Column Base Plate Design

COLUMN BASE PLATE DESIGN (BS5950-1:2000)

Base plate reference;

Compresso! " mome!# : Bo$#s ! #e!so!; #e!so! ;

Des%! &or'es ! mome!#s  Axial force;

F c = ;*+,5 ;*+,5;; kN (Compression)

Bending moment;

M = +,2 kNm; +,2 kNm;

Shear force;

Fv = 51,. kN 51,. kN

Co$/m! e#$s Colmn section;

UC .05.059 .05.059 (Gre S25)

!epth;

! = .0,9 mm .0,9 mm

Breadth;

B = .05,. mm .05,. mm

Flange thickness;

" = 15, mm 15, mm

#e$ thickness;

t = 9,9 mm 9,9 mm

!esign strength;

p%c = 25 N&mm 25 N&mm'

Colmn flange to $ase plate eld;

+ mm 3; 3;

Colmn e$ to $ase plate eld;

+ mm 3; 3;

Bsep$#e e#$s Steel grade;

S25

!epth;

!p = 550 mm 550 mm

Breadth;

Bp = 550 mm 550 mm

"hickness;

t p = .5 mm .5 mm

!esign strength;

p%p = 2*5 N&mm 2*5 N&mm'

S#&&e!er e#$s Steel grade;

S25

"hickness;

t s = 22 mm

!esign strength;

p%s = 2*5 N&mm'

!epth at edge of $aseplate;

ds = 5 mm

!epth at colmn flange;

ds' = 150 mm

Stiffener to $ase plate eld;

+ mm 3

Stiffener to colmn flange eld;

+ mm 3 ( !o, 4er#'$ e$s per s#&&e!er)

6o$!% o! 7o$# ! !'8or p$#e e#$s "otal nm$er of $olts;

+ No, M2 Gre +,+

*nner $olt spacing;

s$olt = 150 mm

Spacing $eteen inner and oter $olt;

s$olt' = 150 mm

+dge distance;

e = 50 mm

 Anchor plate steel grade;

S25

 Anchor plate dimension (s,are);

$ap = 150 mm

 Anchor plate thickness;

t ap = 20 mm

!esign strength;

p%ap = 2*5 N&mm'

+m$eddment to top of anchor plate;

+ = 550 mm

Characteristic strength of concrete;

f c   = 50 N&mm'

Co!'re#e 'ompresso! &or'e ! 7o$# #e!so! &or'e "o determine approximate effective idth of $aseplate- assme $ase plate $eteen stiffeners acts as a fixed ended $eam./ Moment at stiffener;

Ms = ls'&' ' c

here ls is span $eteen stiffeners

Moment de to cantilever otside stiffeners;

Mc = l &'

here lc is cantilever span $e%ond stiffeners

+,ating the to moments;

lc = ls&√0

!istance $eteen stiffener centrelines;

ls = B 1 t s = ;.2,.; mm

Cantilever span;

lc = ls&√(0) = 1..,* mm

+ffective idth of plate;

B pc = min(Bp- ls 1 '×lc) = 550,0 mm

!istance from $olts to compression edge;

h = ! p / e = 500 mm

 Assming a rectanglar compression $lock of idth $ pc- length x and intensit% 230f c then./ From static e,ili$rim;

M = 230f c Bpcx(h/x&') / Fc(h/!p&')

4earranging the ,adratic e,ation;

235fc  Bpcx' / 230f cBpchx 1 Fc(h/!p&') 1 M = 2

Factor a;

a = 235 × f c × Bpc = +250,0 N&mm

Factor $;

$ = /230 × f c × Bpc × h = -+250000,0 N

Constant c;

c = F c × (h/!p&') 1 M = 102*12500,0 Nmm

!epth of compression $lock;

x = 6/32 ×$ / √($' / 7×a×c)8&('×a) = 12,* mm

Compression force in concrete;

Cf  = 230 × f c × Bpc × x = 20,+ kN

"ension force in $olts;

"f  = Cf  / Fc = 1.9,. kN Therefore the bolts are in tension

Compresso! se 7e!!% #8 s#&&e!er  Center of force for trianglar plate = 235 x 9 = 52mm

