Chapter 7.docx

CHAPTER7 STIRRUPS & LATERAL TIES

101 | R e i n f o r c e d C o n c r e t e

2nd Floor 3rd Floor 4th Floor Roof deck

For 3m length Fy.i Fy.j -116.058 -180.705 -92.8214 -157.002 -43.3879 -107.372 30.836 -47.7588

For 5m length Fy.i Fy.j 119.6972 -158.328 118.8608 -151.802 118.9183 -136.364 83.6707 -87.2842

For lengths 3m 2ND FLOOR b = 295 mm f’c = 28 MPa

d = 407.5 mm fy= 230 MPa

0.66√f ′ c bw d = 0.66 √28 (295) (407.5) = 419829.14 N = 419.83 kN> Vs (OK!) 𝜋

Av = 2Ab = 4 (10)2 (2) Use 10 mm Ø RSB

Av = 50 𝜋 mm2

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(407.5) = Vs 115340 N S = 127.64 mm “say” 120 mm

Solution:

S=

For maximum spacing, 0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(295)(407.5) = 209.91 kN> VS d

So, 𝑠𝑚𝑎𝑥 = 2 = 180.705 − 116.058 Vu − 116.058 = 3 3 − (0.4075 + 0.2) Vu = 167.61 kN VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (295) (407.5) Vc = 108137 N = 108.14 kN ØVc = 0.75 (108.14) ØVc = 81.11 kN< Vu STIRRUPS ARE NECESSARY Vs = VS =

Vu − Vc ϕ 167.61 − 0.75

108.14= 115.34 kN

(verify if the section is adequate to carry the shear)

407.5 2

= 203.75 𝑜𝑟 600mm

Therefore, use s = 120 mm @ d+200 407.5+200 = 607.5 mm ≈ 610mm N=

0.61 = 5 stirrups @ 120mm 0.12

5 @ 120 mm forward, 2 @ 120mm = 240 mm @ 0.9m 180.705 − 116.058 Vu − 116.058 = 3 3 − (0.9) Vu = 161.31 kN

102 | R e i n f o r c e d C o n c r e t e

= 0.17𝜆√f ′ c bd VC Vc = 0.17 (1) √28 (295) (407.5) Vc = 108137 N = 108.14 kN

3RD FLOOR

ØVc = 0.75 (108.14) ØVc = 81.11 kN< Vu

Use 10 mm Ø RSB

b = 285 mm f’c = 28 MPa

d = 402.5 mm fy= 230 MPa

Solution:

STIRRUPS ARE NECESSARY Vu − Vc ϕ 161.31 VS = 0.75 − 108.14 = 106.94 kN Vs =

(verify if the section is adequate to carry the shear) 0.66√f ′ c bw d = 0.66 √28 (295) (407.5) = 419829.14 N = 419.83 kN> Vs (OK!) 𝜋 Av = 2Ab = 4 (10)2 (2) Av = 50 𝜋 mm2

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(407.5) = Vs 106940 N S = 137.67 mm “say” 130 mm S=

For maximum spacing,

= 209.91 kN> VS d

407.5 2

= 203.75 𝑜𝑟 600mm

Therefore, use s = 130 mm So, 4 @ 130 mm = 520 mm SPACING

LENGTH

7 @ 120mm 840 mm 4 @ 130mm 520mm Rest@140mm

Vu = 144.11 kN VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (285) (402.5) Vc = 103190.25 N = 103.19 kN ØVc = 0.75 (103.19) ØVc = 77.39 kN< Vu STIRRUPS ARE NECESSARY

0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(295)(407.5)

So, 𝑠𝑚𝑎𝑥 = 2 =

157.002 − 92.8214 Vu − 92.8214 = 3 3 − (0.4025 + 0.2)

