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

ExAmpLE 7.6
A
A
A
y
A y
y
12 kips 10 kips
B
4 ft 8 ft 9 ft
14 kips 8 kips
14 kips
V
212
y
b
a
(1)
12 kips 10 kips
B
4 ft 8 ft 9 ft
12 kips 10 kips
B
4 ft 8 ft 9 ft
c
d
(2)
2 kips
f
C
C
C
e
(3)
h
g
D
D
D y
D
x
x
x
–8 kips
Draw the shear-force and bending-moment diagrams for the
simply supported beam shown. Determine the maximum
bending moment that occurs in the span.
Plan the Solution
Complete the load diagram by calculating the reaction forces
at pin A and roller D. Since only concentrated loads act on this
beam, use Rule 1 to construct the shear-force diagram from
the load diagram. Construct the bending-moment diagram
from the shear-force diagram, using Rule 4 to calculate the
change in bending moments between key points.
SolutioN
Support Reactions
An FBD of the entire beam is shown. Since no loads act in the
horizontal direction, the equilibrium equation ΣF x = 0 is
trivial and will not be considered further. The nontrivial
equilibrium equations are as follows:
Σ Fy = Ay + Dy
−12 kips − 10 kips = 0
Σ M = −(12kips)(4 ft) − (10 kips)(12 ft) + D (21 ft) = 0
A
The following beam reactions can be computed from these equations:
A
y
= 14 kips and D = 8 kips
Construct the Shear-Force Diagram
On the load diagram, show the reaction forces acting in their proper directions. Draw a series
of vertical lines beneath key points on the beam, and draw a horizontal line that will define
the axis for the V diagram. Use the steps outlined next to construct the V diagram. (Note: The
lowercase letters on the V diagram correspond to the explanations given for each step.)
y
a V(0 − ) = 0 kips (zero shear at end of beam).
b V(0 + ) = 14 kips (Rule 1: V diagram jumps up by an
amount equal to the 14 kip reaction).
c V(4 − ) = 14 kips (Rule 2: Since w = 0, the area under
the w curve is also zero. Hence, there is no change in
the shear-force diagram).
d V(4 + ) = 2 kips (Rule 1: V diagram jumps down by
12 kips).
e V(12 − ) = 2 kips (Rule 2: The area under the w curve is
zero; therefore, ∆V = 0).
f V(12 + ) = –8 kips (Rule 1: V diagram jumps down by
10 kips).
g V(21 − ) = –8 kips (Rule 2: The area under the w curve
is zero; therefore, ∆V = 0).
h V(21 + ) = 0 kips (Rule 1: V diagram jumps up by an
amount equal to the 8 kip reaction force and returns
to V = 0 kips).
Notice that the V diagram started at V a = 0 and finished at V h = 0.
y