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

y
120 kN.m
130 kN
V
–130 kN
M
–120 kN.m
60 kN/m
140 kN.m
A B C D
3 m 1 m 1 m
a
b
h
i
(1)
2.167 m
g
j
50 kN
x
it to intersect the V = 0 axis. Use the known slope and the
required ∆V to find ∆x:
∆ V 130 kN
x = ∆ = + =
w + 60 kN/m 2.1667m
Construct the Bending-Moment Diagram
Starting with the V diagram, use the following steps to
construct the M diagram:
c d e h M(0 – ) = 0.
50 kN
(2) (3) (4) i M(0 + ) = −120 kN ⋅ m (Rule 6: For a counterclockwise
f external moment, the M diagram jumps down by an
amount equal to the 120 kN ⋅ m reaction).
j M(2.1667) = −260.836 kN ⋅ m (Rule 4: ∆M = area
under V diagram). Area (1) = −140.836 kN ⋅ m;
therefore, ∆M = −140.836 kN ⋅ m.
n
Use Rule 5 (slope of M diagram = shear force V)
m
to sketch the M diagram between i and j. Since V b =
–50 kN.m
–130 kN, the M diagram has a large negative slope at
l
i. As x increases, the shear force becomes less
k
–190 kN.m
negative until it reaches zero at g. As a result, the
slope of the M diagram will be negative between i
and j but will flatten as it reaches point j.
k M(3) = −240 kN ⋅ m (Rule 4: ∆M = area under V diagram). Area (2) = +20.833
kN ⋅ m; hence, ∆M = +20.833 kN ⋅ m. Adding ∆M to the −260.836 kN ⋅ m moment
at j gives M k = −240 kN ⋅ m at x = 3 m.
Use Rule 5 (slope of M diagram = shear force V) to sketch the M diagram
between j and k. Since V g = 0, the M diagram has zero slope at j. As x increases, the
shear force becomes increasingly positive until it reaches its largest positive value at
point c. This situation means that the slope of the M diagram will be positive between
j and k, curving upward more and more as x increases.
l M(4 – ) = −190 kN ⋅ m (Rule 4: ∆M = area under V diagram). Area (3) = +50 kN ⋅ m. Adding
∆M = +50 kN ⋅ m to the –240 kN ⋅ m moment at k gives M l = −190 kN ⋅ m at x = 4 m.
m M(4 + ) = −50 kN ⋅ m (Rule 6: For a clockwise external moment, the M diagram jumps
up by an amount equal to the 140 kN ⋅ m external concentrated moment).
n M(5) = 0 kN ⋅ m (Rule 4: ∆M = area under V diagram). Area (4) = +50 kN ⋅ m.
50 kN
–260.83 kN.m –240 kN .m
The maximum bending moment is –260.8 kN ⋅ m, and it occurs at x = 2.1667 m.
mecmovies
ExAmpLES
m7.1 Six rules for constructing shear-force and bendingmoment
diagrams.
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