Project 2 - Design following connections of a factory building

Design following connections of a factory building

• Secondary Beam to primary beam connection – Fin plate shear connection

• Primary beam to column connection – Extended end plate connection

• Column to base plate connection - Pinned connection
  • Secondary beam size – ISMB 200
  • Primary beam size – ISMB 300
  • Column size – ISMB 500

• • Column size – ISMB 500

Member end forces (Factored)

	Shear Force	Moment	Axial force
Secondary beam	120 kN	-	15 kN (C)
Primary beam	450 kN	280 kN-m	-
Column	150 kN	-	1250 kN (C)

Consider the yield strength of the member/plate and ultimate strength of member/plate .

Consider 4.6 grade ordinary grade bolts.

AIM AND INTRODUCTION : Following connections are required to be designed-

• Secondary Beam to primary beam connection – Fin plate shear connection

• Primary beam to column connection – Extended end plate connection

• Column to base plate connection - Pinned connection

First part-

• Secondary Beam to primary beam connection – Fin plate shear connection

Inputs and Assumptions :

• Secondary beam size – ISMB 200
• Primary beam size – ISMB 300
• Factored Shear Force on secondary beam- 120kN
• Axial force on secondary beam - 15kN (C)
• `f_y`= 250NPa
• `f_u`= 410MPa
• Bolts - 4.6 grade
• `f_(ub)`= 400MPa
• `f_(yb)`= 240 MPa

PROCEDURE IN DETAIL :

Shear force on secondary beam = 120 kN

Axial force on secondary beam = 15 kN ( C)

Shear capacity of bolt -

Assuming bolt of diameter - `d=`20 mm

Diameter ofbolt hole - d_0 = 21.5 mm

Thickness of fin plate = 16 mm ( Assumed)

`A_(nb) = pi/4xxd^2` = (3.14/4)x `d^2`= 314 `mm^2`

`A_(sb)= 0.78xxpi/4xxd^2` = 0.78x314 = 245 `mm^2`

Shear strength of bolt

Single shear case

Since web of of ISMB 200 is connected to fin plate by bolts, the shear force on bolt act on single shear plane on the bolts. Therefore its a single shear case

Design shear strength og bolt is given by

`V_(sdb)= f_u/sqrt3(n_n*A_(nb) + n_s*A_(sb))/gamma_(mb)`

= 400/1.732 ( 1x245)/1.25 x1000= 42.27 kN

Bearing strength of bolt -

`V_(dpb)= 2.5*k_b*d*t*f_u//gamma_(mb)`

Where `k_b` is least of the follwing-

`k_b= e/(3d_0)` , `p/(3d_0)` , `f_(ub)/f_u` , 1

e = 1.5`d_0` = 32.25 mm say 40 mm ,

p = 2.5 d = 2.5 x 20 50 mm say 60 mm

`k_b` = 0.62 , 0.68 , 1.025 , 1

Thus `k_b` = 0.62

`V_(dpb)` = 2.5x0.62x20x10x400/1.25x1000 = 99.20 kN

Therefore bolt strength or bolt value = 42.27 kN

Number of bolts required = Shear force/bolt value = 120/42.27 =2.84, provide 4 bolts

Shear strength of bolt provided = 4x 42.27 = 169.08 > Shear force of 120 kN ( Safe)

Providing 4 bolts in 2 columns

Weld thickness = 6mm

Strength of weld

Fin plate

Depth, `l_w` = 200 mm

Thickness, t = 10 mm

Thickness of weld = 8 mm

Effective throat thickness , `t_t` = ks = 0.7 t = 5.6 mm

Design strength of the weld,

`p_(dw) = (Lwxxt_txxf_u)/(1.73xxgamma_(mw))`

= 200x5.6x250/1.73x1.25x1000

= 129.48 kN > Shear force of 120 kN ( Safe)

Fin plate dimension-

Min Length of fon plate - p+2e = 60 + 2x40 = 140 , provide l= 200 mm

Width of plate 2x40 + 60 = 140 mm , provide width = 150 mm

RESULT : Thus Pimary to the secondary beam Fin plate connection is designed. Silent features of design are as follows-

1) 4 bolt of 20 mm dia with total shear capacity 169.08 > Shear force of 120 kN ( Safe)

2) Fin plate 200x 150x 10 mm

3) weld thickness 8 mm

4 ) weld length 200 mm

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Second part

• Primary beam to column connection – Extended end plate connection

Vertical Shear force `F_y` on primary beam - 450 kN

Moment on Primary beam `M_z` - 280 kNm

Prismatic section	Overall Depth ( mm)	Flange width `b_f` ( mm)	Flange thickness `t_f`(mm)	Web thickness `t_w`(mm)	Yield strength `(f_y)` `N//(mm^2)`	Ultimate strength(f_u) `N//(mm^2)`
Primary beam : ISMB 300	300	140	12.4	7.5	250	410
Column: ISMB 500	500	180	17.2	10.2	250	410

