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

Boundary Conditions
Boundary conditions for the cantilever beam are
x = 0, v = 0 and x = 0, dvdx / = 0
Evaluate Constants
Substitute the boundary condition dv/dx = 0 at x = 0 into Equation (c) to evaluate the
constant C 1 :
EI dv
dx
w w wL
= L − x + C ⇒ EI = L − + C ∴ C =−
6 ( ) 3
(0) 6 ( 0) 3
1
1 1
6
Next, substitute the value of C 1 and the boundary condition v = 0 at x = 0 into Equation (d),
and solve for the second constant of integration C 2 :
3
w w wL
EIv =− L − x + Cx + C ⇒ EI =− L − − + C
24 ( ) 4
(0) 24 ( 0) 4
1 2
6 (0)
wL
∴ C2
=
24
Elastic Curve Equation
Substitute the expressions obtained for C 1 and C 2 into Equation (d) to complete the elastic
curve equation:
4
3
2
3 4 2
w wL wL
wx
EIv =− L − x − x + v = − L − Lx + x
24 ( ) 4
6 24 , which simplifies to 2 2
(6 4 ) (e)
24EI
Similarly, the beam slope equation from Equation (c) can be completed with the expression
derived for C 1 :
EI dv
dx
3
w wL dv wx
= L − x − = − L − Lx + x
6 ( ) 3
6 , which simplifies to 2 2
(3 3 ) (f)
dx 6EI
Beam Deflection at B
At the tip of the cantilever, x = L. Substituting this value into Equation (e) gives
EIv
B
3 4 4
w wL wL
wL
=− L − L − L + ∴ v =−
24 [ ( )] 4
6 ( ) B
24 8EI
Ans.
Beam Rotation Angle at B
If the beam deflections are small, the rotation angle θ is equal to the slope dv/dx. Substituting
x = L into Equation (f) gives
3
⎛ ⎞
θ
⎝
⎜
⎠
⎟ = − − ∴ ⎛ 3
dv w wL
⎝ ⎜ dv ⎞ wL
EI
L L
⎠
⎟ =− =
dx 6 [ ( )] 3
6 dx 6EI
B
B
B
Ans.
404