anch.bolt design

8/9/2019 Anch.bolt Design

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CPS M312MB1A-128 TO 134

300mm

Width = 300mm Z

Loading Data :-

Shear force Fx = 0 kNShear force Fz = 13!"1 kN

Axial force Fy = "2"0 kN

Moment My = 0 kN-m 300 #

Moment Mx = 000 kN-m

Moment Mz = 000 kN-m

300

Mz is the moment @ major axis and Mx is the moment @ minor axis.

Now e = Mz = 0 m

Fy

Now !" = 0.0#0 m and !$ = 0.%00 m

&ase - ' e ( !" small eccentricity

&ase - '' !"(e(!$ Medi)m eccentricity

&ase - ''' e * !$ ar+e eccentricity

Thi$ i$ th% &a$% o' m%di(m %&&%nt)i&it*

,istance of olts from the end of the late = "!mm

/idth of the memer alon+ -axis = 120mm

/idth of the memer alon+ 1-axis = +0mm

2rade of the concrete to e )sed = M2!

,' ind o) $%i$mi& .oad i$ a /a)t o' th% .oading in th% d%$ign th% /%)mi$$i.% $t)%$$%$ $ha..

% in&)%a$%d * 33 'o) a.. $t%%.$ and 2! 'o) a.. Bo.t$

Ass)me the size of the ase late3L%ngth =


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DS,5 O6 BOLTS 6O7 BAS PLAT CAS - ,,9

4he ase has een ass)med as Fixed.

%50 60

55#

int = 7.86

= 8.69

eccentricity e = 0 m

Ass)min+ no tension in the farther olts3 contact width a =

0.7# m

:ase ress)re = where c = safe earin+ caacity of concrete

= " x %.$$ ;since wind is the art of loadin+)ire any desi+n and nominaldiameter of olts will s)ffice.

?roide 8 5o$ of M1+ Anchor :olts of 4ye - '

N!mm5

N!mm5

a=3x(L2e )P=C=

cab

2


3/8

DS,5 O6 BOLTS 6O7 BAS PLAT CAS - ,,, 9

Max comressie )nder ase late =

"! :ase late

";"

T C =

1;84

,istance of olt from the end of the late = "!mm

alon+ the section considered

?ermissile stress in concrete can e increased y $$ and allowale tension in olts

can e increased y 5# as wind has een considered in desi+n.

?ermissile stress in concrete = + < 108 5mm2

m = mod)lar ratio = 10;8

n = deth of ne)rtal axis '.e. the ortion of the len+th of the concrete area which is effectie.

d = distance of the farther side olts from the nearer ed+e

Accordin+ to concrete eam theory deth of the ne)tral axis BnB is +ien y

D%$ign o' An&ho) Bo.t$

Ca.&(.ating T%n$ion d(% to M>>

Size of the str)ct)ral memer = +0 mm

distance of the farther side olts from the nearer ed+e = 22! mmSo 3 c = + 5mm2 and 4 = 120 5mm2

4herefore 3 n = and eer arm jd =

4akin+ moments ao)t line of action of tensile force 4 3 & x A = M C ? x

So & = 28 ?5

4ension 4 in the windward olts d)e to Mzz is 4 = & - ?

No.of anchor olts for Mzz i.e. alon+ - axis on one side = 2 5o$

4herefore axial tension d)e to Mzz er olt = -223!1! ?5

t= allowale tension in olts = 67.# N!mm5 x %.5# = %%9.%5# N!mm5 ; increasin+ y 5# )illiri)m we hae P @ T = C

n=d

1+tb

mc


4/8

"!

";"

T C = +

N!mm5

2"3

Simi.a).* T%n$ion d(% to M =Size of the str)ct)ral memer = 120 mm

alon+ the section considered

d = distance of the farther side olts from the nearer ed+e = 22! mm

4herefore 3 n = and eer arm jd =

4akin+ moments ao)t line of action of tensile force 4 3 & x A = M C ? x

So & = 20 ?5

D>)atin+ ertical forces we +et ? C 4 = & 4 =

4herefore Net axial tension d)e to Mx and Mz = -97.406 kN,ia of :olts

No.of anchor olts for Mxx i.e.alon+ 1 - axis on one side = 2 5o$

4herefore axial tension er olt d)e to Mxx = -2+3!1! ?5 %5 -

%" %9.9#4herefore axial tension er olt = -48"03 ?5 50 56

?roide 4 5o$ olts of M1+ 57 75

$0 "8

Shear in an extreme olt d)e to Fx = 0 ?5 $" 69

Shear in an extreme olt d)e to Fz = 33;2"! ?5 75 %$0

Ees)ltant shear in extreme olt = 33;2"! ?5 79 %8"

#" 577

"7 $5%

Now3

-48.70 @ 33; 12!18 !

4h)s n)mer of olts and its diameter are ade>)ate

4herefore roide 4 no$ o' M 1+ T*/%-, An&ho) Bo.t$

Normal4ension

25.1P

F

P

F

s

s

t

t

+


5/8

Thi&?n%$$ o' Ba$% P.at% :- for Mzz

Net rojection on either side of the section = 120 mm

Stress ordinate )nder the flan+e of the section = 40120034 5mm2

&onsiderin+ %mm wide stri :endin+ Moment @ face of the col)mn = ;+33;"8144 N-mm

484+; mm

4h)s3roide ase late of en+th = 300mm

of :readth = 300mm

thickness = 1+mm

Now M = f 1 = %9# x t5!" 4herefore 3

t= Mx 6

185x1.33xb


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anch.bolt design

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