Rc Stirrups Design
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RC Vertical Stirrups (Stage 1: Evaluation) 8m Span Floor Beam @ 2nd Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
3.749546511 m
stirrups needed for =
53.409 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 45.890 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535 m span = 8 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
d = 0.535m
3.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 53.409 kN = 3.265 m 3.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 53.409 kN Vu (design for stirrups)
0 kN
after cross-multiplying
3.265 m
d
Vu'= 45.890 kN substitute, Vu = 45.890 kN Vu = 45.890 kN
0.065 m
3.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
53.409 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 3.265 m 3.8 m
CONCLUSION: Since Vu = 45.890 kN , is less than ½ØVc = 52.700 kN , then, stirrups aren't necessary
RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 2nd Floor REFERENCES:
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
Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW
DESIGN OF PLASTIC HINGE ZONE
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.1)
d 4
535
=
a.2) 6ø =
=
4 6
25
133.75 mm governs! = 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
Use whichever is least HENCE, use s =
for main Longitudinal Bar
130 mm
C. STIRRUP SPACING TO BE LEAST OF: 535 d c.1) = = 2 2
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
267.5 mm 267.5 mm to be the least
c.2) 600mm Use whichever is least HENCE, use s =
265 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
Values for Detailing @ @ REST @ TOTAL
Spacing 50 mm 130 mm 265 mm 3475 mm [SATISFACTORY]
No. Of Bars Remarks symmetrical both sides B.W. To CL 8 symmetrical both sides B.W. To CL B.W. To CL 9 symmetrical both sides 18 B.W. To CL symmetrical both sides USE 10mm ø STIRRUPS (Minimum size for stirrups) 1
RC Vertical Stirrups (Stage 1: Evaluation) 8m Span Floor Beam @ 3rd Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
3.749546511 m
stirrups needed for =
53.409 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 45.890 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535 m span = 8 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
d = 0.535m
3.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 53.409 kN = 3.265 m 3.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 53.409 kN Vu (design for stirrups)
0 kN
after cross-multiplying
3.265 m
d
Vu'= 45.890 kN substitute, Vu = 45.890 kN Vu = 45.890 kN
0.065 m
3.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
53.409 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 3.265 m 3.8 m
CONCLUSION: Since Vu = 45.890 kN , is less than ½ØVc = 52.700 kN , then, stirrups aren't necessary
RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 3rd Floor REFERENCES:
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
Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW
DESIGN OF PLASTIC HINGE ZONE
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.1)
d 4
535
=
a.2) 6ø =
=
4 6
32
133.75 mm governs! = 192 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
32 mm
Use whichever is least HENCE, use s =
for main Longitudinal Bar
130 mm
C. STIRRUP SPACING TO BE LEAST OF: 535 d c.1) = = 2 2
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
267.5 mm 267.5 mm to be the least
c.2) 600mm Use whichever is least HENCE, use s =
265 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
Values for Detailing @ @ REST @ TOTAL
Spacing 50 mm 130 mm 265 mm 3475 mm [SATISFACTORY]
No. Of Bars Remarks symmetrical both sides B.W. To CL 8 symmetrical both sides B.W. To CL B.W. To CL 9 symmetrical both sides 18 B.W. To CL symmetrical both sides USE 10mm ø STIRRUPS (Minimum size for stirrups) 1
RC Vertical Stirrups (Stage 1: Evaluation) 8m Span Floor Beam @ 4th Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
3.749546511 m
stirrups needed for =
53.409 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 45.890 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535 m span = 8 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
d = 0.535m
3.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 53.409 kN = 3.265 m 3.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 53.409 kN Vu (design for stirrups)
0 kN
after cross-multiplying
3.265 m
d
Vu'= 45.890 kN substitute, Vu = 45.890 kN Vu = 45.890 kN
0.065 m
3.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
53.409 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 3.265 m 3.8 m
CONCLUSION: Since Vu = 45.890 kN , is less than ½ØVc = 52.700 kN , then, stirrups aren't necessary
RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 4th Floor REFERENCES:
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
Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW
DESIGN OF PLASTIC HINGE ZONE
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.1)
d 4
535
=
a.2) 6ø =
=
4 6
36
133.75 mm governs! = 216 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
