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

Since C 1 = C 3 ,
C
2 2
Pb( L − b )
= C = −
6L
1 3
Elastic Curve Equations
Substitute the expressions obtained for the constants of integration [i.e., Equations (k)
and (l)] into Equations (e) and (h) to complete the elastic curve equations:
(l)
3 2 2
Pbx Pb( L − b )
EIv = −
x, 6L
6L
which simplifies to
Pbx
v =− − −
LEI L 2 b 2
[
6
x 2] (0 ≤ ≤ )
(m)
and
3
2 2
Pbx P Pb L − b
EIv = − x − a −
x
6L
6 ( ) 3
( )
,
6L
which simplifies to
3
Pbx
−
v =− − − −
LEI L 2 b 2 x 2
Px ( a)
[ ]
6
6EI
( a ≤ x ≤ L)
(n)
The slopes for the two portions of the beam can be determined by substituting the values
for C 1 and C 3 into Equations (d) and (g), respectively, to obtain
Also,
EI dv
dx
2 2 2
Pbx Pb( L − b )
= −
2L
6L
dv Pb
∴ = − − − ≤ ≤
dx LEI L 2 b 2
( 3
6
x 2) (0 x a )
EI dv
dx
2
2 2
Pbx P Pb L − b
= − x − a −
2L
2 ( ) 2
( )
6L
dv Pb
−
∴ = − − − −
dx LEI L 2 b 2 x 2
Px ( a)
( 3 )
6
2EI
2
( a ≤ x ≤ L)
(o)
(p)
The deflection v and slope dv/dx can be computed for any location x along the beam span
from Equations (m), (n), (o), and (p).
Beam Slope at Supports
The slope of the beam can be determined at each support from Equations (o) and (p).
At pin support A, the beam slope is found from Equation (o), with x = 0 and a = L − b:
⎛ dv ⎞ Pb
+
⎝
⎜
⎠
⎟ =− − = − − + = −
dx LEI L 2 b 2
Pb
LEI L b L b Pab( L b)
( ) ( )( )
Ans.
6
6
6LEI
A
At roller support C, the beam slope is found from Equation (p), with x = L:
⎛ dv ⎞ Pb
−
⎝
⎜
⎠
⎟ =− − − −
dx LEI L 2 b 2 L 2
PL ( a)
( 3 )
6
2EI
C
=
2 2
Pb(2L − 3 bL + b )
=
6LEI
Pab( L+
a)
6LEI
2
Ans.
407