6 in × 2 in, 100 lb/ft; length $=10\%$ of total depth $D$
Equivalent derrick load, $F_{de}$
800,000 lbf
Drillpipe breaking strength
600,000 lbf
Available hook horsepower, $HP$
800 HP
Tripping speed, $v$
60 ft/min
Design safety factor, $SF$
2 (applied to every strength-rated component)
Buoyancy and travelling-block/hook weight are explicitly ignored.
Find. The maximum depth $D$ at which the string can still be drilled without exceeding any one of the rig's load-bearing limits.
Fig. 1 — Four independent load ceilings on the hook load; the string's own weight climbs at 28 lb/ft of depth until it hits the lowest ceiling.
Approach. Compute the four independent load ceilings the rig imposes on hook load (derrick, drillpipe tension, hoisting horsepower, drilling-line tension), express the string's own weight as a function of depth, and find the depth at which the governing (lowest) ceiling is reached.
Derrick load ceiling. $W_{derrick,max}=F_{de}/SF=800{,}000/2$, so $\boxed{W_{derrick,max}=400{,}000\ \text{lbf}}$.
Drillpipe tensile ceiling. $W_{DP,max}=600{,}000/SF=600{,}000/2$, so $\boxed{W_{DP,max}=300{,}000\ \text{lbf}}$.
Hoisting-horsepower ceiling. Hook power relates load and tripping speed by $HP=\dfrac{W\,v}{33{,}000}$; solving for $W$ at the rig's rated 800 HP and the selected 60 ft/min tripping speed, $W_{HP,max}=\dfrac{33{,}000(800)}{60}$, so $\boxed{W_{HP,max}=440{,}000\ \text{lbf}}$. This is an available-capacity limit, not a strength margin, so $SF$ is not re-applied to it.
Drilling-line tensile ceiling. With $n=10$ lines and no reeving-efficiency table given, assume frictionless sheaves ($E_n=1$), so the fast-line tension is $T_{fast}=W/n$. The line's own allowable tension is $103{,}400/SF=51{,}700$ lbf, giving $W_{line,max}=n\times51{,}700$, so $\boxed{W_{line,max}=517{,}000\ \text{lbf}}$.
Governing ceiling and string weight vs. depth. The lowest of the four is the drillpipe tensile ceiling, $300{,}000$ lbf (Fig. 1). With drill collars occupying $0.10D$ and drill pipe the remaining $0.90D$: $$W(D)=100(0.10D)+20(0.90D)=28D\ \text{lb}$$
Solve for maximum depth. $28D=300{,}000 \Rightarrow D=300{,}000/28$, so $\boxed{D_{max}=10{,}714\ \text{ft}}$ (to the nearest foot).
Check: (1) the design safety factor is applied to strength-rated components (derrick, drillpipe, wire line) but not to the horsepower-based load/speed relation, which is an available-capacity limit rather than a failure margin — this is the standard convention in hoisting-system design problems of this type; (2) frictionless sheaves ($E_n=1$) are assumed for the fast-line tension since no reeving-efficiency table is supplied when such a table is absent.
