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

ExAmpLE 7.5
Draw the shear-force and bending-moment diagrams for the
cantilever beam shown.
y
19 kN
Plan the Solution
Initially, determine the reactions at fixed support A. Three sections
will need to be considered, one each for the intervals between
AB, BC, and CD. For each section, draw the appropriate
free-body diagram, develop the equilibrium equations, and solve
the equations for functions describing the variation of V and M
with location x along the span. Plot these functions to complete
the shear-force and bending-moment diagrams.
A
3 kN/m
B
C
2 m 2 m 2 m
D
x
SolutioN
Support Reactions
An FBD of the entire beam is shown. Since no forces act in the x
direction, the reaction force A x = 0 will be omitted from the FBD.
The nontrivial equilibrium equations are as follows:
M A
19 kN
3 kN/m
6 kN
Σ F = A + 19 kN − 6kN = 0
y
y
A
B
C
D
Σ M = − M + (19kN)(2 m) − (6 kN)(5 m) = 0
A
A
2 m 2 m 1 m 1 m
A y
From these equations, the beam reactions are found to be
A
y
=− 13 kN and M = 8 kN⋅m
Since A y is negative, it really acts downward. The correct direction of this
reaction force will be shown in subsequent free-body diagrams.
A
8 kN.m
A
a
a
V
M
Interval 0 ≤ x < 2 m: Section the beam at an arbitrary distance x between A and
B. The FBD for this section is shown. From the equilibrium equations for this
FBD, determine the desired functions for V and M:
13 kN
x
Σ F = −13 kN − V = 0
y
∴ V = −13 kN
(a)
Σ M = (13kN) x − 8kNm ⋅ + M = 0
a−
a
∴ M = − (13kN) x + 8kNm ⋅
(b)
Interval 2 m ≤ x < 4 m: From a section cut between B and C, determine the
desired shear and moment functions:
19 kN
Σ F = − 13 kN + 19 kN − V = 0
y
∴ V = 6kN
(c)
8 kN.m
2 m
b
M
Σ M = (13kN) x − (19kN)( x − 2 m) − 8kNm ⋅ + M = 0
b−
b
∴ M = (6 kN) x − 30kNm ⋅
(d)
13 kN
A
x
B
b
V
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