Structures Assignment: AS3600-2009 Reinforced Concrete Solutions

This assignment provides detailed solutions to several problems related to reinforced concrete structures, adhering to the Australian standard AS3600-2009. The problems cover topics such as determining the neutral axis parameters and critical second moment of inertia for a given concrete section, designing a flanged beam for bending under specified loading conditions, analyzing a box beam for bending, and designing shear reinforcement for a concrete beam subjected to shear force and torsion. Calculations are performed to determine the required areas of steel reinforcement, spacing of shear reinforcement, and longitudinal checks for balance. The solutions include references to specific clauses and equations from the AS3600-2009 code and detailing considerations are also addressed. The document includes calculations for ultimate moment capacity, shear capacity, and torsion resistance, ensuring that the structural designs meet the required safety factors and code provisions. The assignment concludes with a bibliography of relevant literature on reinforced and prestressed concrete design.

Structures 1
STRUCTURES
By (Name)
Course
Professor’s name
University name
City, State
Date of submission

Structures 2
Question 1
As= Ast+ Asc where Asc ( 2 N 28 ) , Ast ( 8 N 28)
4960+ 1240=6200 m m2
b=300 mm
Es=200000 MPa , Ec=26000 MPa
d= D−60=740 mm
Neutral axis parameters,
k = √ ( pn ) 2+2 pn− pn
Where, n= Es
Ec
200000
26000 =7.69
p= As
bd = 6200
300× 740 =0.028
k = √ ( 0.028 ×7.69 )2 +2× 0.028 ×7.69−0.028× 7.69
k =0.512
Thereforekd =0.512× 740=378.88 mm,

Structures 3
Critical 2nd moment of inertia, Icr =b d2
12 [ 4 k2+12 pn ( 1−k )2 ]
¿ 300× 7402
12 [ 4 ×0.5123 +12 ×0.028 ×7.69 ( 1−0.512 )2 ]
¿ 4.812 ×1010 mm4
Question 2
L=8 m B1 exposure t=70 mm a=3000 mm
bw=300 mm
D=950 mm
g=25 kN /m q=100 kN /m
bef =bw +0.1 a 300+0.1 ×3000=600 mm
d= D−cover − ∅
2 −tie
d=950−25− 32
2 −10=904 mm
Loading
ULS=1.2 g+1.5 q
¿ 1.2 ×25+1.5 ×100=180 kN /m

Structures 4
Moment
M ¿= w l2
8 = 180 ×82
8 =1440 kNm
Mu=Mu 1+ Mu 2
Ultimate moment, M u=∅ ∝2 f '
c bef t ( d− t
2 )
0.85 ×08.5 × 50× 600× 70 ( 905− 70
2 )=1318.5 kNm
Flanged-beam
M u 2=∝2 f '
c b' t (d− t
2 )
0.85 ×50 ×70 ( 600−300 ) × (905− 70
2 )×10−6=776.475 kNm
As 2= Mu 2
f sy (d−t/2)= 776.475× 106
500(905− 70
2 )
=1785 m m2
Therefore, provide 3 N 32(2400 mm2)
Web-beam
Mu 1=Mu−Mu 2
¿ 1318.5−776.476=663.525 kNm

Structures 5
ξ= α2 f '
c
f sy
= 0.85 ×50
500 =0.085
pt 1 =ξ− √ξ2− 2 ξM
ϕb d2 f sy
¿ 0.085− √0.0852− 2× 0.085× 663.525× 106
0.8× 300× 9052 ×500 =7.05 ×10−3
As1 =pt 1 bd=7.05 ×10−3 ×300 × 905
¿ 1914.1 mm2
Therefore provide, 3N32 (2400 m m2 )
Question 3
Box Beam
b=3000 mm , bw=1200 mm , bw1 =900 mm , t=150 mm
d= D− ∅
2 −cover where, cover is exposed B1, ∅ =N 32
Loading
g=75 kN /m q=35 kN / m
ULS=1.2 g+1.5 q

Structures 6
1.2 ×75+1.5 ×35=142.5 kN / m
Moment
M ¿= w l2
8 = 142.5 ×202
8 =7125 kNm
M u=M u 1+ Mu 2
Ultimate moment, M u=∅ ∝2 f '
c bef t ( d− t
2 )
0.85 ×08.5 × 32×3000 × 150 (959−150
2 )=9197.14 kNm
Flanged-beam
Mu 2=∝2 f '
c b' t (d− t
2 )
¿ 0.85 ×32 ×150 ( 3000−900 ) (959−150
2 )=757.4 kNm
As 2= M u 2
f sy (d−t/2)
¿ 757.4 × 106
500(959−75) =1714 mm2
Therefore, provide 3 N 32(2400 mm2)