125m

L

125m

Cantilever moment in plate along diagonal plane

mcc = 230 × f c × (':)'&' × 52 = 0.1250 Nmmmm

!esign moment;

mc = 0.1250 Nmm&mm

late thickness re,ired;

tpc = √(0 × mc&p%p$p) = .2,* mm

Te!so! se 7e!!% #8 s#&&e!er  !istance from inner $olt to face of stiffener;

l $ = B&' / s$olt&' = ;,; mm

!istance from oter $olt to face of stiffener;

l $' = s$olt&' 1 s$olt' / (B&' 1 t s) = ;50,.; mm

*nner $olt lever arm;

m = l$ / 23<×ssp = 1,. mm

ter $olt lever arm;

m' = l$' / 23<×ssp = .,9 mm

Moment in plate (inner $olt);

mt = ('×"f  × m) & N$olt = 2+209 Nmm

Moment in plate (oter $olt);

mt' = ('×"f  × m') & N$olt = 15.10.2 Nmm

+ffective idth resisting moment (inner $olt);

Bpt = min(e- m&tan(52)) 1 min((!p/!)&'/e-m&tan(52)) = ;

121,1; mm +ffective idth resisting moment (oter $olt);

B pt' = min(e- m'&tan(52)) 1 m'&tan(52) = 12*,1 mm

late thickness re,ired;

tpt = max( √(7 × mt&(p%p×Bpt))- √(7 × mt'&(p%p×Bpt'))) = 1,* mm

P$#e #8'!ess late thickness re,ired;

tp>re, = max(tpc- tpt) = ;.2,* mm;

late thickness provided;

t p = .5 mm PASS - Plate thickness provided is adequate (0.931

S#&&e!er se'#o! '$ss&'#o! &or 7e!!% arameter ε;

ε = ('?: N&mm' & p%s)23: = 1,019

4atio for section classification;

4 = d s'&ts = *,+

Section classification ("a$le );

P$s#'; (@sing stem of "/section)

S8er 'p'#; # 'o$/m! &'e (pos#o! o& mm/m mome!#) v = 230 × p%s × 23 × ts × ds' = 2,2 kN

Shear capacit%; S#&&e!er s8er o! 'ompresso! se Shear force;

c = Cf &' = ;10.,9 N

!epth of stiffener at critical location;

d vc = ds 1 (ds' / ds) × '×x&(!p/!) = ;+2,+ mm

Shear capacit%;

vc = 230 × p%s × 23 × ts × dvc = 2*0, kN PASS - Shear capacit! on co"pression side is adequate (0.399 The stiffener is in #$% shear at the colu"n face on the co"pression side (0.&&0

S#&&e!er 7e!!% o! 'ompresso! se 9ever arm;

lac = (!p/!)&' / x&' = ;11,+; mm

Moment in stiffener;

msc = Cf &' × lac = ;11,9.; kNm

For effective length of cantilever ("a$le 7)./  At spport;

Co!#!/o/s #8 $#er$ ! #orso!$ res#r!# (')

 At tip;

L#er$ res#r!# #o #e!so! &$!%e (2)

9oading condition;

Norm$

+ffective length;

9 + = 23 × (!p/!)&' = ;10+,9; mm

β = select(Class-lastic-32-Compact-32-Semi/compact-2300?-2) = 1,000 β = 1,00 4atio β (cl3 735303); +,ivalent slenderness (B3'3?);

λ9" = '3< × (β×9+×ds'&ts')23: = 1*,.

Bending strength ("a$le 0);

p$ = 2*5,0 N&mm'

Bending capacit% (cl3 73'3:);

mcsc = min(3'×p%s×ts×ds''&0- p$×ts×ds''&7)

mcsc = 2*,2 kNm PASS - Stiffener bendin' capacit! on co"pression side is adequate (0.)) S#&&e!er s8er o! #e!so! se Shear force;

t = "f &' = *9, kN

!epth of stiffener at critical location;

d vt = ds 1 (ds' / ds) × '×e&(!p/!) = ;10*,0; mm

Shear capacit%;

vt = 230 × p%s × 23 × ts × dvt = ...,* kN PASS - Shear capacit! on tension is adequate (0.&09 The stiffener is in #$% shear at the colu"n face on the tension side (0.1*

S#&&e!er 7e!!% o! #e!so! se 9ever arm;

lat = (!p/!)&' / e = ;1,1; mm

Moment in stiffener;

mst = "f &' × lat = ,95 kNm

Bending capacit%;

mcst = min(3'×p%s×ts×ds''&0- p%s×ts×ds''&7) mcst = 2*,2 kNm PASS - Stiffener bendin' capacit! on tension side is adequate (0.1*9

S#&&e!er #o 7se p$#e e$ #eld force per mm de to colmn shear force;

f sl = Fv&6' × ('×(!p/!) 1 !)8 = ;0,0.2; kN&mm

9ongitdinal capacit% of < mm fillet eld;

p sl = 1,2.2 kN&mm; (Cl3 03<3?35))