Using bolt dia 24 mm

Resolving moment for the lever arm from center to center of flanges-

`2 (F_1+F_2)xx(300-12) = 280xx10^3` = 280x`10^3`/(300-12.4)

`2(F_1+F_2)` = 973.57

Also `F_1=F_2=F` then

2F = 973.57

F = 486.78 kN

Shear strength of bolt `V_(dsb)`= `f_u/sqrt3*A_(nb)/gamma_(mb)` ( As per cl 10.3.3 , IS 800)

= 400x352/1.732x1.25

= 65034.64 N = 65.04 kN

Bearing strength of bolt -

`V_(dpb)= 2.5*k_b*d*t*f_u//gamma_(mb)` ( As per Cl. 10.3.4, IS 800 )

Where `k_b` is least of the follwing-

`k_b= e/(3d_0)` , `p/(3d_0)` , `f_(ub)/f_u` , 1

e = 1.5`d_0` = 39 mm say 50 mm ,

p = 2.5 d = 2.5 x 24= 60 mm mm say 70 mm

`k_b` = 0.641 , 0.647 , 1.025 , 1

Thus `k_b` = 0.641

`V_(dpb)` = 2.5x0.641x24x10x410/1.25x1000 = 126.15 kN

Therefore bolt strength or bolt value = 65.04 kN

Numbwe of bolts required = Shear force/Strength og one bolt

= 450/65.04 = 6.92 , provide 8 bolts

Total strength of 8 bolts = 65.04x8 = 520 kN > Shear force of 450 kN ( Safe)

Tension capacity of bolt-

Bolt dia `d`= 20 mm

Grade of bolt = 4.6

Hole dia `d_0`= 22 mm

Yield strength of bolt material = 240 `N/(mm^)`

Ultimate tensile strength of bolt material `f_(ub)`=400 `N//(mm^2)`

Net area of bolt `A_(nb)`= 245 `mm^2`

Shank area of bolt `A_(sb)=`314 `mm^2`

`T_(nf)=0.9*f_(ub)*A_(nb)lef_(yb)*A_(sb)(gamma_(m1)/gamma_(m0))`

Where `gamma_(m1)` =1.15

`gamma_(m0)` =1.1

`gamma_(mf)`= 1.25

`T_(nf)` = 0.9x400x245 = 88.20 kN `le`240*314*1.15/1.1 = 78.79 kN ( NOT TRUE)

Assuming bolt of diameter - `d=`24 mm

Grade of bolt : 4.6

`A_(sb) = pi/4*d^2`= 452.16 `mm^2`

`A_(nb) = pi/4xxd^2` = 0.78x452.16 = 352 `mm^2`

Tension capacity of bolt `T_(nf)`= 0.9f_(ub)A_(nb)/1.25`lef_(yb)*A_(sb)*gamma_(m1)/gamma_(m0)`

= 0.9x 400x352/1.25

= 101376 N = 101.38 kN

`f_(yb)*A_(sb)* gamma_(m1)/gamma_(m0)` = 240*452.16*1.15/1.1 = 112949.24 N

= 112.94 kN > 101.38 (OK)

Tensile strength of bolt ,

`T_(db)=T_(nf)/gamma_(mf)`

= 101.38/1.25 = 81.1 kN

Tension on each bolt = Total tension/Nos of bolt = 486.78/8 = 60.85 kN < Tensile strength of bolt ( 81.1 kN) , Hence Safe

Permissible prying force Q = Tension capacity- F

= 101.38-486.78 = - 385.32 kN

Priying force is negative so Q = 0

Evaluation of weld at flanges and Web=

i) Weld size around the weld-

Length of weld available in flange 2x `b_f-t_w` = 2x140 -7.5 = 272.5 mm

Design strength of fillet weld `f_(wd)=f_u/(sqrt3*gamma_(mw)`

`f_u` is smaller of ultimate stress of weld or of parent metal = 410 MPa ( weld material is same as parent metal) and

`gamma_(mw)` is partial safety factor = 1.25 for shop weld and 1.5 for field weld

Design strength of shop fillet weld = 410/1.731x1.25 = 189.37 `N//(mm^2)`

Design strength of field fillet weld = 410/1.731x1.5 = 157.8`N//(mm^2)`

Check for tension in top or bottom flange-

T = M/(D-`t_f`) = 280/(300-12.4)

