CE 318N DESIGN STRENGTH OF BEARING BOLTS Clause 103 examples Design lap joint between two plates each 10mm thick as shown in figure to transmit a factored load of 180 KN using M20 bolts of grade 46 Given ID: 916591
Download Presentation The PPT/PDF document "L-5" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
L-5
CE – 318N
DESIGN STRENGTH OF BEARING BOLTS
Clause
10.3
examples
Slide2Design lap joint between two plates each 10mm thick as shown in figure to transmit a factored load of 180 KN using M20 bolts of grade 4.6. Given
that steel plates of grade Fe410 are used. consider tension in bolt. Find the strength of joint
Slide3Given Data:
Property class 4.6 fub= Ultimate strength of bolt= 400MPa fyb= Yield strength of bolt= 400×0.6= 240MPa fu= Ultimate strength of plate= 410MPa fy= yeild strength of plate= 250 MPaBolt M20
d
= diameter of bolt=
20mm;
d
0
= diameter of hole=
20+2=22mm
Shank
area of bolt =
A
sb
=
π
/4* d
2
=
π
/4*
20
2
=314mm
2
Net
tensile stress area of bolt =
A
nb
= 0.78*
Asb
=.78*314= 245mm
2
edge distance e =1.5d
0
= 1.5*22= 33mm;
pitch p =2.5d= 2.5*20= 50 mm;
; thickness of plate t=min of thickness of connecting members = 10mm
Slide4LAP JOINT
Bolts will be in single shear and bearinga. Strength on shearing All three reduction factor =1 Vdsb
=(f
ub /√3) ( nnAnb+ nsAsb) / γmb ; Number of shear plane with thread intercepting the shear plane nn = 1 Number of shear plane with shank intercepting the shear plane ns = 0 thickness of plate ‘t’= 10mmγmb =Partial safety factor for material of bolt= 1.25
Slide5Design Strength of Bolts in single shear=
Vdsb =(f
ub
/√3) ( nnAnb + 0)/ γmb ; Vdsb = 400*1*245/√3 ×1.25×1000 Vdsb = 45.64KN
Check for reduction factorReduction factor for long Joints :
Length of joint =50mm < 15d
ie
300mm so
rf
=1
Reduction Factor for Large Grip
Lengths: grip
lenth
= 20mm<5d
ie
100mm
so
rf
=1
Reduction
Factor for
packing
No packing material used
so
rf
=1
Slide6LAP JOINT
b) Strength in bearing Vdpb = 2.5 kbd
t
f
u / γmb ; kb is minimum of e/(3do), p/(3do)-0.25, fub/ fu and 1.0, kb= min ( 33/3*22, 50/3*22-0.25, 400/410, 1.0)
kb= min (
0.5, 0.5, 0.97,
1.0
)
k
b
= 0 .5
substituting values in
V
dpb
= 2.5
k
b
d
t
f
u
/
γ
mb
;
V
dpb
= 2.5 * 0.5* 20* 10*410/1.25*1000 = 82KN
V
dpb
=
82KN
Slide7LAP JOINT
c) Tensile Strength of Bolt Tdb= 0.9 fub
A
nb
/ γmbAn=245 mm2Tdn = 0.9*245* 400 / 1.25*1000= 70.56 kNBolt Value= MIN ( 45.64, 82, 70.56) = 45.64 KNRequired no of bolts= 180/ 45.64 =3.94=4 boltsStrength of joint = 4* 45.64 = 183 KN
Slide8b = 50+33+33= 116
Slide9Check for strength of plate at critical section
Tdn = 0.9 An fu / γm1
Where
fu= Ultimate stress of material An= Net effective area at the critical sectionγm1 = partial safety factor at ultimate stress = 1.25 There are two holes at critical sectionTdn = 0.9* (116 -2*22)*10*410/1.25*1000= 212 kNStrength of joint = minimum of strength of bolt and strength of plate = min(183, 212)= 183kN which is more than 180kn hence safe
Slide10Also Find efficiency of joint assuming
a) width of plate given as 116mmb) width of plate not given/ per pitch length
