Question

The A-36 steel drill shaft of an oil well extends 12000 ft into the ground. The pipe used to drill the well is suspended freely from the derrick at A. (Figure 1) The shaft consists of three different sizes of pipe, AAB=2.25 in.2, ABC=1.40 in.2, and ACD=1.15 in.2, each having the length and weight per unit length indicated. please do part D

Homework 4 Problem 4.6 5 of R The A-36 steel dill shaft of an ail wel extends 12000 ft into the graund. The plpe used to arll the well is auspended freely trom the derrlck at A. (Figure 1) The shaft consists of three different eizes of pipe. A 2.25 in. An1.40 in2, and Ac1.15 in2, each having the length and weight per unit length indoated. Correct Significant Figures Foedback: Your answer 12.85 ksi was either rounded differently or used a different numbar of signiticant figures than required for this par Part C Driermine the maximum average normal stress in pipe string CD Exprees your answer to three signilicant figures and include appropriate unita 3.48 ksi Submit Provious Answers Correct Significant Figures Feedback: Your answer 3.4T8 ksi was either rounded differently or used a different rumber af signilicant figuress than required far this par Part D Figure 1 af Detemine the elongstlon of Ita end D wth reepect to the ted end at A. Express your answer to three significant figures and include appropriate units ? ИА Aa WAR 3.2 lb/t 5000 ft Value Units бр Авс waC2.8 lb/ 5000 ft Request Answer Submit C AcD wcD- 2.0 lb/ft 2000 ft D Provide Feedback

Answer 1

Consider the elongation in small element of each section of pipe. ſ dx dx = L x H = LAB TMP + LBC TMP + Lcd 2011

Calculate the elongation in the pipe AB due to self-weight and weight force acting on it. 44B Jo 41°) Jo ARETO AB=P** d (A) = P14 P • dx = pt« (w.xp+X+Wacłac + wcp-co)• da A RE _p500 ((3.2)x+(2.8)(5000) +(2.0)(2000)). dx (2.25)(29x109) _ p500 ((3.2)x+18000). da 1 6 =1"**T65725410") ** (0.525310|32.+ 1802 (6.525x107) [16x + 1800x75 = (6.525x107) [1.6(5000)? +18000(5000) –o-o] =1.99234 ft Activate Windows

Calculate the elongation in the pipe BC due to self-weight and weight force acting on it. Agc = {** (A) = pisc Pz • dx = ptc (WbcX + wcpłcd) • dx " To A E do =p5000 (2.8)x+(2.0)(2000)) . dx Jo (1.40)(29x109) ABCE - Do (209.7654905) * (106-10)(25.2008 * (4.06x107)[148° +4000x}" * (4.06x10²)L..4 and [1.4(5000) + 4000(5000) –0-o] =1.35468 ft Activate Windows Go to PC settings to activate Wi

Calculate the elongation in the pipe CD due to self-weight. 400 = 60 d (1)= P600 Pz - dx = pico (w cox). dx Aco = Jo 4 (2) Jo AE JO AcDE _p2000 (2.0)x. dx Jo (1.15)(29x109) - 22000 (2.0)x - dx = 5" (3.335x10°) (3.335410) ? (3.335x107[**]* * 3.335x103)| (2000)? -07 = 0.11994 ft Calculate the elongation of its end D with respect to the fixed end at A. A2 = 448 +Agc +Ac =1.99234 +1.35468+0.11994 = 3.46696 ft Activate Win Go to PC setting