Transcript
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
dead load
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
LIVE LOAD
