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^2Please show how to get units in kip ftFor part C:Please solve 3. A 24 ft long, simply supported beam carries only a uniform live load (W). The beam has thefollowing properties:f,= 60 ksif= 3,000 psi• b = 14 ind = 26 in• h = 30 inSteel: contains five (5) # 8 bars.a)Compute the depth of the compressive stress block (a) and the actual percentage ofreinforcement (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 theuniform live load (WL) in units of k/ft without the live load service factor (1.6).

Answer We are given Beam span = 24 ft fy = 60 ksi f'c = 3 Ksi b = 14 in d = 26 in h = 30 in Steel…

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).

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Connections

In steel structures, connections are the members that join two or more steel members. The connections are the most important members of any steel structure, and hence, they are designed more carefully. The majority of the construction cost of steel structures is incurred due to the connections. The design of connections is done as per the recommendations and specifications of the American Institute of Steel Construction (AISC).

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

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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