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

The sum of forces in the vertical direction yields the desired
function for V:
P P
Σ Fy
= − P − V = 0 ∴ V =− (c)
2
2
The equilibrium equation for the sum of moments about section
b–b gives the desired function for M:
A
y
B
P
C
x
Σ =
⎛ L
−
⎞ P
Mb−b
P
⎝
⎜ x
⎠
⎟ − x + M = 0
2 2
P PL
∴ M = − x +
2 2
(d)
P—
2
—
L
2
—
L
2
P—
2
Again, the internal shear force V is constant and the internal
bending moment M varies linearly in the interval L/2 ≤ x < L.
V
M
– P—
2
P—
2
PL
4
Plot the Functions
Plot the functions given in Equations (a) and (b) for the interval
0 ≤ x < L/2, and the functions defined by Equations (c) and
(d) for the interval L/2 ≤ x < L, to create the shear-force and
bending-moment diagram shown.
The maximum internal shear force is V max = ±P/2. The
maximum internal bending moment is M max = PL/4, and it
occurs at x = L/2.
Notice that the concentrated load causes a discontinuity
at its point of application. In other words, the shear-force diagram
“jumps” by an amount equal to the magnitude of the
concentrated load. The jump in this case is downward, which
is the same direction as that of the concentrated load P.
ExAmpLE 7.2
y
M 0
Draw the shear-force and bending-moment diagrams for the
simple beam shown.
A
y
A
A y
L—
2
L—
2
B
B
M 0
L—
2
L—
2
C y
C
C
x
x
Plan the Solution
The solution process outlined in Example 7.1 will be used to
derive V and M functions for this beam.
SolutioN
Support Reactions
An FBD of the beam is shown. The equilibrium equations are
as follows:
Σ F = A + C = 0
y y y
Σ M = − M + CL=
0
A
From these equations, the beam reactions are
M
M
Cy
= and Ay
= −
L
L
0
0 0
y
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