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

of M diagram = shear force V). The slope of the M diagram is zero at l. As x increases,
V becomes increasingly negative; consequently, the slope of the M diagram becomes
more and more negative until it reaches its most negative slope at x = 20 ft.
The maximum positive bending moment is +122.5 kip ⋅ ft, and it occurs at x = 13 ft. The
maximum negative bending moment is –150 kip ⋅ ft, and this bending moment occurs at x = 0.
ExAmpLE 7.9
Draw the shear-force and bending-moment diagrams for the
cantilever beam shown. Determine the maximum bending
moment that occurs in the beam.
y
60 kN/m
140 kN.m
50 kN
Plan the Solution
The effects of external concentrated moments on the V and M
diagrams are sometimes confusing. Two external concentrated
moments act on this cantilever beam.
A
B C
3 m 1 m 1 m
D
x
SolutioN
Support Reactions
An FBD of the beam is shown. For the purpose of calculating
external beam reactions, the distributed loads are replaced by
their resultant forces. The equilibrium equations are as follows:
Σ F = A + 180 kN − 50 kN = 0
y
y
Σ M = (180 kN)(1.5 m) − (50 kN)(5 m)
A
−140 kN⋅m − M = 0
From these equations, the beam reactions are A y = −130 kN
and M A = −120 kN ⋅ m.
Construct the Shear-Force Diagram
Before beginning, complete the load diagram by noting
the reaction forces and using arrows to indicate their
proper directions. Use the original distributed loads—
not the resultant forces—to construct the V diagram.
a V(0 − ) = 0 kN.
b V(0 + ) = −130 kN (Rule 1).
c V(3) = +50 kN (Rule 2). The area under the w curve
between A and B is +180 kN; therefore, ∆V = +180 kN
between b and c.
d V(4) = +50 kN (Rule 2: ∆V = area under w curve).
There is zero area under the w curve between B and
C; therefore, no change occurs in V.
e V(5 − ) = +50 kN (Rule 2: ∆V = area under w curve).
There is zero area under the w curve between C and
D; therefore, no change occurs in V.
f V(5 + ) = 0 kN (Rule 1).
A
g To complete the V diagram, locate the point between b and c at which V = 0.
The slope of the V diagram in this interval is +60 kN/m (Rule 3). At point b,
V = −130 kN; consequently, the shear force must change by ∆V = +130 kN in order for
y
M A
A y
120 kN.m
130 kN
V
y
–130 kN
1.5 m
180 kN
60 kN/m
140 kN.m
A B C D
3 m 1 m 1 m
60 kN/m
140 kN.m
A B C D
3 m 1 m 1 m
a
b
(1)
2.167 m
g
50 kN
c d e
(2) (3) (4)
50 kN
x
50 kN
f
x
50 kN
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