Hi,
Good question, here are my solutions:
Problem 518: You need to compute the moment at Point A.
Given: tieback tension T= 10 kips; Tieback; Landslide pressure p = 1.0 ksf; the depth of the active landslide is 18ft; slide plane and ground slope are at 3:1; the tieback acts at a 2:1 angle from the top of the caisson. And from the diagram the caissons are on 5 ft centers and therefore each retain a 5 ft wide section of the active landslide.
Compute horiz. tension, Th x cos (angle) from the 2:1 triangle the hypotenuse is the square root of 5.
So Th = (10 kips)(2/(5^.5)=8.94 kips; the moment arm r= 18’ +2’+1’
Now find P = p(Area active landslide) P= (1.0ksf)(5’ x 18’) =90 kips. P horiz. = Ph
Ph = Ph cos(angle) = (90 kips)(3/(10^.5)) =85.38 kips. R= (1’+2’+18’/2)
Ma=0 so M(caisson) = Ph® - Th® = (85.38 kips)(12’) - (8.94 kips)(21’) = 836.8 ft-kips, answer C.
Problem 519: Find: Le (length extended into rock)
Given: fictional resistance, fr=10 psi; dia. of grouted hole = 6”; FS =1.5, and T = 10 kips.
T= Le(perimeter of the tieback)(fr)/FS The perimeter = (pi)(dia)
10 kips (1.5) = Le(pi)(6”)(10 psi) Le = 15,000 lbs/(188.5 lbs/in) =79.6 inches or 6.6 ft. Answer (B) 7
I had difficulty with all the extra information, and also forgot to add the 2' space between the active slide and the rock. I like this problem, you don't need any references, but there are plenty of places to miss something, so it is good practice.