Chapter 6 - Design Of Beams.docx

CHAPTER 6 DESIGN OF BEAMS

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

Moment Results from Grasp:

Exterior Beams Midspan Support 45.1587 -92.7698 80.8547 -179.847 139.8988 -233.931 156.3688 -254.288

Roof Deck 4th Floor 3rd Floor 2nd Floor

Interior Beams Midspan support 54.6783 -100.902 93.9771 -200.572 126.8842 -242.67 141.9795 -260.629

Solving for 𝝆b; f’c = 28 Mpa; fy = 414 Mpa; db = 25 mm 𝝆b =

0.85f′cβ600 fy(600+fy)

0.8(28)(0.85)600

=

414(600+414)

𝝆b = 0.02891 Solving for 𝝆, 𝝆max = 0.75 𝝆b = 0.75(0.02891) = 0.02168 𝝆min =

1.4 fy

=

1.4 414

= 0.003381

𝝆 = 0.6 𝝆max = 0.6 (0.02168) = 0.01301 So, use 𝝆 = 0.01301 Solving for ω, ω=

ρfy f′c

=

(0.01301)(414) 28

ω = 0.19236 Solving for 𝝎𝒎𝒂𝒙, 𝝎𝒎𝒂𝒙 =

ρ𝑚𝑎𝑥fy f′c

=

(0.02168)(414) 28

𝝎𝒎𝒂𝒙 = 0.32055

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

Designing for Exterior Beams:

Roof Deck Mumidspan = 45.1587 kN - m Musupport = -92.7698 kN - m Midspan Mu = 45.1587 kN – m 𝝆 = 0.01301 ω = 0.19236

fy = 414 MPa f’c = 28 MPa

Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd

Solution: Mu = ∅f′cωbd2(1 − 0.59ω) 45.1587×106 = 0.9 (28) (0.19236) bd2 (1 – 0.59 (0.19236)) bd2 = 10508570.9 mm Assume d = 1.75b b(1.75b)2 = 10508570.9 mm b = 150.83 mm “say” 220 mm d = 263.95 mm “say” 300 mm As = ρbd = 0.01301 (150.83) (263.95) = 517.95 mm2 Using 25 mm ∅ main bars: N=

517.95 π(25)2

Investigate:

= 1.06 “say” 2 bars

𝜋(25)2

(2) = ρ (220) (327.5) ρ = 0.0136 (ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

(0.0136)(414) 28

ω = 0.2011 Mu(capacity) = 0.9 (28) (0.2011) (220) (327.5)2 (1 – 0.59 (0.2011)) Mu(capacity) = 105.39 kN – m Mu(capacity) > Mu(design) = 45.1587 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY!

4

S = 220 – 100 – 2(25) = 70 mm > 50 mm (OK!)

Support

2 A ȳ = 2 A (62.5) ȳ = 62.5 mm

Mu = -92.7698 kN – m 𝝆max = 0.02168 ωmax = 0.32055

D = 300 + 62.5 = 362.5 mm “say” 390 mm dnew = 390 – 62.5 = 327.5 mm

Solution:

fy = 414 MPa f’c = 28 MPa

Mumax = ∅f′cωmaxbd2(1 − 0.59ωmax) Mumax = 0.9 (28) (0.32055) (220) (327.5)2 (1 – 0.59 (0.32055)) Mumax = 154.56 kN – m > Mu(design) 116 | R e i n f o r c e d C o n c r e t e

𝜋(25)2

DESIGN AS SINGLY REINFORCED Mu = ∅f′cωbd2(1 − 0.59ω) 92.7698 x 106 = 0.9 (28) (ω) (220) (327.5)2 (1 – 0.59 (ω)) ω = 0.1738 ω=

ρfy f′c

0.1738 =

ρ(414) 28

ρ = 0.0118 (ρmin < ρ < ρmax) (OK!) As = ρbd = 0.0118 (220) (327.5) = 850.19 mm2

(2) = ρ (220) (327.5) ρ = 0.0136 (ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

(0.0136)(414) 28

ω = 0.2011 Mu(capacity) = 0.9 (28) (0.2011) (220) (327.5)2 (1 – 0.59 (0.2011)) Mu(capacity) = 105.39 kN – m Mu(capacity) > Mu(design) = 92.7698 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY!

