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

e V(10) = +15 kips (Rule 2: ∆V = area under w curve). The area under the w curve
between B and C is −20 kips. Since w is constant in this region, the slope of the
V diagram is also constant (Rule 3) and equal to −2 kips/ft between d and e.
f V(20 − ) = −35 kips (Rule 2: ∆V = area under w curve). The area under the w curve
between C and D is −50 kips. The slope of the V diagram is constant (Rule 3) and
equal to −5 kips/ft between e and f.
g V(20 + ) = 0 kips (Rule 1).
h To complete the V diagram, locate the point between e and f where V = 0. The slope of the
V diagram in this interval is −5 kips/ft (Rule 3).
At point e, V = +15 kips; consequently, the shear force must change by ∆V = −15 kips in
order for it to intersect the V = 0 axis. Use the known slope and the required ∆V to find ∆x:
V 15 kips
∆ x = ∆ = − = 3.0 ft
w −5kips/ft
Since x = 10 ft at point e, point h is located at x = 13 ft.
Construct the Bending-Moment Diagram
Starting with the V diagram, the steps that follow are used to construct the M diagram:
30 kips
V
A
–30 kips
a
b
5 ft
y
35 kips
(1)
c
M i Constant
negative
slope
j
2 kips/ft
B
C
5 kips/ft
10 ft 10 ft
65 kips 35 kips
d
(2)
1— (35 kips + 15 kips)(10 ft) = +250 kip.ft
2
15 kips
e
(3)
3 ft
1— (–35 kips)(7 ft) = –122.5 kip.ft
2
100 kip.ft
Positive slope
Positive slope
Positive slope
Large positive slope
–150 kip.ft
k
h
l
1—
2
(4)
g
f
D
(15 kips)(3 ft) = +22.5 kip.ft
122.5 kip.ft
x
–35 kips
Small negative slope
Negative slope
Large negative
slope
m
i M(–5) = 0 (there is zero moment at the free end
of a simply supported beam).
j M(0) = −150 kip ⋅ ft (Rule 4: ∆M = area under V
diagram). The area of region (1) is (−30 kips)
(5 ft) = −150 kip ⋅ ft; therefore, ∆M = −150 kip ⋅ ft.
The M diagram is linear between points i and j,
having a constant negative slope of −30 kips.
k M(10) = +100 kip ⋅ ft (Rule 4: ∆M = area under
V diagram). The area of trapezoid (2) is +250
kip ⋅ ft; hence, ∆M = +250 kip ⋅ ft. Adding ∆M =
+250 kip ⋅ ft to the −150 kip ⋅ ft moment at j gives
M k = +100 kip ⋅ ft at x = 10 ft.
Use Rule 5 (slope of M diagram = shear
force V) to sketch the M diagram between
j and k. Since V d = +35 kips, the M diagram
has a large positive slope at j. As x increases,
the shear force stays positive but decreases to a
value V e = +15 kips at point e. As a result, the
slope of the M diagram will be positive between
j and k but will flatten as it nears point k.
l M(13) = +122.5 kip ⋅ ft (Rule 4: ∆M = area under
V diagram). Area (3) under the V diagram is
+22.5 kip ⋅ ft; thus, ∆M = +22.5 kip ⋅ ft. Add
+22.5 kip ⋅ ft to M k = +100 kip ⋅ ft to compute
M l = +122.5 kip ⋅ ft at point l. Since V = 0 at this
location, the slope of the M diagram is zero at
point l.
m M(20) = 0 kip ⋅ ft (Rule 4: ∆M = area under
V diagram). The area of triangle (4) is −122.5
kip ⋅ ft; therefore, ∆M = −122.5 kip ⋅ ft.
The shape of the bending-moment diagram between
l and m can be sketched from Rule 5 (slope
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