1. a flat of size 250 x 10 mm is connected to the gusset ...nptel.vtu.ac.in/vtu-nmeict/dss1/tension...

1. A flat of size 250 X 10 mm is connected to the gusset plate, what is the minimum net area

for tension member at section x-x and at x, b, c, d & e as shown below. Assume 20mm bolt.

Compute the design tensile strength of the plate.

DATA :

b (Breadth of Plate) = 210 mm

n ( No. of bolts in a zigzag line) = 3 no.

d (gross dia of bolt) = 20 mm

P1 (staggered pitch) = 60 mm

P2 (staggered pitch) = 80 mm

g (gauge distance) = 50mm

I. Calculation of Minimum Net Area:

i. Path x-x :

Anet = [(210-1*20)*10]

Anet = 1900 mm2

][ *)( tndbAnet −=

x

FLATE 60 80

50 210

b x

c d

e

50

GUSSETE PLATE

40

40

of 24

ii. Path x,b,c,d and e

Bolts are connected in different staggered pitches P1 & P2

Steinman’s formula:

Anet = Effective width X Thickness

Effective width =

++−

2

22

1

21

44)(

gP

gPndb

Effective width =

++−

50*480

50*460)20*3210(

22

Effective width = 200 mm

Anet = 200 x 10

Anet = 2000 mm2

Net area is the least of above two values i.e., Anet = 1900 mm2

II. Calculation of Design tensile strength of the plate

Design tensile strength of the plate is least of the following

i. Yield strength of plate and

ii. Rupture of plate.

(i) Design tensile strength of the plate due to Yield, P-32 Cl: 6.2

=dgT 568.18 kN

mo

gdg

yfATγ

*=

1000*1.1250*10*250

=dgT

of 24

(i) Design tensile strength of the plate due to Rupture, P-32 Cl: 6.3.1

=dnT 560.88 kN

Therefore, Design tensile strength of the plate = 560.88 kN

2. An angle section ISA 200 X 100 X 10 mm is used as tension member and is connected to gusset plate by 16mm dia both the leg as shown below. Compute net area of the section along1-a-b-2, 3-d-f-4, 1-a-c-b-2, & 3-d-e-f-4 and also calculate the net area along other possible lines of failure.

ISA 200X100X10

1 3

2 4

a

b

c

d

e

f

g*

90

Fig(a). Angle with both position details

60 60 60

60 60 60

200

100

80

ga

gb

90

60

1m

undn

0.9ATγ

f=

1000*25.1410*1900*9.0

=dnT

of 24

g* = ga + gb – thickness

= 80 + 60 – 10

= 130 mm

For an M16 bolt, dh = 18 mm

i. Along section 1-a-b-2 and 3-d-f-4

Anet = [(2903-2*18)*10]

Anet = 2543 mm2

ii. Along section 1-a-c-b-2 and 3-d-e-f-4

Anet =

++∑−

2

2

1

1

44)(

22

*g

Pg

PtdA hg t*

Anet =

++−

130*460

90*460)10*)18*3(2903(

22

10*

Anet = 2532.23 mm2

Net area is the least of above two values i.e., Anet = 2532 mm2

(The probability crack propagation through 1-a-b-2, 3-d-f-4, 1-a-c-b-2, & 3-d-e-f-4 and other alternatives need to worked out, the minimum & the path defines the ultimate strength of the material)

3. The long leg of ISA 200 X 100 is connected to gusset plate by 22 mm dia bolts in two rows with gauge space of 90 mm and staggered pitch of 50mm as shown fig. determine suitable thickness of the angle to transmit a pull of 350 kN.

][ *)( tndbAnet −=

Fig(b). Angle Flattened into one plane

of 24

Pitch = 2.5 x 22= 55 mm

Edge distance =1.7 x dia of hole =1.7x24= 40.8

Net area An = t4gP)nd(b X

1

12

h

+−

i) Net area along section (a-b-c):

An(abc) = [ ] t 024)1(200 XX +− = 176t mm

ii) Net area along section (a-b-d-e):

An(abde)= t 4x905024)2(200 XX

2

+−

Anet = 160t mm2

Net area is the least of above values i.e., Anet = 160t mm2

Design strength due to rupture of critical section:

8mm7.41mmt

kN 3501.25

410160t x x 0.9γ

fA 0.9XTm1

undn

≈=

==

=

Design strength due to yielding of gross section:

mm 8mm 7.7t

kN 3501.1

250xtx200γ

fAT

m0

ygdg

≈=

==

=

a

b

c d

e

90

50 50

100

55

65

1-ISA 200 X 100 X t

of 24

4. An angle section ISA 50 X 30 X 6 mm is used as a tension member with its longer leg connected by 12 mm dia bolts. Then what is its strength? And also check the efficiency of member if the Gusset is connected to 30-mm leg of angle. Assume thickness of the gusset plate is 6mm.

