NCEES Civil Sample Questions and Solutions (2008), prob 110

[bcjones15]

the solution is summing five resultant forces around point b, but i can only see four. i guess i don't understand how the book is breaking down the uniform load. can anyone help?

 

[layad]

The solution is summing moments about point B. Consider the forces: reactions at A and B, forces of 1 kip at two places, and components of 0.45 kip/ft. Therefore, moment about point B will have reaction component at A, 1 kip force component at A, 1 kip force component on the incline, and two components of 0.45 kip/ft distributed force. I hope that helps. I would have posted my answer earlier had I seen your question.

 

[bcjones15]

"two components of 0.45 kip/ft distributed force"

I guess I don't fully understand the part of your reply indicated above. It seems to me that the distributed force only contains a "y" component and can be equated to a concentrated force acting in the middle (16 ft orthogonally from point B) of the staircase. How is this approach incorrect?

 

[layad]

I guess I don't fully understand the part of your reply indicated above. It seems to me that the distributed force only contains a "y" component and can be equated to a concentrated force acting in the middle (16 ft orthogonally from point B) of the staircase. How is this approach incorrect?

Yes, you are right. It'll have just one component. Sorry for my ambiguous statement earlier. However, since o.45 kips/ft acts on a section that is inclined, we'll need to consider the total force along the incline, which would be (0.45 kips/ft)/(cos39.5)*20 ft

 

[CaltransPEHopeful]

For that distributed force I used 11.67 kips (got this from length of inclined section compliments of Pythagorus) mulitplied by the distance (16 ft) from Point B.

 

[Sue1409]

...since o.45 kips/ft acts on a section that is inclined, we'll need to consider the total force along the incline, which would be (0.45 kips/ft)/(cos39.5)*20 ft

I think I might be having the same trouble bcjones did. If the vertical force of 0.45 kips/ft is acting perpendicular to the 20-ft moment arm, why should the force be divided by cos39.5? The resulting "total force along the incline" would be acting perpendicular to the incline, not perpendicular to the 20-ft distance.

 

[mac17]

...since o.45 kips/ft acts on a section that is inclined, we'll need to consider the total force along the incline, which would be (0.45 kips/ft)/(cos39.5)*20 ft

I think I might be having the same trouble bcjones did. If the vertical force of 0.45 kips/ft is acting perpendicular to the 20-ft moment arm, why should the force be divided by cos39.5? The resulting "total force along the incline" would be acting perpendicular to the incline, not perpendicular to the 20-ft distance.

I know this is way late, but I thought I would post in case anyone else stumbles across this thread like I did.

I think the confusion is over where the cos(39.5) is placed in the solution. It would make more sense to see it under the 20ft as in:

(0.45 kips/ft)*20 ft/(cos39.5)

The reason for this is because the uniform load is over the length of (20/(cos39.5))ft, not 20ft. Obviously this expression will get the same answer as the one in the solutions...

 

[icreep]

@mac17

Thank you. I'm reviewing for my 10/25/13 PE Exam and this is exactly the problem I stumbled over so thanks for having the foresight to post.