Why we need to consider only 3 cases when we want to find the max positive shear ? IMO , We know that the max shear will only occur when the 67.5kN act on B , am i right ? There is no need to consider 3 different cases .....Or my concept is incorrect ? If so , can anyone show me the example where the max shear will occur at B although the max concentrated load doesn't act on the particular point ? ggg gggsfsf


Sure. Change your wheel loads to (from left to right): 99kN, 99kN, 100kN, 1kN.

When the first wheel has just reached B (Case 1), you have its 99kN at max influence, the second wheel's 99kN at a high positive influence, and the third 100kN wheel at a small positive influence.

Compare to Case 3 when the third (and maximum) wheel reaches B: the 100kN wheel at max influence only provides slightly more positive effect than the 99kN wheel did in Case 1, but instead of having two big wheels with positive influence we instead have two big wheels with negative influence.

It's clear from inspection, without running the numbers, that Case 1 is more onerous. If you change the wheel loads a bit, you can make it less obvious, and need to run the numbers to confirm which is more onerous.

  • $\begingroup$ Do you mean for the load which they are very closer to each other ( the difference in load between them are too small) , we have to do 3 analysis to determine which load give the max shear ? Am i right? For the cases that i asked in original post , it's actually no need to do 3 analysis , because the difference in load between them are not small $\endgroup$ – kelvinmacks Feb 2 '17 at 0:45
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    $\begingroup$ @kelvinmacks - It's not directly "how big is the largest one compared to the second largest". It's "how big is each load relative to the size of the influence it might be placed on". Try doing some investigation yourself: for the example in the question, keeping all the wheel loads the same except the front one, how big would the front wheel need to be to make Case 1 more critical than Case 3? You'll find it's a lot less than 67.5kN. $\endgroup$ – AndyT Feb 2 '17 at 9:32
  • $\begingroup$ what do you mean by make case 1 more critical than Case 3? $\endgroup$ – kelvinmacks Feb 2 '17 at 10:19
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    $\begingroup$ @kelvinmacks - I know what I said. I also know what you said. So I asked you another question. But instead of answering it, you've just repeated stuff. Shall I try and rephrase my question? Here goes: Is it "make" that you're struggling to understand? Or "Case 1"? Maybe it was "Case 3"? The word "more"? Or "critical"? Was it the words "more" and "critical" together in "more critical" which didn't make sense to you? Or does each individual word make sense, but you can't understand the whole phrase "make Case 1 more critical than Case 3"? $\endgroup$ – AndyT Feb 2 '17 at 11:53
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    $\begingroup$ The whole point I was making was that you should do the calculation yourself in order to understand it! I shall try and rephrase: In the question, there is an 18kN load, a 40.5kN load, a 67.5kN load and a 40kN load. Try increasing the 18kN to 25kN, and redo all the calculations from the question. Try increasing it again to 32.5kN. Try increasing it again to 40kN. You will find that with 40kN, Case 1 gives a higher shear than Case 3 does. Hence you will realise that you can't just say "67.5kN is the biggest load, putting it at B is the most onerous". $\endgroup$ – AndyT Feb 2 '17 at 12:08

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