a. Bolted connection :

Properties of 1-ISA 50 X 30 X 6

A = 447 mm2

(ii) Design tensile strength of the plate due to Yield, P-32 Cl: 6.2

=dgT 101.60 kN

Design of end connections:

Nominal diameter of the bolt = 12 mm

p-75, Cl: 10.3.3

1. For single shear of bolts

Hole diameter = 12+1 = 13mm

Where, 1.5 mm for the bolts less than 12 mm &

2 mm for the bolts greater than 12 mm.

Minimum pitch = 2.5 x d = 2.5 x 12 = 30mm, Minimum edge distance = 1.7 x d = 22.1mm

Say 25mm

Assuming shank interfering the shear plane;

.boltstheofofcapacityShear∴

( )sbsnbnunsb AnAnfV += 3/

mo

ygdg

fATγ

*=

1000*1.1250*447

=dgT

of 24

2/400 mmNfu =∴ i.e ultimate tensile strength.

1=sn

( )24

dAsbπ

=

( )2124π

=

2sb mm 113.1A =

( )1.113*13

400=∴ nsbV

N 26119.32Vnsb =∴

mbnsbdsb VV γ/=∴ = 32.26119 /1.25 = 20895.21 N

kN 20.89Vdsb =

2. Strength of bolt in Bearing:

Nominal bearing strength of the bolt

ubnpb ftdkV ****5.2= ,p-76

= 4006125.2 ×××× bk


Say 25mm

0.1,,25.03

,3 u

ub

oob f

fdp

dek −=

,1.0410400 0.25,

3x1330 ,

3x1325

−=

1.00 0.98, 0.52, 0.64,=

52.0=∴ bk

of 24

400 6 12 0.52 2.5 XXXX=

N37440= kNVdpb 952.29=

∴Bolt value = 20.89 kN, ∴No. of bolts = .864.489.206.101 nos= Say 5 nos.

Detailing

2. Check for design Strength due rupture of critical section(Tdn) :

The rupture of an angle connected through one leg is affected by shear lag.

The design strength, Tdn as governed by rupture at net section is given by

Where,

β =

Anet =

−−

2tdoA * t

Anet =

−−

261350 *6 =204 mm2

25 30 25 30 30 30

Lc

1-ISA 50X30X6

5-BOLTS 12mm Dia

1-ISA 50X30X6

mo

ygo

m

undn r

fAr

fAT ***

1

*9.0 β+=

×

×

×−

Lb

ff

tw s

u

y076.04.1

of 24

Ago =

−

2tB * t

Ago =

−

2630 * 6 = 162 mm2

(Assume gauge g=28mm)

W=30mm, bs =30+28-6=52mm,

Lc = 4*30 =120mm.

β =

β = 1.3 0.7> β <1.44 where

1*

*

mu

moy

rfrf

=dnT 108.10 kN

Check for block shear: P-33, Cl: 6.4.1

It’s the minimum of the following

1

***1

9.0*3 m

utn

mo

yvgdb r

fArfAT +=−

mo

ytg

m

uvndb r

fAr

fAT ***2

1*39.0

+=−

×××−120

52

410

250

6

30076.04.1

1.12501623.1

25.1410*204*9.0 **

+=dnT

of 24

Lv = 25+4*30=145 mm

Avg = Lv * t = 145 * 6 = 870 mm2

Avn = Avg – 4.5*do*t = 870 – 4.5*13*6 = 519 mm2

Atg = Lt * t = 22*6 = 132 mm2

Atn = Atg - 0.5 do X t = 132 - 0.5 X 13 X 6 = 93 mm2

1m

u tn

mo

y vg1db γ

fA 0.9γ 3fAT X X

X

X+=−

1.25410 93 0.9

1.1 3250 870T X X

X

X1db +=−

=−1dbT 141.61 kN

mo

y tg

m

uvn 2db

fA 3

fA 0.9T X

X

X X

1γγ

+=−

1.1250 X 132

1.25 X 3410 X 519 X 0.9

2dbT +=−

2−dbT = 118.46 kN

dbT is the least of above two values i.e., 118.46 kN

Therefore, Design strength of the member =Least of Tdg, Tdn, Tdb

Design strength of the member =101.6 kN

Efficiency of tension member(the efficiency of the material based on the yield strength the material) = 101.6*1000*100/(447*250/1.1) = 100%

22

Lv

of 24

Gusset is connected to 30-mm leg of angle a. Bolted connection :


A = 447 mm2

(iii)Design tensile strength of the plate due to Yield, P-32 Cl: 6.2

=dgT 101.60 kN

Design of end connections:

As per above problem

∴Bolt value = 20.89 KN, ∴No. of bolts = .864.489.206.101 nos= Say 5 nos.