Using 25 mm ∅ main bars: N=

850.19 π(25)2

= 1.73 “say” 2 bars

4th Floor

4

S = 220 – 100 – 2(25) = 70 mm > 50 mm (OK!)

Mumidspan = 80.8547 kN - m Musupport = -179.847 kN - m Midspan

2 A ȳ = 2 A (62.5) ȳ = 62.5 mm dnew = 390 – 62.5 = 327.5 mm

Mu = 80.8547 kN – m 𝝆 = 0.01301 ω = 0.19236

fy = 414 MPa f’c = 28 MPa

Solution: Mu = ∅f′cωbd2(1 − 0.59ω) 80.8547×106 = 0.9 (28) (0.19236) bd2 (1 – 0.59 (0.19236)) bd2 = 18815141.89 mm Assume d = 1.75b b(1.75b)2 = 18815141.89 mm b = 183.15 mm “say” 270 mm d = 320.51 mm “say” 350 mm Investigate:

As = ρbd = 0.01301 (183.15) (320.51) = 763.71 mm2

Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd

Using 25 mm ∅ main bars:

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

N=

763.71 π(25)2

= 1.56 “say” 2 bars

4

S = 270 – 100 – 2(25) = 120 mm > 50 mm (OK!) 2 A ȳ = 2 A (62.5) ȳ = 62.5 mm D = 350 + 62.5 = 412.5 mm “say” 430 mm dnew = 430 – 62.5 = 367.5 mm

Support Mu = -179.847 kN – m 𝝆max = 0.02168 ωmax = 0.32055

fy = 414 MPa f’c = 28 MPa

Solution: Mumax = ∅f′cωmaxbd2(1 − 0.59ωmax) Mumax = 0.9 (28) (0.32055) (270) (367.5)2 (1 – 0.59 (0.32055)) Mumax = 238.85 kN – m > Mu(design) DESIGN AS SINGLY REINFORCED Mu = ∅f′cωbd2(1 − 0.59ω) 179.847 x 106 = 0.9 (28) (ω) (270) (367.5)2 (1 – 0.59 (ω)) ω = 0.2258 ω=

ρfy f′c

0.2258 =

ρ = 0.01527

Investigate: Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd 𝜋(25)2

(2) = ρ (270) (367.5) ρ = 0.00989 (ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

ρ(414)

(0.00989)(414) 28

ω = 0.14623 Mu(capacity) = 0.9 (28) (0.14623) (270) (367.5)2 (1 – 0.59 (0.14623)) Mu(capacity) = 122.78 kN – m

28

(ρmin < ρ < ρmax) (OK!)

As = ρbd = 0.01527 (270) (367.5) = 1515.17 mm2 Using 25 mm ∅ main bars: 1515.17 N = π(25)2 = 3.09 “say” 4 bars 4

S = 270 – 100 – 2(25) = 120 mm > 50 mm (OK!) 4 A ȳ = 2 A (62.5) + 2 A (112.5) ȳ = 87.5 mm dnew = 430 – 87.5 = 342.5 mm

Mu(capacity) > Mu(design) = 80.969 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY! 118 | R e i n f o r c e d C o n c r e t e

Solution: Mu = ∅f′cωbd2(1 − 0.59ω) 139.8988×106 = 0.9 (28) (0.19236) bd2 (1 – 0.59 (0.19236)) bd2 = 32554888.85 mm Assume d = 1.75b b(1.75b)2 = 32554888.85 mm b = 219.88 mm “say” 290 mm d = 384.79 mm “say” 390 mm Investigate: As = ρbd = 0.01301 (219.88) (384.79) = 1100.75 mm2

Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd

Using 25 mm ∅ main bars:

𝜋(25)2

(4) = ρ (270) (342.5) ρ = 0.02123 (ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

(0.02123)(414)

N=

1100.75 π(25)2

= 2.24 “say” 3 bars

4

290−100−3(25)

S= (OK!)

28

ω = 0.3139 Mu(capacity) = 0.9 (28) (0.3139) (270) (342.5)2 (1 – 0.59 (0.3139)) Mu(capacity) = 204.14 kN-m Mu(capacity) > Mu(design) = 179.145 kN – m

2

= 57.5 mm > 50 mm

3 A ȳ = 3 A (62.5) ȳ = 62.5 mm D = 390 + 62.5 = 452.5 mm “say” 470 mm dnew = 470 – 62.5 = 407.5 mm

THEREFORE, THE DESIGN IS SAFE AND OKAY!