Slide11Soln
Strength of joint= 183knThis strength of solid plate given by Tdn = 0.9 Ag fu / γm1
Or =
fy Ag/γm0 which ever minimumWhere fu= Ultimate stress of material =410KN/mm2 fy= yield stress of material =250KN/mm2 Ag= (116*10)= 1160 mm2Γm1 = partial safety factor for material of bolt= 1.25Γm0 = partial safety factor at yeild = 1.1
Tdn
= 250*1160/
1.10*1000= 263
kN
Efficiency = 183/263*100= 69%
Slide12EFFICIENCY PER PITCH LENGTH
There are two bolts per pitch lengthShearing Strength of bolt per pitch length= 2* 45.64=91.2 KnBearing Strength of bolt per pitch length= 2* 82=164 KnTensile strength of plate per pitch length= 0.9 (p-dh
)t
f
u /γm1 = 0.9*(50-22)10*410/1.25*1000= 82.65KNStrength of joint= 82.65knThis strength of SOLID PLATE per pitch length given by Tdn = fy Ag/γm0 which ever minimumWhere fu= Ultimate stress of material =410KN/mm2 fy= yield stress of material =260KN/mm2 Ag=p*t= (50*10)= 500 mm2Γm1 = partial safety factor for material of bolt= 1.25Γ
m0 = partial safety factor at ultimate stress = 1.1
T
dn
= 250*500/
1.10*1000= 113
kN
Efficiency = 82.65/113*100= 73.14%
Slide13Slide14Design butt joint between two plates each 10mm thick as to transmit a factored load of 180 KN using M20 bolts of grade 4.6. Given
that steel plates of grade Fe410 are used. (a) single cover butt joined, cover plate 8mm thick
(b)
double cover butt joint, each cover plate 6 mm
thick
Slide15Given Data:
Property class 4.6 fub= Ultimate strength of bolt= 400MPa fyb= Yield strength of bolt= 400×0.6= 240MPa fu= Ultimate strength of plate= 410MPa fy= yeild strength of plate= 240 MPaBolt M20
d
= diameter of bolt=
20mm; d0= diameter of hole= 20+2=22mm Shank area of bolt =Asb = π/4* d2 = π/4* 202 =314mm2 Net tensile stress area of bolt =Anb = 0.78* Asb =.78*200.9= 245mm2 edge distance e =1.5d0= 1.5*22= 33mm; pitch p =2.5d= 2.5*20= 50 mm;; m
Slide16BUTT JOINT
Bolts will be in single shear and bearingcover plate 8mm thickStrength of Bolts in single shear= All three reduction factor =1
V
dsb =(f ub /√3) ( nnAnb+ nsAsb) / γmb ; Here number of shear plane with thread intercepting the shear plane n
n = 1
Number
of shear plane
with shank
intercepting the shear plane n
s
= 0
V
dsb
= 400*1*245/√3 ×1.25×1000
V
dsb
= 45.64KN
Slide17b) Strength in bearing
Vdpb = 2.5 kbd t fu / γmb
;
t = (min( 10or 8)= 8mm kb is minimum of e/(3do), p/(3do)-0.25, fub/ fu and 1.0, kb= min ( 33/3*20, 50/3*20-0.25, 400/410, 1.0) kb= min ( 0.5, 0.5, 0.97, 1.0) k
b= 0 .5 substituting values in
V
dpb
= 2.5
k
b
d
t
f
u
/
γ
mb
;
Vdpb
= 2.5 * 0.5* 20* 8*410/1.25*1000 = 65.6KN
Slide18Bolt Value= MIN (
45.64, 65.6) = 45.64 KNRequired no of bolts= 180/45.64= 4 bolts
Slide19c) Check Tensile Strength of plate at critical section
Tdb= 0.9fu An
/
γ
mbAn= Net effective area at the critical section= (220-4*22)*10= 1320mm2Tdn = 0.9*1320* 410 / 1.25*1000= 389 kNBolt Value= MIN ( 4*45.64, 389) = 45.64 KNStrength of joint= 4* 45.64 = 183 KN > 180KN hence safe
Slide20double cover butt joint with cover plates of 6mm thickness.
