Zeal Polytechnic, Pune.: Third Year (Ty) Diploma in Civil Engineering Scheme: I Semester: V
Zeal Polytechnic, Pune.: Third Year (Ty) Diploma in Civil Engineering Scheme: I Semester: V
Zeal Polytechnic, Pune.: Third Year (Ty) Diploma in Civil Engineering Scheme: I Semester: V
Page 1 of 10
Scheme – I
Sample Question Paper
Program Name : Civil Engineering Program Group
Program Code : CE/CR/CS
Semester : Fifth 22502
Course Title : Design of Steel and RCC Structures
Max. Marks : 70 Time : 4 Hrs.
Instructions:
1) All questions are compulsory.
2) Illustrate your answers with neat sketches wherever necessary.
3) Figures to the right indicate full marks.
4) Assume suitable data, if necessary.
5) Preferably, write the answers in sequential order.
6) Use of non-programmable electronic pocket calculator is permissible.
7) Mobile phone, pager and any other electronic communication devices are not
permissible in examination hall.
1
c) ( i) Define cover provided for reinforcement in RC section and state it’s
recommendations as per IS 456:2000.
(ii) Define over reinforced section and state two reasons due to which they are avoided
in actual practice.
2
Scheme – I
Sample Test Paper - I
Program Name : Civil Engineering Program Group
Program Code : CE/CR/CS
Semester : Fifth 22502
Course Title : Design of Steel and RCC Structures
Max. Marks : 20 Time : 1.15 Hrs.
Instructions:
1) All questions are compulsory.
2) Illustrate your answers with neat sketches wherever necessary.
3) Figures to the right indicate full marks.
4) Assume suitable data, if necessary.
5) Preferably, write the answers in sequential order.
6) Use of non-programmable electronic pocket calculator is permissible.
7) Mobile phone, pager and any other electronic communication devices are not
permissible in examination hall.
3
Scheme – I
Sample Test Paper - II
Program Name : Civil Engineering Program Group
Program Code : CE/CR/CS
Semester : Fifth 22502
Course Title : Design of Steel and RCC Structures
Max. Marks : 20 Time : 1.15 Hrs.
Instructions:
1) All questions are compulsory.
2) Illustrate your answers with neat sketches wherever necessary.
3) Figures to the right indicate full marks.
4) Assume suitable data, if necessary.
5) Preferably, write the answers in sequential order.
6) Use of non-programmable electronic pocket calculator is permissible.
7) Mobile phone, pager and any other electronic communication devices are not
permissible in examination hall.
4
22502
11920
4 Hours / 70 Marks Seat No.
Marks
P.T.O.
22502 [2]
Marks
2. Attempt any THREE of the following: 12
a) Explain the limit state of serviceability applicable to steel
structures.
b) In steel constructions bolts of grade 4.6 are generally used.
What do you mean by grade 4.6?
c) Define over reinforced sections and state two reasons due to
which they are avoided.
d) Diameter of steel bar is 20 mm. Use Fe415 steel and design
bond stress is 1.2 MPa. For plain bars in tension. Find
development length in tension and compression.
P.T.O.
22502 [4]
Marks
6. Attempt any TWO of the following: 12
a) Design a square column to carry an axial load of 1500 kN. The
unsupported length of the column is 3.5 m. Use M20 concrete
& 1% Fe500 steel for longitudinal reinforcement. Use MS bar
for lateral ties. Apply the check for minimum eccentricity.
b) Design a circular column to carry an axial load of 1500 kN.
using MS Lateral ties. Use M25 concrete and Fe415 steel.
The unsupported length of column is 3.75 m.
c) Design on R.C. column footing with following data.
Size of column = 400 mm × 400 mm.
Safe bearing capacity of soil = 200 kN/m2.
Load on column = 1400 kN.
Concrete M20 and steel Fe 415 is used.
Calculate depth of footing from B.M. Criteria.
No shear check is required.
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
WINTER-19 EXAMINATION
Subject Name: DESIGN OF STEEL & RCC STRUCTURES Subject Code : 22502
Model Answer
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more Importance
(Not applicable for subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in the figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values
6) may vary and there may be some difference in the candidate’s answers and model answer
7) In case of some questions credit may be given by judgement on part of examiner of relevant
answer
8) For programming language papers, credit may be given to any other program based on equivalent
concept.
