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

— 1 w 0 L
v
2
M A
A x
A
B
2L
A y 3
L—
3 B y
A x
448
— 1 w 0
v
2
M A
A y
A
2x
3
x
L
x
will require that six equations be solved. In addition to the three equilibrium equations,
three more equations will be obtained from three boundary conditions. These six equations
will be solved to yield the constants of integration and the unknown beam reactions.
SolutioN
Equilibrium
Consider an FBD of the entire beam. The equation for the sum of forces in the horizontal
direction is trivial, since there are no loads in the x direction:
x
3
a
a
w 0
V
x
L
M
x
w 0
x
Σ F = A = 0
The sum of forces in the vertical direction yields
integration
Equation (d) will be integrated twice to give
and
EI dv
dx
x
x
1
Σ F = A + B − w L = 0
2
y y y 0 (a)
The sum of moments about roller support B gives
Σ M 1
= w L ⎛ L ⎞
B 0
⎝ ⎠ − AL y − M A
2 3
= 0 (b)
Next, cut a section through the beam at an arbitrary
distance x from the origin and draw a free-body diagram,
taking care to show the internal moment M acting
in a positive direction on the exposed surface of
the beam. The equilibrium equation for the sum of
moments about section a–a is
Σ M 1
= w ⎛ x ⎞ ⎛ ⎞
⎝ L ⎠
x x 0
⎝ ⎠ − Ax y − MA
+ M
2 3
= 0
From this equation, the bending-moment equation can
be expressed as
w0 M =− + + ≤ ≤
6L x 3 A y x M A (0 x L ) (c)
Substitute the expression for M into Equation (10.1)
to obtain
EI d 2
v w0 =− + +
dx 6L x 3 A y x M
2
A (d)
w0 L x A
4 y
x
2
=− + + M A x + C 1
(e)
24 2
w0 EIv
L x A
5 y M x
3 A x
2
=− + + + C 1 x + C 2
(f)
120 6 2
Boundary Conditions
For this beam, Equation (c) is valid for the interval 0 ≤ x ≤ L. The boundary conditions,
therefore, are found at x = 0 and x = L. From the fixed support at A, the boundary conditions
are x = 0, dv/dx = 0 and x = 0, v = 0. At roller support B, the boundary condition is
x = L, v = 0.