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8/12/2019 1401871257_Design and Drawing of Steel Structures 3-2 Set-1 A
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S.1Design and Drawing of Steel Structures (April/May-2013, Set-1) JNTU-Anantapur
B.Tech. III-Year II-Sem. ( JNTU-Anantapur )
Code No.: 9A01601/R09
III B.Tech. II Semester Regular & Supplementary Examinations
April/May - 2013DESIGN AND DRAWING OF STEEL STRUCTURES
( Civil Engineering )
Time: 3 Hours Max. Marks: 70
Use of IS 800-2007, IS 875 Part III, Steel Tablets and
Railway Design Standards Code are permitted in the examination hall.
- - -
Part A
(Answer any ONE question, 1 28 marks)
1. A column is subjected to an axial load of 750 kN. The beam connected to the flange of the column has an eccentric
load of 150 kN and the beam connected to the web of the column has an eccentric load of 60 kN. Design the column
for an effective length 4 m. Draw:
(a) Side view of the column
(b) Plan view of the column. (Unit-III, Topic No. 3.3)
2. A simply supported beam of effective span 8 m carries a U.D.L of 40 kN/m. Taking fy = 250 N/mm2and E = 2 105N/
mm2. Design the beam as laterally supported. Draw elevation and sectional details of the beam. (Unit-II, Topic No. 2.3)
Part B
(Answer any THREE questions, 3 14 marks)
3. (a) Write about major types of tension members. (Unit-III, Topic No. 3.1)
(b) The tension member of a roof truss carries a maximum axial tension of 250 kN. Design the section. Diameter of
connecting riveters 20 mm, safe stress in tension = 150 N/mm2. (Unit-III, Topic No. 3.1)
4. Design a built up column with four angles laced together. The effective length of the column is 7.20 m and it supports
a load of 1200 kN. (Unit-IV, Topic No. 4.1)
5. A steel column consists of ISHB 300 with cover plate 300 mm 25 m for each flange. The column carries an axial load
of 2300 kN. Design a gusted base plate for the column. Use 18 mm diameter rivets. (Unit-V, Topic No. 5.1)
6. Design an I section purlin with an without sag bars for trussed roof from the following data,
Span of roof = 10 m, spacing of purlin along slope of truss 1.8 m
Spacing of truss = 4 m slope of roof truss = 1 vertical 2 horizontal
Wind load on roof surface normal to roof = 1200 N/mm2.
Vertical loads from roof sheets = 200 N/mm2. (Unit-VI, Topic No. 6.2)
7. Design the cross section of a welded plate girder for a U.D.L of 80 kN/m. The effective span is 16 m.
(Unit-VII, Topic No. 7.1)
Set-1Solutions
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B.Tech. III-Year II-Sem. ( JNTU-Anantapur )
S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2014
SOLUTIONS TO APRIL/MAY-2013, SET-1, QP
Part A
(Answer any ONE question, 1 28 marks)
Q1. A column is subjected to an axial load of 750 kN. The beam connected to the flange of the columnhas an eccentric load of 150 kN and the beam connected to the web of the column has an eccentricload of 60 kN. Design the column for an effective length 4 m. Draw:
(a) Side view of the column
(b) Plan view of the column.
Answer : April/May-13, Set-1, Q1
Given that,
Axial load on column, P= 750 kN
Beam connected to the flange with an eccentric load of W1= 150 kN
Beam connected to the web with an eccentric load of W2= 60 kN
Effective length of column,Leff
= 4 m = 4000 mm
The total vertical loads acting on the column,
W= P+ W1+ W
2
W= 750 + 150 + 60
W= 960 kN
Assume,
Slenderness ratio of column, = 80Yield stress for steel,f
y= 260 N/mm2
Forfy= 260 N/mm2and = 80, the permissible stress in axial compression
bcis equal to 103 N/mm2.
Required area of section,
Areq
=bc
W
Areq
=103
109603
Areq
= 9320.388 mm2
Assume effective section are required,
A = 1.5Areq
A = 1.5 9320.388
A = 13980.582 mm2.
