maximum load P
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![The aluminum W 310 x 202 (wide flange) column is subjected to an eccentric axial
load P as shown. The maximum compressive stress in the column is given by the
so-called secant formula:
where
--- [¹ + ( √5)]
ec
o 1+ sec
max
E
ō = P/A = average stress
A = 25800 mm² = cross-sectional area of the column
e = 85 mm = eccentricity of the load
c = 170 mm = half depth of the column
r = 142 mm = radius of gyration of the cross section
L = 7100 mm = length of the column
E = 71 x 10⁹ Pa = modulus of elasticity
Determine the maximum load P that the column can carry if the maximum stress
is not to exceed 120 × 10^6 Pa. Terminate the computation when Σa <
0.0005%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc0b8fd4-83f5-4ce5-b54d-512021d4fb00%2Fe9bc29ba-90f2-43a3-91bc-56d25235b574%2F865jhud_processed.png&w=3840&q=75)
Transcribed Image Text:The aluminum W 310 x 202 (wide flange) column is subjected to an eccentric axial
load P as shown. The maximum compressive stress in the column is given by the
so-called secant formula:
where
--- [¹ + ( √5)]
ec
o 1+ sec
max
E
ō = P/A = average stress
A = 25800 mm² = cross-sectional area of the column
e = 85 mm = eccentricity of the load
c = 170 mm = half depth of the column
r = 142 mm = radius of gyration of the cross section
L = 7100 mm = length of the column
E = 71 x 10⁹ Pa = modulus of elasticity
Determine the maximum load P that the column can carry if the maximum stress
is not to exceed 120 × 10^6 Pa. Terminate the computation when Σa <
0.0005%.
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