3rd Floor Mumidspan = 139.8988 kN - m Musupport = -233.931 kN - m Midspan Mu = 139.8988 kN – m 𝝆 = 0.01301 ω = 0.19236

fy = 414 MPa f’c = 28 MPa

Investigate: Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd 119 | R e i n f o r c e d C o n c r e t e

𝜋(25)2

(3) = ρ (290) (407.5) ρ = 0.01246 (ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

Using 25 mm ∅ main bars: N=

(0.01246)(414)

1772.63 π(25)2

= 3.61 “say” 4 bars

4

28

ω = 0.18423

S = 290 – 100 – 2(25) = 140 mm > 50 mm (OK!)

Mu(capacity) = 0.9 (28) (0.18423) (290) (407.5)2 (1 – 0.59 (0.18423)) Mu(capacity) = 199.27 kN – m

4 A ȳ = 2 A (62.5) + 2 A (112.5) ȳ = 87.5 mm

Mu(capacity) > Mu(design) = 139.8988 kN – m

dnew = 470 – 87.5 = 382.5 mm

THEREFORE, THE DESIGN IS SAFE AND OKAY!

Support Mu = -233.931kN – m 𝝆max = 0.02168 ωmax = 0.32055

fy = 414 MPa f’c = 28 MPa

Solution: Mumax = ∅f′cωmaxbd2(1 − 0.59ωmax) Mumax = 0.9 (28) (0.32055) (290) (407.5)2 (1 – 0.59 (0.32055)) Mumax = 315.43 kN – m > Mu(design) DESIGN AS SINGLY REINFORCED Mu = ∅f′cωbd2(1 − 0.59ω) 233.931 x 106 = 0.9 (28) (ω) (290) (407.5)2 (1 – 0.59 (ω)) ω = 0.22179

Investigate: Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd 𝜋(25)2

(4) = ρ (290) (382.5) ρ = 0.0177 (ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

(0.0177)(414) 28

ω = 0.26171 ω=

ρfy f′c

0.22179 = ρ = 0.015

ρ(414) 28

(ρmin < ρ < ρmax) (OK!)

As = ρbd = 0.015 (290) (407.5) = 1772.63 mm2

Mu(capacity) = 0.9 (28) (0.26171) (290) (382.5)2 (1 – 0.59 (0.26171)) Mu(capacity) = 236.61 kN – m Mu(capacity) > Mu(design) = 233.038 kN – m

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

THEREFORE, THE DESIGN IS SAFE AND OKAY!

2nd Floor Mumidspan = 156.3688 kN - m Musupport = -254.288 kN - m Midspan Mu = 156.3688 kN – m 𝝆 = 0.01301 ω = 0.19236

fy = 414 MPa f’c = 28 MPa

Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd

Solution: Mu = ∅f′cωbd2(1 − 0.59ω) 156.3688×106 = 0.9 (28) (0.19236) bd2 (1 – 0.59 (0.19236)) bd2 = 36387509.43 mm Assume d = 1.75b b(1.75b)2 = 36387509.43 mm b = 228.19 mm “say” 300 mm d = 399.33 mm “say” 400 mm As = ρbd = 0.01301 (228.19) (399.33) = 1185.51 mm2 Using 25 mm ∅ main bars: N=

Investigate:

1185.51 π(25)2

= 2.42 “say” 3 bars

𝜋(25)2

(3) = ρ (300) (402.5) ρ = 0.0122 (ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

(0.0122)(414) 28

ω = 0.18039 Mu(capacity) = 0.9 (28) (0.18039) (300) (402.5)2 (1 – 0.59 (0.18039)) Mu(capacity) = 197.42 kN – m Mu(capacity) > Mu(design) = 156.3688 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY!

4

300−100−3(25)

S= (OK!)