The bolts will be in double shear and bearing
All RF=1
Vdsb =(f ub /√3) ( nnAnb+ nsAsb) / γmb ; Here number of shear plane with threat intercepting the shear plane nn =
2Number of shear plane
with shank
intercepting the shear
plane
n
s
=
0
V
dsb
= 400 × (245+245) /√3×
1.25
×
1000
V
dsb
=
90.63 KN
Slide21b) Strength in bearing
Vdpb = 2.5 kbd t fu / γ
mb
;
t= least of aggregate thickness of cover plate and minimum thickness of the main plate joined. ( 10, (6+6) )=10mm kb is minimum of e/(3do), p/(3do)-0.25, fub/ fu and 1.0, kb= min ( 33/3*20, 50/3*20-0.25, 400/410, 1.0) kb= min ( 0.5, 0.5, 0.97, 1.0) = 0 .5 substituting values in Vdpb = 2.5 kbd t fu / γmb ; Vdpb = 2.5 * 0.5* 20* 10*410/1.25*1000 = 65.59KN
Slide22Bolt Value
= MIN (90.63, 65.59) = 65.59 KNRequired no of bolts= 180/65.59 =2.86= 3 bolts
Slide23Tdn
=0.9Anfu/γm1= 0.9×( 220-22) *10*410/1.25*1000 = 584knCheck for the strength of plate
Slide24Find the efficiency and the
maxm force which can be transferred through the double covered butt joint shown below. Also find net tensile strength of main and cover plate. Given bolts properety class 4.6. Fe410 steel plate used
Slide25Given Data
Property Class 4.6fub= Ultimate strength of bolt= 400MPafyb= Yeild
strength of bolt= 400×0.6= 240MPa
f
u= Ultimate strength of plate= 410MPaBolts diameterApp diameter of bolt= 6.04 * SQRT(t) ; t is thickness of thinner plateApp diameter of bolt=d= 6.04*SQRT(10)= 19.1mm ≂20mmd0=22mm Edge/end distance = 40mm(given) pitch= 60mm (given) t= min(20 and 16) = 16mm (for bearing)γmb = Partial safety factor for material of bolt= 1.25
Slide26The bolts will be in double shear and bearing
STRENGTH IN SHEARING:-Assume that thread cut both the shearing planes ie nn=2 ns=0Strength of a bolt in double shear= Vdsb
=
fub (nn An) /√3× γmb Vdsb = 2 × 400 × 245/√3× 1.25× 1000 = 90.534KN
Slide27Strength of Bolts in Bearing=
Vdpb ≤ Vnpb / γmb ; (γmb = 1.25) where
V
npb = 2.5 kbd t fu t= least of aggregate thickness of cover plate and minimum thickness of the main plate joined. ( 20, 16) =16mmEnd/edge distance e =40mmPitch= 60mm kb is minimum of e/(3do), p/(3do)-0.25, fub
/ fu and 1.0,
Slide28k
b= min ( 40/3×22, 60/3×22-0.25, 400/410, 1.0) = .6061Design strength in bearing Vdpb = 2.5 kb
d
t
fu / γmb = (2.5×.6061×20×16×410) /1.25×1000= 159. 06KNThe strength of the bolt will be minimum of shearing and bearing = min( 90.53 and 159.06)Design strength of joint in double shear=6 × 90.314=543.187KNSo Strength of bolts in joint= 543.18 KN
Slide29Strength of Plate
Net tensile strength at 1-1 (critical for main plate) is given by Tdn = 0.9 An fu / γm1 An= (b-dh)t
=0.9
×
410× (200-22) ×16/1.25×1000=840.73 KNNet tensile strength at 3-3 (critical for cover plate) is given by= Tdn = 0.9 An fu / γm1 = =0.9×410× (200-3×22) ×20/1.25×1000 =989.66 KN(2tc)Strength of plate in joint= 840.73 KNStrength of joint= 543.18KNMaxm design force that can be transferred= = 543.18KN
Slide30Design strength of solid plate=
0.9 Ag fby / γm0 =0.9×250×200
×
16/1.25
×1000=727.27KNEfficiency of joint= 543.18/727.27 ×100=74.4%
Slide31Determine the strength and efficiency of the lap joint shown in
fig . The bolts are of 20mm diameter and of grade 4.6 The two Plates to be joined are 10 mm and 12 mm thick (Steel is of grade Fe 410).
