# NCEES Sample Questions & Solutions, #110



## Sue1409 (Oct 19, 2010)

I'm having trouble understanding the solution for Problem #110 in the NCEES Sample Questions and Solutions book. Why is the 0.45 kips/ft distributed load being divided by cos39.5? I understand that gives the value of the force acting perpendicular to the incline, but the distributed load is already acting perpendicular to the 20-ft length (i.e., the 16-ft moment arm). Can anyone explain this?


----------



## Eddierizzle (Oct 20, 2010)

Sue1409 said:


> I'm having trouble understanding the solution for Problem #110 in the NCEES Sample Questions and Solutions book. Why is the 0.45 kips/ft distributed load being divided by cos39.5? I understand that gives the value of the force acting perpendicular to the incline, but the distributed load is already acting perpendicular to the 20-ft length (i.e., the 16-ft moment arm). Can anyone explain this?


Once I get home I'll take a look at that problem to see if I can explain it to you. Unfortunately I don't have the problem in front of me to look at.


----------



## Ble_PE (Oct 20, 2010)

Sue1409 said:


> I'm having trouble understanding the solution for Problem #110 in the NCEES Sample Questions and Solutions book. Why is the 0.45 kips/ft distributed load being divided by cos39.5? I understand that gives the value of the force acting perpendicular to the incline, but the distributed load is already acting perpendicular to the 20-ft length (i.e., the 16-ft moment arm). Can anyone explain this?


The cos39.5 is in the denominator to account for the length of the stair beam. In other words, the distributed load is acting over a length of 20/cos39.5, not along the length of 20 ft.

A more straightforward way of completing the question is to calculate the length of the stair beam and use that in the moment equation. So the term you would have would be 0.45*25.93*16. Does that help?


----------



## Sue1409 (Oct 21, 2010)

Ble_PE said:


> A more straightforward way of completing the question is to calculate the length of the stair beam and use that in the moment equation. So the term you would have would be 0.45*25.93*16. Does that help?


Yes, it does. Thanks for the reply!


----------



## armenterosjl (Oct 21, 2010)

Sue1409 said:


> I'm having trouble understanding the solution for Problem #110 in the NCEES Sample Questions and Solutions book. Why is the 0.45 kips/ft distributed load being divided by cos39.5? I understand that gives the value of the force acting perpendicular to the incline, but the distributed load is already acting perpendicular to the 20-ft length (i.e., the 16-ft moment arm). Can anyone explain this?


I had the same problem, I found the solution:

The stair concentrated load is: 0.45 kips x L inc, and L inc = 20 ft / cos (angle), cos (angle) = 20 ft / 25.9278 ft = 0.7714


----------



## Sue1409 (Oct 24, 2010)

armenterosjl said:


> I had the same problem, I found the solution:The stair concentrated load is: 0.45 kips x L inc, and L inc = 20 ft / cos (angle), cos (angle) = 20 ft / 25.9278 ft = 0.7714


Thank you, too, for the reply. It makes perfect sense to me now. So much so, that I feel a little stupid for asking the question to begin with! I think I've been cramming too much....


----------

