Exercise 3: Design of doubly reinforced concrete section and shear design Questions 1: A doubly reinforced beam section is shown in the figure below. Calculate the nominal flexural strength Mn and the design flexural strength øMn . b = 275 mm 6.5 d = 510 mm h == 600 mm 2 No. 25 bars A = 1000 mm² Design parameters: f'e=280 kgf/cm² (or 28 MPa) fy=4200 kgf/cm² (or 420 MPa) 6 No. 25 bars As = 3000 mm² 5.3 m 6.7 m 6.7 m 5.3 m B Strip 5.3 m Questions 2: 2a. What is the typical criterion defining the ultimate state of concrete members? 2b. What are under-reinforced concrete section and over-reinforced concrete section? 2c. What are the three major components of concrete shear? 2d. Why should the concrete shear Vc not be taken greater than the allowable limit (e.g. the following code-specified values)? 22.5.5.1.1 V shall not be taken greater than 52 √fb,d. 22.5.5.1.1 V1.33√fbd [0.42 √ƒb„d] (ACI 318-19) Design- Slab thickness = 150 mm (+401-112) 2e. What is the purpose of limiting the maximum spacing of shear reinforcement along the beam length? Beam 2 (cross-section =300x600 mm) Questions 3: A 5.3 m B 5.3 m C 5.3 m D 3a. Based on the results (design shear) from Exercise 1, design the shear reinforcement for the continuous beam. Write down the detailed steps for at least one section. Structural Framing Plan 3b. Locate the regions where only minimum shear reinforcement is required (V< øsVc). 3c. Draw the reinforcement details, including the longitudinal and shear reinforcement, of the whole continuous beam. Design parameters: f'e=280 kgf/cm² (or 28 MPa) fy=4200 kgf/cm² (or 420 MPa) for both longitudinal and shear reinforcement Concrete cover = 4 cm or (40 mm) Longitudinal reinforcement diameter DL= 24 mm Shear reinforcement diameter Dv= 12 mm Shear reinforcement design (unit: mm) Span AB Span BC Left end Mid. (V
Exercise 3: Design of doubly reinforced concrete section and shear design Questions 1: A doubly reinforced beam section is shown in the figure below. Calculate the nominal flexural strength Mn and the design flexural strength øMn . b = 275 mm 6.5 d = 510 mm h == 600 mm 2 No. 25 bars A = 1000 mm² Design parameters: f'e=280 kgf/cm² (or 28 MPa) fy=4200 kgf/cm² (or 420 MPa) 6 No. 25 bars As = 3000 mm² 5.3 m 6.7 m 6.7 m 5.3 m B Strip 5.3 m Questions 2: 2a. What is the typical criterion defining the ultimate state of concrete members? 2b. What are under-reinforced concrete section and over-reinforced concrete section? 2c. What are the three major components of concrete shear? 2d. Why should the concrete shear Vc not be taken greater than the allowable limit (e.g. the following code-specified values)? 22.5.5.1.1 V shall not be taken greater than 52 √fb,d. 22.5.5.1.1 V1.33√fbd [0.42 √ƒb„d] (ACI 318-19) Design- Slab thickness = 150 mm (+401-112) 2e. What is the purpose of limiting the maximum spacing of shear reinforcement along the beam length? Beam 2 (cross-section =300x600 mm) Questions 3: A 5.3 m B 5.3 m C 5.3 m D 3a. Based on the results (design shear) from Exercise 1, design the shear reinforcement for the continuous beam. Write down the detailed steps for at least one section. Structural Framing Plan 3b. Locate the regions where only minimum shear reinforcement is required (V< øsVc). 3c. Draw the reinforcement details, including the longitudinal and shear reinforcement, of the whole continuous beam. Design parameters: f'e=280 kgf/cm² (or 28 MPa) fy=4200 kgf/cm² (or 420 MPa) for both longitudinal and shear reinforcement Concrete cover = 4 cm or (40 mm) Longitudinal reinforcement diameter DL= 24 mm Shear reinforcement diameter Dv= 12 mm Shear reinforcement design (unit: mm) Span AB Span BC Left end Mid. (V
Chapter2: Loads On Structures
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![Exercise 3:
Design of doubly reinforced concrete section and shear design
Questions 1:
A doubly reinforced beam section is shown in the figure below. Calculate the nominal
flexural strength Mn and the design flexural strength øMn .
b = 275 mm
6.5
d = 510 mm
h == 600 mm
2 No. 25 bars
A = 1000 mm²
Design parameters:
f'e=280 kgf/cm² (or 28 MPa)
fy=4200 kgf/cm² (or 420 MPa)
6 No. 25 bars
As = 3000 mm²
5.3 m
6.7 m
6.7 m
5.3 m
B
Strip
5.3 m
Questions 2:
2a. What is the typical criterion defining the ultimate state of concrete members?