2

= 62.5 mm > 50 mm

3 A ȳ = 3 A (62.5) ȳ = 62.5 mm D = 400 + 62.5 = 462.5 mm “say” 465 mm dnew = 465 – 62.5 = 402.5 mm

Support Mu = -254.288 kN – m 𝝆max = 0.02168 ωmax = 0.32055

fy = 414 MPa f’c = 28 MPa

Solution: Mumax = ∅f′cωmaxbd2(1 − 0.59ωmax) Mumax = 0.9 (28) (0.32055) (300) (402.5)2 (1 – 0.59 (0.32055)) 121 | R e i n f o r c e d C o n c r e t e

Mumax = 318.35 kN – m > Mu(design)

As = ρbd 𝜋(25)2

DESIGN AS SINGLY REINFORCED Mu = ∅f′cωbd2(1 − 0.59ω) 254.288 x 106 = 0.9 (28) (ω) (300) (402.5)2 (1 – 0.59 (ω)) ω = 0.24224 ω=

ρfy f′c

0.24224 =

ω=

ρfy f′c

=

(0.02139)(414) 28

ω = 0.31627 Mu(capacity) = 0.9 (28) (0.31627) (300) (382.5)2 (1 – 0.59 (0.31627)) Mu(capacity) = 284.54 kN – m

ρ(414)

ρ = 0.01638

(5) = ρ (300) (382.5) ρ = 0.02139 (ρmin < ρ < ρmax) (OK!) 4

28

(ρmin < ρ < ρmax) (OK!)

As = ρbd = 0.01631 (300) (402.5) = 1969.43 mm2

Mu(capacity) > Mu(design) = 254.288 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY!

Using 25 mm ∅ main bars: N=

1969.43 π(25)2

= 4.01 “say” 5 bars

4

300−100−3(25)

S= (OK!)

2

= 62.5 mm > 50 mm

5 A ȳ = 3 A (62.5) + 2 A (112.5) ȳ = 82.5 mm dnew = 465 – 82.5 = 382.5 mm

Designing for Interior Beams:

Roof Deck Mumidspan = 54.6783 kN - m Musupport = -100.902 kN - m Midspan Mu = 54.6783 kN – m fy = 414 MPa 𝝆 = 0.01301 f’c = 28 MPa ω = 0.19236 Solution: Mu = ∅f′cωbd2(1 − 0.59ω) 54.6783×106 = 0.9 (28) (0.19236) bd2 (1 – 0.59 (0.19236)) bd2 = 12723811.64 mm

Investigate: Mu = ∅f′cωbd2(1 − 0.59ω) where:

Assume d = 1.75b b(1.75b)2 = 12723811.64 mm b = 160.76 mm “say” 250 mm d = 281.33 mm “say” 300 mm

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

As = ρbd = 0.01301 (160.76) (281.33) = 588.40 mm2 Using 25 mm ∅ main bars: N=

588.40 π(25)2

= 1.20 “say” 2 bars

4

S = 250 – 80 – 2(25) = 120 mm > 50 mm (OK!) 2 A ȳ = 2 A (52.5) ȳ = 52.5 mm D = 300 + 52.5 = 352.5 mm “say” 380 mm dnew = 380 – 52.5 = 327.5 mm

Mu(capacity) > Mu(design) = 54.6783 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY!

Support Mu = -100.902 kN – m 𝝆max = 0.02168 ωmax = 0.32055

fy = 414 MPa f’c = 28 MPa

Solution: Mumax = ∅f′cωmaxbd2(1 − 0.59ωmax) Mumax = 0.9 (28) (0.32055) (250) (327.5)2 (1 – 0.59 (0.32055)) Mumax = 175.64 kN – m > Mu(design) DESIGN AS SINGLY REINFORCED Mu = ∅f′cωbd2(1 − 0.59ω) 100.902 x 106 = 0.9 (28) (ω) (250) (327.5)2 (1 – 0.59 (ω)) ω = 0.16548 ω=

ρfy f′c

0.16548 =

ρ(414)

ρ = 0.01119

Investigate: Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd 𝜋(25)2

(2) = ρ (250) (327.5) 4 ρ = 0.01199 (ρmin < ρ < ρmax) (OK!)

28

(ρmin < ρ < ρmax) (OK!)

