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

6.92 ksi
z
6.92 ksi
H
A
H
y
11.54 ksi
8.49 ksi
8.49 ksi
(1)
K
K
The shear stress created by the 36 kip · in. torque is computed from the elastic torsion
formula:
Tc (36 kip⋅in.)(2 in.)
τ = =
= 6.92 ksi
4
J 10.4004 in.
The normal and shear stresses acting at a point should be summarized on a stress
element before one begins the stress transformation calculations. Often, it is more
efficient to calculate the stress magnitudes from the appropriate formulae but to determine
the proper direction of the stresses by inspection.
x
The axial stress acts in the same direction as the 25 kip force; therefore, the 8.49 ksi
axial stress acts in compression in the x direction.
The direction of the torsion shear stress at the point of
interest can be confusing to determine. Examine the illustration
of the pipe, and note the direction of the equivalent
x 4.24 ksi
torque acting at H. The shear stress arrow on the +x face of
29.2°
the stress element acts in the same direction as the torque;
8.12 ksi therefore, the 6.92 ksi shear stress acts upward on the +x face
of the stress element. After the proper shear stress direction
has been established on one face, the shear stress directions
12.36 ksi on the other three faces are known.
3.88 ksi
8.49 ksi
60 kip·in.
x
25 kips
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.
Equivalent Forces at K
The pipe will be sectioned just to the right of a stress element
at K, and the equivalent forces and moments acting at the
section of interest will be determined. While the equivalent
force is simply the 25 kip axial force, the equivalent torque at K
is the sum of the torques applied to the pipe shaft at B and C.
The equivalent torque at the section of interest is 60 kip · in.
Normal and Shear Stresses at K
The normal and shear stresses at K can be calculated from the
equivalent forces shown in the accompanying figure. The 25 kip
axial force creates a compressive normal stress of 8.49 ksi. The
60 kip · in. equivalent torque creates a shear stress given by
11.54 ksi
K
4.24 ksi
12.29 ksi
8.49 ksi
34.9°
x
16.54 ksi
8.05 ksi
Tc (60 kip⋅in.)(2 in.)
τ = =
= 11.54 ksi
4
J 10.4004 in.
As at H, the 8.49 ksi axial stress acts in compression in the
x direction. The equivalent torque at K creates a shear stress
that acts downward on the +x face of the stress element. The
proper stress element for K is shown.
Stress transformation Results at K
The principal stresses and the maximum shear stress at K are
shown in the accompanying figure.
618