Slide32For
Fe 410 grade of steel : fu =410 MpaFor bolts of grade 4.6 : fub = 400MPa
f
yb
= Yeild strength of bolt= 400×0.6= 240MPaFor 20 mm diameter bolt Anb = 245 mmYmb = partial safety factor for material of bolt = 1.25Ym1 = partial safety factor for resistance governed by ultimate stress = 1.25
Slide33The
bolts will be in single shear and bearing. The bearing strength of the bolt will be governed by the thickness of the thinner plate. Hence t =10 mm.Strength of the bolt in single shear, Vdsb
=
fub×Anb/√3× γmbVdsb = 400×245/√3× 1.25× 1000 = 45.26 KNStrength of joint per pitch length in shear= 1×45.26 (one bolts in one pitch)= 45.26KN
Slide34Strength
of bolt in bearing=Vdpb = 2.5 kbd
t
f
u /γmbt= least of the main plates jointed= min(12, 10)= 10mmThe end distance e has been assumed to be for rolled edge from Table 5.3 kb is minimum of e/(3do), p/(3do)-0.25, fub/ fu and 1.0, kb
= min ( 33/3×22, 50/3
×
22-0.25
,
400/410
, 1.0)
k
b
= min ( 0.5, 0.51, 1.95, 1) =0.5
V
dpb
= 2.5
k
b
d
t
f
u
= 2.5×0.5×20×10×410/1.25×1000=
82.0KN
Strength of the joint per pitch in bearing
=
1×82= 82KN (one
bolts in one pitch
)
Slide35Net tensile strength of plate per pitch length at 1-1
Tdb= 0.9fu An / γm1Tdb= 0.9f
ub
(p-ndo)t/ γm1 (one bolt at 1-1)= 0.9× 410×(50-22) ×10/1.25×1000=82.65KNThe strength of the joint per pitch length ( minimum of shearing, tearing and bearing ) is 45.6KN.tensile strength of solid plate per pitch lengthTdb= 0.9fu A/ γm1 = 0.9× 410×100
×10/1.25×1000=147.6kn
Efficiency of joint = 45.52/147.6*100 = 30.66%
Slide36Web Angle Connection
Framed Connections
Slide37Two framing angles ISA150mm×150mm×10mm are used to make beam to column connection. One angle is placed on either side of web of beam . four bolts of 20mm
dia and grade 4.6 are used to connect the angle legs to the beam web. Determine the reaction that can be transferred through the jointColumn section ISHM300 @ 618.03N/m t
f
=10.6mm
Beam section ISMB 350 @514.04N/m tw=8.1mm
Slide38Slide39Given Data
Thickness of web of the beam= 8.1mmThickness of flange of the column= 10.6mmThikness of angle= 10mm fub
= Ultimate strength of bolt=
400MPa
fyb= Yeild strength of bolt= 400×.6= 240MPafu= Ultimate strength of plate= 410MPad=20mm d0=22mmγm0 =Partial safety factor at yeild= 1.1γm1 =Partial safety factor at ultimate= 1.25
Slide40Strength
Calculations:=The bolts connecting the angle legs with the web of the beam are in double shear and bearing. For 20 mm diameter bolt Anb = 245 mm
Design strength of bolts in double shear,
Vdsb = 2×fub×Anb/√3× γmb Vdsb = 2 × 400 × 245/√3× 1.25× 1000 = 90.52 KNStrength of joint= 4×90.52 = 362.08 KN
Slide41Strength of bolt in bearing=
Vdpb = 2.5 kbd t fu /γmb
k
b
is minimum of e/(3do), p/(3do)-0.25, fub/ fu and 1.0, assume p= 60mm and e=40mm kb= min ( 40/3×22, 60/3×22-0.25, 400/410, 1.0) kb= min ( .61, .659, 97, 1.0) = 0.61 t= last of aggregate thickness of cover plate and the minimum thickness of the main plates jointed = min(20, 8.1)= 8.1mmVdpb = 2.5 k
bd t fu
= 2.5×0.61×20×8.1×410/1.25×1000= 81.03KN
Slide42The strength of the bolt will be minimum of shearing and bearing = 81.03KN. Bearing strength= 4× 81.03KN. = 324.12 KN The Maxm reaction
that can be transferred through the
joint=
324.12 KN
Slide43In this type of connection, the end shear, i.e., the reaction of the beam is transferred through web angles either to the column flange or to the column web.
