Question

1. Complete the table below for the welded connection samples shown. ETNICES Ful enetration But Weld Partial Penetration Burt Weld Flat/Plate Grade 300 Steel t>17 fyp 280 MPa up 440 MPa A-A B-B 10mm Weld Electrode: E48XX/W50X Quality: SP w480 MPa 1 Determine the nominal tension capacity of the welded connection as per AS4100 Nominal capacity is the design capacity without relevant capacity reduction factors Assume the design throat thickness can be taken to the full depth of penetration. 2 Actual yield load observed from experiment 3 Actual ultimate load observed from experiment 4 Comment on the failure mode and the difference between the actual and nominal capacity 2. Complete the questions below for the bolted splice connection shown. Grade 300 Steel t

Answer 1

Answer 1

To determine the nominal strength of welded connection we are using following formula from AS4100....

$D.S.=B*t*fy/\gamma mw$

Where

D.S.= design strength of the welded connection

B= effective width of the main plate

t= effective throat thickness of weld

fy= smaller of yield strength between weld and parent metal

$\gamma mw$= factor of saftey=1.25 for workshop weld

=1.5 for field work

given data:

B=25mm

t=20 for full pentration of weld

fy=280 MPa

$\gamma mw$=1.25 as work in lab.

Then

D.S.= (20*25)*(280/1.25)

D.S.= 112 KN

And

As there is no experimental result is given so we can't predict about these and due to this we can't comment on both result untill we can't get the actual strength of the weld

Again.....fp

Foe partial penetration

t= throat thickness= (7/8)*thickness of plate

=(7/8)*20

=17.5 mm

Then

Design strength of the weld is

D.S.= (25*17.5)*(280/1.25)

D.S.= 98 KN

Answer for 2

mode of failure for bolted connection...

1. Bolt may fail in shear

2. Bolt may fail in bearing

3. If bolt are enough strong then plate may fail in tension.

(Plate will be either main plate or cover plate)

And the tensile strength of the bolted connection is minimum of the above three discussed strength.

Nominal tensile strength of the bolted connection.....

(A)Bolt may fail in shear...

To calculate the strength of plate in shear we use the following formula

S.S. = (n_{​​​​​​n}​​​*A_{​​​​​​n} + n_​​s*A_{​​​​​​s})*(fub)/(1.25*√3)

Where

n_{​​​​​​n}​​​= no.of shearing plane in net area= 1

n_{​​​​​​S}​​​=no.of shearing plane in shank area=1

As= shank area of bolt = (π*12*12/4) = 113.097 mm^{​​​​2}​.

An= net area of bolt = 0.78*As= 0.78*113.097=88.215

fub=ultimate strength of bolt=400 MPa (given)

As there is given to consider any reduction factor

So

S.S. = (1*88.2+1*113.1)*(400)/(1.25*√3)

S.S.= 37.19 KN

(B) Bolt may fail in bearing

bearing strength of the bolted connection is..

B.S.= 2.5*Kb*d*t*Fup/1.25

Where....

Kb=minimum of

1. (end distance)/(3* hole diameter) = (50/3*13)=1.28

2. (Pitch)/(3*dia. Of hole)-0.25= (40/3*13)-0.25 =0.77

3.(Fub/Fup) = 400/440=0.91

4. =1

Then Kb= 0.77

d =dia of bolt 12 mm

d' = dia of hol= 12+1=13 mm(as per code)

t= minimum thickness = 6+6=12 <20

Fup= 440 MPa

Then

B.S. = (2.5*0.77*12*12*440/1.25)

B.S.=97.57 KN

(C) tensile strength of the bolted connection is.....

T.S. = minimum of (1,2)

TS1 = Ag*fy/1.1

TS2 =An*Fup*0.9/1.25

Where

Ag= gross area of plate=20*100=2000

An= net area of plate= 20*(100-2*13) =1480

Fy= yield strength of the plate=320 MPa

Fup= 440

Then

TS1= (20*100*320/1.1)

= 581.81KN

TS2= (1480*440*0.9/1.25)

= 468.86KN

T.S. = minimum of (581.81,468.86)

T.S. =468.86 KN

then final value of tensile strength of the bolted connection is minimum of shear, bearing and tension value

Bolt strength=min.(37.19,97.57,468.86)

Bolt strength= 37.19KN<------- answer

Actual strength can only be find out by experiment which is not available at this time and due to this we can't conclude any conclusions about the connection.

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