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

Since KL/ry > 133.7, the column is classified as a long column, so Equation (16.24) is
used to calculate the critical compression stress:
σcr
= 0.877 σ = 0.877(10.81 ksi) = 9.48 ksi
The allowable compression stress is determined from Equation (16.25):
σ
allow
e
σ cr 9.48 ksi
= = = 5.68 ksi
1.67 1.67
The allowable axial load for KL = 12 ft can be calculated from the allowable stress:
2
P = σ A = (5.68 ksi)(2.38 in. ) = 13.52 kips
Ans.
allow
allow
ExAmpLE 16.6
z
y
70 mm
3 mm
70 mm
A square tube made of 6061-T6 aluminum alloy has the cross-sectional dimensions
shown. Use the Aluminum Association column design formulas to determine
the allowable axial load P allow that may be supported by the tube if the
effective length of the compression member is
(a) KL = 1,500 mm.
(b) KL = 2,750 mm.
Plan the Solution
After computing the section properties of the square tube, the Aluminum Association
design formulas [Equations (16.26) through (16.28)] will be used to
calculate the allowable loads for the specified effective lengths.
SolutioN
Section Properties
The centroid of the square tube is found from symmetry. The cross-sectional area of the
tube is
2 2 2
A = (70 mm) − (64 mm) = 804 mm
The moments of inertia about both the y and z centroidal axes are identical:
I
y
4 4
(70 mm) (64 mm)
= Iz
= − = 602,732 mm
12 12
Similarly, the radii of gyration about both centroidal axes are the same:
4
r
y
4
602,732 mm
= rz
= = 27.38 mm
2
804 mm
(a) Allowable axial load P allow for KL = 1,500 mm: For an effective length KL = 1,500 mm,
the effective-slenderness ratio for the square tube member is
KL
r
= 1,500 mm
27.38 mm = 54.8
Since this slenderness ratio is greater than 9.5 and less than 66, Equation (16.27) must be
used to determine the allowable compression stress. The SI version of this equation can
be used to give σ allow :
702
σ = [139 − 0.868( KL/ r)] MPa = [139 − 0.868(54.8)] = 91.43 MPa
allow