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

j M(6) = 0 kN ⋅ m (Rule 4: ∆M = area under the V diagram). The area under the
V diagram between x = 4 m and x = 6 m is simply the area of rectangle (3):
(−2 kN) × (2 m) = −4 kN ⋅ m. Adding ∆M = −4 kN ⋅ m to the bending moment at point i
(M i = +4 kN ⋅ m) gives the bending moment at point j: M j = 0 kN ⋅ m. This result is
correct, since we know that the bending moment at roller C must be zero.
The maximum shear force is V = 4 kN. The maximum bending moment is M =
+5.333 kN ⋅ m, at x = 2.667 m, where the shear-force diagram crosses V = 0 (between
points b and c).
ExAmpLE 7.8
Draw the shear-force and bending-moment diagrams for the
simply supported beam shown. Determine the maximum
positive bending moment and the maximum negative bending
moment that occur in the beam.
Plan the Solution
The challenges of this problem are
(a) to determine both the largest positive and largest
negative moments and
(b) to sketch the proper shape of the M curve as it goes
from negative to positive values.
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 = B + D − 30 kips − 20 kips − 50 kips = 0
y y y
Σ M = (30 kips)(5 ft) − (20 kips)(5 ft)
B
− (50 kips)(15ft) + D (20 ft) = 0
From these equations, the beam reactions are B y =
65 kips and D y = 35 kips.
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(−5 – ) = 0 kips.
b V(−5 + ) = −30 kips (Rule 1).
c V(0 − ) = −30 kips (Rule 2).
There is zero area under the w curve between
A and B; therefore, ∆V = 0 between b and c.
d V(0 + ) = +35 kips (Rule 1).
y
30 kips
V
A
–30 kips
a
b
30 kips
A
30 kips
5 ft
y
35 kips
(1)
c
A
5 ft
5 ft
2 kips/ft
B
y
y
2 kips/ft
B
2 kips/ft
B y
B
C
10 ft 10 ft
20 kips 50 kips
5 ft 10 ft
C
10 ft 10 ft
C
5 kips/ft
5 kips/ft
5 kips/ft
10 ft 10 ft
65 kips 35 kips
d
(2)
2 kips/ft
∆V = –20 kips
15 kips
e
5 kips/ft
(3)
g
3 ft
h
(4)
f
D
x
–35 kips
D y
D
D
x
x
∆V = –50 kips
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