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

ExAmpLE 9.9
71 mm
89 mm
280 mm
V
10 mm 110 mm
z
190 mm
c y b a
d
35 mm
z
e
45 mm
V
f
g h i
z
y
y
10 mm
V
40
mm
30
mm
a
60 mm
10 mm
60 mm
20 mm
66 mm
10 mm
140 mm
10 mm
66 mm
84 mm
160 mm
A beam with the cross section shown is subjected to a vertical
shear force V = 450 kN. The location of the neutral axis appears
on the sketch, and the moment of inertia of the cross
section about the neutral axis is I = 30,442,500 mm 4 .
(a) Determine the shear stresses in the cross section at points
a through i in the accompanying sketch.
(b) Plot the distribution of shear stress in the cross section.
Plan the Solution
SolutioN
(a) Shear stress at point a
Begin by making a vertical cut through the top flange of the
cross section at point a. For our calculation of Q, we consider
the flange area between a and the free surface at the rightmost
edge of the flange. The vertical distance from the z axis to the
centroid of this area is 71 mm – 10 mm/2 = 66 mm. Accordingly,
for the area selected,
Q = (66 mm)(60 mm × 10mm) = 39,600 mm 3
The shear stress at point a is
3
VQ (450,000 N)(39,600 mm )
τ a = = = 58.5 MPa (a)
4
It (30,442,500 mm )(10 mm)
While this approach is simple and easy to understand for
point a, other points in the cross section are not so directly
isolated. For this cross section, approach 3 from Example 9.8
might be a more effective method for calculating shear stresses.
For any specific point at which we wish to calculate
shear stress, we will also consider a mirror-image point located
on the opposite side of the vertical axis of symmetry of
the cross section (i.e., the y axis in this figure). Accordingly,
in addition to point a, we will consider point a′ when making
our shear stress calculations. The specific areas are shown in
the accompanying figure. Owing to symmetry, points a and a′
must have the same shear stress magnitude.
For the highlighted area,
Q = 2 × (66 mm)(60 mm × 10 mm) = 79,200 mm 3
60 mm
a'
z
y
10 mm
V
a
60 mm
66 mm
Two vertical cuts have been made through the upper flange to
isolate these areas; therefore, t = 2 × (10 mm) = 20 mm. The
shear stress at point a is
3
VQ (450,000 N)(79,200 mm )
τ a = = = 58.5 MPa (b)
4
It (30,442,500 mm )(20 mm)
Plainly, the results obtained in Equations (a) and (b) are the
same. Note that values used in Equation (a) for the shear force V
and the moment of inertia I are respectively identical to those
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