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

obtained from the continuity conditions at x = L/2. Finally, two nontrivial equations will
be derived from equilibrium for the entire beam. These eight equations will be solved to
yield the constants of integration and the unknown beam reactions.
SolutioN
Equilibrium
Draw an FBD of the entire beam. Since no loads act in the horizontal direction, the reactions
A x and C x will be omitted. Write two equilibrium equations:
M A
v
3L
4
w
wL
2
L—
4
M C
wL
Σ F = A + C − = 0
2
y y y (a)
Σ M wL
=
⎛ L ⎞
C
⎝ ⎠ − AL y − MA + MC
2 4
= 0
(b)
A y
v
A
L—
2
B
L—
2
C
C y
x
For this beam, two equations are required to describe the
bending moments for the entire span. Draw two FBDs:
one FBD that cuts through the beam between A and B,
and the second FBD that cuts through the beam between
B and C. From these two FBDs, derive the bendingmoment
equations and, in turn, the differential equations
of the elastic curve.
M A
M A
A y
v
A y
A
A
x
L—
2
a
a
V
x
M
B
x
w
x –
w x –
L—
2
L—
2
b
b
V
M
x
For the interval 0 ≤ x ≤ L/2 Between A and B,
M = A x + M
which gives the differential equation
y
EI d 2
v = Ax y + M
2
dx
integration
Integrate Equation (c) twice to obtain
and
EI dv
dx
A
A
(c)
Ay
= x
2
+ M A x + C 1 (d)
2
Ay
EIv = x
3
M A
+ x
2
+ C 1 x + C 2 (e)
6 2
For the interval L/2 ≤ x ≤ L Between B and C,
which gives the differential equation
w
=−
⎛ L
M x −
⎞
2 ⎝ 2⎠
2
+ Ax+
M
y
A
EI d 2
v
2
dx
w
=−
⎛ L x −
⎞
2 ⎝ 2⎠
2
+ Ax+
M
y
A
(f)
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