As = ρbd = 0.01119 (250) (327.5) = 916.18 mm2 Using 25 mm ∅ main bars: N=

916.18 π(25)2

= 1.87 “say” 2 bars

4

ω=

ρfy f′c

=

(0.01199)(414) 28

ω = 0.17728 Mu(capacity) = 0.9 (28) (0.17728) (250) (327.5)2 (1 – 0.59 (0.17728)) Mu(capacity) = 107.26 kN – m

S = 250 – 80 – 2(25) = 120 mm > 50 mm (OK!) 2 A ȳ = 2 A (52.5) ȳ = 52.5 mm dnew = 380 – 52.5 = 327.5 mm 123 | R e i n f o r c e d C o n c r e t e

Solution: Mu = ∅f′cωbd2(1 − 0.59ω) 93.9771×106 = 0.9 (28) (0.19236) bd2 (1 – 0.59 (0.19236)) bd2 = 21868765.46 mm Assume d = 1.75b b(1.75b)2 = 21868765.46 mm b = 192.57 mm “say” 270 mm d = 337 mm “say” 380 mm Investigate: As = ρbd = 0.01301 (192.57) (337) = 844.30 mm2

Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd

Using 25 mm ∅ main bars:

𝜋(25)2

(2) = ρ (250) (327.5) ρ = 0.01199 (ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

(0.01199)(414)

N=

844.30 π(25)2

= 1.72 “say” 2 bars

4

S = 270 – 80 – 2(25) = 140 mm > 50 mm (OK!)

28

ω = 0.17728 Mu(capacity) = 0.9 (28) (0.17728) (250) (327.5)2 (1 – 0.59 (0.17728)) Mu(capacity) = 107.26 kN – m

2 A ȳ = 2 A (52.5) ȳ = 52.5 mm D = 380 + 52.5 = 432.5 mm “say” 450 mm dnew = 450 – 52.5 = 397.5 mm

Mu(capacity) > Mu(design) = 100.902 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY! 4th Floor Mumidspan = 93.9771 kN - m Musupport = -200.572 kN - m Midspan Mu = 93.9771 kN – m 𝝆 = 0.01301 ω = 0.19236

fy = 414 MPa f’c = 28 MPa

Investigate: Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd 124 | R e i n f o r c e d C o n c r e t e

𝜋(25)2

(2) = ρ (270) (397.5) ρ = 0.00915 (ρmin < ρ < ρmax) (OK!)

Using 25 mm ∅ main bars:

4

ω=

ρfy f′c

=

N=

1549.77 π(25)2

= 3.16 “say” 4 bars

4

(0.00915)(414) 28

S = 270 − 80 − 2(25) = 140 mm > 50 mm (OK!)

ω = 0.13529 Mu(capacity) = 0.9 (28) (0.13529) (270) (397.5)2 (1 – 0.59 (0.13529)) Mu(capacity) = 133.84 kN – m

4 A ȳ = 2 A (52.5) + 2 A (102.5) ȳ = 77.5 mm dnew = 450 – 77.5 = 372.5 mm

Mu(capacity) > Mu(design) = 93.9771 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY!

Support Mu = -200.572 kN – m 𝝆max = 0.02168 ωmax = 0.32055

fy = 414 MPa f’c = 28 MPa

Solution: Investigate: Mumax = ∅f′cωmaxbd (1 − 0.59ωmax) Mumax = 0.9 (28) (0.32055) (270) (397.5)2 (1 – 0.59 (0.32055)) Mumax = 279.44 kN – m > Mu(design)

Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd

DESIGN AS SINGLY REINFORCED

(4) = ρ (270) (372.5) ρ = 0.01952 (ρmin < ρ < ρmax) (OK!)

2

Mu = ∅f′cωbd2(1 − 0.59ω) 200.572 x 106 = 0.9 (28) (ω) (270) (397.5)2 (1 – 0.59 (ω)) ω = 0.21345 ω=

ρfy f′c

0.21345 =

ρ(414)

ρ = 0.01444

28

(ρmin < ρ < ρmax) (OK!)

As = ρbd = 0.01444 (270) (397.5) = 1549.77 mm2

𝜋(25)2 4

ω=

ρfy f′c

=

(0.01952)(414) 28

ω = 0.28862 Mu(capacity) = 0.9 (28) (0.28862) (270) (372.5)2 (1 – 0.59 (0.28862)) Mu(capacity) = 226.09 kN – m Mu(capacity) > Mu(design) = 200.572 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY! 125 | R e i n f o r c e d C o n c r e t e

3rd Floor Mumidspan = 126.8842 kN - m Musupport = -242.67 kN - m Midspan Mu = 126.8842 kN – m 𝝆 = 0.01301 ω = 0.19236

fy = 414 MPa f’c = 28 MPa Investigate:

Solution: Mu = ∅f′cωbd (1 − 0.59ω) 126.8842×106 = 0.9 (28) (0.19236) bd2 (1 – 0.59 (0.19236)) bd2 = 29526350.68 mm 2

Assume d = 1.75b b(1.75b)2 = 29526350.68 mm b = 212.84 mm “say” 260 mm d = 372.47 mm “say” 375 mm As = ρbd = 0.01301 (212.84) (372.47) = 1031.39 mm2 Using 25 mm ∅ main bars: N=

1031.39 π(25)2

(3) = ρ (260) (372.5) ρ = 0.0152(ρmin < ρ < ρmax) (OK!) 4

ω=

ρfy f′c

=

(0.0152)(414) 28

ω = 0.2247 Mu(capacity) = 0.9 (28) (0.2247) (260) (372.5)2 (1 – 0.59 (0.2247)) Mu(capacity) = 177.20 kN – m

= 2.10 “say” 3 bars

260−80−3(25) 2

𝜋(25)2

Mu(capacity) > Mu(design) = 126.8842 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY!

4

S=

Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd

= 52.5 mm > 50 mm (OK!)

3 A ȳ = 3 A (52.5) ȳ = 52.5 mm D = 375 + 52.5 = 427.5 mm “say” 425 mm dnew = 425 – 52.5 = 372.5 mm

Support Mu = -242.67 kN – m 𝝆max = 0.02168 ωmax = 0.32055

fy = 414 MPa f’c = 28 MPa

Solution: Mumax = ∅f′cωmaxbd2(1 − 0.59ωmax) Mumax = 0.9 (28) (0.32055) (260) (372.5)2 (1 – 0.59 (0.32055)) Mumax = 236.31 kN – m < Mu(design)

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

DESIGN AS DOUBLY REINFORCED As1 = ρbd = 0.02168 (260) (372.5) = 2099.71 mm2 N1 =

2099.71 π(25)2

= 4.28 “say” 5 bars

4

Mu1 = Mumax = 236.31 kN – m Mu2 = Mu – Mu1 Mu2 = ∅As2fy(d-d’) 242.67x106 – 236.31x106 = 0.9 (As2) (414) (372.5 – 52.5) As2 = 53.34 mm2 N=

53.34 π(25)2

a = β1c 140.48 = 0.85c c = 165.27 mm f’s = 600

𝑐−𝑑′ 𝑐

= 600

165.27−52.5 165.27

f’s = 409.40 MPa < fy = 414 MPa compression steel does not yield 𝑓𝑦

A’s = As2 𝑓′𝑠

= 0.11 “say” 2 bars

414

A’s = 53.34 409.40 = 53.94 mm2

4

S=

0.85f’c ab = As1fy 0.85(28) a (260) = 2099.71 (414) a = 140.48 mm

260−80−3(25) 2

= 52.5 mm > 50 mm (OK!)

Investigate:

5 A ȳ = 3 A (52.5) + 2 A (102.5) ȳ = 72.5 mm

Assume compression steel yields (f’s = fy)

dnew = 425 – 72.5 = 352.5 mm

As =

𝜋 (25)2 4

(5) = 2454.37 mm2

As2 = A’s =

𝜋 (25)2 4

As1 = As – As2 = 1472.62 mm2

(2) = 981.75 mm2

𝜋 (25)2 4

(5) -

𝜋 (25)2 4

(2) =

Solve for a and c: C1 = T1 0.85f’c ab = As1fy 0.85(28) a (260) = 1472.62(414) a = 98.52 mm Check if compression steel yields: f’s = 600 where: C1 = T1

𝑐−𝑑′ 𝑐

a = β1c 98.52 = 0.85c c = 115.91 mm Solve for the stress in compression steel

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

f’s = 600

𝑐−𝑑′ 𝑐

= 600

115.91−52.5 115.91

f’s = 328.24 MPa < fy = 414 MPa compression steel does not yield Investigate: f’s = 600

𝑐−𝑑′ 𝑐

= 600

𝑐−52.5

Solution: Mu = ∅f′cωbd2(1 − 0.59ω) 141.9795×106 = 0.9 (28) (0.19236) bd2 (1 – 0.59 (0.19236)) bd2 = 33039074.26 mm

