rc stirrups design

RC Vertical Stirrups (Stage 1: Evaluation)8m Span Floor Beam @ 2nd Floor

REFERENCES:

Is Shear Reinforcing NecessaryACI Code 11.4.6.1 Refer to the diagram shown belowConditions:

stirrups needed for = 3.74954651123883 m

53.409 kN (SEE STAAD RESULTS) Parameters:

Vu, for design = 45.890 kNVu diagram

ØVc =105.400 kN In Metric Units,

Ø Vc

stirrups needed to here b = 0.3 m52.700 kN h = 0.6 m

cc = 0.065 m

d = 0.535 m

span = 8 m

d = 0.535mf'c = 27.6 MPafy = 414 MPa

3.8 m 275 MPaface of support l = 1.0

Vu = Vu' + 0 kNwhere, 53.409 kN

Vu'=

53.409 kN0 kN

3.265 m 3.8 mafter cross-multiplying 3.265 m

Vu'=45.890 kN 3.8 m

substitute, Vu = 45.890 kN + 0 kN

Vu = 45.890 kNTriangular Shape for Vu' Calculation

* For ØVc, 53.409 kN

ØVc = = 0.751.0 27.6 MPa

300 5356

where, 3.265 m

Ø = 0.75 (new ACI value) 3.8 m

l = 1.0 (see the Parameters)

f'c = 27.6 MPa (see the Parameters)

bw = 0.3 m (see the Parameters)

d = 0.535 m (see the Parameters)

substitute,CONCLUSION:ØVc =105,399.75 N

or Since Vu = 45.890 kN ,105.400 kN is less than

52.700 kN , then, stirrups aren't necessary

Reinforced Concrete Mechanics and Design by James K. Wight and James G. MacGregor

Design of Reinforced Concrete 9th Edition by Jack C. McCormac and Russell H. Brown

If Vu ≤ ½ØVc then stirrups aren't necessaryIf Vu > ½ØVc then stirrups are necessary, proceed to the design stage

Ø Vs

zone where concrete carries

fyt =

Calculation : See the diagram shown at the right* For Vu @ distance "d" from left end, Shear Diagram to Beam Centerline

in Trapezoidal Shape

Vu' (by Ratio and Proportion)

½ØVc =

CL

dh

b

Vu (design for stirrups)

d

Vu'∅((𝜆√(𝑓^′ 𝑐 ))/6) 𝑏_𝑤 𝑑

A3

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D12

Insert Value here

AF18

input your value

AF19

input your value

AF20

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AF21

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AG22

Input Value

AF23

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AF24

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AF25

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AF26

insert value

AG33

insert value

RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 2nd Floor

REFERENCES:

Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW

NOTESA. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH

2h TO BE LEAST OF:

a.1) d

= 535 = 133.75 mm governs!4 4

a.2) 6 25 = 150 mm

a.3) 150 mm

--------------> 133.75 mm is the least

B. ø = DIA. OF MAIN LONGITUDINAL BAR FOR

DISTANCE UNDER CONSIDERATION IN TOP OR

BOTTOM WHICHEVER IS THE SMALLER.using 25 mm for main Longitudinal Bar

Use whichever is leastHENCE, use s = 130 mm

C. STIRRUP SPACING TO BE LEAST OF:

c.1) d

= 535 = 267.5 mm2 2

267.5 mm to be the least

c.2) 600mm


Values for DetailingSpacing No. Of Bars Remarks

@ 50 mm 1 B.W. To CL symmetrical both sides


REST @ 265 mm 9 B.W. To CL symmetrical both sidesTOTAL 3475 mm 18 B.W. To CL symmetrical both sides

[SATISFACTORY]



6ø =

(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)


USE 10mm ø STIRRUPS (Minimum size for stirrups)

DESIGN OF PLASTIC HINGE ZONE

PLASTIC HINGE ZONE

H34

insert Value

RC Vertical Stirrups (Stage 1: Evaluation)8m Span Floor Beam @ 3rd Floor

REFERENCES:






