Structures: Solved Questions and Answers

This document contains solutions to five questions related to structures, including calculations for neutral axis parameters, ultimate moment, shear reinforcement, and torsion. The bibliography lists relevant research articles on the topic.

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
Bernardo, L.F.A. and Lopes, S.M.R., 2004. Neutral axis depth versus flexural ductility in high-
strength concrete beams. Journal of Structural Engineering, 130(3), pp.452-459.
Chan, C.L., Khalid, Y.A., Sahari, B.B. and Hamouda, A.M.S., 2002. Finite element analysis of
corrugated web beams under bending. Journal of Constructional Steel Research, 58(11),
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
El-Sayed, A.K., El-Salakawy, E.F. and Benmokrane, B., 2006. Shear strength of FRP-reinforced
concrete beams without transverse reinforcement. ACI Structural Journal, 103(2), p.235.
Foster, S., 2010, December. 12 Design of FRC beams for shear using the VEM and the draft
Model Code approach. In Shear and Punching Shear in RC and FRC Elements: Technical
Report: Proceedings of a Workshop Held on 15-16 October 2010, in Salò, Lake Garda, Italy
(Vol. 57, p. 195). fib Fédération internationale du béton.
Loo, Y.C. and Chowdhury, S.H., 2010. Reinforced and Prestressed Concrete: Analysis and
Design with Emphasis on Application of AS3600-2009. Cambridge University Press.
Marčiukaitis, G., Jonaitis, B. and Valivonis, J., 2006. Analysis of deflections of composite slabs
with profiled sheeting up to the ultimate moment. Journal of Constructional Steel Research,
62(8), pp.820-830.
Tsavdaridis, K.D. and D'Mello, C., 2011. Web buckling study of the behaviour and strength of
perforated steel beams with different novel web opening shapes. Journal of Constructional Steel
Research, 67(10), pp.1605-1620.
Tang, Y.Q., Chen, L.Q. and Yang, X.D., 2008. Natural frequencies, modes and critical speeds of
axially moving Timoshenko beams with different boundary conditions. International Journal of
mechanical sciences, 50(10-11), pp.1448-1458.
Zararis, P.D., 2003. Shear strength and minimum shear reinforcement of reinforced concrete
slender beams. Structural Journal, 100(2), pp.203-214.