I just treated this like a statics problem and did the concrete load of (145 lb/ft^3)(3ft)(12ft) =5220 lb/ft as a point load. From there, I set up my right triangle and did sum of forces to get my forces in x and y to equal 7382 lb/ft each. So total forces are 14764 lb/ft aka 14.76 kips.
The way I did it was a bit different (but I ended up with the same answer). I drew it like a beam with a triangle distributed load and a 45 degree support strut (with the 3ft distribution O.C.). Then I drew my free body diagram, converted it into a point load using the centroid, and then found the moment about the ground connection. This gave me the perpendicular force at the strut connection, then divided that reaction force by sin(45) and came out with 14.75 kips. There's a couple ways to do the problem of course but I got lost in getting the "pcf" into a point load until I realized you assume the distributed load is a 45-45-90 triangle. Solution was a little bit confusing but the question is definitely something that can be on the exam. I really appreciate these video, keep up the good work guys!
while I appreciate the free practice problem....this is very difficult to follow along with. Not sure how he developed his FBD triangle at the bottom. The lateral reaction he calculated at the top of the wall is 3.84klf, but in the right triangle he uses the full 10.44klf value?
For anyone who needs and easier explanation. This problem is solved using the retaining wall equations in NCEES. Then taking the moment at the base to solve for the Fy force in the brace then converting Fy to F with Fy=Fsin(theta).
Correct. That is also what is shown in the video. When he does Pmax * H / 2, that is calculating the equivalent force for the distributed load. When he sums moments you see F (h/3) to show the force acting 2/3 down the wall.
I just treated this like a statics problem and did the concrete load of (145 lb/ft^3)(3ft)(12ft) =5220 lb/ft as a point load.
From there, I set up my right triangle and did sum of forces to get my forces in x and y to equal 7382 lb/ft each. So total forces are 14764 lb/ft aka 14.76 kips.
this was extremely confusing to follow
The way I did it was a bit different (but I ended up with the same answer). I drew it like a beam with a triangle distributed load and a 45 degree support strut (with the 3ft distribution O.C.). Then I drew my free body diagram, converted it into a point load using the centroid, and then found the moment about the ground connection. This gave me the perpendicular force at the strut connection, then divided that reaction force by sin(45) and came out with 14.75 kips. There's a couple ways to do the problem of course but I got lost in getting the "pcf" into a point load until I realized you assume the distributed load is a 45-45-90 triangle.
Solution was a little bit confusing but the question is definitely something that can be on the exam. I really appreciate these video, keep up the good work guys!
while I appreciate the free practice problem....this is very difficult to follow along with. Not sure how he developed his FBD triangle at the bottom. The lateral reaction he calculated at the top of the wall is 3.84klf, but in the right triangle he uses the full 10.44klf value?
The instructor can improve upon explaining the Why instead of focusing just on the What
yes i was coming to the comments to say this
Agree. He just ran through the numbers, closed his books, and went home. Very little commitment.
For anyone who needs and easier explanation. This problem is solved using the retaining wall equations in NCEES. Then taking the moment at the base to solve for the Fy force in the brace then converting Fy to F with Fy=Fsin(theta).
Isn't Fr 2/3 distance down the wall and not 1/2?
Correct. That is also what is shown in the video. When he does Pmax * H / 2, that is calculating the equivalent force for the distributed load. When he sums moments you see F (h/3) to show the force acting 2/3 down the wall.
Ncees pe hb 1.1, p. 24, 80.
Couldn't solve it despite many attempts. Perhaps you might be fortunate to do what I couldn't ?
Why you did not divide by 2 for Rt (the question didn't say it's a continuous wall)
Also thanks for the video you do it is really refreshing.
No explanation or drawing of tributary area?
A = 3'x12' = 35 sf.
Why this never was an aspect of the solution?