Detailing


Say 25mm

1-ISA 50X30X6

25 30 30 30 30 25

1-ISA 50X30X6

5-BOLTS 12mm Dia

Lv

mo

ygdg

fATγ

*=

1000*1.1250*447

=dgT

of 24




Where,

β =

Anet =

−−

2tdoA * t

Anet =

−−

261330 *6 =84 mm2

Ago=

−

2tB * t

Ago=

−

2650 * 6 = 282 mm2

(Assume gauge, g=19mm)

W=50mm, bs=50+19-6=63 mm,

Lc = 4*30 =120mm.

β =

β = 1.01 0.7> β <1.44 where

1*

*

mu

moy

rfrf

=dnT 97.10 kN

Check for block shear: P-33, Cl: 6.4.1

mo

ygo

m

undn

fAfAT

γ

β

γ

****9.0

1

+=

×××−L

b

f

f

t

w s

u

y076.04.1

×××−120

63

410

250

6

50076.04.1

1.1

250*282*01.1

25.1

410*84*9.0+=dnT

of 24


1

***1

9.0*3 m

utn

mo

yvgdb r

fArfAT +=−

mo

ytg

m

uvndb r

fAr

fAT ***2

1*39.0

+=−

Lv = 25+4*30=145mm

Avg = Lv * t = 145 * 6 = 870mm2

Avn = Avg – 4.5*do*t = 870 – 4.5*13*6 = 519 mm2

Atg = Lt * t =11*6 = 66 mm2

Atn = Atg - 0.5 do*t = 66 - 0.5*13*6 = 27 mm2

1

***1

9.0*3 m

utn

mo

yvgdb r

fArfAT +=−

25.1410*27*9.0

1.1*3250*870

1 +=−dbT

=−1dbT 122.13 KN

mo

ytg

m

uvndb r

fAr

fAT ***2

1*39.0

+=−

1.1250*66

25.1*3410*519*9.0

2 +=−dbT

2−dbT = 103.45 kN

22

Lv

of 24

dbT is the least of above two values i.e., 103.45 kN

Therefore, Design strength of the member =Least of Tdg, Tdn, Tdb

Design strength of the member = 97.107 kN

Efficiency of tension member(the efficiency of the material based on the yield strength the material) = 97.107*1000*100/(447*250/1.1) = 95.58%

Hence, in this case, by connecting the short leg, the efficiency is reduced by about 4% DESIGN PROBLEMS

Design a tension member 3.4m between c/c of intersection using double angle section carrying a factored pull of 700kN. The member is subjected to reversal of stresses.

Solution:

Factored load = 700kN

Strength governed by yielding(Cl: 6.2, P-32)

Tension capacity of the member =

Area of 2 section required,

Ag =3080mm2

The section is chosen based on the area and the min radius of gyration required.

Slenderness ratio: (P-20. Table 3)

minr =18.88

Choose the section for an area greater than cm2 and minr greater 18.88

Try ISA 100X75X10

A=33.00cm2=3300mm2

Strength governed by yielding(Cl: 6.2, P-32)

Tension capacity of the member = >700kN Hence Safe

mo

ygdg

fATγ

*=

2501.1*310*700 Ag =

180min

≤=rLeffλ

kNfATmo

ygdg 750*

==γ

of 24

Connection Details:

Taking 20mm bolts of grade 4.6

Dia of hole(do) = 20+2=22mm

1) Strength of one bolt in Single shear: P-75, Cl:10.3.3:

+=

b

sbsnbnu

m

AnAnXfVγ3

dsb

Assuming both Shank and tread interfering the shear plane;

nn=1 ns=1, 25.1=bmγ

( )24

dAnbπ

= = ( ) 22 46.254184

mm=π

( )24

dAsbπ

= = ( ) 22 16.314204

mm=π



=

b

bdpb

m

ftdkVγ

****5.2 , P-76

0.1,,25.03

,3 u

ub

oob f

fdp

dek −=

0.1,410400,25.0

22*350,

22*340

−=

00.1,98.0,51.0,61.0=

51.0=∴ bk

kNX 3.1031000*25.1

16.314*146.254*1

3

400=

+=

of 24

=

1000*25.1400*10*20*51.0*5.2

dpbV

kNVdpb 60.81=

∴Bolt value = 81.60 kN

∴No. of bolts = .23.960.81

750 nos= Say 10 nos.