Ø Vc


cc = 0.065 m

d = 0.535 m

span = 8 m

d = 0.535mf'c = 27.6 MPafy = 414 MPa



Vu'=

53.409 kN0 kN


Vu'=45.890 kN 3.8 m




ØVc = = 0.751.0 27.6 MPa

300 5356

where, 3.265 m












Ø Vs


fyt =




½ØVc =

CL

dh

b


d

Vu'∅((𝜆√(𝑓^′ 𝑐 ))/6) 𝑏_𝑤 𝑑

A3

Input your value

D12

Insert Value here

AF18

input your value

AF19

input your value

AF20

input value

AF21

Input Value

AG22

Input Value

AF23

Input Value

AF24

Input Value

AF25

Input Value

AF26

insert value

AG33

insert value

RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 3rd Floor

REFERENCES:



2h TO BE LEAST OF:

a.1) d

= 535 = 133.75 mm governs!4 4

a.2) 6 32 = 192 mm

a.3) 150 mm

--------------> 133.75 mm is the least






c.1) d

= 535 = 267.5 mm2 2


c.2) 600mm






[SATISFACTORY]



6ø =





PLASTIC HINGE ZONE

H34

insert Value

RC Vertical Stirrups (Stage 1: Evaluation)8m Span Floor Beam @ 4th Floor

REFERENCES:






Ø Vc


cc = 0.065 m

d = 0.535 m

span = 8 m

d = 0.535mf'c = 27.6 MPafy = 414 MPa



Vu'=

53.409 kN0 kN


Vu'=45.890 kN 3.8 m




ØVc = = 0.751.0 27.6 MPa

300 5356

where, 3.265 m












Ø Vs


fyt =




½ØVc =

CL

dh

b


d

Vu'∅((𝜆√(𝑓^′ 𝑐 ))/6) 𝑏_𝑤 𝑑

A3

Input your value

D12

Insert Value here

AF18

input your value

AF19

input your value

AF20

input value

AF21

Input Value

AG22

Input Value

AF23

Input Value

AF24

Input Value

AF25

Input Value

AF26

insert value

AG33

insert value

RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 4th Floor

REFERENCES:



2h TO BE LEAST OF:

a.1) d

= 535 = 133.75 mm governs!4 4

a.2) 6 36 = 216 mm

a.3) 150 mm

--------------> 133.75 mm is the least






c.1) d

= 535 = 267.5 mm2 2


c.2) 600mm






[SATISFACTORY]



6ø =





PLASTIC HINGE ZONE

H34

insert Value

RC Vertical Stirrups (Stage 1: Evaluation)8m Span Floor Beam @ 5th Floor

REFERENCES:






Ø Vc


cc = 0.065 m

d = 0.535 m

span = 8 m

d = 0.535mf'c = 27.6 MPafy = 414 MPa



Vu'=

53.409 kN0 kN


Vu'=45.890 kN 3.8 m




ØVc = = 0.751.0 27.6 MPa

300 5356

where, 3.265 m












Ø Vs


fyt =




½ØVc =

CL

dh

b


d

Vu'∅((𝜆√(𝑓^′ 𝑐 ))/6) 𝑏_𝑤 𝑑

A3

Input your value

D12

Insert Value here

AF18

input your value

AF19

input your value

AF20

input value

AF21

Input Value

AG22

Input Value

AF23

Input Value

AF24

Input Value

AF25

Input Value

AF26

insert value

AG33

insert value

RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 5th Floor

REFERENCES:



2h TO BE LEAST OF:

a.1) d

= 535 = 133.75 mm 4 4

a.2) 6 20 = 120 mm governs!

a.3) 150 mm

--------------> 120 mm is the least






c.1) d

= 535 = 267.5 mm2 2


c.2) 600mm






[SATISFACTORY]



6ø =





PLASTIC HINGE ZONE

H34

insert Value

RC Vertical Stirrups (Stage 1: Evaluation)6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor

REFERENCES:






Ø Vc


cc = 0.065 m

d = 0.535 m

span = 6 m

d = 0.535mf'c = 27.6 MPafy = 414 MPa



Vu'=

###2.265 m 2.8 m

after cross-multiplying 2.265 m

Vu'=80.960 kN 2.8 m



* For ØVc, 100.083 kN

ØVc = = 0.751.0 27.6 MPa

300 5356

where, 2.265 m







or Since Vu = 80.960 kN ,105.400 kN is greater than

52.700 kN , then, proceed to the design stage




Ø Vs


fyt =




½ØVc =

CL

dh

b


d

Vu'∅((𝜆√(𝑓^′ 𝑐 ))/6) 𝑏_𝑤 𝑑

A3

Input your value

D12

Insert Value here

AF18

input your value

AF19

input your value

AF20

input value

AF21

Input Value

AG22

Input Value

AF23

Input Value

AF24

Input Value

AF25

Input Value

AF26

insert value

AG33

insert value

RC Vertical Stirrups (Stage 2: Designing Part One)6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor

REFERENCES:

Conditions and Calculations to be Satisfied: ACI Code Requirements


2h TO BE LEAST OF:

a.) d

= 535 m = 133.75 mm governs!4 4

b.) 6 25 = 150 mm

c.) 150 mm

--------------> 133.75 mm is the least





Use 12 mm ø stirrups

ACI Code 11.4.7.9

Vs = 562,132.013 N or 562.13 kN

For Vs, ; Vs = 80.96 kN - 105.40 kN0.75



6ø =


Step 1

ACI Code 12.13.2.1 allows No. 25M and smaller stirrups to be anchored by a standard 90⁰ or 135⁰ hook stirrup hook around a longitudinal bar. Provide a No.10M or larger bar in each of the upper

corners of the stirrup to anchor them.

Step 2 Under no circumstance may Vs be allowed to exceed 2/3 √(𝑓^′ 𝑐) 𝑏_𝑤 𝑑 𝑉_𝑠=(𝑉_𝑢−∅𝑉_𝑐)/∅

2/3 √(𝑓^′ 𝑐) 𝑏_𝑤 𝑑

DESIGN OF PLASTIC HINGE ZONE (Refer to the Figure below)

PLASTIC HINGE ZONE

H91

insert Value

E100

insert Value

Vs = 32.586 kN < (the section is large enough)

RC Vertical Stirrups (Stage 2: Designing Part Two)6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor

REFERENCES:

ACI Equation 11-13

94.922 < 95.400

therefore use 95.400

ACI Code 11.4.5.1The maximum spacing of vertical stirrups permitted by the code is the lesser of d/2 or 600mm

265 mm < 600 mm (ok)

( )* using 12 mm double-leg stirrups,

theoretical use s = 1020 mmwhere,

Av = 2 π 122

= 226.195 sq.mm > (governs!)4

substitute, S = 226.195 275 535 = 1021 mm >32.586 x10^3

Values for DetailingSpacing No. Of Bars Remarks@ 50 mm 1 B.W. To CL symmetrical both sides@ 130 mm 8 B.W. To CL symmetrical both sides


[SATISFACTORY]



Step 3

mm2 mm2

mm2 for Avmin

Step 4

Avmin

Smax(exceeds the maximum limit)

𝐴_(𝑣 𝑚𝑖𝑛)=1/16 √(𝑓^′ 𝑐) (𝑏_𝑤 𝑠_𝑚𝑎𝑥)/𝑓_𝑦𝑡 ≥(0.33𝑏_𝑤 𝑠_𝑚𝑎𝑥)/𝑓_𝑦𝑡

𝑆_𝑚𝑎𝑥=𝑑/2=

1/16 √(𝑓^′ 𝑐) (𝑏_𝑤 𝑠_𝑚𝑎𝑥)/𝑓_𝑦𝑡 =

=(0.33𝑏_𝑤 𝑠_𝑚𝑎𝑥)/𝑓_𝑦𝑡

2/3 √(𝑓^′ 𝑐) 𝑏_𝑤 𝑑

𝑠=(𝐴_𝑣 𝑓_𝑦𝑡 𝑑)/𝑉_𝑠 ,

rc stirrups design

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