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18.15 Use Euler's formula and a factor of safety of 2.5 to design

a W14 structural steel wide-flange column to support an
axial load of 350 kips. The length of the column is 34 ft and
its ends are pin-connected.
20

User Mishbah
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Answer:

To design the column, we need to calculate the maximum compressive stress that the column can withstand.

Euler's formula states that the critical compressive stress is given by:

Pcr = (π² * E * I) / L²

where:

Pcr = critical compressive load

E = modulus of elasticity of steel

I = moment of inertia of the cross-sectional area of the column

L = effective length of the column

From the AISC steel manual, we can find the properties of a W14x74 beam:

- Area (A) = 21.8 in²

- Moment of inertia (I) = 735 in⁴

- Modulus of elasticity (E) = 29,000 ksi (kips/in²)

First, we need to calculate the effective length factor, K, for the column. Since the ends of the column are pin-connected, K = 1.0.

Next, we can calculate the critical load:

Pcr = (π² * 29,000 ksi * 735 in⁴) / (34 ft * 12 in/ft)²

Pcr = 859.6 kips

To find the maximum compressive stress, we divide the axial load by the cross-sectional area of the column:

σmax = (2.5 * 350 kips) / (21.8 in²)

σmax = 45.36 ksi

Finally, we check if the maximum stress is less than the allowable stress for the material. From the AISC steel manual, the allowable stress for a W14x74 column is 50 ksi. Since σmax is less than 50 ksi, the design is safe.

Therefore, a W14x74 structural steel wide-flange column is suitable for this application with pin-connected ends, a length of 34 ft, and a factor of safety of 2.5 to support an axial load of 350 kips.

Step-by-step explanation:

User Harry Mallon
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