Where,

β =

Anet =

−−

2tdoA * t

Anet =

−−

21022100 *10 =730 mm2

Ago=

−

2tB * t

Ago=

−

21075 * 10 = 700 mm2

W=75mm, bs=60+75-10=125mm,

Lc = 9*50 =450 mm.

β =

+=

mo

ygo

m

undn r

fAr

fAT ***

1

*9.0*2 β

×

×

×−

Lb

ff

tw s

u

y076.04.1

×

×

×−

450125

410250

1075076.04.1

of 24

β = 1.3 0.7> β <1.44 where

1*

*

mu

moy

rfrf

=dnT 844.63 KN

Check for block shear : P-33, Cl:6.4.1


1

***1

9.0*3 m

utn

mo

yvgdb r

fArfAT +=−

mo

ytg

m

uvndb r

fAr

fAT ***2

1*39.0

+=−

Lv = 40+9*50=490 mm

Avg = Lv * t = 490 * 10 = 4900 mm2

Avn = Avg – 4.5*do*t = 4900 – 9.5*22*10 = 2810 mm2

Atg = Lt * t = 40*10 = 400 mm2

Atn = Atg -0 .5 do*t = 400 - 0.5*22*10 = 290 mm2

1

***1

9.0*3 m

utn

mo

yvgdb r

fArfAT +=−

+=− 25.1

410*290*9.01.1*3

250*4900*21dbT

=−1dbT 1457.13 kN

mo

ytg

m

uvndb r

fAr

fAT ***2

1*39.0

+=−

+=− 1.1

250*40025.1*3

410*2810*9.02dbT

+=

1000*1.12507003.1

1000*25.1730*410*9.0*2 **

dnT

of 24

2−dbT = 1139.65kN

dbT is the least of above two values i.e., 1139.6 kN >700kN ∴ Safe

Provide ISA 100X75X10 as Tension member.

(If required next lower section can be tried to make the section economical)

Design a tension member 3.4m between c/c of intersection using unequal angles with short leg back to back and long legs connected to same side of gusset plate (Welded) carrying a factored pull of 200kN. The member is subjected to reversal of stresses.

Factored load =200kN

Strength governed by yielding (cl.6.2)

mo

y*gdg

γfATmember theofcapacity Tention =

22x3

g cm 8.8mm 880250

1.1200x10A requiredsection twoof Area ===

18.881803400r

180rL

λ ratio sslendernes

min

min

eff

==∴

≤=

Choose the section for an area greater than 8.8 cm2 and rmin greater than 18.88

Try 125x75x10ISA 2

follows as are 125x75x10ISA of Properties 22 mm 3804cm 38.04=A = Area =

mm 20.7cm 2.07= rzz = mm 58.1 mm 5.81=ryy =

safe kN 200kN 864.551.1x10003804x250

γfATmember theofcapacity Tention

mo

y*gdg >===

Welded Connection:

Size of weld

of 24

mm 3= size Minimum

mm 6=Ssay mm 9x1243=t

43 size Maximum =

≤

2

umw N/mm 410=f ; 1.50= γ weldfield as assumingBy

N/mm. L 662.80=x1.53

x410}{0.7x6xL=xγ3

}xf{0.7xSxL =)(P bottomat weldofStrength 11

mw

u11

N/mm. L 662.80=x1.53

x410}{0.7x6xL=xγ3

}xf{0.7xSxL =)(P bottomat weldofStrength 21

mw

u12

P=P+P 21

2Pabout moment By taking .section angle theof with thatcoincides weld theofgravity ofCenter Px125=x250P1

( ) ( )x125220x10=x250662.80L 31

mm 155saymm 150.87662.80x250

x125200x10=L3

1 =

Since the load is acting exactly at the centre of its connection therefore L1=L2=155 mm

Check for design Strength due rupture of critical section(Tdn) :



Where,

β =

Anc = ttA

−

2= 2120010

210125 mmx =

−

Ago=

−

2tB * t

mo

ygo

m

undn r

fAr

fAT ***

1

*9.0 β+=

×

×

×−

Lb

ff

tw s

u

y076.04.1

of 24

Ago= 102

1075 x

− = 700 mm2

Lc = 4*30 =120mm.

