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

and the combined normal stress on side K is
σK = σaxial + σ bend =−0.5 ksi − 2.4 ksi = − 2.9 ksi = 2.9 ksi(C) Ans.
Neutral-Axis location
For an eccentric axial load, the neutral axis (i.e., the location with zero stress) is not
located at the centroid of the cross section. Although not requested in this example, the
location of the axis of zero stress can be determined from the combined stress distribution.
By the principle of similar triangles, the combined stress is zero at a distance of 3.958 in.
from the left side of the structural member.
ExAmpLE 8.9
The C-clamp shown is made of an alloy that has a yield strength of 324 MPa in both
tension and compression. Determine the allowable clamping force that the clamp can
exert if a factor of safety of 3.0 is required.
Plan the Solution
The location of the centroid for the tee-shaped
cross section must be determined at the outset.
P P
Once the centroid has been located, the eccentricity
40 mm
e of the clamping force P can be deter-
mined and the equivalent force and moment
a
acting on section a–a established. Expressions
for the combined axial and bending
a
stresses, written in terms of the unknown P,
can be set equal to the allowable normal stress. From these expressions, the
maximum allowable clamping force can be determined.
H
12 mm
4 mm
12 mm
4 mm
12 mm
Section Properties
The centroid for the tee-shaped cross section is located as shown in the sketch
on the left. The cross-sectional area is A = 96 mm 2 , and the moment of inertia
about the z centroidal axis can be calculated as I z = 2,176 mm 4 .
z
6 mm
10 mm
y
4 mm
12 mm
Allowable Normal Stress
The alloy used for the clamp has a yield strength of 324 MPa. Since a
factor of safety of 3.0 is required, the allowable normal stress for this
material is 108 MPa.
K
4 mm
P
internal Force and Moment
A free-body diagram cut through the clamp at section a–a is shown.
The internal axial force F is equal to the clamping force P. The internal
bending moment M is equal to the clamping force P times the eccentricity
e = 40 mm + 6 mm = 46 mm between the centroid of section
a–a and the line of action of P.
40 mm
y
M
F
x
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