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

integration
Integrate Equation (f) twice to obtain
and
EI dv
dx
3
w
=−
⎛ L
−
⎞ Ay
x + x
2
+ M A x + C 3
6 ⎝ 2⎠
(g)
2
4
w
=−
⎛ L
−
⎞ Ay
EIv x + x
3
M A
+ x
2
+ C 3 x + C 4
24 ⎝ 2⎠
(h)
6 2
Boundary Conditions
There are four boundary conditions for this beam. Substituting x = 0, dv/dx = 0 into Equation
(d) gives C 1 = 0, and substituting the value of C 1 and x = 0, v = 0 into Equation (e)
gives C 2 = 0. Next, substitute the boundary condition x = L, dv/dx = 0 into Equation (g) to
obtain the following expression for C 3 :
C
3
3 2
wL AL y
= − − MAL
48 2
Finally, substitute the boundary condition x = L, v = 0 and the expression obtained for C 3
into Equation (h) to obtain the following expression for C 4 :
C
4
7wL AL y MAL
= − + +
384 3 2
4 3 2
Continuity Conditions
The beam is a single, continuous member. Consequently, the two sets of equations
must produce the same slope and the same deflection at x = L/2. Consider slope
equations (d) and (g). At x = L/2, these two equations must produce the same slope;
therefore, set the two equations equal to each other and substitute the value L/2 for
each variable x:
This equation reduces to
2
A ⎛ L ⎞
+ M
⎛ L ⎞
⎝ ⎠ ⎝ ⎠ =− w + A ⎛ L 2
y
⎞
+ M
⎛ L ⎞
⎝ ⎠ ⎝ ⎠ + C
2 2 2 6 (0) 3 y
A
A
2 2 2
3 2 2
wL AL
AL
3
y
y
wL
0 = C3
= − − MAL
∴ + MAL
= (i)
48 2 2 48
Similarly, deflection equations (e) and (h) must produce the same deflection at x = L/2:
A y 3 2
L M A L w A L M L
C L C
6 2 2 2 24 (0) y 3 2
⎛ ⎞
4
+
⎛ ⎞
=− +
⎛ ⎞
+
A ⎛ ⎞
+
⎛ ⎞
⎝ ⎠ ⎝ ⎠
6 ⎝ 2⎠
2 ⎝ 2⎠
⎝ 2 ⎠ +
This equation reduces to
3
3 4
C C L wL 4 AL 3 M L 2 wL 3
7
AL 2
y A y
L
4 =−
⎛ ⎞
⎡
⎤
3 ∴ − + + =− MAL
⎝ 2 ⎠
⎢ − − ⎥ ⎛ ⎞
384 3 2 ⎣ 48 2 ⎦
⎝ 2⎠
(j)
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