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

integration
Equation (b) will be integrated twice. The first integration gives a general equation for the
slope dv/dx of the beam:
EI dv
dx
2
Px
=− + C1 (c)
2
Here, C 1 is a constant of integration. A second integration gives a general equation for the
elastic curve v:
3
Px
EIv =− + Cx 1 + C2 (d)
6
In this equation, C 2 is a second constant of integration. The constants C 1 and C 2 must be
evaluated before the slope and elastic curve equations are complete.
Boundary Conditions
Boundary conditions are values of the deflection v or slope dv/dx that are known at particular
locations along the beam span. For this beam, the bending-moment equation M in
Equation (a) is valid in the interval 0 ≤ x ≤ L. The boundary conditions, therefore, are
found at either x = 0 or x = L.
Consider the interval 0 ≤ x ≤ L for this beam and loading. At x = 0, the beam is unsupported.
The beam will deflect downward, and as it deflects, the slope of the beam will no
longer be zero. Consequently, neither the deflection v nor the slope dv/dx is known at x = 0.
At x = L, the beam is supported by a fixed support. The fixed support at B prevents
deflection and rotation; therefore, we know two bits of information with absolute certainty
at x = L: v = 0 and dv/dx = 0. These are the two boundary conditions that will be
used to evaluate the constants of integration C 1 and C 2 .
Evaluate Constants
Substitute the boundary condition dv/dx = 0 at x = L into Equation (c) to evaluate the
constant C 1 :
EI dv
dx
2
2
Px
PL ( )
PL
=− + C1
⇒ EI(0)
=− + C1 ∴ C1
=
2
2 2
Next, substitute the value of C 1 and the boundary condition v = 0 at x = L into Equation (d),
and solve for the second constant of integration C 2 :
3
3 2
Px
PL ( ) PL PL
EIv =− + Cx 1 + C2
⇒ EI(0)
= − + L + C ∴ C =−
6
6 2 ( ) 2 2
3
Elastic Curve Equation
Substitute the expressions obtained for C 1 and C 2 into Equation (d) to complete the elastic
curve equation:
3 2 3
Px PL PL
P
EIv =− + x − v = − x + L x − L
6 2 3 , which simplifies to [ 3 2 ]
6EI
3 2 3
(e)
Similarly, the beam slope equation from Equation (c) can be completed with the expression
derived for C 1 :
EI dv
dx
2 2
Px PL dv P
=− + = −
2 2 ,which simplifies to [
dx 2EI L 2 x 2
] (f)
2
3
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