Structures 7
Mu 1=Mu−Mu 2
¿ 9197.14−757.4=8439.74 kNm
ξ= α 2 f ' c
f sy
=0.85 × 32
500 =0.0544
pt 1 =ξ− √ξ2− 2 ξM
ϕb d2 f sy
¿ 0.0544− √0.05442 −2 ×0.0544 × 8439.74 ×106
0.8× 900 ×9592 ×500 =0.041
As1 =pt 1 ( area of concrete section )
0.041 ( 1200× 959−900 ×610 ) =24673.8 mm2
Provide 4 N 32( As=32000 m m2)
Question 4
Shear reinforcement
Ag=2000 ×100+360 × 900=5.24 × 105 mm2
Asf =12600 mm2
g=25 kN /m
concecrete loads P=405 kN

Structures 8
Shear force, V = 1.2 G+1.5Q
2
¿ 1.2× 25× 1.5+405 ×2
2 =630 kN
V umax =0.2 f ' bw d
¿ 0.2 ×30 ×360 × 940=2030.4 kN
ϕ V umax=0.7 ×2030.4=1421.28
V ¿
ϕ = 630
0.7 =900 kN <V umax
Concrete shear capacity,
V uc=β1 β2 β3 bd 3
√ f ' c
3
√ Ast
bd × 10−3
β1=1.1 (1.6− 940
1000 )=0.7 use 1.1
β2=1
β3=2 × do
av
=2 × 0.94
2.5 =0.65 say 1
3
√f 'c=3
√32=3.17< 4 MPa acceptable
V uc=1.1× 1× 1× 360× 940 ×3.17 × 3
√ 12600
360 ×940 × 10−3 =394.03 kN

Structures 9
∅ V uc=0.7V uc=0.7× 394.03=275.821 kN
V umin=V uc+0.6 bd ×10−3
275.82+0.6 ×360 × 940× 10−3 =478.86 kN
ϕV umin=0.7 ×478.86 335.2 kN <V ¿
∴ shear reinforcement required
Shear reinforcement
V us= ( V ¿−ϕ V uc
ϕ )= 630−275.82
0.7 =505.97 kN
θv=30o +15o × V ¿−ϕ V umin
ϕ V umax−ϕ V umin
¿ 30o +15o × 630−335.2
1421.28−335.2 =34.1o
Say, Asv=220 mm2
S= Asv f sy do cot θv
V us
= 220 ×500 ×940 cot 34.1
505.97 ×103 =301.8 mm
Say spacing of 300mm
Longitudinal check
For balance

Structures 10
12600 ×500=0.85× 32(200 ×100+360( γ ku d −100))
γ ku d=187.83 mm
Distance between the C.O.G of compression
Ζ=
2000× 100× 100
2 +( 87.83× 360)×(100+ 87.83
2 )
2000× 100+87.83 ×360 =20.08 mm
Ζ=d−Ζ=920 mm
τ ¿= V ¿
Ζ b = 630× 103
920 ×360 =1.9 MPa
Asf
S = b
f syf
[ ( τ¿
∅ −k co f 'ct )
μ − gp
bf
]
From transverse reinforcement, Asv
S =220
300 =0.73 mm2 /mm
Where, μ=0.9 kco=0.5
f ' ct =0.36 √f 'c=0.36 √32=2.04 MPa

Structures 11
360
500 [ ( 1.9
0.7 −0.5× 2.04 )
0.9 − 3.6
360 ]=1.35 mm2 /mm
S= Asf
1.35
220
1.35 =163 mm Not okay for longitudinal reinforcement
Question 5
Similar to question 4 but design for torsion
T ¿=10 kNm
Ultimate torsion maximum, τumax =0.2 f '
c J t
Where, Jt =0.33 x2 y=0.33× 3602 × 1000=42768000
Therefore, τumax =0.2 ×32 × 42768000=273.73 kNm
τ ¿=ϕ τ umax [1− V ¿
ϕ V umax
]
¿ 0.7 × 273.73 [ 1− 630
142.28 ] =106.67 kNm> τ¿

Structures 12
τuc =Jt ( 0.3 √f '
c )=4276800 ( 0.3 √32 )=72.58 kNm
ϕ τuc [0.5− V ¿
ϕV uc ]
Since, ϕV uc=275.82kN
ϕ τuc [0.5− V ¿
ϕV uc ]=0.7 × 72.58 [0.5− 630
275.82 ]=−90.65 kNm
Therefore, it is okay
θv=34.1°
Asw
Sw = τ¿
∅ f sy 2 At cot θv
= 10 × 106
0.7 ×500 ×2 × ( 280 × 920 ) cot34.1
¿ 0.0375 m m2 /m m
Min reinforcement required, Asv min
Sv
= 0.35 bw
f sy
= 0.35 ×360
500 =0.252 m m2 /m
Therefore, Asw
Sw < Asvmin
Sv
ok

Structures 13
Bibliography
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pp.1391-1406.
Eltaher, M.A., Alshorbagy, A.E. and Mahmoud, F.F., 2013. Determination of neutral axis
position and its effect on natural frequencies of functionally graded macro/nanobeams.
Composite Structures, 99, pp.193-201.

Structures 14
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