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

320
SHEAR STRESS IN bEAMS
y
9 kN
(2)
x
A
B
C
D
(1)
E
300
mm
150
mm
4.5 kN
1 m 1 m
4.5 kN
4.5 kN
V
4.5 kN·m
−4.5 kN
M
1.350 kN·m
2.025 kN·m
FIGURE 9.1
Simply supported beam with concentrated load applied at midspan.
be designated as member (1), and the upper board will be designated as member (2). The
cross-sectional dimensions of the beam are shown in Figure 9.2.
The objective of this investigation is to determine the forces acting on member (1) at
sections B and C.
From the bending-moment diagram, the internal bending moments at sections B and
C are M B = 1.350 kN · m and M C = 2.025 kN · m, respectively. Both moments are positive;
thus, beam segment BC will be deformed as shown in Figure 9.3a. Compressive normal
stresses will be produced in the upper half of the beam cross section, and tensile normal
stress will be produced in the lower half. The bending stress distribution over the depth of
the cross section at these two locations can be determined from the flexure formula with the
use of a moment of inertia I z = 33,750,000 mm 4 about the z centroidal axis. The distribution
of bending stresses is shown in Figure 9.3b.
To determine the forces acting on member (1), we will consider only those normal
stresses acting between points b and c (on section B) and between points e and f (on
section C). At B, the bending stress varies from 1.0 MPa (T) at b to 3.0 MPa (T) at c. At C,
the bending stress varies from 1.5 MPa (T) at e to 4.5 MPa (T) at f.
From Figure 9.2, the cross-sectional area of member (1) is
A = (50 mm)(120 mm) = 6,000 mm
1
2
z
y
120 mm
(2)
(1)
100 mm
50 mm
FIGURE 9.2 Cross-sectional
dimensions of beam.
To determine the resultant force at section B that acts on this area, the stress distribution can
be split into two components: a uniformly distributed portion having a magnitude of 1.0 MPa
and a triangular portion having a maximum intensity of (3.0 MPa − 1.0 MPa) = 2.0 MPa. By
this approach, the resultant force acting on member (1) at section B can be calculated as
2 2
1
2 2
Resultant FB
= (1.0 N/mm )(6,000 mm ) + (2.0 N/mm )(6,000 mm )
2
= 12,000 N = 12kN
Since the bending stresses are tensile, the resultant force acts in tension on section B.
Similarly, the stress distribution on section C can be split into two components:
a uniformly distributed portion having a magnitude of 1.5 MPa and a triangular portion having