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d) Write the expression for minimum shear reinforcement giving the meaning of 1 2
the terms involved
0.87𝑓𝑦𝐴𝑠𝑣
Sv =
0.4𝑏
1
Sv – Spacing of stirrups
fy – Characteristic strength of steel,
Asv – Area of Stirrup bar,
b – width of the beam
e) Define the aspect ratio in case of slab and state its importance. 2
The ratio of Ly/Lx is know as aspect ratio of the slab, where Ly is longer side and 1
Lx is the shorter side of the slab.
The importance of this is that, if the ratio of Ly/Lx is greater than 2, then the
should be designed as a one way slab and if the ratio is less than 2 it should be
1
designed as a two way slab.
f) Write the two IS specifications for longitudinal reinforcement of an axially 1 M any2 2
loaded short column
IS specifications for longitudinal reinforcement of an axially
loaded short column:
i) Minimum diameter of bar in column = 12 mm
ii) Minimum number of bars in square/rectangle column = 4 Nos
iii) Minimum number of bars in circular column = 6 Nos
iv) Cover of the column = 40 mm
v) Minimum and maximum steel in column
vi) Max % of steel = 6 % of gross cross-sectional area of column
vii) Min % of steel = 0.8 % of gross cross-sectional area of column
g) Enlist two loads to be considered as per IS 875 -1987 while designing steel 1M any2 2
structure
Loads to be considered as per IS 875 -1987 while designing steel structures
1.DEAD LOAD ----IS 875-PART-1 -1987
2. LIVE LOAD -- IS 875-PART-2 -1987
3. WIND LOAD --- IS 875-PART-3 -1987
4. SNOW LOAD --- IS 875-PART-4 -1987
Q2 Attempt Any THREE of the following 12 M
a) Explain the limit state of serviceability applicable to steel structures. 4M
The acceptable limit for safety and serviceability of the structure before
failure occurs is called as Limit state. To assure the serviceability of the
2
P a g e 2 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2013 Certified)
c) Define over reinforced sections and state two reasons due to which they are 4M
avoided.
When xu > xmax or pt > pt lim The section is called a over reinforced section, It
is avoided due to the following reasons:
1) In over reinforced section, percentage of steel is more than critical
2
percentage, Due to this, the concrete crushes and reaches its ultimate
stress before steel reaches its yield point . In this case, the beam will fail
initially due to overstress in the concrete, suddenly without giving any
warning by way of large deformations and cracks as it does in the case
of under reinforced section. So, there is a huge loss of life and property. 2
2)The moment of resistance of the section does not increase more than
that of balanced section even if the steel is increased as compared to
balanced section
d) Diameter of steel bar is 20 mm. Use Fe415 steel and design bond stress is 1.2 4M
MPa. For plain bars in tension. Find development length in tension and
compression.
d= 20mm
fy= 415N/mm2
ꞇ bd= 1.2N/mm2 1
Development Length is given by
0.87𝑓𝑦Ø)
Ld=( 4 ꞇbd
For deformed bars, ꞇbd = 1.2*1.6
In tension 1
0.87∗415∗20)
Ld =(
4∗1.6∗1.2
= 940.23 mm 1
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2013 Certified)
In compression
ꞇbd = 1.6*1.25* ꞇbd
0.87∗415∗20) 1
Ld =(
4∗1.6∗1.25∗1.2
= 752.19 mm
Q3 Attempt any TWO of the following 12M
a) Design the lap joint for plates 100x10mm and 80x10mm thick connected to
transmit 120kN factored load using single row of 18mm dia bolts of 4.6 grade 6M
and plates of 415 grades.
For bolts of grade 4.6
fub = 400 N/mm2
For Fe 415 grade steel fu = 415
(d) Dia of bolts = 18mm 1
Dia of bolt hole = 18+2 = 20mm.