36 mm
Use whichever is least HENCE, use s =
for main Longitudinal Bar
130 mm
C. STIRRUP SPACING TO BE LEAST OF: 535 d c.1) = = 2 2
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
267.5 mm 267.5 mm to be the least
c.2) 600mm Use whichever is least HENCE, use s =
265 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
Values for Detailing @ @ REST @ TOTAL
Spacing 50 mm 130 mm 265 mm 3475 mm [SATISFACTORY]
No. Of Bars Remarks symmetrical both sides B.W. To CL 8 symmetrical both sides B.W. To CL B.W. To CL 9 symmetrical both sides 18 B.W. To CL symmetrical both sides USE 10mm ø STIRRUPS (Minimum size for stirrups) 1
RC Vertical Stirrups (Stage 1: Evaluation) 8m Span Floor Beam @ 5th Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
3.749546511 m
stirrups needed for =
53.409 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 45.890 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535 m span = 8 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
d = 0.535m
3.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 53.409 kN = 3.265 m 3.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 53.409 kN Vu (design for stirrups)
0 kN
after cross-multiplying
3.265 m
d
Vu'= 45.890 kN substitute, Vu = 45.890 kN Vu = 45.890 kN
0.065 m
3.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
53.409 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 3.265 m 3.8 m
CONCLUSION: Since Vu = 45.890 kN , is less than ½ØVc = 52.700 kN , then, stirrups aren't necessary
RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 5th Floor REFERENCES:
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
Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW
DESIGN OF PLASTIC HINGE ZONE
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.1)
d 4
535
=
a.2) 6ø =
4 6
= 20
133.75 mm = 120 mm
governs!
a.3) 150 mm -------------->
120 mm
is the least
B. ø = DIA. OF MAIN LONGITUDINAL BAR FOR DISTANCE UNDER CONSIDERATION IN TOP OR
BOTTOM WHICHEVER IS THE SMALLER. using
20 mm
Use whichever is least HENCE, use s =
for main Longitudinal Bar
120 mm
C. STIRRUP SPACING TO BE LEAST OF: 535 d c.1) = = 2 2
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
267.5 mm 267.5 mm to be the least
c.2) 600mm Use whichever is least HENCE, use s =
265 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
Values for Detailing @ @ REST @ TOTAL
Spacing 50 mm 120 mm 265 mm 3515 mm [SATISFACTORY]
No. Of Bars Remarks symmetrical both sides B.W. To CL 9 symmetrical both sides B.W. To CL B.W. To CL 9 symmetrical both sides 19 B.W. To CL symmetrical both sides USE 10mm ø STIRRUPS (Minimum size for stirrups) 1
RC Vertical Stirrups (Stage 1: Evaluation) 6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
1.474372805 m
stirrups needed for =
100.083 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 80.960 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535m
2.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 100.08 kN = 2.265 m 2.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 100.083 kN Vu (design for stirrups)
after cross-multiplying
2.265 m
d
Vu'= 80.960 kN substitute, Vu = 80.960 kN Vu = 80.960 kN
0.065 m
d = 0.535 m span = 6 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
2.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
100.083 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 2.265 m 2.8 m
CONCLUSION: Since Vu = 80.960 kN , is greater than ½ØVc = 52.700 kN , then, proceed to the design stage
RC Vertical Stirrups (Stage 2: Designing Part One)
RC Vertical Stirrups (Stage 2: Designing Part One) 6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor REFERENCES:
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
Conditions and Calculations to be Satisfied: ACI Code Requirements
DESIGN OF PLASTIC HINGE ZONE (Refer to the Figure below)
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.)
d = 4
b.) 6ø = c.)
535 m
4 6
=
25
133.75 mm governs! = 150 mm
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 least HENCE, use s =
Step 1 Use
130 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
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.
12 mm ø stirrups
Step ACI Code 11.4.7.9 Under no circumstance may Vs be allowed to exceed 2
2 𝑓 ′ 𝑐 𝑏𝑤 𝑑 3 For Vs,
Vs =
𝑉𝑠 =
562,132.013 N
𝑉𝑢 − ∅𝑉𝑐 ∅
;
Vs = 32.586 kN
Vs = <
or 80.96 kN
562.13 kN 0.75
105.40 kN
2 𝑓 ′ 𝑐 𝑏𝑤 𝑑 (the section is large 3 enough)
RC Vertical Stirrups (Stage 2: Designing Part Two)
6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor 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
REFERENCES:
Step ACI Equation 11-13 3
𝐴𝑣 1 16
𝑓′𝑐
𝑚𝑖𝑛
𝑏𝑤 𝑠𝑚𝑎𝑥 𝑓𝑦𝑡
=
1 𝑏𝑤 𝑠𝑚𝑎𝑥 0.33𝑏𝑤 𝑠𝑚𝑎𝑥 𝑓 ′𝑐 ≥ 16 𝑓𝑦𝑡 𝑓𝑦𝑡
= 94.922 mm2
therefore use
<
95.400
95.400 mm2 = mm
2
0.33𝑏𝑤 𝑠𝑚𝑎𝑥 𝑓𝑦𝑡
for Avmin
Step ACI Code 11.4.5.1 4 The maximum spacing of vertical stirrups permitted by the code is the lesser of d/2 or 600mm 𝑑 265 mm 𝑆𝑚𝑎𝑥 = = < 600 mm (ok) 2
* using
12 mm
) double-leg stirrups,
(
theoretical where, Av = substitute,
𝑠=
𝐴𝑣 𝑓𝑦𝑡 𝑑 , 𝑉𝑠 π 4
2
S=
12
use s = 1020 mm 2
= 226.195 sq.mm > Avmin (governs!)
