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

mecmovies
ExAmpLE
m9.2 Derivation of the shear stress formula.
ExAmpLE 9.2
A 10 ft long simply supported laminated wooden beam consists of eight 1.5 in. by 6 in.
planks glued together to form a section 6 in. wide by 12 in. deep, as shown. The beam
carries a 9 kip concentrated load at midspan. Determine
(a) the average horizontal shear stress in the glue joints at b, c, and d, and
(b) the maximum horizontal shear stress in the cross section,
at section a–a, located 2.5 ft from pin support A.
12 in.
1.5 in.
(typ)
z
y
b
c
d
Plan the Solution
The transverse shear force V acting at section a–a can be
determined from a shear-force diagram for the simply
supported beam. To determine the horizontal shear stress
in the indicated glue joints, the corresponding first moment
of area, Q, must be calculated for each location. The
average horizontal shear stress will then be determined by
the shear stress formula given in Equation (9.2).
A
y
2.5 ft
9 kips
a
a
B
5 ft 5 ft
6 in.
C
x
SolutioN
internal Shear Force at Section a–a
The shear-force and bending-moment diagrams can readily
be constructed for the simply supported beam. From the
shear-force diagram, the internal shear force V acting at
section a–a is V = 4.5 kips.
Section Properties
The centroid location for the rectangular cross section can
be determined from symmetry. The moment of inertia of
the cross section about the z centroidal axis is equal to
y
A
4.5 kips
V
2.5 ft
9 kips
a
a
B
5 ft 5 ft
4.5 kips
x
C
4.5 kips
I
z
3 3
bh (6 in.)(12in.)
= = = 864 in.
12 12
4
− 4.5 kips
(a) Average Horizontal Shear Stress in Glue Joints
The shear stress formula is
τ =
VQ
It z
M
11.25 kip·ft
22.5 kip·ft
333