𝜋
Gross Area of bolt = 4 𝑥182 = 254.47mm2
Net area of bolts = 0.78*Gross Area
=0.78x254.47
= 198.49mm2
Single shear strength of bolt 1
Vdsb = Vnsb
γm1
𝑓𝑢𝑏
Vnsb = (nn.Anb + ns Asb)
√3
400
= (1x198.49+0)
√3
= 45838.33N
45838.33
Vdsb = = 36670.67N
1.25
Vdsb = 36.67 kN
Vnpb = 2.5kb.d.t.fu
Kb is least of the below
𝑒 𝑃 𝑓𝑢𝑏
[ , − 0.25, , 1.0]
3𝑑𝑜 3𝑑𝑜 𝑓𝑢
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Assuming e = 1.5do
= 1.5x20 = 30mm
P = 2.5d = 45 mm
30 45 400
= 0.5 , (3𝑥20 − 0.25) = 0.5 , 415 = 0.964, 1.0
3𝑥20
30 ₮ 45₮ 45 ₮ 45 ₮ 45
Alternatively
Single shear strength of bolt
Vdsb = Vnsb
1
γm1
𝑓𝑢𝑏
Vnsb = (nn.Anb + ns Asb)
√3
400
= (1x198.49+0)
√3
= 45838.33N
45838.33
Vdsb = = 36670.67N
1.25
Vdsb = 36.67 kN
120
Therefore no of bolts: = 3.27 ≈ 4 𝑁𝑜 ′ 𝑠 1
36.67
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Vnpb = 2.5kb.d.t.fu
1
Kb is least of the below
𝑒 𝑃 𝑓𝑢𝑏
[ , − 0.25, , 1.0]
3𝑑𝑜 3𝑑𝑜 𝑓𝑢
35 65 400
= 0.583 , (3𝑥20 − 0.25) = 0.833 , 415 = 0.964 , 1.0
3𝑥20
Vnpb = 2.5x0.583x18x10x415
= 108875.25 N
𝑉𝑛𝑝𝑏 108875.25
𝑉𝑑𝑝𝑏 = =
1.25 1.25
= 87100.2N
= 87.1 kN
30 ₮ 65 ₮ 65 ₮ 65 ₮ 30 ₮
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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b) Design a suitable fillet welded connection for ISA 80x50x8 with its longer leg 6M
connected to gusset plate of thickness 8mm. The angle is subjected to
factored load of 270kN. Cxx = 27.5mm. Assume weld applied to all 3 edges
and shop weld. Take fy = 250 MPa & fu = 410 MPa.
Given Pu = 270 kN
Minimum size of weld = 3 mm 1
3
Maximum size of weld = 4 (8) = 6𝑚𝑚
Provide a weld size of 4 mm.
Note: - Students may assume any other size of weld between 3 mm to 6 mm.
The answer will vary examiner needs to check as per the size of weld
considered by the student.
Pdw = fwd * L * t
270∗103
Length of weld required =
530.24
= 509.21 ≈ 510 mm
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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27.5
1
0.87x415x804.25
Xu = = 161.32 mm
0.36x20x250
Check Xu max 1
Xumax = 0.48d = 0.48x450 = 216mm
Since Xu < Xumax section is under reinforced
MR = 0.87 fy Ast (d – 0.42Xu)
= 0.87x 415 x 804.25 (450 – 0.42 x 161.32)
= 110994360 N-mm 1
MR = 110.99 kN-m
1
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Effective span
0.3 0.3
Le = 6 + + = 6.3𝑚. 1
2 2
wl2 20∗6.32
BM for simply supported beam = = = 99.23𝑘𝑁𝑀
8 8
1
Factored BM = 1.5x99.23 = 148.84 kNM
Mulim = 0.133*fck*b*d2
𝟏
As b = 𝟐 𝑑
Calculate effective depth d and width b 1
𝑑3
Mulim = 0.133 fck 2
𝑑3
148.84x106 = 0.133x20 2 = 481.89mm ≈ 490 mm.
