0 at x = L/2 into
Equation (c), and solve for the constant of integration C 1 :
EI dv
dx
2
wLx 0 w0x
4 2
4
wLL 0 ( /2) w0
( L/2)
= − + C1
⇒ EI(0)
= − + C
8 12L
8 12L
∴ C
1
5w0L
3
= −
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