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

1.216 ksi
H
1.216 ksi
H
1.373 ksi
30.3°
x
Bending stress due to M: The 154.2 kip · in. internal bending moment acting as shown
creates tensile normal stresses above the z centroidal axis in the HSS. To compute the
bending stress by the flexure formula, the bending moment has a value M = −154.2 kip · in.
and the y coordinate for point H is y = 2.5 in. The bending stress is then
My ( 154.2 kip in.)(2.5 in.)
s bend =− =− − ⋅ = 3.60 ksi ( T)
I
107.04 in.
4
z
Shear stress due to V: The shear stress at H associated with the 5.71 kip shear force can
be calculated from the shear stress formula:
4.98 ksi
x
0.687 ksi
3.60 ksi
1.396 ksi
3
VQ (5.71 kips)(11.391 in. )
τ H = =
= 1.215 ksi
4
It (107.04 in. )(2 × 0.25 in.)
z
Stress element: The normal and shear stresses at H are shown on the
stress element. The normal stresses due to both the axial force and the
bending moment act in the x direction.
The direction of the shear stress on the stress element can be determined
from the FBD at H. The internal shear force at H acts upward on
the right face of the FBD. The shear stress due to V = 5.71 kips acts in the
same direction—that is, upward on the right face of the stress element.
Stress transformation Results at H
The principal stresses and the maximum shear stress at H can be
determined from the stress transformation equations and procedures
detailed in Chapter 12. The results of these calculations are shown in
the accompanying figure.
0.710 ksi
2.080 ksi
mecmovies
ExAmpLES
m15.2 The inverted tee shape is subjected to a transverse shear
force V and a bending moment M, each acting in the direction
shown. Determine the bending stress, the transverse shear stress
magnitude, the principal stresses, and the maximum shear stress
acting at location H.
FIGURE m15.2
630