A clearance of 2mm is provided between the beam and the column. Web angles are usually connected in the compression zone of beam to provide lateral support to the compression flange. The length of the web angles varies from 0.6 to 0.75 depth of the member; upper limit is the clear depth of the web
Slide44the thickness of the angle should be 8mm
upto beam depth of 450mm and 10mm for deeper beams. From flexibility point of view, the gauge on column flange should be as large as possible. Usually it is 90 < g < 140; however 100 or 140 mm gauge is commonly used.
Slide45Design a web angle connection with one angle on each side of web of the beam between a beam MB350 and a Column ISHB 400 for a reaction of 300
kN using 20mm dia black bolts grade 5.6. Grade of rolled steel section is Fe410.
Slide46Thickness of web of the beam= 8.1mm
Thickness of flange of the column= 12.7mmfub= Ultimate strength of bolt= 500MPafyb= Yeild strength of bolt= 500×.6= 300MPaf
u
= Ultimate strength of plate= 410MPa
d=20mm d0=22mmγm0 =Partial safety factor at yeild= 1.1Strength CalculationsConnectiong of angles with beam The bolts connecting the anglelegs with the web of the beam are in double shear and bearing. Consider web angles of thickness 8mm while the web thickness of beam is 8.1 mmDesign strength of bolts in double shear, Vdsb = 2×fub×Anb/√3× γmb
Vdsb
= 2 × 500 × 245/√3×
1.25
×
1000 = 113.16 KN
Slide47Strength of bolt in bearing=
Vdpb = 2.5 kbd t fu /γmb
k
b
is minimum of e/(3do), p/(3do)-0.25, fub/ fu and 1.0, assume p= 60mm and e=40mm kb= min ( 40/3*22, 60/3*22-0.25, 500/410, 1.0) kb= min ( .61, .659, 1.22, 1.0) = 0.61 t= last of aggregate thickness of cover plate and the minimum thickness of the main plates jointed= min(16, 8.1)= 8.1mmVdpb = 2.5 kbd t fu = 2.5×0.61×20×8.1×410/1.25×1000= 81.03KN
The strength of the bolt will be minimum of shearing and bearing and is 81.03KN.No of bolts=300/81.03=3.7 app 4 bolts
Slide48Connection of angles with column
: The bolt connecting the angle legs with the flange of the column are in single shear and bearing against angle of thickness 8mm while the flange thickness of column is 12.7 mm. Thus,Strength of bolts in single shear = Vdsb = fub×Anb/√3× γ
mb
V
dsb = 500 × 245/√3× 1.25× 1000 = 56.58 KNBearing strength of the bolt on the angles,Vdpb = 2.5 kbd t fu = 2.5×0.61×20×8×410/1.25×1000= 80.903KNThe strength of the bolt will be minimum of shearing and bearing and is 56.58KN.No of bolts=300/56.58=5.3
Slide49Slide50A bracket is bolted to a vertical column as shown below. If M20 bolts of grade 4.6 are used Determine the
maxm value of factored load P which can be carried safely. Given that thickness of web of ISMC 300 IS 7.6mm.