226.195 275 535 = 32.586 x10^3
1021 mm
(exceeds the maximum
> Smax limit)
Values for Detailing Spacing No. Of Bars @ 50 mm B.W. To CL 1 @ 130 mm B.W. To CL 8 REST @ 265 mm B.W. To CL 6 2680 mm 15 B.W. To CL TOTAL [SATISFACTORY]
Remarks
symmetrical both sides symmetrical both sides symmetrical both sides
symmetrical both sides
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
3.749546511 m
stirrups needed for =
53.409 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 45.890 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535 m span = 8 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
d = 0.535m
3.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 53.409 kN = 3.265 m 3.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 53.409 kN Vu (design for stirrups)
0 kN
after cross-multiplying
3.265 m
d
Vu'= 45.890 kN substitute, Vu = 45.890 kN Vu = 45.890 kN
0.065 m
3.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
53.409 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 3.265 m 3.8 m
CONCLUSION: Since Vu = 45.890 kN , is less than ½ØVc = 52.700 kN , then, stirrups aren't necessary
RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 2nd Floor REFERENCES:
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
Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW
DESIGN OF PLASTIC HINGE ZONE
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.1)
d 4
535
=
a.2) 6ø =
=
4 6
25
133.75 mm governs! = 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
Use whichever is least HENCE, use s =
for main Longitudinal Bar
130 mm
C. STIRRUP SPACING TO BE LEAST OF: 535 d c.1) = = 2 2
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
267.5 mm 267.5 mm to be the least
c.2) 600mm Use whichever is least HENCE, use s =
265 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
Values for Detailing @ @ REST @ TOTAL
Spacing 50 mm 130 mm 265 mm 3475 mm [SATISFACTORY]
No. Of Bars Remarks symmetrical both sides B.W. To CL 8 symmetrical both sides B.W. To CL B.W. To CL 9 symmetrical both sides 18 B.W. To CL symmetrical both sides USE 10mm ø STIRRUPS (Minimum size for stirrups) 1
RC Vertical Stirrups (Stage 1: Evaluation) 8m Span Floor Beam @ 3rd Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
3.749546511 m
stirrups needed for =
53.409 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 45.890 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535 m span = 8 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
d = 0.535m
3.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 53.409 kN = 3.265 m 3.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 53.409 kN Vu (design for stirrups)
0 kN
after cross-multiplying
3.265 m
d
Vu'= 45.890 kN substitute, Vu = 45.890 kN Vu = 45.890 kN
0.065 m
3.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
53.409 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 3.265 m 3.8 m
CONCLUSION: Since Vu = 45.890 kN , is less than ½ØVc = 52.700 kN , then, stirrups aren't necessary
RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 3rd Floor REFERENCES:
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
Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW
DESIGN OF PLASTIC HINGE ZONE
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.1)
d 4
535
=
a.2) 6ø =
=
4 6
32
133.75 mm governs! = 192 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
32 mm
Use whichever is least HENCE, use s =
for main Longitudinal Bar
130 mm
C. STIRRUP SPACING TO BE LEAST OF: 535 d c.1) = = 2 2
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
267.5 mm 267.5 mm to be the least
c.2) 600mm Use whichever is least HENCE, use s =
265 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
Values for Detailing @ @ REST @ TOTAL
Spacing 50 mm 130 mm 265 mm 3475 mm [SATISFACTORY]
No. Of Bars Remarks symmetrical both sides B.W. To CL 8 symmetrical both sides B.W. To CL B.W. To CL 9 symmetrical both sides 18 B.W. To CL symmetrical both sides USE 10mm ø STIRRUPS (Minimum size for stirrups) 1