0.5𝑥𝑓𝑐𝑘 4.6𝑀𝑢𝑙𝑖𝑚
Ast = [1 − √1 − ] bd
𝑓𝑦 𝑓𝑐𝑘𝑏𝑑2
P a g e 9 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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0.5𝑥20 4.6𝑥148.84𝑥106
= [1 − √1 − ] 245x490 = 848.6mm2 1
500 20𝑥245𝑥4902
𝐴𝑠𝑡 848.6
No. of bars = = = 2.7 ≈ 3 No’s 1
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑟 314.16
Alternatively,
Or
Ast can also be determined as
1
𝐴𝑠𝑡
Pt,lim = ∗ 100
𝑏𝑑
0.76∗𝑏𝑥∗𝑑 0.76∗245∗490
Ast = = = 912.38mm2
100 100
912.38
= 2.9 ≈ 3 No’s.
314.16
1
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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𝑤𝑑∗𝑙 60∗5
Factored shear force (Vu) = = = 150𝑘𝑁
2 2
1
# Calculate Shear stress حv
𝑉𝑢 150𝑥10 3
حv= = 350𝑥500 = 0.857N/mm2 < حc max hence OK
𝑏𝑑
حc value is 0.5
Check if shear reinforcement is required
As حv > حc Shear reinforcement is required
1
Calculation of Balance shear
Vus = Vu - حc x bd
3
= 150x10 -0.5x350x500 = 62500 N.
𝜋
Area of bent up bar = [ 4 𝑥202] x 1 = 314.16mm2
Assuming 450 bend
# Shear resisted by Bent-up bar 1
Vusb = shear resisted by bent-up bar
= 0.87 x fy x Asb x Sin45
= 0.87 x 415 x 314.16 x Sin45 = 80205.33 N
𝑉𝑢
Vusb > hence OK.
2
# Shear Resisted by vertical shear stirrups will be 1
𝑉𝑢
Vusv = = 31.25 𝑘𝑁
2
2∗𝜋82
For two legged vertical stirrups Asv = = 100.53 mm2
4
S = 0.87 ∗ 𝑓𝑦 ∗ 𝐴𝑠𝑣 ∗ 𝑑
Vusv
= 0.87x415x100.53x500 = 580.74 mm
31.25x103
Maximum spacing is given as 300 mm
0.75d = 0.75x500=375 mm
1
580.74 mm
Least value of the above is 8 mm stirrups @ 300 mm c/c
Alternatively
Check for minimum stirrups if capable of taking the shear force resisted by OR
minimum stirrups
Vusv (min) = 0.4 x b x D = 70 kN >Vusv
Minimum stirrups are sufficient
2∗𝜋∗82
Asv = = 100.53 mm2
4
P a g e 11 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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0.87∗𝑓𝑦∗𝐴𝑠𝑣 0.87∗415∗100.53
Sv = = = 259.25 ≈ 250 mm
0.4∗𝑏 0.4∗350
0.87∗𝑓𝑦∗𝐴𝑠𝑣∗𝑑
Vus = (𝑆𝑖𝑛 𝛼 + 𝐶𝑜𝑠 𝛼)
𝑆𝑣
1) 𝛼 – angle between the inclined stirrup and the axis of member not
less than 450
iii) Bent up bars. 2
Vus = 0.87fy Asb.Sin α
1) Asb – area of bentup bar
2) Vus - balance shear
3) Fy - Grade of steel/characteristic strength of stirrup
4) α - angle between the inclined stirrup and the axis of member not less
than 450
P a g e 12 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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= 3000 = 107. 14 mm
20 x 1.4 1
# Factored max BM 1
Md = wd x le2 / 8
12.563 𝑥 3.112
Md= = 15.19 kN-m
8
#Check for required depth 1
d= SQRT(15.19 x 106/0.138 x 20 x 1000)
d = 74.19mm
davail = 110mm > dreqd -: ok
Provide D = 135mm
davail 110mm
#Area of main steel and its spacing
4.6 𝑀𝑑
Ast = 0.5 fck/fy [1 − √1 − ] 𝑥 𝑏𝑑
𝑓𝑐𝑘𝑏𝑑2
1
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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P a g e 14 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2013 Certified)
P a g e 15 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
P a g e 16 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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# Required Depth 1
0.138 fckbd2 = Md
12.64∗ 106
dreq = √ = 67.67mm
0.138∗ 20 ∗ 1000
P a g e 17 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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P = 1500 kN
Factored load = 1500 x 1.5 = 2250kN
Lo = 3.5 m.