TOTAL DISTANCE FROM SUPPORT 840 mm 1360 mm

Vu − Vc ϕ 144.11 VS = 0.75 − 103.19 = 88.96 kN Vs =

(verify if the section is adequate to carry the shear) 0.66√f ′ c bw d = 0.66 √28 (285) (402.5) =400620.99 N = 400.62 kN> Vs (OK!) 𝜋

Av = 2Ab = 4 (10)2 (2) Av = 50 𝜋 mm2

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(402.5) = Vs 88960 N S = 163.46 mm “say” 130 mm S=

103 | R e i n f o r c e d C o n c r e t e

For maximum spacing, 0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(285)(402.5) = 200.31 kN> VS So, 𝑠𝑚𝑎𝑥 =

d 2

=

402.5 2

= 201.25 𝑜𝑟 600 mm

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(402.5) = Vs 80480 N S = 180.67 mm “say” 140 mm S=

For maximum spacing, 0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(285)(402.5) = 200.31 kN> VS

Therefore, use s = 130 mm d

So, 𝑠𝑚𝑎𝑥 = 2 =

@ d+200 402.5+200 = 602.5 ≈ 610mm 0.61 N= = 4 stirrups @ 130 mm 0.13 1 @ 90mm & 4 @ 130 mm forward, 2 @ 130 mm = 260 mm @ 0.9m 157.002 − 92.8214 Vu − 92.8214 = 3 3 − (0.9) Vu = 137.75 kN VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (285) (402.5) Vc = 103190.25 N = 103.19 kN

402.5 2

= 201.25 𝑜𝑟 600mm

Therefore, use s = 140 mm So, 5 @ 140 mm SPACING

LENGTH

1 @ 90 mm 90mm 6 @ 130 mm 520mm 5 @ 140 mm 700 mm Rest @150mm

TOTAL DISTANCE FROM SUPPORT 90mm 610 mm 1310 mm

4TH FLOOR b = 255 mm f’c = 28 MPa

d = 367.5 mm fy= 230 MPa

Use 10 mm Ø RSB Solution:

ØVc = 0.75 (103.19) ØVc = 77.39 kN< Vu STIRRUPS ARE NECESSARY Vu − Vc ϕ 137.75 VS = 0.75 − 103.19 = 80.48 kN Vs =

(verify if the section is adequate to carry the shear) 0.66√f ′ c bw d = 0.66 √28 (285) (402.5) = 400620.99 N = 400.62 kN> Vs (OK!) 𝜋

Av = 2Ab = 4 (10)2 (2) Av = 50 𝜋 mm2

107.372 − 43.3879 Vu − 43.3879 = 3 3 − (0.3675 + 0.2) Vu = 95.27 kN VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (255) (367.5) Vc = 84299.59 = 84.30 kN 104 | R e i n f o r c e d C o n c r e t e

ØVc = 0.75 (84.30) ØVc = 63.23 kN< Vu

Solution:

STIRRUPS ARE NECESSARY Vu − Vc ϕ 95.27 VS = − 84.30 = 42.73 kN Vs =

0.75

(verify if the section is adequate to carry the shear) 47.7588 30.836 = 𝑥 3−𝑥

0.66√f ′ c bw d = 0.66 √28 (255) (367.5) = 327280.76N = 327.28 kN> Vs (OK!)

x = 1.82

𝜋

Av = 2Ab = 4 (10)2 (2) Av = 50 𝜋 mm2

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(367.5) = Vs 42730 N S = 310.72 mm ≈ 300 mm S=

For maximum spacing, 0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(255)(367.5) = 163.64 kN> VS d

367.5

So, 𝑠𝑚𝑎𝑥 = = = 183.75 ≈ 2 2 200 mm 𝑜𝑟 600 mm Therefore, use s = 200 mm

SPACING

LENGTH

TOTAL DISTANCE FROM SUPPORT

47.7588 𝑉𝑢 = 1.82 1.82 − (0.3125 + 0.2) Vu = 34.31 kN VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (205) (312.5) Vc = 57627.77 N = 57.63 kN ØVc = 0.75 (57.63) ØVc = 43.22 kN> Vu but, 1 1 ØVc = 2 (43.22) = 21.61 kN 2 then,

