1401871257_Design and Drawing of Steel Structures 3-2 Set-1 A

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


2/18


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


3/18



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


ISWB 550 @ 112.5 kg/m

17.6 mm

250 mm


Axial load

750 kN

550

mm 10.5 mm


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


4/18



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.


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


5/18



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.


6/18



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)


7/18



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.


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


8/18



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.


9/18



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


10/18



(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


11/18



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.


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


12/18



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


13/18



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


14/18



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)


15/18



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.


16/18



Q7. Design the cross section of a welded plategirder for a U.D.L of 80 kN/m. The effectivespan is 16 m.


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


17/18



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


18/18



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.

Documents

1401871257_Design and Drawing of Steel Structures 3-2 Set-1 A