# A Rectangular Beam Has b

July 11, 2017 | Author: Marco Ramos Jacob | Category: Beam (Structure), Strength Of Materials, Structural Engineering, Building Engineering, Civil Engineering

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11.) A rectangular beam has b = 300 mm and d =490 mm . Concrete compressive strength f’c = 27.6 MPa and steel yield strength fy = 276 MPa . calculate the required tension steel area if the factor moment Mu is (a) 20 Knm (b) 140 Kn-m (c) 485 Kn-m (d) 620 Kn-m.

SOLUTION: Solve for ρ max and Mu max : '

¿

Ρb

¿

.85 f c β 1 600 fy (600+ fy)

pb

0.85 (27.6)(0.85)(600) 276(600+276) Pb Ρmax= 0.75 ρb

= 0.0495 ρmax = .75(0.0495)

Ρmax=0.0371 ωmax

¿

¿

pmax fy f 'c

ωmax

0.0371( 276) 27.6 ωmax =

0.371

Rn max = f’c ω(1-0.59ω) 0.59(.371)]

Rn maax =27.6( .371)[1Rn max =8.001 MPa

Mn max = Rn max b d2 (490)

Mn max = 8.001(300)

2

Mn max = 576.279 x 106 n-mm Mn max = 576.279 kN-mm Mu max =Ø Mn max

Mu max =0.90 x 576.279 Mu max =518.65 kN-m

A) Mu = 20 kN-m < Mu max (singly reinforced) Mu = Ø Rn b d2 20 X 106 = .90 Rn ( 300)(490)2 Rn = 0.309 MPa ρ

¿

0.85 f ' c fy

1-

1-

2 Rn 0.85 f ' c

0.85 (27.6) 276

¿

ρ

1-

1–

2(0.309) 0.85(27.6)

ρ =0.00113 < ρ min

f'c ¿√ 4 fy

ρmin

ρ min =

1.4 fy

if f’c > 31.36 MPa otherwise ρ min =

1.4 fy

= .005072

As =ρ b d

As = 0.005072(300)(490) As = 746 mm2

b) Mu = 140 kN-m < Mu max

(singly reinforced)

Mu = Ø Rn b d2

140 x 106 = .90 Rn (300)

(490) Rn = 2.16 MPa

ρ

ρ=

¿

0.85 f ' c fy

¿

0.85 (27.6) 276

1-

2 Rn 0.85 f ' c

1-

1-

1–

2(2.16) 0.85(27.6)

ρ = 0.00822

As = ρ b d

ρ= 0.00822 > ρ min As= 0.00822(300)(490) As = 1,209 mm2

c) Mu =485 kN –m < Mu max

(Singly Reinforce)

Mu =Ø Rn b d2

485 x106 Rn (300)(490)2

ρ

¿

0.85 (f ' c ) fy

1-

1–

2 Rn 0.85(f ' c )

ρ=

¿

0.85 (27.6) 276

1-

1–

2(7.48) 0.85(27.6)

ρ =0.03384> ρ min As = ρ b d

As =0.03384(300)(490) As = 4,975 mm2

d) Mu = 620 kN –m > Mu max The beam will be doubly reinforced. 12.) A rectangular beam has b = 310 mm and d=460 mm. The beam will be design to carry a service dead load of 230 kN-m and service live load of 190 kN-m .Compression reinforcement if necessary will have its centroid 70 mm from extreme concrete fiber . Determine the required steel area . Use f’c = 30 MPa and fy = 415 MPa.