RC Vertical Stirrups (Stage 1: Evaluation) 8m Span Floor Beam @ 4th Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
3.749546511 m
stirrups needed for =
53.409 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 45.890 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535 m span = 8 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
d = 0.535m
3.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 53.409 kN = 3.265 m 3.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 53.409 kN Vu (design for stirrups)
0 kN
after cross-multiplying
3.265 m
d
Vu'= 45.890 kN substitute, Vu = 45.890 kN Vu = 45.890 kN
0.065 m
3.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
53.409 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 3.265 m 3.8 m
CONCLUSION: Since Vu = 45.890 kN , is less than ½ØVc = 52.700 kN , then, stirrups aren't necessary
RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 4th Floor REFERENCES:
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
Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW
DESIGN OF PLASTIC HINGE ZONE
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.1)
d 4
535
=
a.2) 6ø =
=
4 6
36
133.75 mm governs! = 216 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
36 mm
Use whichever is least HENCE, use s =
for main Longitudinal Bar
130 mm
C. STIRRUP SPACING TO BE LEAST OF: 535 d c.1) = = 2 2
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
267.5 mm 267.5 mm to be the least
c.2) 600mm Use whichever is least HENCE, use s =
265 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
Values for Detailing @ @ REST @ TOTAL
Spacing 50 mm 130 mm 265 mm 3475 mm [SATISFACTORY]
No. Of Bars Remarks symmetrical both sides B.W. To CL 8 symmetrical both sides B.W. To CL B.W. To CL 9 symmetrical both sides 18 B.W. To CL symmetrical both sides USE 10mm ø STIRRUPS (Minimum size for stirrups) 1
RC Vertical Stirrups (Stage 1: Evaluation) 8m Span Floor Beam @ 5th Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
3.749546511 m
stirrups needed for =
53.409 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 45.890 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535 m span = 8 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
d = 0.535m
3.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 53.409 kN = 3.265 m 3.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 53.409 kN Vu (design for stirrups)
0 kN
after cross-multiplying
3.265 m
d
Vu'= 45.890 kN substitute, Vu = 45.890 kN Vu = 45.890 kN
0.065 m
3.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
53.409 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 3.265 m 3.8 m
CONCLUSION: Since Vu = 45.890 kN , is less than ½ØVc = 52.700 kN , then, stirrups aren't necessary
RC Vertical Stirrups Stage 2: Designing 8m Span Floor Beam @ 5th Floor REFERENCES:
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
Since the result of the evaluation of this beam is no stirrups needed, REFER TO THE FIGURE BELOW
DESIGN OF PLASTIC HINGE ZONE
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.1)
d 4
535
=
a.2) 6ø =
4 6
= 20
133.75 mm = 120 mm
governs!
a.3) 150 mm -------------->
120 mm
is the least
B. ø = DIA. OF MAIN LONGITUDINAL BAR FOR DISTANCE UNDER CONSIDERATION IN TOP OR
BOTTOM WHICHEVER IS THE SMALLER. using
20 mm
Use whichever is least HENCE, use s =
for main Longitudinal Bar
120 mm
C. STIRRUP SPACING TO BE LEAST OF: 535 d c.1) = = 2 2
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
267.5 mm 267.5 mm to be the least
c.2) 600mm Use whichever is least HENCE, use s =
265 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
Values for Detailing @ @ REST @ TOTAL
Spacing 50 mm 120 mm 265 mm 3515 mm [SATISFACTORY]
No. Of Bars Remarks symmetrical both sides B.W. To CL 9 symmetrical both sides B.W. To CL B.W. To CL 9 symmetrical both sides 19 B.W. To CL symmetrical both sides USE 10mm ø STIRRUPS (Minimum size for stirrups) 1
RC Vertical Stirrups (Stage 1: Evaluation) 6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor 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
REFERENCES:
Is Shear Reinforcing Necessary?
