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

Similarly, the moment of inertia I z about the z centroidal axis is calculated as follows:
A i
(in. 2 )
y i
(in.)
I zi
(in. 4 )
| d i |
(in.)
d i2 A i
(in. 4 )
I z
(in. 4 )
(1) 4.00 4 21.3333 1.1410 5.2075 26.5408
(2) 1.75 0.25 0.0365 2.6090 11.9120 11.9485
38.4893
The product of inertia I yz about the centroid is calculated thusly:
A i
(in. 2 )
y i
(in.)
z i
(in.)
_
y – yi
(in.)
_
z – zi
(in.)
I yz = ( y _ – y i )(z _ – z i )A i
(in. 4 )
(1) 4.00 4 −0.25 −1.1410 −0.6090 2.7795
(2) 1.75 0.25 −2.25 2.6090 1.3910 6.3510
9.1304
Coordinates of Points H and K
The (y, z) coordinates of point H are
y
H
= 8in. − 2.859 in. = 5.141 in. and z = 0.859 in.
H
and the coordinates of point K are
y
K
=− 2.859 in. and z = 0.859 in.
K
Moment Components
The bending moment acts about the –z axis; therefore,
M
z
=−9,000 lb⋅ in. and M = 0
y
Bending Stresses at H and K
Since the angle shape does not have an axis of symmetry, the bending stresses at points
H and K must be computed from Equation (8.21) or Equation (8.22). Because M y = 0,
Equation (8.22) is the more convenient of the two equations in this instance. The bending
stress at point H is therefore
⎛ Iz z − Iyz
y⎞
Iyy
Iyzz
σ H = ⎜ M y
M
2 2
⎝ II − I ⎠
⎟ + ⎛ − + ⎞
⎜
⎝ II − I
⎟
⎠
y z
yz
y z
yz
⎡
4 4
(6.7393 in. )(5.141 in.) (9.1304 in. )(0.859 in.)
= 0 + − + ⎤
⎢
( 9,000 lb in.)
4 4 4 2 ⎥ − ⋅
⎣ (6.7393 in. )(38.4893 in. ) − (9.1304 in. ) ⎦
= 1,370 psi = 1,370 psi (T)
z
299