2b. What are under-reinforced concrete section and over-reinforced concrete section?
2c. What are the three major components of concrete shear?
2d. Why should the concrete shear Vc not be taken greater than the allowable limit (e.g.
the following code-specified values)?
22.5.5.1.1 V shall not be taken greater than 52 √fb,d.
22.5.5.1.1 V1.33√fbd [0.42 √ƒb„d]
(ACI 318-19)
Design-
Slab thickness
= 150 mm
(+401-112)
2e. What is the purpose of limiting the maximum spacing of shear reinforcement along
the beam length?
Beam 2 (cross-section =300x600 mm)
Questions 3:
A
5.3 m
B
5.3 m
C
5.3 m D
3a. Based on the results (design shear) from Exercise 1, design the shear reinforcement
for the continuous beam. Write down the detailed steps for at least one section.
Structural Framing Plan
3b. Locate the regions where only minimum shear reinforcement is required (V< øsVc).
3c. Draw the reinforcement details, including the longitudinal and shear reinforcement,
of the whole continuous beam.
Design parameters:
f'e=280 kgf/cm² (or 28 MPa)
fy=4200 kgf/cm² (or 420 MPa) for both longitudinal and shear reinforcement
Concrete cover = 4 cm or (40 mm)
Longitudinal reinforcement diameter DL= 24 mm
Shear reinforcement diameter Dv= 12 mm
Shear reinforcement design (unit: mm)
Span AB
Span BC
Left
end
Mid.
(V<os Vc)
Right Left
end end
Mid.
(V<os Vc)
Right
end
Left
end
Span CD
Mid.
(V<s Vc)
Right
end
Required
Av/s
Design
spacing
Sdesign](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14edd9a8-fad3-4b3f-b234-2144164e926a%2Fd660effb-0241-4992-b1b4-fb709046b460%2Fiylab_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise 3:
Design of doubly reinforced concrete section and shear design
Questions 1:
A doubly reinforced beam section is shown in the figure below. Calculate the nominal
flexural strength Mn and the design flexural strength øMn .
b = 275 mm
6.5
d = 510 mm
h == 600 mm
2 No. 25 bars
A = 1000 mm²
Design parameters:
f'e=280 kgf/cm² (or 28 MPa)
fy=4200 kgf/cm² (or 420 MPa)
6 No. 25 bars
As = 3000 mm²
5.3 m
6.7 m
6.7 m
5.3 m
B
Strip
5.3 m
Questions 2:
2a. What is the typical criterion defining the ultimate state of concrete members?
2b. What are under-reinforced concrete section and over-reinforced concrete section?
2c. What are the three major components of concrete shear?
2d. Why should the concrete shear Vc not be taken greater than the allowable limit (e.g.
the following code-specified values)?
22.5.5.1.1 V shall not be taken greater than 52 √fb,d.
22.5.5.1.1 V1.33√fbd [0.42 √ƒb„d]
(ACI 318-19)
Design-
Slab thickness
= 150 mm
(+401-112)
2e. What is the purpose of limiting the maximum spacing of shear reinforcement along
the beam length?
Beam 2 (cross-section =300x600 mm)
Questions 3:
A
5.3 m
B
5.3 m
C
5.3 m D
3a. Based on the results (design shear) from Exercise 1, design the shear reinforcement
for the continuous beam. Write down the detailed steps for at least one section.
Structural Framing Plan
3b. Locate the regions where only minimum shear reinforcement is required (V< øsVc).
3c. Draw the reinforcement details, including the longitudinal and shear reinforcement,
of the whole continuous beam.
Design parameters:
f'e=280 kgf/cm² (or 28 MPa)
fy=4200 kgf/cm² (or 420 MPa) for both longitudinal and shear reinforcement
Concrete cover = 4 cm or (40 mm)
Longitudinal reinforcement diameter DL= 24 mm
Shear reinforcement diameter Dv= 12 mm
Shear reinforcement design (unit: mm)
Span AB
Span BC
Left
end
Mid.
(V<os Vc)
Right Left
end end
Mid.
(V<os Vc)
Right
end
Left
end
Span CD
Mid.
(V<s Vc)
Right
end
Required
Av/s
Design
spacing
Sdesign
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