Adopt ISWB 550 @ 112.5 kg/m: with following properties:
Area provided,A= 14334 mm2
Depth of the section,D= 550 mm
Width of flange, bf= 250 mm
Thickness of flange, tf= 17.6 mm
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S.3Design and Drawing of Steel Structures (April/May-2013, Set-1) JNTU-Anantapur
B.Tech. III-Year II-Sem. ( JNTU-Anantapur )
Thickness of web, tw= 10.5 mm
Moment of inertia,Ixx
= 74906.1 104mm4
Moment of inertia,Iyy= 3740.6 10
4
mm
4
Radius of gyration, rxx
= 228.6 mm
Radius of gyration, ryy
= 51.1 mm
Modulus of section,Zxx
= 2723.9 103mm3
Modulus of section,Zyy
= 299.2 103mm3
17.6 mm
250 mm
Eccentric load 150 kN
Axial load
750 kN
550
mm 10.5 mm
Eccentric load 60 kN
ISWB 550 @ 112.5 kg/m
17.6 mm
250 mm
Eccentric load 150 kN
Axial load
750 kN
550
mm 10.5 mm
Eccentric load 60 kN
ISWB 550 @ 112.5 kg/m
Figure: Showing the Section
Determine the Slenderness Ration ()
=minr
l
rmin
= ryy
= 51.1 mm
Length, l= 4 m = 4000 mm
=1.51
4000
= 78.278
Determine Average Compressive Stress, bc, cal
bc, cal
=providedArea
loadaxialTotal
=14334
109603
bc, cal
= 66.974 N/mm2
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For,T
D=
6.17
550= 31.25 and
minr
l= 78.278
Evaluate ac
value from IS 800-1984
ac= 132 N/mm2
Assume an angle section ISA 150 mm 75 mm 10
mm @ 0.169 kg/m for an eccentric load about X-X & Y-Y axis.
Moment in yy-axis =
+
2
75
2
D 150
Myy
=
+
2
75
2
550 150
Myy
= 46875 kN-mm
Assume minimum eccentricity = 100 mm
Moment in plane of xx-axis = Load of beam connectedto web Eccentricity
Mxx
= 60 100
Mxx
= 6000 kN-mm
Bending stress about xx-axis,
bc,xx
=xx
yy
Z
M
=3
3
109.2723
1046875
bc,xx
= 17.209 N/mm2
Bending stress about yy-axis,
bc,yy
=yy
xx
Z
M
=3
3
102.299
106000
bc,yy
= 20.053 N/mm2
Allowable bending compressive stress,
bc
= 0.66fy
= 0.66 260
bc
= 171.6 N/mm2
Check for Safety of Interaction Expression
1,,cal,
+
+
bc
yybcxxbc
ac
bc
++
6.171
053.20209.17
132
974.66
= 0.725 < 1
Hence the section of safe.
Provide ISWB 550 @ 112.5 kg/m with angles of ISA150 mm 75 mm 10 mm @ 0.169 kg/m.
Q2. A simply supported beam of effective span 8m carries a U.D.L of 40 kN/m. Taking f
y= 250
N/mm2and E = 2 105N/mm2. Design thebeam as laterally supported. Draw elevationand sectional details of the beam.
Answer : April/May-13, Set-1, Q2
Given that,
Effective length,L= 8 m
Uniformly distributed load, W= 40 kN/m
fy= 250 N/mm2
E= 2 105N/mm2
Self weight =350
LW =350
840 = 0.914 kN/m
Total uniformly distributed load = 40 + 0.914
W= 40.914 kN/m
Bending Moment, M
M=8
2WL
=8
8914.402
M= 327.312 106N-mm
M= 327.312 kN-m
Shear Force, V
V=2
WL
=2
8914.40
V= 163.656 kN
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S.5Design and Drawing of Steel Structures (April/May-2013, Set-1) JNTU-Anantapur
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Calculate the Section Modulus Required Zreq
Zreq
=bc
M
bc=
bt= 0.66f
y
= 0.66 250
bc
= bt
= 165 N/mm2
Required sectional modulus,Zreq
=165
10312.3276
Zreq
= 1983.709 103mm3
Adopt ISWB 500 @ 95.2 kg/m (from steel tables)
Zxx
= 2091.6 103mm3
Ixx
= 52209.9 104mm4
d= 500 mm, bf= 250 mm
tf
= 14.7 mm
tw
= 9.9 mm
Check for Shear
Calculate shear stress, v, cal
=wtd
V
=9.9500
10656.1633
v, cal
= 33.062 N/mm2
Permissible shear stress, v = 0.4fy= 0.4 250
= 100 N/mm2
v, cal
is less than v
The design is safe in shear.Check for Deflection
Maximum deflection, Ymax
=EI
WL
384
54
Ymax
=45
43
109.52209102384
)8()10914.40(5
(1000)3
Ymax
= 20.897 mm
Allowable deflection,Y=325
Span
=325
8000
Y = 24.615 mm
Yis greater than Ymax
Hence the beam is safe in deflection.