1

Vu >2ØVc s=

𝑑 2

=

312.5 2

= 156.25 mm “say” 120 mm

Rest@200mm

Solving for minimum area of stirrup:

ROOF DECK

Av =

b = 205 mm f’c = 28 MPa

d = 312.5 mm fy= 230 MPa

Av=

𝑏𝑤 𝑠 3 𝑓𝑦 205 (120) 3 (230)

= 35.65 mm2

Therefore, use s = 120 mm Use 10 mm Ø RSB SPACING

LENGTH

TOTAL DISTANCE FROM SUPPORT

Rest @120mm

105 | R e i n f o r c e d C o n c r e t e

For lengths 5m 2ND FLOOR

= 419829.14 N = 419.83 kN> Vs (OK!)

b = 295 mm f’c = 28 MPa

d = 407.5 mm fy= 230 MPa

Use 10 mm Ø RSB Solution:

𝜋

Av = 2Ab = 4 (10)2 (2) Av = 50 𝜋 mm2

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(407.5) = Vs 57970 N S = 253.96 mm “say” 250 mm S=

For maximum spacing, 0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(295)(407.5) = 209.91 kN> VS d

407.5

So, 𝑠𝑚𝑎𝑥 = = = 203.75 ≈ 2 2 200 mm 𝑜𝑟 600mm Therefore, use s = 200 mm 158.328 119.6972 = 𝑥 5−𝑥 x = 2.85 158.328 Vu = 2.85 2.85 − (0.4075 + 0.2) Vu = 124.58 kN VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (295) (407.5) Vc = 108137.81 N = 108.14 kN ØVc = 0.75 (108.14) ØVc = 81.11 kN< Vu

SPACING

LENGTH

TOTAL DISTANCE FROM SUPPORT

Rest @200mm 3RD FLOOR b = 285 mm f’c = 28 MPa

d = 407.5 mm fy= 230 MPa

Use 10 mm Ø RSB Solution:

STIRRUPS ARE NECESSARY Vu − Vc ϕ 124.58 VS = 0.75 − 108.14 = 57.97 kN Vs =

(verify if the section is adequate to carry the shear) 0.66√f ′ c bw d = 0.66 √28 (295) (407.5) 106 | R e i n f o r c e d C o n c r e t e

151.802 118.8608 = 𝑥 5−𝑥 x = 2.80 151.802 Vu = 2.80 2.80 − (0.4075 + 0.2) Vu = 118.87 kN

Therefore, use s = 200 mm SPACING

LENGTH

TOTAL DISTANCE FROM SUPPORT

Rest@200mm

VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (285) (407.5) Vc = 104472.12 N = 104.47 kN

4TH FLOOR

ØVc = 0.75 (104.47) ØVc = 78.35 kN< Vu

b = 255 mm f’c = 28 MPa

STIRRUPS ARE NECESSARY

Use 10 mm Ø RSB

Vu − Vc ϕ 118.87 VS = − 104.47 = 54.02 kN

Solution:

d = 367.5 mm fy= 230 MPa

Vs =

0.75

(verify if the section is adequate to carry the shear) 0.66√f ′ c bw d = 0.66 √28 (285) (407.5) = 405597.64 N = 405.60 kN> Vs (OK!) 𝜋

Av = 2Ab = 4 (10)2 (2) Av = 50 𝜋 mm2

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(407.5) = Vs 54020 N S = 272.53 mm “say” 250 mm

136.364 118.9183 = 𝑥 5−𝑥

S=

x = 3.07

For maximum spacing,

118.9183 Vu = 3.07 3.07 − (0.3675 + 0.2) Vu = 96.94 kN

0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(285)(407.5) = 202.80 kN> VS d

