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

ExAmpLE 8.11
8 in.
8 in.
z
H
K
(1)
y
4 in.
(2)
9,000 lb·in.
0.50 in.
(typ)
4 in.
An unequal-leg angle shape is subjected to a bending moment M = 9,000 lb · in.,
oriented as shown. Calculate the bending stresses at points H and K, and determine
the orientation of the neutral axis.
Plan the Solution
To begin the calculation, we must first locate the centroid of the angle shape.
Then, the area moments of inertia I y and I z and the product of inertia I yz must be
computed with respect to the centroid location. The bending stresses at points H
and K will be computed from Equation (8.21), and the orientation of the neutral
axis will be computed from Equation (8.23).
SolutioN
Section Properties
The angle shape will be subdivided into two areas, (1) and (2), as shown. (Note:
The fillets will be neglected in this calculation.) The corner of the angle will be
used as the reference location (as indicated in the sketch) for calculations in both
the horizontal and vertical directions. The location of the centroid in the vertical
direction is calculated in the following manner:
Ref.
0.50 in.
3.50 in.
0.50 in.
(typ)
A i
(in. 2 )
y i
(in.)
y i A i
(in. 3 )
(1) 4.00 4 16.00
0.859 in.
(2) 1.75 0.25 0.4375
5.75 16.4375
8 in.
(1)
z
2.859 in.
y
(2)
4 in.
yA i i
y = Σ Σ A
i
3
16.4375 in.
= = 2.859 in.
2
5.75 in.
Similarly, the location of the centroid in the horizontal direction is calculated as
follows:
Ref.
0.50 in.
3.50 in.
0.50 in.
(typ)
A i
(in. 2 )
z i
(in.)
z i A i
(in. 3 )
(1) 4.00 −0.25 −1.00
(2) 1.75 −2.25 −3.9375
5.75 −4.9375
z
zA i i
= Σ Σ A
i
3
4.9375 in.
= − =−0.859 in.
2
5.75 in.
The location of the centroid for the angle shape is shown in the sketch. Next, the moment
of inertia I y is calculated for the angle shape about its y centroidal axis:
A i
(in. 2 )
z i
(in.)
I yi
(in. 4 )
| d i |
(in.)
d i2 A i
(in. 4 )
I y
(in. 4 )
(1) 4.00 −0.25 0.0833 0.609 1.4835 1.5668
(2) 1.75 −2.25 1.7865 1.391 3.3860 5.1725
6.7393
298