Slide51Sol:- Given M20 bolts of grade 4.6,
Thus fub= 400MPa fyb= 240MPa fu= 410 MPa
(for plate and ISMC)
γ mb = 1.25Bolts are in single shearShear Capacity of Bolts:-Assume shear plane cuts the root of thread ns=0 and nn=1Anb= 0.78 ×π/4 ×20 ×20 = 245mm2
Slide52Shear capacity of Bolt
Vnsb= 400 (1× 245)/ (√3 × 1000) = 56.58 Vdsb= V
nsb
/
γmb = 1/ 1.25 = 45.27 KNNo REDUCTION FACTOR FOR SHEAR CAPACITY OF BOLTS is required as
Slide53Bearing strength of bolt
Since the thickness of channel web less than that of plate hence bearing capacity shall be governed by channel web.Bearing strength of bolt= Vnpb = 2.5 k
b
d
t fu where kb is smaller of e/(3do), p/(3do)-0.25, fub/ fu and 1.0,Or kb= min (33/3 ×22), 80/3 ×22-.25, 400/410, 1)= 0.5d=20mm ; t= 7.6 Vnpb= 2.5 ×0.5 ×20 ×7.6 × 410/1000Vdbp= Vnpb/ γmb = 62.32knDesign strength of bolt is minimum of bearing and shearing ie
45.27KN
Slide54Direct Shear F
1= P/n= P/5= 0.2PCenter of gravity of the bolted connection is at the center of central bolt.F2= Pe r/ Σ r
2
For outer four bolts r= √ 80
2+ 602 =100mmFor cental bolt r=0Σ r2 = 4 ×1002F2= P × 250 × 100/4 ×100 ×100= 0.625 PLet Angle between F1 and F2= θ cos θ = 60/100=0.6Total Shear force on extreme bolt= F= √ ( F1 2 + F2 2
+ 2 F1F
2
Cos
θ
)
= 0.76199P
Equating it to strength of bolt we get
0.76199P = 45.27 or P= 59.41KN
FORCE IN EXTREME BOLT:
Slide55If F given
Than F1= shear component= 3/5 FF2= tension component= 4/5 F
Slide56These are the bolts made of high tensile steel which are
pretentioned and then provided with nuts. The nuts are clamped also. Hence resistance to shear force is mainly by friction. The slip will occur when load overcomes the frictional resistance provided by the preload of the bolt. After slip occurs, the behaviour is similar to the normal bolts.
There are two types of HSFG bolts namely (
i
). Parallel shank and (ii) waisted shank.In this case also, it is commonly assumed that equal size bolts share the loads equally in transferring the external force. HSFG Bolts
Slide57Nut is tighten to develop a clamping force on the plate which is indicated as the tensile force T in the bolt. T should be about 90% of proof load.
For slip critical connection the horizontal force F is induced in the joints which is equal to the Tensile force T multiplied by coefficient of friction.µf This friction resistance to slip between the plate surfaces should exceed the slip caused by externally applied shear.
Slide58Parallel shank bolt are designed for no slip at serviceability loads. Hence they slip at higher loads and slip into bearing at ultimate load. Hence such bolts should be checked for their bearing strength at ultimate load.
Waisted shank HSFG bolts are designed for no slip even at ultimate load and hence there is no need to check for their bearing strength.
