Answer :
The center line deflection of the section is 0.0513 ft. As per the maximum permissible center line deflection of 0.0583 ft, the center line deflection of this section is satisfactory for the service live load.
W24 x 55 (Ix = 1350 in ) is selected for a 21 ft simple span to support a total service live load of 3 k/ft (including beam weight).
Use E = 29000 ksi.
The maximum permissible value of center line deflection is 1/360 of the span.
The maximum permissible center line deflection can be computed as;
[tex]$$\Delta_{max} = \frac{L}{360}$$[/tex]
Where, [tex]$$L = 21\ ft$$[/tex]
The maximum permissible center line deflection can be computed as;
[tex]$$\Delta_{max} = \frac{21\ ft}{360}$$$$\Delta_{max} = 0.0583\ ft$$[/tex]
The total service live load is 3 k/ft. So, the total load on the beam is;
[tex]$$W = \text{Load} \times L
= 3\ \text{k/ft} \times 21\ \text{ft}
= 63\ \text{k}$$[/tex]
The moment of inertia for the section is;
[tex]$$I_x = 1350\ in^4$$$$= 1.491 \times 10^{-3} \ ft^4$$[/tex]
The moment of inertia can be converted to the moment of inertia in SI units as follows;
[tex]$$I_x = 1.491 \times 10^{-3} \ ft^4$$$$= 0.0015092 \ \text{m}^4$$$$\Delta_{CL} = 0.0513\ ft$$[/tex]
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