500 mm
14.7
mm
14.7
mm
9.9 mm
250 mm
470.6
mm
500 mm
14.7
mm
14.7
mm
9.9 mm
250 mm
470.6
mm
Figure: Showing the Sectional Details of I-rolled beam
Part B
(Answer any THREE questions, 3 14 marks)
Q3. (a) Write about major types of tensionmembers.
Answer : April/May-13, Set-1, Q3(a)
The major tension members can be divided into,
1. Wires and cables
2. Bars and rods
3. Single structural shapes and plates
4. Build up sections.
1. Wires and Cables
The use of wires as structural tension members are
normally cold-drawn from hot-rolled rods. Generally the size
of wire is specified by the gauge number. Cables or wire
ropes are used for rigging slings, hoists, derricks, hangers
for suspension bridges, guy wires etc.
2. Bars and Rods
The members carrying small magnitude tensile force
may be made from round rods or flat bars or hot-rolled squarerods. The ends of the round bars are threaded and used with
nuts. Flat bars may be bolted, welded or riveted to an adjoin-
ing part or forged into eyebar or loop ends and connected to
a pin.
3. Single Structural Shapes and Plates
Single structural shapes that are commonly used are
tee-sections, angle sections and channel sections. The main
use of single angles are for light truss tension members and
bracing. Single channels and plates can also be used as
tension members.
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4. Built-up Section
Built-up section comprising of two or more plates are
used, when heavy tensile load is required for a member. This
type of built up section provides,
(a) Greater rigidity by way of greater moment of
inertia,
(b) Greater area, and
(c) Suitable dimension.
(b) The tension member of a roof trusscarries a maximum axial tension of 250kN. Design the section. Diameter ofconnecting riveters 20 mm, safe stressin tension = 150 N/mm2.
Answer : April/May-13, Set-1, Q3(b)
Given that,
Axial tension, P= 250 kN = 250 103N
Diameter of rivets, d= 20 mm
Gross diameter of rivets = d+ 1.5 = 20 + 1.5 = 21.5 mm
Safe stress in tension, at= 150 N/mm2
Calculate the Required Area
Area required,Areq =at
tensionAxial
=150
102503
Areq
= 1666.667 mm2
Provide an angle section of ISA 125 95 12 mm @
19.6 kg/m
Area = 2498 mm2
Determine the Net Area, Anet
Anet
=A1+ k A
2
Where,
A1= (125 Diameter of rivet t/2) t
= (125 21.5 12/2) 12
A1= 1170 mm
A2= (95 t/2) t
= (95 12/2) 12
A2= 1068 mm2
k=21
1
3
3
AA
A
+
=1068)11703(
11703
+
= 0.767
Anet
= 1170 + 0.767 1068
Net area = 1989.156 1990 mm2
Check for Load Capacity
Load capacity = Area provided (Anet
) at
= 1990 150
= 298500 N
= 298.5 kN greater than 250 kN
Hence the section provided is safe.