β = 1.24 0.7> β <1.44 where

1*

*

mu

moy

rfrf

Check for block shear : P-33, Cl:6.4.1


1

***1

9.0*3 m

utn

mo

yvgdb r

fArfAT +=−

mo

ytg

m

uvndb r

fAr

fAT ***2

1*39.0

+=−

Avg =Avn =thickness of gusset plate x shearing length of weld =2x160x10=3200mm2

Atn = Atg =thickness of gusset plate x angle length =10x250 =2500mm2

1

***1

9.0*3 m

utn

mo

yvgdb r

fArfAT +=−

100025.1410*2500*9.0

10001.1*3250*3200

1 xxTdb +=−

=−1dbT 1158 kN

mo

ytg

m

uvndb r

fAr

fAT ***2

1*39.0

+=−

10001.1250*2500

100025.1*3410*3200*9.0

2 xxdbT +=−

2−dbT = 1114 kN

24.1160

75

410

25075076.04.1

10=×××−=

β

kNkNxx

Tdn 200110310001.1

25070024.1100025.1

410*1200*9.02 **>=

+=

of 24

dbT is the least of above two values i.e., 1114 kN >200kN

Provide 2- ISA 125x75x10 as tension member.

(If required next lower section can be tried to make the section economical)

Design end connection for an angle ISA 125X75X10mm thick using a lug angle with 18mm dia bolts. Take fy=250 N/mm2, thickness of the gusset plate is 8mm. Show all details in sketch. Use bolts of Grade 4.6


Ag = 1902 mm2

Assuming that the 125mm leg (longer leg) is connected to the gusset plate,

Area of connected leg = 10*2

10125

− =1200mm2,

Area of outstanding leg = 10*2

1075

− =700mm2,

Load in the connected leg of main angle = kN27327.432*7001200

1200=

+

Load in the outstanding leg of main angle = kN27.15927327.432 =−

P-83, Cl:10.12.2:

Load in the outstanding leg of lug angle =20% more than load in the outstanding leg of main angle

mo

fyAgdgT

γ*

=

1.1250*1000*1902

=dgT

kNdgT 27.432=

of 24

Load in the outstanding leg of lug angle = 1.2 X 159.27 = 191.12 kN

Load in the connected leg of lug angle =40% more than load in the outstanding leg of main angle

Load in the connected leg of lug angle = 1.4 X 159.27 = 22.98 kN

Area of Lug angle required

Try Lug angle of ISA75X50X8

Area a = 9.36 2cm =936 2mm

Strength of lug Angle in rupture :

Safe

Bolt value:

d=18 mm, dia of bolt hole(d0)= 18+2=20mm

P-75, Cl:10.3.3:

1) Strength of one bolt in Single shear:

Assuming Shank is interfering the shear plane;

nn=0 ns=1, 25.1=bmγ

( )24

dAsbπ

= = ( ) 22 46.254184

mm=π



241.8295.840250

1.1*31012.191*cmmm

X

fymoTdg

Ag ====γ

kNkNXm

Antfu12.1913.27631025.1

410*)85075(*9.0

1

*9.0>=

−+==

γ

+=

b

sbsnbnu

m

AnAnXfγ3

kNX 471000*25.1

46.254*1

3

400==

of 24

=

b

bdpb

m

ftdkVγ

****5.2 , P-76

0.1,,25.03

,3 u

ub

oob f

fdp

dek −=

0.1,410400,25.0

20*360,

20*340

−=

00.1,98.0,75.0,67.0=

67.0=∴ bk

=

1000*25.1400*8*18*67.*5.2

dpbV

kNVdpb 18.77=

∴Bolt value = 47 kN

No. of bolts required to connect:

1. outstanding leg of lug angle with gusset plate = .06.447

12.191 nos=

Hence provide 4 bolts with a pitch of 60mmc/c.

2. Connecting leg of lug angle with outstanding leg of main angle = .74.447

98.222 nos=

Hence provide 5 bolts with a pitch of 60mmc/c.

3. Connecting leg of main angle with gusset plate = .81.547273 nos=

Hence provide 6 bolts with a pitch of 60mmc/c

of 24

125

Gusset plate

Lug Angle ISA 75X50X8

75

Main Angle- ISA 125X75X10 mm c/c 6-18 mm dia bolts at 60 mm c/c

5-18 mm dia bolts at 60 mm c/c 4-18 mm dia bolts at 60 mm c/c

of 24

1. a flat of size 250 x 10 mm is connected to the gusset ...nptel.vtu.ac.in/vtu-nmeict/dss1/tension...

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