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

ExAmpLE 7.7
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
A
This example focuses on finding the maximum moment in a
beam that has a uniformly distributed load. To calculate the
maximum moment, we must first find the location where V = 0.
To do this, we will determine the slope of the shear-force diagram from the intensity of
the distributed loading, using Rule 3. Once the location where V = 0 is established, the
maximum bending moment can be calculated from Rule 4.
SolutioN
Support Reactions
An FBD of the beam is shown. For the purpose of calculating
external beam reactions, the −1.5 kN/m distributed load can be
replaced by its resultant force of (1.5 kN/m)(4 m) = 6 kN
acting downward at the centroid of the loading. The equilibrium
equations are as follows:
Σ F = A + C − 6kN = 0
y y y
Σ M = − (6 kN)(2 m) + C (6m) = 0
A
From these equations, the beam reactions are
A
y
= 4kN and C = 2kN
y
y
A
y
y
A y
2 m
1.5 kN/m
4 m 2 m
6 kN
1.5 kN/m
4 m 2 m
B
B
C
C
C y
x
x
Construct the Shear-Force Diagram
On the load diagram, show the reaction forces acting in their proper directions. The original
distributed load—not the 6 kN resultant force—should be used to construct the V
diagram. The resultant force can be used to determine the external beam reactions; however,
it cannot be used to determine the variation in shear force in the beam.
The steps that follow are used to construct the V diagram. (Note: The lowercase
letters on the diagram correspond to the explanations given for each step.)
a V(0 – ) = 0 kN (zero shear at end of beam).
b V(0 + ) = 4 kN (Rule 1: V diagram jumps up by an amount
equal to the 4 kN reaction force).
c V(4) = −2 kN (Rule 2: The change in shear force, ∆V, is
equal to the area under the w curve). The area under the
w curve between A and B is (−1.5 kN/m) (4 m) = −6 kN;
therefore, ∆V = −6 kN. Since V b = +4 kN, the shear force
at c is V b = +4 kN – 6 kN = −2 kN.
Because w is constant between A and B, the slope of
the V diagram is also constant (Rule 3) and equal to
−1.5 kN/m between b and c. Consequently, the V diagram
is linear in this region.
d V(6 − ) = −2 kN (Rule 2: The area under the w curve is
zero between B and C; therefore, ∆V = 0).
A
y
1.5 kN/m
4 m 2 m
4 kN 2 kN
4 kN
V
b
a
(1)
1
2.667 m
1.5 kN/m
e V(6 + ) = 0 kN (Rule 1: V diagram jumps up by an amount equal to the 2 kN reaction
force and returns to V = 0 kN).
f
B
(2) (3)
c
e
d
C
x
–2 kN
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