Design of Connection
(i) Strength of rivet in single shear,
= 100 4
2d
= 100 4
5.212
= 36305.030 N
= 36.305 kN
(ii) Strength of rivet in bearing = 300 d t
= 300 21.5 12
= 77400 N = 77.4 kN
Rivet value = Least strength = 36.305 kN
Number of rivets =valueRivet
Load=
305.36
250
= 6.886 6
Provide 6 rivets at a pitch of 2.5 20 = 50 mm
Pitch = 50 mm
Edge distance = 35 mm (Assume)
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S.7Design and Drawing of Steel Structures (April/May-2013, Set-1) JNTU-Anantapur
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35 mm50 mm50 mm50 mm50 mm50 mm
35 mm
125 mm
6-20 mm
rivets
ISA 125 95 12
Gusset Plate
35 mm50 mm50 mm50 mm50 mm50 mm
35 mm
125 mm
6-20 mm
rivets
ISA 125 95 12
Gusset Plate
Figure
Q4. Design a built up column with four angles laced together. The effective length of the column is7.20 m and it supports a load of 1200 kN.
Answer : April/May-13, Set-1, Q4
Given that,
Axial load, P= 1200 kN
Effective length, l= 7.20 m
4 MS angle sections are used which are laced together to form a square column.
Assume slenderness ratio, = 50;fy= 250 N/mm2Forf
y= 250 N/mm2and = 50
Adopt ac
= 132 N/mm2(from IS 800 : 1984)
Determine the Required Area
Area required,Areq
=acloadAxial
Areq
=132
1012003
Areq = 9090.909 mm2
Area required for each flange =4
909.9090
= 2272.727 mm2
Adopt ISA 110 110 15 mm @ 24.2 kg/m with following properties,
Area,A= 3081 mm2
Ixx
=Iyy
= 337.4 104mm4
Cxx
=Cyy
= 32.7 mm
rxx
=ryy
= 33.1 mm
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Minimum radius of gyration, rmin
=provA
Ix
Where,
Ix
= 4 [Ixx
+A(B/2 Cxx
)2]
Ix
= 4 [337.4 104+ 3081 (B/2 32.7)2]
Aprov
= 4 A
= 4 3081
Aprov
= 12324 mm2
rmin
=
21
30814
])7.322(3081104.337[424
+ B
rmin = [1095.099 + (0.5B 32.7)2]1/2 ... (1)
Compressive stress bc
=Area
loadAxial=
30814
1012003
bc
= 97.371 N/mm2
From IS 800 : 1984
Forfy= 250 N/mm2and
bc= 97.371 N/mm2, we get,
= 90 +x= 90 7.371
= 82.629
We know that, =minr
l
82.629 =21
])7.325.0(099.1095[
1020.7
2
3
B+
1095.099 + (0.5B 32.7)2=
23
629.82
1020.7
On simplification, we get,
B = 226.616 250 mm
Provide a square column of size 250 mm 250 mm.Check
Minimum radius of gyration = [1095.099 + (0.5 250 32.7)2]1/2
rmin
= 98.053
=minr
l=
053.98
1020.73
= 73.43
From IS 800 : 1984,
For = 73.43 andfy= 250 N/mm2, we get,
bc
= 108.570 N/mm2.
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Check for safe capacity of section,
Psafe
=Aprov
bc
= (4 3081) 108.57P
safe= 1338016.680 N
Psafe
= 1338.017 kN greater than 1200 kN
Hence the section provided is safe.
Design of Lacing Bars
Assume single lacing system is inclined at 45 to the axis of the column.
Gauge distance, g= 60 mm (Assume)
Spacing = 4[rmin
Cyy
]
= 4[98.053 32.7]
= 261.412 mm
260 mm c/c
Distance, d= Spacing + 2(g)
d= 260 + 2(60)
d= 380 mm
Length of the lacing bar is given as,
Lb
= 2dcot
= 2 380 cot 45
= 760 1
Lb
= 760 mm
rmin= rxx= ryy= 33.1 mm
e=
minr
Lb
=1.33
760
= 22.961
23
Note
e should be equal to or less than 0.7
(or) 50 which ever is less.
e
= 23 < (0.7 73.43) > 50
= 23 < 51.401
Hence safe (adequate).
Finding the Angle
e=
minr
lb
e=
1.33
cot2 d
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(0.7 73.43) 33.1 = 2dcot
cot =3802
1.33)43.737.0(
cot = 2.239
tan =239.2
1
= 24.070
= 25
Hence provide = 25 for convenience in solving.
