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

The 8.45 kN · m (i.e., 8.45 × 10 6 N · mm) torque acting about the
z axis creates shear stress at K. The magnitude of this shear stress can
be calculated from the elastic torsion formula:
y
6
Tc (8.45 × 10 Nmm)(100 ⋅ mm)
τ = =
= 13.438 MPa
4
J 62,879,706 mm
The 1,500 kPa internal fluid pressure creates tensile normal stresses in
the 12 mm thick wall of the pipe. The longitudinal stress in the pipe wall is
pd (1,500 kPa)(176 mm)
s long = = = 5,500 kPa = 5.500 MPa (T)
4t 4(12 mm)
13.438 MPa
8.45 kN·m
H
13.438 MPa
K
x
and the circumferential stress is
pd (1,500 kPa)(176 mm)
s hoop = = = 11,000 kPa = 11.000 MPa ( T)
2t 2(12 mm)
Note that the longitudinal stress acts in the z direction. Furthermore,
the circumferential direction at point K is the y direction.
Combined Stresses at K
The normal and shear stresses acting at point K are summarized on
a stress element. Note that, at point K, the torsion shear stress acts
in the +y direction on the +z face of the stress element. The transverse
shear stress associated with the 9 kN shear force acts in the
opposite direction.
Torsion shear
13.438 MPa
Beam shear
2.533 MPa
z
y
Hoop stress
11.000 MPa
K
Longitudinal stress
5.500 MPa
Bending stress
49.619 MPa
Stress transformation Results at K
The principal stresses and the maximum shear stress at K 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.
The absolute maximum shear stress at K is 29.64 MPa.
10.91 MPa
z
y
11.00 MPa
44.12 MPa
K
10.8°
16.56 MPa
46.2 MPa
29.6 MPa
13.08 MPa
mecmovies
ExAmpLES
m15.6 A 12 kN force is applied to the component shown.
Determine the internal forces acting at section a–a.
651