= 973.57 kN

Tension capacity of flange = `A_g*f_y//1.1`

= 140x12.4x250/1.1 =395.55 kN < Tension ( 973.57 kN) --- Not safe

Increasing the thickness of flange using cover plate of 20 mm on top and bottom flange

Then Tension capacity of flange = `A_g*f_y//1.1`

= 140x(12.4+20)x250/1.1 =1031 kN > Tension ( 973.57 kN) --- Safe

Shear capacity of flange = `A_v*f_y//sqrt3*1.1`= 1031/1.732 = 595 kN

`T/T_d+ V/V_d` = 973.57/1031+ 0/595 = 0.94 --- Safe

Evaluation of weld at flanges and Web=

i) Weld size around the web-

Length of weld available in flange 2x `b_f-t_w` = 2x140 -7.5 = 272.5 mm

Design strength of fillet weld `f_(wd)=f_u/(sqrt3*gamma_(mw)` ( cl. 10.5.7.1.1, IS 800: 2007)

`f_u` is smaller of ultimate stress of weld or of parent metal = 410 MPa ( weld material is same as parent metal) and

`gamma_(mw)` is partial safety factor = 1.25 for shop weld and 1.5 for field weld

Design strength of shop fillet weld = 410/1.731x1.25 = 189.37 `N//(mm^2)`

Assume size of weld 10 mm-

Length available around flange for Fillet eeld = `2b_f+2(b_f-t_w)`

= 2*140+2(140-7.5)

= 545 mm

Shear force = design strength of welf *Effective length* throat thickness

450 *1000 = 189.37*10*l

Required Effective length = 237.63 mm < Available length for weld ( safe)

Factored design tensile force `T_f= T_(nf)/gamma_(mf)` = 88.2/1.25 = 70.56 kN

Tension capacity of flange = `A_g*f_y//1.1`

= 140x12.4x250/1.1 =434 kN < Tension ( 869.27 kN) --- Not safe

Increasing the thickness of flange using cover plate of 6 mm on top and bottom flange

Let us assume Cover plate of 6 mm

Then Tension capacity of flange = `A_g*f_y//1.1`

= 140x(12.4+16)x250/1.1 =903 kN > Tension ( 869.27 kN) --- Safe

Shear capacity of flange = `A_v*f_y//sqrt3*1.1`= 903/1.732 = 521.36 kN

`T/T_d+ V/V_d` = 869.27/903 + 0/521.36 = 0.96 --- Safe

Using bolt dia 24 mm

Width of end plate = 2e+p = 2*50+70 = 170 mm say 200 mm

Length of end plate = 2e+ 200 +2p = 2*50+200+2*70 = 440 mm

Thickness of plate = 20 mm

RESULT : Thus Extended end plate connection is designed.

1) Size of End plate 200x440x20 mm

2) 8 bolts of dia 24 mm 4.4 grade

3) Cover plate on top and bottom flange 20 mm thick

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Part-3 : Column to base plate connection - Pinned connection

• Column size – ISMB 500
• Shear force on primary beam - 150kN
• Axial force - 1250 kN ( C)

Property of ISMB 500 section-

Flange width = 180 mm

Overall depth = 500 mm

Flange thickness = 17.2 mm

Web thicness = 10.2 mm

Axial compressive load = 1250 kN

Shear force = 150 kN

Assuming Fe 415 Grade of steel

Ultimate strength `( f_u)` = 415 Mpa

The yield strength `(f_y)` = 250 Mpa

`gamma_(mw)` partial safety factor for weld = 1.5

Grade of concrete (M20) , `f_(ck)` = 20 Mpa

`gamma_(m0_` = 1.1

`gamma_(mw)` = 1.25 ( for shop welding)

Bearing strength of concrete = `0.45 f_(ck)`

= 0.45x 20

= `9 N//mm^2`

Area of base plate required = 1250x `10^3`/ 9

= 138889 `mm^2`

= 13.89x`10^4``mm^2`

Now finding length and width of base plate , A = LXB

L = h+2a , B = `b_f`+2b

Assume a = b = x say

13.89x `10^4` = (500+2x)(180+2x)

= 90000+`1000x`+`360x^2`+ `4x^2`

`138900` = `4x^2`+`1360x`+90000

`x^2`+ `340x`-22500=0

`x`= 57 mm

therefore a = 57 mm and b= 57 mm

L = 500+2X57 = 614 mm

B = 180+2x 57 = 294 mm

Area to be provided = 614x 294 =180516 `mm^2`

The intensity of pressure for concrete pedestal ( w) = P/A < `0.45f_(ck)`

= 1250x1000/180516

= 6.92 mm `N//mm^2`< 0.45`f_(ck) =``( 9 N//mm^2)` SAFE

Minimum thickness of plate (`(t_s)`= [ 2.5 w ( a^2- 0.3b^2 `gamma_(mo)/f_y`]`^0.5`> `t_f` ( as per cl.7.4.3.1 of IS 800:2007)