Length of the lacing bar,Lb= 2dcot
Lb
= 2 380 cot (25)
Lb = 1629.825
1630 mm
rmin
= ryy
= rxx
= 33.1 mm
e=
minr
lo
=1.33
1630
e= 49.245
50
e< 0.7
50 < 51.401
Hence safe.
For single lacing length,
l =2
bLsec
=2
1630sec 25
l = 899.253
900 mm
Effective length of lacing, le= l
Minimum thickness of lacing /r
le 12 145
min
=
24
12900
min
= 129.904 >/ 145Hence safe.
For = 129.904,fy= 250 N/mm2(from table)
ac
= 57 N/mm2
Transverse shear, V = 0.025W
= 0.025 1200
V= 30 kN
For single lacing system, force acting in lacing is
given as,
F= sinn
V
=25sin4
30
= 17.747 kN
Number of rivets =VR
F
.
cos2
=305.36
25cos747.172
= 0.886 1 rivet
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Compressive stress =tb
F
=)2460(
10747.173
= 12.324 < ac
57 N/mm2
Hence safe.
Tensile stress=tdb
F
o )(
=24)5.2160(
10747.17
3
= 19.207 N/mm2< ac57 N/mm2
Hence safe.
A single lacing system comprises of 60 mm 24 mm
flats connected by one power driven rivet of 20 mm ateach end.
120 mm
260 mm
25
25
25
120 mm
260 mm
25
25
25
Figure: Single Lacing Systems Column
Cxx
X X
250mm
Y
Y
Lacing
Cxx
X X
250mm
Y
Y
Lacing
Figure: 4 Angle Section Laced Together
Q5. A steel column consists of ISHB 300 withcover plate 300 mm 25 m for each flange.The column carries an axial load of 2300 kN.Design a gusted base plate for the column.Use 18 mm diameter rivets.
Answer : April/May-13, Set-1, Q5
Given that,
Load on column, P= 2300 kN
Section used for column - ISHB 300
1. Depth of the section, h= 300 mm
2. Width of the flange, bf= 250 mm
Diameter of rivets = 18 mm
Assume
1. Permissible bearing pressure on concrete
bc= 4 N/mm2
2. Safe bearing capacity of soil = 300 kN/m2
Area of base plate,A =concreteonpressureBearing
columnonloadAxial
A =4
1023003
= 575 103mm2
Provide gusset plate of 15 mm
Gusset angle of 150 115 12
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Minimum width of base plate,
b = 300 + 2(25) + 2(15) + 2(115)
b = 610 mm
Length of base plate, d=b
Area=
610
105753
d= 942.623 950 mm
Provide base plate of dimension = 950 610 mm
The arrangement of base plate, gusset plate, angles and column section is shown in the figure.
Cover plate
300 25
ISHB 300
115 mm
15 mm
350 mm
15 mm
115 mm
610 mm
P/2
950 mmy
y
P/2
W = 4 N/mm2
300 mm
Cover plate
300 25
ISHB 300
115 mm
15 mm
350 mm
15 mm
115 mm
610 mm
P/2
950 mmy
y
P/2
W = 4 N/mm2
300 mm
Figure
Upward Pressure by Soil
w =plateofArea
loadAxial
w =610950
1023003
w = 3.969 4 N/mm2
For unit width; w= 4 N/mm2
Project of cantilever, l = 115 12
l = 103 mm
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Cantilever moment =2
2lw
=2
10342
= 21218 N-mm
Now moment of resistance = Cantilever Moment
fbZ= 21218
185 6
12
t= 21218 (Qfb= 185 N/mm
2)
t2 = 688.151 mm
t= 26.233 mm
Thickness of base plate required, tb= t Gusset angle thickness
tb
= 26.233 12
tb
= 14.233 14 mm
Bending moment at the centre of 1 mm wide strip
=2
115
8
38022 ww
=2
1154
8
380422
Design moment,M= 45750 N-mm
Equating moment of resistance = Design moment
fbZ = 45750
185 6
12
t= 45750
t= 38.52 mm
tb 40 mm
Thickness of base plate, tb= 40 mm (or) 14 mm
Provide base plate of size 950 610 40 mm
Design of Connection Between Gusset Base and Column
Assuming that the column ends are faced well to take up the bearing stresses. The connection existing between
gusset plate and column will be designed for half of the axial load.