fck = 20 N/mm2
fy = 500 N/mm2 1
As Asc is 1% of gross area , Asc = 0.01 Ag
Ac = Ag – Asc
= Ag – 0.01 Ag = 0.99 Ag
Pu = 0.4 fck. Ac + 0.67 fy Asc
2250 x 103 = 0.4 x 20 x 0.99 Ag + 0.67 x 500 x 0.01 Ag
Ag = 199645.08 mm2
#Size of Footing
√𝐴𝑔 1
𝐷=
= 446.82 ~ = 450mm
P a g e 18 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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P a g e 19 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Asc = 0.0 1 A g
Ac = Ag – A S C
= 0.99 Ag
Pu = 0.4 fck.Ac + 0.67 fy ASC
2250x103 = 0. 4 × 25 × 0.99 A g + 0.67 × 415 × 0.01 Ag
A g = 177 437.8 mm2
#Diameter of column
4 𝑥 177437.8 1
𝐷= √
3.14
= 475.43 mm D = 480 mm
D should not be less than 0.12× L
0.12 × 3 750 = 450mm -ok
D should not be less than 400 mm -ok
D=480mm
#Check for min eccentricity
emin = Lo/500 + D/30 1
= 3750/500 + 480/30
=23.5mm
emax = 0.05 x D
= 0.05 x 480 = 24mm
emin < emax :- OK
#Check for slenderness ratio
L/D = 3750/480 = 7.8 < 12 :- short column
Asc = 0.01 x (π x 4802/4) 1
= 1808.64 mm2
Assume 20mm dia of bar
Provide 6 bars of 20mm
Transverse steel
Diameter = Ø t = ¼ x Ø L max or 6mm whichever is greater
a) = ¼ x 20 or 6mm 1
5mm or 6mm
Ø t = 6mm
a) Spacing
Min of the following
1) D = 480mm 1
2) 16 Ø L = 16 x 20 = 320mm
3) 300mm
P a g e 20 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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c) Design on R.C. column footing with following data. Size of column = 400 rnm 6M
x 400 mm. Safe bearing capacity of soil = 200 kN/m2. Load on column =
1400 kN. Concrete M20 and steel Fe 415 is used. Calculate depth of footing
from B.M. Criteria. No shear check is required.
Column = 400mm x 400mm
SBC = 200KN/m2
P = 1400KN
Factored Load = 1400 x 1.5 = 2100KN
fck = 20N/mm2
fy = 415N/mm2
Ultimate Bearing Capacity = 2 x SBC = 400KN/m2
Size of footing
1
Wf = 2100KN
Af = 1.05 x Wf/SBC
= 1.05 x 2100/400 = 5.51m2
L = B = √𝐴𝑓 = √5.51 = 2.35m
Adopt footing of size 2.35m x 2.35m
#Calculation of Upward soil pressure (q)
1
q = Wf/( L x B ) = 2100 / (2.35 x 2.35 )= 380.27KN/m2
# Calculation of Depth from flexure 1
Lx = Ly= (2.35-0.4)/2 = 0.975 m
M = q*Lx2/2
Mx= My = 380.27* 0.975*0.975/2 = 180.75 KN m
1
# Check for depth required
𝑀
d-reqd = √[ ]
𝑞𝑓𝑐𝑘𝑏
180.75 𝑥 106
= √ = 255.91mm = 260mm
0.138 𝑥 20 𝑥 1000
D = 260 + 50 = 310mm 1
#Calculation of Steel
4.6 𝑀𝑑
𝐀𝐬𝐭𝐱 = 𝐀𝐬𝐭𝐲 = 0.5 fck/fy [1 − √1 − ] ∗ 𝑏𝑑
𝑓𝑐𝑘𝑏𝑑2
= 0.5 x 20 / 415
4.6∗ 180.75∗ 106
[1 − √1 − 20∗1000∗2602
] ∗ 1000 ∗ 260 = 2377.58 mm2
P a g e 21 | 22
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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1
Using 20mm bar (Aᶲ = 314mm2)
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