Post on 18-Jun-2020
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
Page 1 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
Page 2 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
Page 3 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
Page 4 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
Page 5 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
Page 6 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
Minimum pitch = 2.5 x d = 2.5 x 12 = 30mm, Minimum edge distance = 1.7 x d = 22.1mm
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
Page 7 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
Page 8 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
Page 9 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
Page 10 of 24
Gusset is connected to 30-mm leg of angle a. Bolted connection :
Properties of 1-ISA 50 X 30 X 6
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
Minimum pitch = 2.5 x d = 2.5 x 12 = 30mm, Minimum edge distance = 1.7 x d = 22.1mm
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
Page 11 of 24
3. 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 =
−−
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
Page 12 of 24
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
+=−
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
Page 13 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
Page 14 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
3. Strength of bolt in Bearing:
Nominal bearing strength of the bolt
=
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=
+=
Page 15 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.
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 =
−−
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
Page 16 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
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
+=−
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
Page 17 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
Page 18 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) :
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 =
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
Page 19 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
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
+=−
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 **>=
+=
Page 20 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
Properties of 1-ISA 125 X 75 X 10
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=
Page 21 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
4. Strength of bolt in Bearing:
Nominal bearing strength of the bolt
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==
Page 22 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
Page 23 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
Page 24 of 24