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

wx 2
L
A
7
24 wL x
—
x
3
a
a
V
2w
L x
M
load that is acting on this FBD is (1/2)x [(2w/L)x] = (wx 2 /L), and it acts at a
distance x/3 from section a–a.
V and M functions applicable for the interval 0 ≤ x < L/2 can be derived from
the equilibrium equations for the FBD:
2 2
7 wx
wx 7
Σ Fy
= wL − − V = 0 ∴ V =− + wL
24 L
L 24
2 3
7 wx
Σ = − +
⎛ x ⎞
wx 7
Ma−a
wLx
⎝
⎜
⎠
⎟ + M = 0 ∴ M = − + wLx
24 L 3
3L
24
(a)
(b)
L—
3
wL
4
A
L—
7
24 wL 2
x
w
B
x –
x –
L—
3
L—
2
x –
L—
2
w
b
b
1—
2
V
x –
M
L—
2
The shear-force function is quadratic (i.e., a second-order
function), and the bending-moment function is cubic (i.e., a
third-order function).
Interval L/2 ≤ x < L: Section the beam at an arbitrary distance
x between B and C. Make sure that you replace the original
distributed loads on the FBD before deriving the V and M
functions.
On the basis of this FBD, the equilibrium equations can
be written as follows:
y
w
7 wL
Σ = − −
⎛ L
−
⎞
⎝
⎜
⎠
7
− wL
F wL w x V V wL − ⎛
⎝
⎜ − L
w x
⎞
y 0
⎠
⎟
24 4 2
24 4 2
(c)
7 wL
Σ = − +
⎛ L
−
⎞
⎝ ⎠ + ⎛
⎝
− L ⎞ 1 ⎛ L
M
−
⎞
a−
a wLx x w x x
⎠ ⎝ ⎠ + M = 0
24 4 3 2 2 2
2
7 wL
∴ M = wLx −
⎛ L x −
⎞
⎝ ⎠ − w⎛
x
⎝
− L ⎞
24 4 3 2 2⎠
(d)
These equations can be simplified to
A
B
7
24 wL 11
L—
L—
24 wL
2
2
7
24 wL 1
24 wL
V
13
24 L
121
5
48 wL2 1152 wL2
C
x
11
– 24 wL
⎛ 13 ⎞
V = w −
⎝
⎜ L x
⎠
⎟
24
and
M = w 2 2
( − 12x + 13 Lx − L )
24
The shear-force function is linear (i.e., a first-order function),
and the bending-moment function is quadratic (i.e., a secondorder
function), between B and C.
Plot the Functions
Plot the V and M functions to create the shear-force and
bending-moment diagram shown.
Notice that the maximum bending moment occurs at a
location where the shear force V is equal to zero.
M
202