ACI Code 11.4.6.1
Refer to the diagram shown below
Conditions: If V u ≤ ½ØV c then stirrups aren't necessary If V u > ½ØV c then stirrups are necessary, proceed to the design stage
CL
1.474372805 m
stirrups needed for =
100.083 kN (SEE STAAD RESULTS)
Parameters: b
Vu, for design = 80.960 kN
d
Vu diagram Ø Vs
In Metric Units,
ØVc =105.400 kN
b = 0.3 m h = 0.6 m
stirrups needed to here
52.700 kN
zone where concrete carries
Ø Vc
cc =
d = 0.535m
2.8 m face of support Calculation : See the diagram shown at the right * For Vu @ distance "d" from left end, Vu = Vu' + 0 kN where, Vu' (by Ratio and Proportion) Vu' 100.08 kN = 2.265 m 2.8 m
Shear Diagram to Beam Centerline in Trapezoidal Shape 100.083 kN Vu (design for stirrups)
after cross-multiplying
2.265 m
d
Vu'= 80.960 kN substitute, Vu = 80.960 kN Vu = 80.960 kN
0.065 m
d = 0.535 m span = 6 m f'c = 27.6 MPa fy = 414 MPa fyt = 275 MPa l = 1.0
2.8 m
+
0 kN Triangular Shape for Vu' Calculation
* For ØVc, ØVc = ∅
100.083 kN
𝜆 𝑓 ′𝑐 𝑏𝑤 𝑑 6
=
0.75
1.0
where, Ø = 0.75 (new ACI value) 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, ØVc = 105,399.75 N or 105.400 kN
27.6 MPa
6
Vu'
300
535 2.265 m 2.8 m
CONCLUSION: Since Vu = 80.960 kN , is greater than ½ØVc = 52.700 kN , then, proceed to the design stage
RC Vertical Stirrups (Stage 2: Designing Part One)
RC Vertical Stirrups (Stage 2: Designing Part One) 6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor REFERENCES:
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
Conditions and Calculations to be Satisfied: ACI Code Requirements
DESIGN OF PLASTIC HINGE ZONE (Refer to the Figure below)
PLASTIC HINGE ZONE
NOTES A. STIRRUP SPACING IN PLASTIC HINGE ZONE OF LENGTH 2h TO BE LEAST OF:
a.)
d = 4
b.) 6ø = c.)
535 m
4 6
=
25
133.75 mm governs! = 150 mm
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 least HENCE, use s =
Step 1 Use
130 mm
(Note: final value for minimum spacing was rounded down to be divisible by 10 or 5 as reality check requires)
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.
12 mm ø stirrups
Step ACI Code 11.4.7.9 Under no circumstance may Vs be allowed to exceed 2
2 𝑓 ′ 𝑐 𝑏𝑤 𝑑 3 For Vs,
Vs =
𝑉𝑠 =
562,132.013 N
𝑉𝑢 − ∅𝑉𝑐 ∅
;
Vs = 32.586 kN
Vs = <
or 80.96 kN
562.13 kN 0.75
105.40 kN
2 𝑓 ′ 𝑐 𝑏𝑤 𝑑 (the section is large 3 enough)
RC Vertical Stirrups (Stage 2: Designing Part Two)
6m Span Floor Beam @ 2nd, 3rd, 4th and 5th Floor 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
REFERENCES:
Step ACI Equation 11-13 3
𝐴𝑣 1 16
𝑓′𝑐
𝑚𝑖𝑛
𝑏𝑤 𝑠𝑚𝑎𝑥 𝑓𝑦𝑡
=
1 𝑏𝑤 𝑠𝑚𝑎𝑥 0.33𝑏𝑤 𝑠𝑚𝑎𝑥 𝑓 ′𝑐 ≥ 16 𝑓𝑦𝑡 𝑓𝑦𝑡
= 94.922 mm2
therefore use
<
95.400
95.400 mm2 = mm
2
0.33𝑏𝑤 𝑠𝑚𝑎𝑥 𝑓𝑦𝑡
for Avmin
Step ACI Code 11.4.5.1 4 The maximum spacing of vertical stirrups permitted by the code is the lesser of d/2 or 600mm 𝑑 265 mm 𝑆𝑚𝑎𝑥 = = < 600 mm (ok) 2
* using
12 mm
) double-leg stirrups,
(
theoretical where, Av = substitute,
𝑠=
𝐴𝑣 𝑓𝑦𝑡 𝑑 , 𝑉𝑠 π 4
2
S=
12
use s = 1020 mm 2
= 226.195 sq.mm > Avmin (governs!)
226.195 275 535 = 32.586 x10^3
1021 mm
(exceeds the maximum
> Smax limit)
Values for Detailing Spacing No. Of Bars @ 50 mm B.W. To CL 1 @ 130 mm B.W. To CL 8 REST @ 265 mm B.W. To CL 6 2680 mm 15 B.W. To CL TOTAL [SATISFACTORY]
Remarks
symmetrical both sides symmetrical both sides symmetrical both sides
symmetrical both sides
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