= [2.5x 6.92 ( 57x57-0.3x57x57)1.1/250]`^0.5`

= [ 17.3 ( 3249- 974.7)x0.0044]`^0.5`

= (3174.8)`^0.5` = (176.54)`^0.5`

= 13.28 mm < `t_f (17.2 mm)`

Adopt minimum thickness of plate (`(t_s)` = `t_f`=17.2 mm

Adopting `t_s` = 18 mm

Provide 625x300x18 mm base plate

DESIGN OF WELDED CONNECTION-

Length awailable for welding around the column profile , `L_a` = 2( h-`2t_f)`+2b_f + 2(b_f- t_w)`

= 2( 500 -17.2) + 2x180 + 2(180-10.2)

= 965.6 +360 + 339.6

= 1665.2 mm

Effective length of Fillet weld = KL = L-2SXEnd return

Where S is size of weld :

Minimum size of weld =`S_(min)`= thickness of thicker part ( IS 800:2007 ,Table 22)

`t_w`= 10.2 mm

`t_f`= 17.2 mm

`t_s`= 18 mm

From the above data maximum thickness = 18 mm,

From table 21 , `S_(min)`= 5 mm for thicker part between 10- 20 mm

Maximum size of weld =`S_(max)` = `t_(min)-1.5` for square weld

= `t_w-1.5` ( as thickness of web is minimum)

= 10.2-1.5 = 8.7 mm

`s_(max)` = 8.7 mm

Size of the Fillet weld = Average of `S_(min)` and `S_(max)` = 5+8.7/2 = 6.85 mm say 7 mm

Thus, S = 7 mm

KL = 1665.2-2*7*12 = 1497.2 mm

Thus effective length of weld = KL = 1497.2 mm

Throat thickness of weld ,`t_t` = 0.7s =0.7*7 = 4.9 mm

Strength of weld/mm of length = `t_t*(f_u)/sqrt3*1/gamma_(mw)`

= 4.9x(410/1.732)x 1/1.25 = 927.95 N

Length of weld required = Shear force / weld strength = 150x1000/927.95 = 161.65 mm < `L_a` 9 1497.2), Hence OK

Providing weld length `L_w`= 200 mm

Therefore weld strength = `A*(f_u)/sqrt3*1/gamma_(mw)`

= `t_t*L_w*(f_u)/sqrt3*1/1.25`

= 4.9* 200*(410)x1.732*1/1.25

= 185588.92 N

= 185.59 kN > Shear force ( 150kN)----------OK

Shearing strength of Bolt, `V_(dsb) = f_u*A_(sb)//(sqrt3*gamma_(mb))`

Using 400 Grade bolt of 22 mm `phi`, `A_(sb)=0.7*pi/4*phi^2`

= 0.7*(3.14/4)*`22^2`

= 265.96`mm^2`

`V_(dsb) = f_u*A_(sb)//(sqrt3*gamma_(mb))`

= 400*265.96/1.1*1.732

= 55838.75

= 55.84 kN

Bearing strength of Bolt ,`V_(dpb) = 2.5*k_b*d*t*f_u`

`k_b` is smaller of `e/(3d_0), p/(3d_0)-0.25 ,f_(ub)/f_u`, 1

`d_0`= d+2 = 22+2 =24 mm

Minimum end & edge distance , e = 1.5`d_0` = 1.5*24 = 36 mm , provide e = 50 mm

minimum pitch distance , p = 2.5*d = 2.5* 22 = 55 mm , provide , p= 70 mm

`e/(3d_0)= 50/3*24 = 0.69

p/(3d_0)-0.25 = 70/3*24 =0.97

f_(ub)/f_u = 400/410 = 0.97

1

Therefore `k_b` = 0.69

`V_(dpb) = 2.5*k_b*d*t*f_u`/`gamma_(mb)`

= 2.5*0.69*22*18*410/1.25 = 224 kN

Bolt strength is lesser of `V_(dsb)` ( 55.84 kN) & `V_(dpb)`( 224 kN)

Therefore Bolt strength OR Bolt value , `B_v` = 55 kN

Number of bolts required = Shear force/Bolt value

= 150/55 = 2.7 , providing 4 Bolts of 22 `phi`grade 415

Total capacity of Bolts = 4* 55 = 220 kN > Shear force ( 150 kN) , Hence ok

RESULT : Thus column base plate connection ( pinned connection) is designed such that the size of Base plate is

1) 625x 300x 18 mm

2) SAize of weld is 7 mm

and 3) Length of weld is 200 mm

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