Engr Waqas

Economy of ASD Vs LRDF

6 posts in this topic

Assalam i alikum,

Can any one explain me this line please? This is from the book of Z.A SIDDIQUE PRC PART 1

''Strength design becomes economical as compared to ASD when dead loads are relatively larger compared with live loads'' 

 

Thanks. 

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yes this is true. u can verify it by solving beam using these two methods. ASD method give large sections as compared to ultimate. but it is good in achieving servicibility criteria.i will try to give u an example solving by these two methods and thier comparison.

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on different steel percentages ultimate design section takes 30 to 80 % more load as compared to ASD design sections.

 

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Thanks brother. I m basically confused that why line stresses on "DEAD LOADS vs LIVE LOADS" ? If live loads are larger as compared to dead ones, will ASD become economical?

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3 hours ago, Engr Waqas said:

If live loads are larger as compared to dead ones, will ASD become economical?

Yes. How? Go through the following paragraphs.

In case of ASD, service loads are used, and both dead and live loads have same load factor of 1. On the other hand, considering the the basic load combination of 1.2 D + 1.6 L, dead load (DL) has a load factor of 1.2 and live load (LL), a load factor of 1.6. Now consider two simple examples.

For the first example, assume that a certain beam has to carry a DL of 30 kips, and a LL of 10 kips. For this beam, total design service load for ASD method will be DL+LL=30+10 = 40 kip. For the Strength Design, basic design load will be 1.2 DL+1.6LL = 1.2(30)+1.6(10) = 52 kips. Strength Design load in this case is  (52-40)/40*100 = 30% larger than the Design Load for ASD.

Now, for the second example, assume that there is another beam, similar to that in the first example, except that the DL & LL values are reverse of those in first example i.,e., now DL = 10 kips, & LL = 30 kips. Total Design Load for ASD method is now  DL+LL=10+30 = 40 kip (same as in first example). However, design load for Strength design changes and it is now 1.2 (10)+1.6(30) =60 kips, which is (60-40)/40*100 = 50 % larger.

Thus, it is very clear that for both the examples, Design load is the same when using ASD method, and resultantly the member size & reinforcement will also be the same. However, since Design load for Strength Design method is larger in second example (when LL is larger than DL), a comparatively larger member size or reinforcement will be required. Thus, ASD method is generally economical (as compared to Strength Design method) when the live load on a member is larger than the dead load.

Regards.

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