bcjones15
New member
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?
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?"two components of 0.45 kip/ft distributed force"
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 ftI 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?
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....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 know this is way late, but I thought I would post in case anyone else stumbles across this thread like I did.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....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
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