In the block shear calculation for welded connection thickness of angle used; not the thickness of gusset plate as in the case of earlier weld problem🤔
Actually your thought is correct but the thing is that the thickness of the plate is not given in the question, so the thickness of the plate is considered equal to that of the angle.
12:00 in Vdpb caculation in last term fu is of plate right and also 13:15 we r providing edge distance = 30 mm then 14 : 30 how we r getting w1 = 40 it should be (75 - 30 = 45 ) plz anyone correct me if i m wrong
@@rajashekarmuttayam8766 it will be 35 even as per Figure 7. Look at minute 19:11 onwards in the video and compare the distance to Figure 7, it will be 35. It's an error in calculation in the video.
@@martindsouza2659 I calculated using 35. I agree with you. In case of block shear failure, the tension failure should happen along the line from bolt to the outer edge of angle and that was 35mm.
@ayyapureddy nandagopal P1 + P2 =Tu Consider it as a case of simply supported beam for instance. P1 and P2 are the forces opposite to Tu. Take moment of the forces about the edge point of the angle section.
@@vishalbarwal6659 no, I just calculated the P1 and P2 value and then the weld lengths Lw1 and Lw2. Then calculated the yeild strength, rupture strength and block shear strength. No need to take average weld length.
@@snehasneha8270 actually when you see the block shear failure in weld connection, you consider that the whole gusset plate which is connected by the angle through weld is torn out due to this failure. So the tensile force is acting along the plane of the gusset plate which is not even welded up.
You can go though a similar problem in Lecture notes---design of steel structures----Uma Shankar Yaligar---pg 110,eg 4. Although here the welding is done in the tension area of the gusset plate and angle.
sir for bearing stress for bolt its fu not fub
sir minimum edge distance is 1.5Xdia.of hole i.e 1.5X22=33mm
It's not the dia of hole it's dia of bolt only 20mm it's correct
1.5×20= 30
@@KRISHNAKrish-ii8vo thank you bro
In the block shear calculation for welded connection thickness of angle used; not the thickness of gusset plate as in the case of earlier weld problem🤔
Actually your thought is correct but the thing is that the thickness of the plate is not given in the question, so the thickness of the plate is considered equal to that of the angle.
Feel free to ask... Maybe I can help for some of the things.
12:00 in Vdpb caculation in last term fu is of plate right and also
13:15 we r providing edge distance = 30 mm then 14 : 30 how we r getting w1 = 40 it should be (75 - 30 = 45 )
plz anyone correct me if i m wrong
Yes it should be 45
Sir Atg value will be =35×10
Not 40×10
tell me how that 40 came there??
it will be 40 only see the figure 7 in IS:800:2007
Yes
@@rajashekarmuttayam8766 it will be 35 even as per Figure 7. Look at minute 19:11 onwards in the video and compare the distance to Figure 7, it will be 35. It's an error in calculation in the video.
@@martindsouza2659 I calculated using 35. I agree with you. In case of block shear failure, the tension failure should happen along the line from bolt to the outer edge of angle and that was 35mm.
He took 165mm as total length of weld then in Avg cal why he multiply by 2
19:00 why don't we consider Atg = 35*10
Actually I have same doubt coz it should be 35*10 ig
Isn't that we need to take the longer weld length running along the toe of the angle in the calculation of the Avg and Avn in the block shear?
Avg = Avn = 2* 164.3 * 10 = 3286 mm^2
Lw1 = 164.3 mm instead of 165 mm i.e average weld length.
@ayyapureddy nandagopal P1 + P2 =Tu
Consider it as a case of simply supported beam for instance. P1 and P2 are the forces opposite to Tu. Take moment of the forces about the edge point of the angle section.
P1 * 75 - Tu *26= 0 solve this to get the load distribution
Tu=200kN Cxx=26mm is the centre from where the forces Tu acts.
Tell me if you got it or not?
Where is the average weld length for beta calculation is given?
Same doubt, where is it written in code? Did you get the answer?
@@vishalbarwal6659 no, I just calculated the P1 and P2 value and then the weld lengths Lw1 and Lw2. Then calculated the yeild strength, rupture strength and block shear strength. No need to take average weld length.
Yes, exactly. But I was curious if it is mentioned, then where can I find it. It's alright, Thanks!
@@vishalbarwal6659 no I didn't found it.
@@shivamsharanlall672 I also didn't get it. But your way of finding Lw is helpful. Thanks!!!
Please mention the name of lecture in file names
Why he didn't calculate the block shear failure in the end with actual weld length?
I didn't get any idea..of block shear pattern in case of weld with unequal length
@@snehasneha8270 we actually need to take the longer weld length for block shear calculation
@@shivamsharanlall672 sir which text book it is?
If there is no weld in tension area.... Then how can we calculate Atn and Atg
@@snehasneha8270 actually when you see the block shear failure in weld connection, you consider that the whole gusset plate which is connected by the angle through weld is torn out due to this failure. So the tensile force is acting along the plane of the gusset plate which is not even welded up.
You can go though a similar problem in Lecture notes---design of steel structures----Uma Shankar Yaligar---pg 110,eg 4.
Although here the welding is done in the tension area of the gusset plate and angle.
At 24:42 how 35 and 40 are divided 🤔?
Exactly i didn't get it how it was divided?
Can anyone help plz??
How to take R min
how can you take the value of cx as 26
It is give in code sp6
U can also check in steel tables ISA75×50×10