Design load for connection =2
loadAxial
=2
2300
= 1150 kN
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Using 18 mm bolts Diameter of bolts = 1.5 + 18
= 19.5 mm
Strength of bolt (in single shear)
= 100 4
(19.5)2
= 29864.765 N
29865 NStrength of bolt in bearing will be higher.
Bolt value = 29865 N
Number of bolts required, n =29865
1011503
n = 38.507 40 Therefore provide 40 bolts of 18 mmto connect thecolumn to the gusset plate. On each gusset plate 20 bolts
are provided.
+ + +
+ + +
+ + +
+ + +
950 mm
+ + +
+ + +
+ + +
+ + +
950 mm
Figure
Q6. Design an I section purlin with an withoutsag bars for trussed roof from the followingdata,
Span of roof = 10 m, spacing of purlin alongslope of truss 1.8 m
Spacing of truss = 4 m slope of roof truss = 1vertical 2 horizontal
Wind load on roof surface normal to roof =1200 N/mm2.
Vertical loads from roof sheets = 200 N/mm2.
April/May-13, Set-1, Q6
Answer :
Given that,
Span of the roof,L= 10 m
Spacing of purlin along slope of truss = 1.8 m
Spacing of truss = 4 m
Slope of roof truss = 1 vertical 2 horizontal
Slope () = 1/2
tan = 1/2
tan = 0.5
= tan1(0.5)
= 26.565
sin = 0.447
cos = 0.894
Wind load on roof surface normal to roof = 1200 N/m2.
Vertical load from roof sheets = 200 N/m2
Calculating the Dead Load (D.L)
Load from roof sheeting = 200 spacing of purlin
= 200 1.8
= 360 N/m2
Assume self weight = 120 N/m2
Total dead load (Wd.L
) = 360 + 120 = 480 N/m2
Calculation of Wind Load
Given that,
Wind load on roof surface = 1200 N/m2
Total wind load = (Ww.l
) = 1200 Spacing of purlin
= 1200 1.8
Ww.l
= 2160 N/m2
Design of I-section Purlin without Sag Bars
The load combination of (dead load + wind load)
creates greater effects on purlin than load combination of
(dead load + live load).
Consider the load combination (dead load + wind
load) for I-section purlin.
Dead load + Wind load
Wd.w.x
= Load normal to the slope
= Ww.l
+ Wd.l
cos
= 2160 + 480 cos (26.565)
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Wd.w.x
= 2589.325 N
Wd.w.y
= Load parallel to slope = Wd.l
sin = 480 sin (26.565)
Wd.w.y
= 214.662 N
Mxx
=10
)(2
.. LW xwd
=10
4)325.2589(2
Mxx
= 4142.920 103N-mm
Myy
=10
)(2
.. LW ywd
=10
4)662.214(2
Myy
= 343.459 103N-mm
Assume,yy
xx
Z
Z= 6, and
bt
= 0.66fy
= 0.66 250
bt
= 165 N/mm2
E= 2 105N/mm2
Finding the required sectional modulus
Zxx, req
=bt
yyyy
xxxx M
Z
ZM
+
=165
]10459.343610920.4142[ 33 +
Zxx, req
= 37.598 103mm3
Select ISMB 100 @ 11.5 kg/m
Zxx= 51.5 103mm3
Zyy
= 10.9 103mm3
Check for Permissible Stress
bt
=yy
yy
xx
xx
Z
M
Z
M+
=3
3
3
3
109.10
10459.343
105.51
10920.4142
+
bt
= 111.955 N/mm2< 165 N/mm2
Hence safe.
