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

f
Before the V diagram is complete, we must locate the point between b and c where
V = 0. To do this, recall that the slope of the shear-force diagram (dV/dx) is equal to
the intensity of the distributed load w (Rule 3). In this instance, a finite length ∆x of
the beam is considered rather than an infinitesimal length dx. Accordingly, Equation
(7.1) can be expressed as
V
Slopeof V diagram = ∆ w
∆ x
= (a)
Given that the distributed load is w = −1.5 kN/m between points A and B, the slope
of the V diagram between points b and c is equal to −1.5 kN/m. Since V = 4 kN at
point b, the shear force must change by ∆V = −4 kN to cross the V = 0 axis. Use the
known slope and the required ∆V to solve for ∆x from Equation (a):
∆ V 4kN
x = ∆ −
=
=
w −1.5 kN/m 2.667m
Since x = 0 m at b, point f is located 2.667 m from the left end of the beam.
Construct the Bending-Moment Diagram
Starting with the V diagram, the steps that follow are used to construct the M diagram.
(Note: The lowercase letters on the M diagram correspond to the explanations given for
each step.)
g M(0) = 0 (zero moment at the pinned end of the simply
y
1.5 kN/m
supported beam).
h M(2.667) = +5.333 kN ⋅ m (Rule 4: The change in bending
x moment, ∆M, between any two points is equal to the area
A
B
C under the V diagram). The V diagram between b and f is a
triangle (1) with a width of 2.667 m and a height of +4 kN.
4 m 2 m
The area of this triangle is +5.333 kN ⋅ m; therefore,
4 kN 2 kN
∆M = +5.333 kN ⋅ m. Since M = 0 kN ⋅ m at x = 0 m and
b
∆M = +5.333 kN ⋅ m, the bending moment at x = 2.667 m
4 kN
1— (4 kN)(2.667 m) = 5.333 kN.m
2
is M h = +5.333 kN ⋅ m.
(1)
The shape of the bending-moment diagram between g
f
e
V
and h can be sketched from Rule 5 (the slope of the M
a
(2) (3)
diagram is equal to the shear force V). The shear force at b is
2.667 m
–2 kN
c
d
+4 kN; therefore, the M diagram has a large positive slope
at g. Between b and f, the shear force is still positive but
1— (–2 kN)(1.333 m) = –1.333 kN.m
2
decreases in magnitude; consequently, the slope of the M
diagram is positive but becomes less steep as x increases.
5.333 kN.m
At f, V = 0, so the slope of the M diagram becomes zero.
h
i
i M(4) = +4 kN ⋅ m (Rule 4: ∆M = area under the V diagram).
The shear-force diagram between f and c is a
4 kN.m
triangle (2) with a width of 1.333 m and a height of
j −2 kN. This triangle has a negative area of −1.333 kN ⋅ m;
M
g
therefore, ∆M = −1.333 kN ⋅ m. At h, M = +5.333 kN ⋅ m.
Adding ∆M = −1.333 kN ⋅ m to this value gives the
bending moment at x = 4 m: M i = +4 kN ⋅ m.
The shape of the bending-moment diagram between h and i can be sketched
from Rule 5 (the slope of the M diagram is equal to the shear force V). The slope of
the M diagram is zero at h, corresponding to V = 0 at f. As x increases, V becomes
increasingly negative; consequently, the slope dM/dx of the M diagram becomes
more and more negative until it reaches −2 kN at point i.
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