SOLUTION: β= 0.85 Mu = 1.4 MD + 1.7 ML

Mu =1.4( 230) +1.7(190) Mu = 645 kN-m

Solve for ØMn max : NOTE: for rectangular beams Cmax = .75 cb C max = 0.75

600 d 600 fy = 203.94 mm

A =β1 C max = 173.35 mm Mn max =0.85 f’c ab [d – a/2 ] Mn max = 0.85 (30) (173.35)(310) [460- 173.35/2] Mn max = 511.58 kN-m ØMn max = 0.90 (511.58) ØMn max = 460.42 kN-m

Since Mu = 645 kN-m > Ø Mn max , Compression steel is necessary b A’s C’s=A's f’s

a

Cc =0.85f’c a b c =

d-d’

d

d- a/2

A’s

+

As

As1

As2 T2= As2 Fy

Mn1 = Mn max = 511.58 kN-m Mn2 =

Mu Ø - Mn1

Mn2 =

645 0.90

- 511.58

Mn2 = 205.088 C = Cmax = 203.94 mm A =173.35 mm

Tension steel: T1 = Cc

As1 fy = 0.85 f’c a b As1 (415)= 0.85(30)((173.35)

(310) As1 = 3,302 mm2 Mn2 = T2 (d –d’)

205.088x 106=As2(415)(460-

70) As2 = 1267 mm2 AS = As1 +As2

As= 3302 + 1267 As = 4569 mm2

Compression Steel: F’s 600

c−d ' c

f’s =600

203.94−70 203.94

f’s =394.06 MPa A1

T = 100 mm c

bf = 1200

a

C1

A1 z

A2

C2

y1 d=500mm

y2 As Bw = 280

A2 = Ac max – A1

T

A2 = 121,759 – 120,000 A2 = 1759 mm2

A2 = bw z

1759= 280 z Z = 6.28 mm

Y1 = d- t /2

y1 = 500 – 100/2 Y1= 450 mm

Y2 = d – t – z/2

y2 = 500 – 100 – 6.28/2 y 2= 396.86 mm

Mn max = C1 y1 + C2 y2

Mn max = 0.85 f'c(A1y1 +A2y2) Mn max -= 0.85(21)(120,00 x 450 +1759 x396.86) Mn m,ax = 976. 36 kN-m

Ø Mn max = 0.90 (976.36) Ø Mn max = 878.72 kN –m

Since Mu = 1080 kN-m > Ø Mn max , the compression reinforcement must be provided.

Bf = 1200 a` A1 C’s

A’s

d’ = 70

C1 A’s A2

z

C2 d= 500mm

500 mm 430

y1

y2

d-d’

As

d’ = 70

As1

As2

T2 Bw = 280 mm

As1 = Asmax Mn1 = Mn max

Mn2

A=t+z

a = 100 + 6.28 a= 106.28 mm

c =a/ β1

c = 106.28/0.85 c = 125.04

f’s = 600

c−d ' c

f’s = 600

125.04−70 125.01 f’s = 264.1

MPa < fy Mn1 = Mn max = 976.36kN-m

As1 = Asmax T1 = C1 +C2

As1 fy = 0.85f’c (A1+A2) As1(415)= 0.85(21)(120,000

+1759) As1 = 5237 mm2 Mn2 = Mn – Mn1

Mn2 =

Mu Ø

Mn2 =

1080 0,90

- Mn1

-

976.36

Mn2 = 223.64 kN-m Mn2 = T2 (d-d’)

Mn2 = As2fy (d-d’) 223.64 x 106 =As2(415)(500-

70) As2 = 1253 mm2 Tension steel area ,As = As1+ As2 = 6,490 mm2 Compression steel area: C’s =T2

A’s f’s = As2 fy A’s(264.1)= 1253(415) A’s =1969 mm2

1.2 m

0.1m 1.2 m

0.47 m 0.28 m

Dead load = weight of concrete: Area = 1.2 (0.1) + .28(0.47) = .2516 m3 Wc

=

Yc x area

Wc = 23.5 (0.2516) Wc = 5.9126 kN/m

Wu A

L= 9 m

B

WuL2/24

- WuL2/12

L= 9m

C

WuL2/24

- WuL2/12

- WuL2/12

Maximum positive moment ( at midspan)

Wu L2 Mu= 24

Wu (9)2 1,080 = 24 Wu = 320

kN/m Wu = 1.4 Wd + 1.7 Wl 1.7 Wl

320 = 1.4(5.9126) + WL =183 .37 kN/m