A 36 ft simply supported beam is loaded with concentrated loads 16 ft inwards from each support. On the left side, the dead load
is 8.0 kips and the live load is 18.0 kips. On the other side, the dead load is 4.0 kips and the live load is 9.0 kips. Self-weight of the beam can be ignored. Lateral supports are provided at the supports and the load points. Determine the least-weight W-shape to carry the load. Use A992 steel, Cb = 1.0 and Table 3.10 of the 15th edition of the Manual. check if a W21×62 will be sufficient considering the correct Cb. Do not use Table 3.10 and work the problem with calculations.
1 answer:
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The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ =
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε =
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:
Thus, modulus of elasticity is 28.6 X 10³ ksi