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The document provides calculations for the maximum allowable force a pinned A992 steel column can support against buckling, applying a factor of safety of 2. It details the critical buckling load along both the stronger and weaker axes, resulting in a maximum allowable force of 512.671 kN. The calculations utilize the modulus of elasticity and moment of inertia for the steel column.

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71 views1 page

Learn On The Go: Open in App

The document provides calculations for the maximum allowable force a pinned A992 steel column can support against buckling, applying a factor of safety of 2. It details the critical buckling load along both the stronger and weaker axes, resulting in a maximum allowable force of 512.671 kN. The calculations utilize the modulus of elasticity and moment of inertia for the steel column.

Uploaded by

eunhojang74
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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KO

Chegg

100% (1 )

Answered by
Civil engineering expert High-quality solutions

Consider the column along the major axis and minor axis as shown.

Use the given formula to calculate the allowable force P against buckling.

π 2 EI
P cr = 2
L eff

Here, P cr is the critical buckling load, E is the modulus of elasticity,I is the moment of
inertia, and L is the effective length of column.

(a) Along the stronger axis

2 EI
π x
(P cr ) x =
L 2eff
2 × 200 × 10 9 × 87.9 × 10 −6
π
(P cr ) x =
12 2
(P cr ) x = 1,204,914.204 N
(P cr ) x = 1,204.914 kN

The critical load above is calculated along the stronger axis.

The modulus of elasticity for A992 steel is


200 GPa = 200 × 10 9 Pa = 200 × 10 9 N
m2

(b) Along the weaker axis

π 2 EI y
(P cr ) y =
L 2eff
2 × 200 × 10 9 × 18.7 × 10 −6
π
(P cr ) y =
62
(P cr ) y = 1,025,342.235 N
(P cr ) y = 1,025.342 kN

Calculate the maximum allowable force.

Min[(P cr ) x , (P cr ) y ]
P allow =
FOS
1,025.342
P allow =
2
P allow = 512.671 kN

The maximum allowable force that the column can support without buckling is
512.671 kN.

? 1 13

Q: The A992 steel column can be considered pinned at its top and bottom and braced
against its weak axis at the mid-height. Part A Determine the maximum allowable force P
that the column can support without buckling. Apply a F.S. = 2 against buckling. Take A =
7.4 (10?3)m2 , Ix = 87.3 (10?6)m4 , and Iy = 18.5 (10?6)m4 .

?
.
3 .

Q: The A992 steel column can be considered pinned at its top and bottom and braced
against its weak axis at the mid-height. a) Determine the maximum allowable force PP
that the column can support without buckling. Apply a F.S. = 2 against buckling. Take A =
7.4 * 10-3 m2 , Ix = 88.4 * 10-6 m4 and Iy = 18.8 * 10-6 m4

Q: The A992 steel column can be considered pinned at its top and bottom and braced
against its weak axis at the mid-height. (Figure 1)Determine the maximum allowable force
P that the column can support without buckling. Apply a F. S. =2 against buckling. Take
A=7.4(10^-3) mi^7 I_x=87.3(10^-6) m^4, and I_y=18.5(10^-6) m^4. Express your answer to
three signi#cant #gures and include the appropriate units.

Q: The A992 steel column can be considered pinned at its top and bottom and braced

allowable force 𝐏 that the column can support without buckling. Apply a F. S .=2 against
against its weak axis at the mid-height. (Figure 1) Part A Determine the maximum

buckling. Take A=7.4(10^-3) m^2, I_x=87.6(10^-6) m^4, and I_y=18.1(10^-6) m^4 Express
your answer to three signi#cant #gures and include the appropriate units.

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