Design of I-Section with Sag Bar
Dead load + Wind load
Wdwx
= Load normal to the slope
= 2589.325 N
Wdwy
= Load parallel to slope
= 214.662 N
Bending Moment
Mxx
=10
)(2
LWdwx
=10
4)325.2589(2
Mxx = 4142.920 103
N-mm
Myy
=10
2)(
2
LWdwy
=10
2
4662.214
2
Myy
= 85.865 103N-mm
Finding the Required Section Modulus
Zxx, req
=bt
yyyy
xxxx M
Z
ZM
+
=165
10865.85610920.414233 +
Zxx, req
= 28.231 103mm3
Select ISLB 100 @ 8.0 kg/m
Zxx
= 33.6 103mm3
Zyy
= 5.1 103mm3
bt
=yy
yy
xx
xx
Z
M
Z
M+
=3
3
3
3
101.5
10865.85
106.33
10920.4142
+
bt
= 140.137 N/mm3< 165 N/mm2
Hence safe.
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Q7. Design the cross section of a welded plategirder for a U.D.L of 80 kN/m. The effectivespan is 16 m.
Answer : April/May-13, Set-1, Q7
Given that,
Length of girder,L= 16 m
Udl = 80 kN/m
Total load on plate girder,
W = 16 80
W = 1280 kN
Reaction at each support =2
W=
2
1280
RA=R
B= 640 kN
16 m
RA
= 640 kN RB
= 640 kN
80 kN/m
16 m
RA
= 640 kN RB
= 640 kN
80 kN/m
Figure
Maximum Bending Moment
M= 640 8
2
8802
M= 2560 kN/m
Maximum Shear Force (S.F)
Shear force = Support reaction = 640 kN
Depth of Web Plate
(i) Provide intermediate stiffeners, ifwt
d> 200
(ii) No stiffeners (i.e., end and intermediate stiffeners)
are required if wt
d
67. But thick webs are requiredfor this condition.
(iii) Thin webs and end stiffeners are required if
K=wt
d= 100
Consider,
K =wt
d= 100
Economical depth,
d=
31
yfMK
d=
31
250
1001025606
(f
y= 250 N/mm2)
d= 1007.937 mm
Provide 1200 mm wide plates.
We know that,
K =wtd
100 =wt
1200
tw
=100
1200
tw
= 12 mm
Size of web plate = 1200 mm 12 mm
Selection of Flange Plate
Assume that the moment is resisted by the flanges,
Area of flange needed is,
1.1
dfA yf M
1.1
1200250fA 2560 106
Af 9386.667 mm2
If flange is kept in semi-compact class, then
ft
b 13.6
Let width of flange, bf= 13.6 t
f
Now,
Af
= 9386.667
bf t
f= 9386.667
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13.6tf t
f= 9386.667
2ft = 690.196
tf
= 26.272 30 mm
Provide 30 mm thick plate
Af
= bf t
f
9386.667 = bf 30
bf
= 312.889 320 mm
Provide flange plate = 320 mm 30 mm
1200 mm
30 mm
12 mm
320 mm
30 mm
Y
Y
ZZ
1200 mm
30 mm
12 mm
320 mm
30 mm
Y
Y
ZZ
Figure
Check for Actual Stresses
Moment of inertia of the section,
I = 12
1200)12320(
12
126032033
I = 8.991 109mm4
Actual flange stress,
f=I
M Y
max
=2
1260
10991.8
1025609
6
f= 179.379 N/mm2
Flange stress is greater than permissible stress i.e.,
165 N/mm2
Redesign the member by considering the depth, d=1400 mm
Thickness of web, tw=
100
1400
tw
= 14 mm
Size of web = 1400 mm 14 mmArea of flange required is,
1.1
dfA yf M
1.1
1400250fA 2560 106
Af 8045.714 mm
2
ft
b 13.6
Af= b
f t
f= 8045.714
13.6 tf t
f= 8045.714
tf
= 24.323 26 mm b
f t
f= 8045.714
bf 26 = 8045.714
bf
= 309.451 320 mm
Provide flange plate of 320 mm 26 mm
1452 mm
14 mm
320 mm
Y
Y
ZZ
26 mm
26 mm
1400 mm1452 mm
14 mm
320 mm
Y
Y
ZZ
26 mm
26 mm
1400 mm
Figure
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Check for Actual Stresses
I= 12
1400)14320(
12
145232033
I= 1.166 1010
Actual flange stress,
f =I
M Y
max
=2
1452
10166.1
10256010
6
f = 159.396 N/mm2less than 165 N/mm2
Hence the design is safe.