𝑐

C1 + C2 = T 0.85f’c ab + A’sf’s = Asfy 𝑐−52.5 0.85(28)(0.85c)(260) + 981.75(600 𝑐 ) = 2454.37(414) c = 127.36 mm a = β1c a = 0.85(127.36) a = 108.26 mm

Assume d = 1.75b b(1.75b)2 = 33039074.26 mm b = 220.96 mm “say” 300 mm d = 386.68 mm “say” 400 mm As = ρbd = 0.01301 (220.96) (386.68) = 1111.58 mm2 Using 25 mm ∅ main bars: N=

f’s = 600

𝑐−𝑑′ 𝑐

= 600

f’c = 28 MPa

𝝆 = 0.01301 ω = 0.19236

127.36−52.5

1111.58 π(25)2

= 2.26 “say” 3 bars

4

127.36

S=

f’s = 352.67 MPa Mu = Mu1 + Mu2 Mu = ∅f′c ab (d – a/2) + ∅A’sf’s (d-d’) Mu = 0.9(28) (108.26)(260)(372.5 – 108.26/2) + 0.9(981.75)(352.67)(352.5-52.5) Mu(capacity) = 319.31 kN - m

300−80−3(25) 2

= 72.5 mm > 50 mm (OK!)

3 A ȳ = 3 A (52.5) ȳ = 52.5 mm D = 400 + 52.5 = 452.5 mm “say” 460 mm dnew = 460 – 52.5 = 407.5 mm

Mu(capacity) > Mu(design) = 242.67 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY!

2nd Floor Mumidspan = 141.9795 kN - m Musupport = -260.629 kN - m Investigate:

Midspan Mu = 141.9795 kN – m

fy = 414 MPa

Mu = ∅f′cωbd2(1 − 0.59ω) where: 128 | R e i n f o r c e d C o n c r e t e

As = ρbd 𝜋(25)2

(3) = ρ (300) (407.5) ρ = 0.01205 (ρmin < ρ < ρmax) (OK!)

As = ρbd = 0.01638 (300) (407.5) = 2002.46 mm2

4

ω=

ρfy f′c

=

Using 25 mm ∅ main bars:

(0.01205)(414)

N=

28

2002.46 π(25)2

= 4.08 “say” 5 bars

4

ω = 0.17817 S=

300−80−3(25) 2

= 72.5 mm > 50 mm (OK!)

Mu(capacity) = 0.9 (28) (0.17817) (300) (407.5)2 (1 – 0.59 (0.17817)) Mu(capacity) = 200.16 kN – m

5 A ȳ = 3 A (52.5) + 2 A (102.5) ȳ = 72.5 mm

Mu(capacity) > Mu(design) = 141.9795 kN – m

dnew = 460 – 72.5 = 387.5 mm

THEREFORE, THE DESIGN IS SAFE AND OKAY!

Support Mu = -260.629 kN – m 𝝆max = 0.02168 ωmax = 0.32055

fy = 414 MPa f’c = 28 MPa

Solution:

Investigate:

Mumax = ∅f′cωmaxbd2(1 − 0.59ωmax) Mumax = 0.9 (28) (0.32055) (300) (407.5)2 (1 – 0.59 (0.32055)) Mumax = 326.31 kN – m > Mu(design)

Mu = ∅f′cωbd2(1 − 0.59ω) where: As = ρbd 𝜋(25)2

(5) = ρ (300) (387.5) ρ = 0.02111 (ρmin < ρ < ρmax) (OK!) 4

DESIGN AS SINGLY REINFORCED Mu = ∅f′cωbd2(1 − 0.59ω) 260.629 x 106 = 0.9 (28) (ω) (300) (407.5)2 (1 – 0.59 (ω)) ω = 0.24223 ω=

ρfy f′c

0.24223 =

ρ(414)

ρ = 0.01638

28

(ρmin < ρ < ρmax) (OK!)

ω=

ρfy f′c

=

(0.02111)(414) 28

ω = 0.31213 Mu(capacity) = 0.9 (28) (0.31213) (300) (387.5)2 (1 – 0.59 (0.31213)) Mu(capacity) = 289.07 kN – m Mu(capacity) > Mu(design) = 260.629 kN – m THEREFORE, THE DESIGN IS SAFE AND OKAY! 129 | R e i n f o r c e d C o n c r e t e