Slide59The design slip resistance or nominal shear capacity of a bolt:
Vnsf = μf ne
K
h
Fo μ = Coefficient of friction (called as slip factor, table 20) ≤0.55. ne = Number of effective interfaces offering friction resistance Kh = 1.0 for fasteners in clearance holes = 0.85 for fasteners in oversized and short slotted holes = 0.7 for fasteners in long slotted holes loaded parallel to the slot Fo = Minimum bolt tension (proof load) at installation ≈ A
sb f
o
A
sb
= Nominal shank area of bolt
f
o
= Proof stress ≈ 0.7
f
ub
f
ub
= Ultimate tensile stress of bolt
Shear Strength of HSFG Bolts (
Cl
10,4
pg
76)
Slide60The factored design force or slip resistance (
Vsf ), should satisfy: Vsf
≤
Vnsf / γmf γmf = 1.10 if slip resistance is designed at service load γmf = 1.25 if slip resistance is designed at ultimate load. The formulae for bearing & tension resistance, and Combined Shear and Tension are similar to those of Black bolts.Commonly used grade 8.8REDUCTION FACTORS FOR THE CAPACITY OF BOLTS IN CASE OF LONG JOINT
Shear Strength of HSFG Bolts
Slide61COEFFICIENT OF FRICTION as per
IS 800:2007 61
Slide62BOLTS SUBJECTED TO COMBINED SHEAR AND TENSION
A Bolt required to resist both design shear force Vsb and design tensile force at the same time shall satisfy: - (𝑽𝒔f
/
V
dsf)𝟐 + (𝑻f/𝑻𝒅f)𝟐≤𝟏.𝟎 Where Vsf = factored shear force on bolt Vdsf = design shear capacity of bolt Tf = factored tensile force on bolt Tdf= design tension capacity of bolt
Slide63Determine the shear capacity of a 20 mm bolt in standard hole of property class 8.8 if it connects plates of 10 mm thickness and is in single shear and
waisted shank friction grip bolting. Use μf =0.33; Given Data: fub= Ultimate strength of bolt= 800MPa
f
yb
= Yeild strength of bolt= 800×0.8= 640MPafu= Ultimate strength of plate= 410MPaKh = 1 (clearance holes)γmf = 1.1 ( for slip resistance at service load) = 1.25 for slip resitance at ultimate loadn c =1Shank area of bolt =Asb = π/4* d2 = π/4* 202 =314.16mm2Net tensile stress area of bolt =Anb = 0.78* Asb=.78*314.16= 245mm2
Slide64The design slip resistance or nominal shear capacity of a bolt:
Vdsf = μf ne
K
h
Fo / γmf Fo = Minimum bolt tension (proof load) at installation ≈ Asb fo Asb = Nominal shank area of bolt
fo = Proof stress ≈ 0.7 f
ub
f
ub
= Ultimate tensile stress of bolt
V
dsf
= 0.33×1×1×.7×800×245/1.1×1000=41.16KN (slip at service load)
V
dsf
= 0.33×1×1×.7×800×245/1.25×1000=36.2 KN (slip at ultimate load)
Slide65PRYING ACTION (10.4.7)
Slide66Prying force is due to flexibility of connecting plates. It occurs only in HSFG bolts
In a tension or hanger connection, the applied load produces tension in the bolts and the bolts are designed as tension members. If the attached plate is allowed to deform, additional tensile forces called prying forces Q are developed in the bolts. which causes the flange of T section to bend in middle portion.ADDITIONAL FORCE IN BOLT
DUE TO PRYING
Slide67PRYING ACTION
Slide68Failure Modes Due to
Prying Forces68
Slide69Additional Force in Bolt due to Prying
69
Slide70Additional Force in Bolt due to Prying
The additional force Q in the bolt due to prying action: WhereQ= Prying Forcel
e
= Minimum of end distance or
lv= distance from bolt center to the toe of fillet weld or half of the root radius for a rolled section.ϒ= 1.5 β= 2 for non-tensioned bolt and 1 for pre-tensioned boltbe = Effective width of flange per pair of bolts, mm fo = Proof stress (KN or KN/mm2)t= thickness of the end plate2Te= Total applied tensile forceCheck Te + Q should be less than tensile capacity of bolt70
Slide71The joint shown in Figure below has to carry a factored load of 180 KN. End plate is of size 160mm×140mm×16mm.The bolts used are M20 HSFG of grade 8.8 weld size is 8mm. Check the safety of bolts
Slide72Slide73Since the bolts are HSFG bolts in tension hence an additional prying force will develop in the bolts in addition to the direct tension.