407.5

So, 𝑠𝑚𝑎𝑥 = 2 = 2 = 203.75 ≈ 200 mm 𝑜𝑟 600 mm

VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (255) (367.5) Vc = 84299.59 N = 84.30 kN ØVc = 0.75 (84.30) ØVc = 63.23 kN< Vu STIRRUPS ARE NECESSARY Vu Vs = − Vc ϕ 96.94 VS = − 84.30 = 44.95 kN 0.75

(verify if the section is adequate to carry the shear) 107 | R e i n f o r c e d C o n c r e t e

0.66√f ′ c bw d = 0.66 √28 (255) (367.5) = 327280.76 N = 327.28 kN> Vs (OK!) 𝜋 4

Av = 2Ab = (10)2 (2) Av = 50 𝜋 mm2 S=

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(367.5) = Vs 44950 N

87.2842 𝑉𝑢 = 2.92 2.92 − (0.3125 + 0.2) Vu = 71.96 kN VC = 0.17𝜆√f ′ c bd Vc = 0.17 (1) √28 (205) (312.5) Vc = 57627.77 N = 57.63 kN

S = 295.38 mm “say” 250 mm

ØVc = 0.75 (57.63) ØVc = 43.22 kN< Vu

For maximum spacing,

STIRRUPS ARE NECESSARY

0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(255)(367.5) = 163.64 kN> VS d

367.5

So, 𝑠𝑚𝑎𝑥 = 2 = 2 = 183.75 ≈ 200 mm 𝑜𝑟 600 mm

Vu − Vc ϕ 71.96 VS = 0.75 − 57.63 = 38.32 kN Vs =

(verify if the section is adequate to carry the shear)

Therefore, use s = 200 mm

SPACING

LENGTH

TOTAL DISTANCE FROM SUPPORT

Rest@200mm

0.66√f ′ c bw d = 0.66 √28 (205) (312.5) = 223731.35 N = 223.73 kN> Vs (OK!) 𝜋 4

Av = 2Ab = (10)2 (2) Av = 50 𝜋 mm2

ROOF DECK b = 205 mm f’c = 28 MPa

d = 312.5 mm fy= 230 MPa

Use 10 mm Ø RSB Solution:

S=

𝐴𝑣 𝑓𝑦 𝑑 50𝜋 (230)(312.5) = Vs 38320 N

S = 294.63 mm ≈ 250 mm For maximum spacing, 0.33√𝑓 ′ 𝑐 𝑏𝑤 𝑑 = 0.33√28(205)(312.5) = 111.87 kN> VS d

312.5

So, 𝑠𝑚𝑎𝑥 = 2 = 2 = 156.25 ≈ 150 mm 𝑜𝑟 600 mm Therefore, use s = 150 mm 87.2842 83.6707 = 𝑥 5−𝑥 x = 2.92

SPACING

LENGTH

TOTAL DISTANCE FROM SUPPORT

Rest @150mm

108 | R e i n f o r c e d C o n c r e t e

LATERAL TIES Spacing of ties: a) 16 x db = 16(25) = 400 mm b) 48 x tie diameter= 48(10) = 480 mm c) Least dimension of column = 400 mm use 10mm ties at 400 o.c. SPACING

LENGTH

5 @50mm 5@100mm 3@150mm Rest @ 400mm

250mm 500mm 450mm

TOTAL DISTANCE FROM SUPPORT 250mm 750mm 1200mm

Based on the results, use this spacing for stirrups from the following lengths: For 3 m length: SPACING

LENGTH

1 @ 90 mm 6 @ 130 mm 5 @ 140 mm Rest @150mm

90mm 520mm 700 mm

TOTAL DISTANCE FROM SUPPORT 90mm 610 mm 1310 mm

For 5 m length: SPACING

LENGTH

TOTAL DISTANCE FROM SUPPORT

All stirrups @200mm

109 | R e i n f o r c e d C o n c r e t e