4.5 in $\times$ 3.64 in ID, 20.0 lb/ft; E-75 (75 ksi), X-95 (95 ksi), G-105 (105 ksi)
Drillpipe wear
Class 3, 37.5% wall loss (62.5% wall remaining)
Find. The drill-collar length and the length & grade(s) of drill pipe needed to reach 12,000 ft safely (tension design, worn pipe) and economically (cheapest grade wherever it suffices).
Fig. 2 — Combination string (schematic, not to scale): cheapest grade (E-75) carries the lower, lightly-loaded section; X-95 covers the top where tension is highest; G-105 is not required.
Approach. Size the drill-collar length from the BHA design factor and buoyed weight-on-bit; derate the drillpipe body area for Class 3 wear; then size each pipe grade, from the bottom (cheapest) up, so the running total of buoyed weight below plus the margin of overpull never exceeds that grade's worn tensile rating — using the LEAST expensive grade that satisfies each successive section.
Drill-collar length. The BHA design factor sizes the collars so their buoyed weight exceeds $DF_{bha}\times WOB$: $$L_{DC}=\frac{DF_{bha}\times WOB}{w_{DC}\times BF}=\frac{1.5(20{,}000)}{72(0.8168)}\Rightarrow\boxed{L_{DC}=510.2\ \text{ft}}$$ Buoyed collar weight $=510.2\times72\times0.8168=30{,}000$ lbf (exactly $DF_{bha}\times WOB$, as intended), leaving $L_{DP,total}=12{,}000-510.2=11{,}489.8$ ft of drill pipe to size.
Worn pipe-body tensile rating. Class 3 wear removes 37.5% of the WALL thickness (62.5% remains): new wall $=(4.5-3.64)/2=0.43$ in, worn wall $=0.43(0.625)=0.26875$ in, worn ID $=4.5-2(0.26875)=3.9625$ in, so $$A_{worn}=\frac{\pi}{4}\big(4.5^2-3.9625^2\big)\Rightarrow\boxed{A_{worn}=3.572\ \text{in}^2}$$ Worn tensile rating $P_t=\text{yield}\times A_{worn}$: E-75 $=267{,}923$ lbf; X-95 $=339{,}372$ lbf; G-105 $=375{,}092$ lbf.
Design load per grade (margin-of-overpull method). The design factor above sizes the BHA only; drillpipe grade is sized by the standard margin-of-overpull criterion, $P_t\ge W_{below,buoyed}+MOP$, i.e. each grade's usable design load is $$P_t-MOP:\quad\boxed{147{,}923\ \text{lbf (E-75)},\ \ 219{,}372\ \text{lbf (X-95)},\ \ 255{,}092\ \text{lbf (G-105)}}$$
Size the bottom (cheapest) section — E-75. Buoyed weight per foot of drill pipe (any grade, same geometry) $=20\times0.8168=16.34$ lb/ft. Running E-75 immediately above the collars until its own design load is reached: $$L_{E75}=\frac{147{,}923-30{,}000}{16.34}\Rightarrow\boxed{L_{E75}=7{,}220.9\ \text{ft}}$$ Weight at the top of the E-75 section $=147{,}923$ lbf (by construction).
Check whether the remaining length needs X-95, or X-95 and G-105. Remaining drill pipe to surface: $11{,}489.8-7{,}220.9=4{,}268.9$ ft. If X-95 alone carries this to surface, the weight at the very top would be $147{,}923+4{,}268.9(16.34)=337{,}690$ lbf. Since $337{,}690<219{,}372$ is FALSE for X-95's own design load taken alone at the top... check instead against each section's OWN top: X-95's design load ($219{,}372$ lbf) is reached after $$L_{X95,max}=\frac{219{,}372-147{,}923}{16.34}=4{,}374.3\ \text{ft}$$ which EXCEEDS the $4{,}268.9$ ft actually remaining — so X-95 alone, run the rest of the way to surface, never reaches its own design limit: $$\boxed{L_{X95}=4{,}268.9\ \text{ft},\quad L_{G105}=0}$$
Confirm the finished string reaches exactly 12,000 ft and stays within all limits. $$510.2+7{,}220.9+4{,}268.9=12{,}000.0\ \text{ft}\ \checkmark$$ Surface tension $=30{,}000+7{,}220.9(16.34)+4{,}268.9(16.34)=217{,}700$ lbf $<219{,}372$ lbf (X-95 design load) — safe, with G-105 not needed anywhere in the string, which is the more economical result.
Check: (1) the "Design Factor for BHA Design" (1.5) is applied ONLY to the drill-collar length calculation; drillpipe grade/length is sized by the margin-of-overpull criterion alone ($P_t\ge W+MOP$, no additional design-factor multiplier), which is the standard convention separating BHA design (design-factor method) from drillpipe tension design (overpull method); (2) Class 3 wear is applied as a 37.5% WALL-THICKNESS loss (not a direct 37.5% strength loss).
Section (bottom to top)
Length
Notes
Drill collars, 6.0 in $\times$ 3.0 in, 72 lb/ft
510.2 ft
buoyed weight $=DF_{bha}\times WOB=30{,}000$ lbf
Drill pipe, E-75
7,220.9 ft
cheapest grade, carries lower string
Drill pipe, X-95
4,268.9 ft
carries to surface; below its own design limit (219,372 lbf)