PASS - #on'itudinal capacit! of stiffener to base plate +eld is adequate (0.0&,  Assme the force of a single $olt is resisted $% the eld on one side of the stiffener3 +ffective eld length;

l s = min(Bpt- Bpt') = ;121,1; mm

#eld force per mm;

fst =   '×"f &(N$olt × ls) = 0,2++ kN&mm

"ransverse capacit% of < mm fillet eld;

p st = 1,50 kN&mm; (Cl3 03<3?35)

PASS - Transverse capacit! of stiffener to base plate +eld is adequate (0.1* ratio s = (f sl&psl)' 1 (f st&pst)' = 0,0.*

4esltant effect (cl3 03<3?35);

PASS - Stiffener to base plate +eld is adequate (0.03, S#&&e!er #o 'o$/m! e$s #eld force per mm de to colmn shear force;

f sct = Fv&(<×ds') = ;0,0.; kN&mm

"ransverse capacit% of < mm fillet eld;

p sct = 1,50 kN&mm; (Cl3 03<3?35)

PASS - Transverse capacit! of colu"n to stiffener +eld is adequate (0.0&* Considering one stiffener and taking moments a$ot the centreline of the colmn flange on the compression side./ ft  = 6Cf &' × (!p/!/x1")&' 1 "f &' × ((!p1!/")&'/e)8&(' × (!/")

#eld force on tension side;

× ds') f t = 0,0 kN&mm #eld force on compression side;

fc   = (Cf &' 1 '×f t× ds' / "f &')&('×ds') = ;0,55; kN&mm

Maximm eld force per mm;

fscl    = max(f t- f c) = 0,55 kN&mm

9ongitdinal capacit% of < mm fillet eld;

p scl = 1,2.2 kN&mm; (Cl3 03<3?35)

PASS - #on'itudinal capacit! of colu"n to stiffener +eld is adequate (0.)0 4esltant effect (cl3 03<3?35);

ratio sc = (f scl&pscl)' 1 (f sct&psct)' = 0,20. PASS - olu"n to stiffener +eld is adequate (0.&03

6o$!% o! 7o$#s Force per $olt;

F$olt = ('×"f )& N$olt = .,+ kN

"ensile area per $olt;

At>$ = .5.,0 mm'

"ensile strength;

pt>$ = 5*0 N&mm'

"ension capacit% (cl3 030);

t>$ = 23< × pt>$ × At>$ = ;15+,1; kN; PASS - /olt capacit! is adequate (0.&&0

A!'8or p$#es Force per anchor plate;

Fap = F$olt = .,+ kN

Bolt hole diameter in anchor plate;

dh = 2 mm

 Anchor plate $earing area; Bearing capacit%;

Aap = $ap' / π×dh'&7 = 2192 mm' ap = 230 × f c × Aap = *5,+ kN PASS - Anchor plate bearin' capacit! is adequate (0.0)3

Bearing pressre on anchor plate;

fap   = Fap & Aap = 1,* N&mm'

#idth of $olt head (across flats);

d$h = .*,0 mm

Maximm cantilever length;

lap = $ap&' × √(') / d$h&' = ++,1 mm

Bending moment in plate;

map = f ap × lap'&' = *,2 Nm&mm

Bending capacit%;

mcap = p%ap × tap'&7 = 2*,5 Nm&mm PASS - Anchor plate bendin' capacit! is adequate (0.&3&

6o$!% o! 7o$# !'8or%e Note / the folloing calclation to check the holding don $olt anchorage into the fondation assmes that the distance from the edge of an anchor plate to the nearest edge of the fondation is at least e,al to the depth of em$edment of the anchor plate3 "ension force to $e resisted;

Ft = "f  = 1.9,. kN

"he clear distance $eteen anchor plates is less than the em$edment (+)3 +ffective concrete plan area;

Aplan>eff  = 6s$olt1'×s$olt'1$ap8×($ap1'×+)1π×+'1'×$ap×+/

(N$olt&')×$ap'  Aplan>eff  = 15..2 mm' For tension failre pll/ot- effective tensile area;

At>eff  = Aplan>eff  = 15..2 mm'

"ensile strength of concrete;

pt = 1,*2 N&mm'

ll/ot capacit% of tension $olts;

 t = pt × At>eff  = 2+*,0 kN PASS - oldin' do+n bolt anchora'e is adequate (0.0*

S8er #r!s&er #o 'o!'re#e  Assmed coefficient of friction;

µ = 2352

 Availa$le shear resistance;

v = Cf  × µ = ;*2; kN PASS - rictional shear capacit! is adequate (0.*&3