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

84.2 MPa (C)
y
126.4 MPa (C)
116.7 kN
6.48 MPa (T)
45.4 MPa (T)
A
x
(2)
B
9.72 MPa (T)
68.0 MPa (T)
174.9 kN
(c) Consider the equilibrium of area (2). In the x direction, the sum of the resultant forces is
Σ Fx
= 174.9 kN - 116.8 kN = 58.2 kN ≠ 0
Area (2) is not in equilibrium. What observations can be drawn from this situation?
Whenever a beam segment is subjected to nonuniform bending—that is, whenever the
bending moments are changing along the span of the beam—portions of the beam
cross section will require additional forces in order to satisfy equilibrium in the longitudinal
direction. Where can these additional forces be applied to area (2)?
116.7 kN
–
1
F H 29.1 kN
2
=
z
y
x
(1)
No shear forces
on these surfaces
(3)
–
1
F H 29.1 kN
2
=
174.9 kN
(2)
The additional force in the horizontal direction F H required to satisfy equilibrium
cannot emanate from the upper and lower surfaces of area (2), since these are
free surfaces. Therefore, F H must act at the boundary between areas (1) and (2), and
at the boundary between areas (2) and (3). By symmetry, half of the horizontal force
will act on each surface. Because F H acts along the vertical sides of area (2), it is
termed a shear force.
mecmovies
ExAmpLE
m9.1 Discussion of the horizontal shear force developed
in a flexural member.
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