3. A 24 ft long, simply supported beam carries only a uniform live load (WL). The beam has the following properties: • f,= 60 ksi f= 3,000 psi • b = 14 in • d = 26 in • h = 30 in • Steel: contains five (5) # 8 bars. Compute the depth of the compressive stress block (a) and the actual percentage of reinforcement (p). b) Compute the R value and use this to compute the moment capacity (M.) in kip ft. c) Use the M; value to solve for the ultimate moment (M.). The use this value to solve for the uniform live load (WL) in units of k/ft without the live load service factor (1.6).
3. A 24 ft long, simply supported beam carries only a uniform live load (WL). The beam has the following properties: • f,= 60 ksi f= 3,000 psi • b = 14 in • d = 26 in • h = 30 in • Steel: contains five (5) # 8 bars. Compute the depth of the compressive stress block (a) and the actual percentage of reinforcement (p). b) Compute the R value and use this to compute the moment capacity (M.) in kip ft. c) Use the M; value to solve for the ultimate moment (M.). The use this value to solve for the uniform live load (WL) in units of k/ft without the live load service factor (1.6).
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
For part A: I got 6.64 in and 0.0108 or 1.08%
For part B:
Please compute the R value: R=ρfy(1-a/2d)
Then compute the Mt value: Mt=Rbd^2
Please show how to get units in kip ft
For part C:
Please solve
Transcribed Image Text:3. A 24 ft long, simply supported beam carries only a uniform live load (W). The beam has the
following properties:
f,= 60 ksi
f= 3,000 psi
• b = 14 in
d = 26 in
• h = 30 in
Steel: contains five (5) # 8 bars.
a)
Compute the depth of the compressive stress block (a) and the actual percentage of
reinforcement (p).
b)
Compute the R value and use this to compute the moment capacity (M.) in kip ft.
c)
Use the M; value to solve for the ultimate moment (M.). The use this value to solve for the
uniform live load (WL) in units of k/ft without the live load service factor (1.6).
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