Video 1

Hello we are going to start the design calculation of tension member where it isconnected with welded joint. So, in last classes we have gone through one workout examplewhere the strength of the tension member has been calculated using bolt connections. Nowsame thing we will be doing for weld connections. In case of weld connection how the neteffective area is going to change with respect to bolt connections, how the shear lag distanceis going to change those things we will come across and we will also see that block shear howto calculate, how to calculate that Atn, Atg, Avn, Avg.So those parameters will be calculated in todays lecture and we will see how what is thedifference between bolt connection and weld connection with respect to strength calculationof tension member.

Example:An ISA 90×60×6 angle of Fe 410 grade steel is connected to a 10 mm thick gusset plate byweld as shown in the figure below. Calculate the design tensile strength of the angle if gussetis connected to the (a) longer leg (b) shorter leg.Solution:(a) Gusset is connected to longer legGross area, Ag for ISA 90 × 60 × 6 = 865 mm2. [From IS hand book: SP: 6(1)-1964]The net area of connected leg, Anc = (90 - 6/2) × 6 = 522 mm2Gross area of outstanding leg, Ago = (60 - 6/2) × 6 = 342 mm2(i) Tensile strength governed by yielding of gross section: [Clause 6.2]

(ii) Tensile strength governed by rupture of net section:Therefore, the length of outstanding leg will be, w = 60 mm.So, the shear lag width, bs = 60 mm.The average length of weld along the direction of load

(iii) Tensile strength governed by block shear:Assuming average length of weld on each side as 75 mmAvg = 2 × 75 × 10 = 1500 mm2 [As gusset plate thickness = 10 mm]Avn = 2 × 75 × 10 = 1500 mm2Atg = 90 × 10 = 900 mm2Atn = 90 × 10 = 900 mm2

Thus, the design tensile strength of the angle = least of (i), (ii) and (iii) = 196.6 kN.

Video 2

(b) Gusset connected to shorter legThe net area of connected leg, Anc = (60 - 6/2) × 6 = 342 mm2Gross area of outstanding leg, Ago = (90 - 6/2) × 6 = 522 mm2Net cross sectional area, An = 522 + 342 = 864 mm2i) Tensile strength governed by yielding of gross section:

(ii) Tensile strength governed by rupture of net section:Here, the section is connected through its shorter leg.Therefore, the length of outstanding leg will be, w = 90 mm.So, the shear lag width, bs = 90 mm.The average length of weld along the direction of load= Lc = (75+75) /2= 75 mm. Thus,

(iii) Tensile strength governed by block shear:

Thus, the design tensile strength of the angle = least of (i), (ii) and (iii) = least of (196.36, 184and 373.9 ) = 184 kN.But in this case we could see that the critical strength are coming due to rapture of the netsection that is 184 kilonewton, right. Moreover, we could see that this critical strength is lessin this case, means, when shorter length is connected with the gusset plate with respect to theearlier one where longer leg is was connected to the gusset plate.So in this example means through this workout example we could see that efficiency of thesame angle can be increased if the connection is done properly connection can be doneproperly in a sense that when the longer leg of the angle section is connected with the gussetplate then we could increase the efficiency of the section. Also from earlier lecture if wecompare with the earlier calculation then we can find that weld connections is taking littlehigher load then the bolt connection.So from these two examples means examples from the strength calculation of bolt connectionand weld connection for the connection with longer leg, connection with shorter leg, fourpermutation combination we could see the best option is connection should be done withlonger leg and if possible with the weld connection. In that case maximum efficiency can beobtained.