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

A i
(mm 2 )
y i
(mm)
y i A i
(mm 3 )
(1) 1,500 90 135,000
(2) 1,800 7.5 13,500
3,300 148,500
10 mm
(1)
150 mm
yA i i
y = Σ Σ A
i
3
148,500 mm
= = 45.0 mm
2
3,300 mm
Ref.
axis
Thus, the z centroidal axis is located 45.0 mm above the reference axis for
the inverted-tee cross section.
Ans.
90 mm
The internal bending moment acts about the z centroidal axis, and consequently,
the moment of inertia must be determined about this same axis for the inverted-tee
cross section. Since the centroids of areas (1) and (2) do not coincide with the z centroidal
axis for the entire cross section, the parallel-axis theorem must be used to
calculate the moment of inertia for the inverted tee shape.
The moment of inertia I ci of each rectangular shape about its own centroid must
be computed for the calculation to begin. For example, the moment of inertia of area
(1) about the z centroidal axis for area (1) is calculated as I c1 = bh 3 /12 = (10 mm)
(150 mm) 3 /12 = 2,812,500 mm 4 . Next, the perpendicular distance d i between the
z centroidal axis for the inverted-tee shape and the z centroidal axis for area A i must
be determined. The term d i is squared and multiplied by A i , and the result is added to
I ci to give the moment of inertia for each rectangular shape about the z centroidal axis
of the inverted-tee cross section. The results for all areas A i are then summed to determine
the moment of inertia of the cross section about its centroidal axis. The complete
calculation procedure is summarized in the following table:
7.5 mm
120 mm
10 mm
(2)
15 mm
See Section A.2 of Appendix A
for a review of the parallel-axis
theorem.
I ci
(mm 4 )
| d i |
(mm)
d i2 A i
(mm 4 )
I z
(mm 4 )
(1) 2,812,500 45.0 3,037,500 5,850,000
(2) 33,750 37.5 2,531,250 2,565,000
8,415,000
The moment of inertia of the cross section about its z centroidal axis
is thus I z = 8,415,000 mm 4 .
Ans.
Ref.
axis
z
120 mm
45 mm
Since the inverted-tee cross section is not symmetric about its z
centroidal axis, two section moduli are possible. [See Equation (8.9).] The distance
from the z axis to the upper surface of the cross section will be denoted c top . The section
modulus calculated with this value is
S
top
4
Iz
8,415,000 mm
= = = 70,125 mm
c 120 mm
top
Let the distance from the z axis to the lower surface of the cross section be denoted
c bot . Then, the corresponding section modulus is
3
40 mm
H
K
(1)
y
120 mm
(2)
150 mm
15 mm
S
bot
4
Iz
8,415,000 mm
= = = 187,000 mm
c 45 mm
bot
3
247