Consider the following simply supported beam. It has normal-weight concrete with a compressive strength of 5,000 psi and Grade-60 bars. Each support is 12 in. wide. The section is a hollow box. The beam is to carry a uniformly distributed dead load of 2.5 kips/ft., including its own weight. The live load is a point load of 60 kips, located at mid-span. Total dead load including self- weight = 2.5 kips/ ft. 12 in. 12 ft. PL = 10 kips 12 ft. Draw the shear diagram for the beam. #3 stirrup. 5 in. 17 in. 5 in. 4 in. 20 in. 12 in. 24 in. 4 in

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Consider the following simply supported beam. It has normal-weight concrete with a
compressive strength of 5,000 psi and Grade-60 bars. Each support is 12 in. wide. The section is
a hollow box. The beam is to carry a uniformly distributed dead load of 2.5 kips/ft., including its
own weight. The live load is a point load of 60 kips, located at mid-span.
Total dead load
including self-
weight = 2.5
kips/ ft.
12 in.
12 ft.
PL = 10 kips
12 ft.
Draw the shear diagram for the beam.
# 3 stirrup
5 in.
17 in.
5 in.
4 in.
20 in.
12 in.
24 in.
4 in
Transcribed Image Text:Consider the following simply supported beam. It has normal-weight concrete with a compressive strength of 5,000 psi and Grade-60 bars. Each support is 12 in. wide. The section is a hollow box. The beam is to carry a uniformly distributed dead load of 2.5 kips/ft., including its own weight. The live load is a point load of 60 kips, located at mid-span. Total dead load including self- weight = 2.5 kips/ ft. 12 in. 12 ft. PL = 10 kips 12 ft. Draw the shear diagram for the beam. # 3 stirrup 5 in. 17 in. 5 in. 4 in. 20 in. 12 in. 24 in. 4 in
Expert Solution
Step 1

Given data, 

Compressive strength of concrete= 5000 psi 

Grade 60 bars 

Uniformly distributed load= 2.5 Kips/ft 

Live load is point load= 60 Kips 

We have to draw the shear diagram. 

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