You are required to carry out the flexural and shear reinforcement design for the simply supported beam shown in Figure 2 below. The beam has a clear span of 25 feet, an overall depth of 20 inches and a width of 12 inches. It is supported by 14 inches deep columns at both ends. The beam carries a uniformly distributed service dead load of 0.82 kip/ft (including its self-weight) and a uniformly distributed service live load of 1.0 kip/ft.

You are required to carry out the flexural and shear reinforcement design for the simply supported beam shown in Figure 2 below. The beam has a clear span of 25 feet, an overall depth of 20 inches and a width of 12 inches. It is supported by 14 inches deep columns at both ends. The beam carries a uniformly distributed service dead load of 0.82 kip/ft (including its self-weight) and a uniformly distributed service live load of 1.0 kip/ft.

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

Members with compression and bending

Compression members are structural elements that carry compressive axial force. They are the vertical members of a building that support beams and their design, which is governed by buckling and strength. Columns are known as struct or compression members, which are subjected to compressive forces. These members provide stiffness to the building and avoid any imperfection regarding the transfer of axial load. For instance, webs in a truss, braces in a framed structure, tiers, and so on.

Question

Say whether shear reinforcement is required and if it is indeed required, carry out the design for the stirrups. Make a sketch of the design.

3) You are required to carry out the flexural and shear reinforcement design for the simply
supported beam shown in Figure 2 below. The beam has a clear span of 25 feet, an overall
depth of 20 inches and a width of 12 inches. It is supported by 14 inches deep columns at
both ends. The beam carries a uniformly distributed service dead load of 0.82 kip/ft
(including its self-weight) and a uniformly distributed service live load of 1.0 kip/ft.
Design Data
●
• fé = 4.35 ksi
Normal weight concrete (wc = 150 lb/ft³)
= 60 ksi
●
25 ft.
Figure 3
fy
Use #9 bars for longitudinal reinforcement
max
● Use #3 bars for stirrups
● Clear cover = 1.5 in.
● Maximum aggregate size a = 0.75 in.

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Step 1: Introduce the problem statement

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Step 2: Compute maximum bending moment and shear force

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Step 3: Design for flexure

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Step 4: Design for flexure

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Step 5: Design for shear

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Step 6: Design for shear

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Step 7: Cross section of beam

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3) You are required to carry out the flexural and shear reinforcement design for the simply supported beam shown in Figure 2 below. The beam has a clear span of 25 feet, an overall depth of 20 inches and a width of 12 inches. It is supported by 14 inches deep columns at both ends. The beam carries a uniformly distributed service dead load of 0.82 kip/ft (including its self-weight) and a uniformly distributed service live load of 1.0 kip/ft. Design Data ● ● ● ● f= = 4.35 ksi Normal weight concrete (wc = = 60 ksi ● Use #3 bars for stirrups ● Clear cover = 1.5 in. fy= Use #9 bars for longitudinal reinforcement ● Maximum aggregate size a 25 ft. max Figure 3 150 lb/ft³) = 0.75 in. a) Carry out the design for the maximum factored bending moment experienced by the beam according to the ACI code requirements and make a dimensioned sketch of the beam showing how the bars are placed in the designed section.
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