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

Construct the Bending-Moment Diagram
Starting with the V diagram, use the steps that follow to
construct the M diagram. (Note: The lowercase letters on
the M diagram correspond to the explanations given for
each step.)
i M(0) = 0 (zero moment at the pinned end of a simply
supported beam).
j M(4) = 56 kip ⋅ ft (Rule 4: The change in bending
moment ∆M between any two points is equal to the area
under the V diagram). The area under the V diagram 14 kips
b c
between x = 0 ft and x = 4 ft is simply the area of
(1)
d 2 kips
rectangle (1), which is 4 ft wide and +14 kips tall. The
V
area of this rectangle is (+14 kips) (4 ft) = +56 kip ⋅ ft
a
(2)
(a positive value). Since M = 0 kip ⋅ ft at x = 0 ft and the
change in bending moment is ∆M = +56 kip ⋅ ft, the
bending moment at x = 4 ft is M j = 56 kip ⋅ ft.
k M(12) = 72 kip ⋅ ft (Rule 4: ∆M = area under the V
diagram). ∆M is equal to the area under the V diagram
between x = 8 ft and x = 12 ft. The area of rectangle (2)
is (+2 kips) (8 ft) = +16 kip ⋅ ft. Therefore, ∆M = +16
M
kip ⋅ ft (a positive value). Since M = +56 kip ⋅ ft at j and
∆M = +16 kip ⋅ ft, the bending moment at k is M k = +56
56 kip.ft
i
j
kip ⋅ ft + 16 kip ⋅ ft = +72 kip ⋅ ft. Even though the shear force decreases from +14 kips to
+2 kips, notice that the bending moment continues to increase in this region.
l M(21) = 0 kip ⋅ ft (Rule 4: ∆M = area under the V diagram). The area under the
V diagram between x = 12 ft and x = 21 ft is the area of rectangle (3), which is
(–8 kips) (9 ft) = −72 kip ⋅ ft (a negative value); therefore, ∆M = −72 kip ⋅ ft. At point k,
M = +72 kip ⋅ ft. The bending moment changes by ∆M = −72 kip ⋅ ft between k and l;
consequently, the bending moment at x = 21 ft is M l = 0 kip ⋅ ft. This result is correct,
since we know that the bending moment at roller D must be zero.
Notice that the M diagram started at M i = 0 and finished at M l = 0. Notice also that the M
diagram consists of linear segments. From Rule 5 (the slope of the M diagram is equal to
the intensity of the shear force V), we can observe that the slope of the M diagram must
be constant between points i, j, k, and l, because the shear force is constant in the corresponding
regions. The slope of the M diagram between points i and j is +14 kips, the M
slope between points j and k is +2 kips, and the M slope between points k and l is –8 kips.
The only type of curve that has a constant slope is a line.
The maximum shear force is V = 14 kips. The maximum bending moment is
M = +72 kip ⋅ ft, at x = 12 ft. Notice that the maximum bending moment occurs where
the shear-force diagram crosses the V = 0 axis (between points e and f ).
A
y
12 kips 10 kips
B
4 ft 8 ft 9 ft
14 kips 8 kips
f
k
C
e
72 kip.ft
(3)
h
g
l
D
x
–8 kips
Relationships Among the Diagram Shapes
Equation (7.3) reveals that the V diagram is obtained by integrating the distributed load w, and
Equation (7.4) shows that the M diagram is obtained by integrating the shear force V. Consider,
for example, a beam segment that has no distributed load (w = 0). For this case, integrating w
gives a constant shear-force function [i.e., a zero-order function f (x 0 )], and integrating a constant
V gives a linear function for the bending moment [i.e., a first-order function f (x 1 )]. If a beam
segment has constant w [i.e., a zero-order function f (x 0 )], then the V diagram is a first-order
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