Fy=250 (for plate)lv= distance from bolt centerto the toe of fillet weld or half of the root radius for a rolled section. =160/2 -8-16/2-40= 24mmβ= 1 for pre-tensioned bolt fo = Proof stress (kN or
kN
/mm
2)=0.7 fub= .7*800=560Mpale= Minimum of end distance or Le= Min (40, 26.34)= 26.3mmϒ= 1.5 be = Effective width of flange per pair of bolts = 140mmt= thickness of the end plate= 16mm2Te= Total applied tensile force =180KNDirect tension in bolt (T)= 180/2= 90KN
Slide74Total tension= 32.43+90=122.43
Q=32430.89N
0r 32.43KN
Design Tension capacity of bolt=
Tdf = 0.9fub Anb /γmb= 0.9×800×.78×/1.25 = 141.12KNtotal capacity > tension (122.43) safe
Slide75PRYING ACTION CONTD.
The thickness of the Tee-flange is determined so that it does not yield.
Slide76PRYING ACTION CONTD.
The thickness of the Tee-flange is determined so that it does not yield Mp = fy/1.1×Ze =fy/1.1 × bet2 / 4 Or t min = √4.4 M
p
/
fy be
Slide77ECENTRIC CONNECTION
Slide78A bracket is bolted to a vertical column as shown below. If M20 bolts of grade 4.6 are used Determine the
maxm value of factored load P which can be carried safely. Given that thickness of web of ISMC 300 IS 7.6mm.
Slide79Sol:- Given M20 bolts of grade 4.6,
Thus fub= 400MPa fyb= 240MPa fu= 410 MPa
(for plate and ISMC)
γ mb = 1.25Bolts are in single shearShear Capacity of Bolts:-Assume shear plane cuts the root of thread ns=0 and nn=1Anb= 0.78 ×π/4 ×20 ×20 = 245mm2
Slide80Shear capacity of Bolt
Vnsb= 400 (1× 245)/ (√3 × 1000) = 56.58 Vdsb= V
nsb
/
γmb = 1/ 1.25 = 45.27 KNNo REDUCTION FACTOR FOR SHEAR CAPACITY OF BOLTS is required as
Slide81Bearing strength of bolt
Since the thickness of channel web less than that of plate hence bearing capacity shall be governed by channel web.Bearing strength of bolt= Vnpb = 2.5 k
b
d
t fu where kb is smaller of e/(3do), p/(3do)-0.25, fub/ fu and 1.0,Or kb= min (33/3 ×22), 80/3 ×22-.25, 400/410, 1)= 0.5d=20mm ; t= 7.6 Vnpb= 2.5 ×0.5 ×20 ×7.6 × 410/1000Vdbp= Vnpb/ γmb = 62.32knDesign strength of bolt is minimum of bearing and shearing ie
45.27KN
Slide82Direct Shear F
1= P/n= P/5= 0.2PCenter of gravity of the bolted connection is at the center of central bolt.F2= Pe r/ Σ r
2
For outer four bolts r= √ 80
2+ 602 =100mmFor cental bolt r=0Σ r2 = 4 ×1002F2= P × 250 × 100/4 ×100 ×100= 0.625 PLet Angle between F1 and F2= θ cos θ = 60/100=0.6Total Shear force on extreme bolt= F= √ ( F1 2 + F2 2 + 2 F
1F2
Cos
θ
)
= 0.76199P
Equating it to strength of bolt we get
0.76199P = 45.27 or P= 59.41KN
FORCE IN EXTREME BOLT:
Slide83If F given
Than F1= shear component= 3/5 FF2= tension component= 4/5 F
Slide84Select nominal
dia of boltsAdopt a pitch of 2.5d to 3d for boltsFind no off bolts to be provided in two or more vertical row.Find resultant force on critical sectionStrength of bolt computed and it should not be less than force on bolt.Steps in design
Slide85No of bolts required ‘n’
If
der
r two vertical lines of bolts and n obtained is bolts required in each row
M is moment in the joint and V is design strength
Slide86END
short questions cont
Slide87APP DIA OF BOLTS= 6.04 * SQRT(t)
t is thickness of thinner plateCover plate thickness= 5/8 * t No of bolts=Load/Bolt ValueImportant