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

1
A
x 1 or x 2
v
m
Draw a free-body diagram around end A of the beam. Place the origin of the x coordinate
system at A when considering segments AB and BC. From the free-body diagram,
derive the following equations for the virtual internal moment m:
= ⎛ ⎝ ⎜ 1 ⎞
m kN⎟ x1 0m≤ x1
≤ 3m
2 ⎠
m
= ⎛ ⎝ ⎜ 1 ⎞
m kN⎟ x2 3m≤ x2
≤ 4.5 m
2 ⎠
v
2 kN E
x 3 or x 4
1
2 kN
Now draw a free-body diagram around end E of the beam. Place the origin of the x
coordinate system at E when considering segments CD and DE. Derive the following
equations for the virtual internal moment m:
= ⎛ ⎝ ⎜ 1 ⎞
m kN⎟ x3 3m≤ x3
≤ 4.5 m
2 ⎠
= ⎛ ⎝ ⎜ 1 ⎞
m kN⎟ x4 0m≤ x4
≤ 3m
2 ⎠
x 2 x 3
x
180 kN
1
45 kN/m
x 4
A
B C D
E
3 m 1.5 m 1.5 m 3 m
300 kN 150 kN
45 kN/m
M
A
V
300 kN
x 1
Real Moment M: Remove the virtual load and reapply
the real loads. Determine the beam reactions, which are
A y = 300 kN and E y = 150 kN, acting in the upward
direction as shown.
Draw a free-body diagram that cuts through segment
AB around support A of the beam. The same x
coordinate used to derive the virtual moment must be
used to derive the real moment; therefore, the origin
of the x coordinate system must be placed at A. From
the free-body diagram, derive the following equation
for the real internal moment M in segment AB of the
beam:
45 kN/m
M =− x + (300 kN) x
2
0 ≤ x ≤ 3 m
1
1 2 1
45 kN/m
180 kN
M
Repeat the process with a free-body diagram cut through segment BC
of the beam. From the free-body diagram, derive the following equation for
the real internal moment M in segment BC of the beam:
A
300 kN
3 m
x 2
B
V
45 kN/m
M =− x − (180 kN)( x − 3m) + (300 kN) x
2
3m≤
x ≤ 4.5 m
2 2 2 2
2
770