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Problem 233 A steel bar 50 mm in diameter and 2 m long is surrounded by a shell of a cast iron 5 mum thick. Compute the load that will compress the combined bar a total of 0.8 mm in the length of 2 m. For steel, E = 200 GPa, and for cast iron, E = 100 GPa. Solution 233 * Click here to expand or collapse this section PL AE 8 = deast iron = Ostet = 0.8mm Peat tron (2000) S cast iron = T 2 2 = 0.8 (+(60? — 507) ](100 000) Pee iren 11 0007 N Peet f2n(502)](200000) os Cast Iron, 4 t=5mm Pte) = 500007 N wi 0002 DFy =0 P = Peast iron + Peteci d= 50mm P = 110007 + 50 0007 P = 610007 N P =191.64kN answer Problem 234 Areinforced concrete column 200 mm in diameter is designed to carry an axial compressive load of 300 KN. Determine the required area of the reinforcing steel if the allowable stresses are 6 MPa and 120 MPa for the concrete and steel, respectively. Use E,, = 14 GPa and E,; = 200 GPa. Solution 234 ¥ Click here to expand or collapse this section Seo = by = 8 ( PL ) (= ) AE),, \AE/ y (= ) (= ) Ee Es Cre on b 14000 ~——-200 000 1006 to = Tost When o,; = 120 MPa 100¢,, = 7(120) Fo = 84 MPa > 6 MPa + (not okay!) When Op = 6 MPa 100(6) = Tos og = 85.71 MPa< 120 MPa} (okay!) Use Og, = 6 MPa and 6; = 85.71 MPa. EFy =0 Py, + Peo = 300 Ost Ast +Oco Aco = 300 85.714 4 + 6 | +n(200)? — A, | = 300(1000) 79.714 ¢ + 60.000m = 300.000 Ag = 1398.9 mm? answer Problem 236 A rigid block of mass M is supported by three symmetrically spaced rods as shown in Fig. P-236. Each copper rod has an area of 900 mm?; E = 120 GPa; and the allowable stress is 70 MPa. The steel rod has an area of 1200 mm’; E = 200 GPa; and the allowable stress is 140 MPa. Determine the largest mass M which can be supported. Copper Steel Copper 160 mm 240 mm 160 mm “a Ye Figure P-236 and P-237 Solution 236 y Click here to expand or collapse this section Sco = Sst (2 ) 7 (2 ) E/., Bl. & oc0(160) ost (240) » 120000 200000 Oo 49 = 90 ap When o,; = 140 MPa 9 (140) Feo co = 126 MPa > 70 MPa (not okay!) When 6c, = 70 MPa ow = 2(70) 77.78 MPa < 140 MPa (okay!) Gat Use Oc, = 70 MPa and Og, = 77.78 MPa. uFy =0 2Peo + Ps = W 2(Gco Aco) + ost Ast = Mg 2[ 70(900) ] + 77.78(1200) = a4(9.81) M = 22358.4 kg answer Problem 237 In Problem 236, how should the lengths of the two identical copper rods be changed so that each material will be stressed to its allowable limit? Solution 237 v Click here to expand or collapse this section Use Og9 = 70 MPa and og, = 140 MPa cq = Sst (=) °C), co TE, 140(240) 120000 200000 Leo = 288 mm answer (b) Condition: 6.; = 20pyr sFy =0 2P4 + Py = 40 2(o Ast) + hr Ap = 40 (Gey lanl tendn = ery (1.0) + op (1.5) = 40 Tp = 7.27 ksi ot = 2(7.27) = 14.54 ksi Sbr = Ost oL 7 (= ( B ) br - E ) st 7.27(1000) L 5, 14.54(1000)(3 x 12) 2x10® 29 x 108 Lip = 29.79in Lip = 2.48 ft answer Problem 239 The rigid platform in Fig. P-239 has negligible mass and rests on two steel bars, each 250.00 mm long. The center bar is aluminum and 249.90 mm long. Compute the stress in the aluminum bar after the center load P = 400 KN has been applied. For each steel bar, the area is 1200 mm2 and E = 200 GPa. For the aluminum har, the area is 2400 mmz2 and E = 70 GPa. Figure P-239 Solution 239 * Click here to expand or collapse this section 06°6bZ ost 6s: = 6ai + 0.10 oL oL CE) & st £ al 250 249.90 t(250) rail ) PA 200 000 70.000 0.00125¢., = 0.00357ea + 0.10 on = 2.8560; + 80 P = 400 kN Py Py Ps DFy =0 2P., + Pa = 400000 Qo Ast +O Aa = 400000 2(2.856c 4) + 80)1200 + 4; (2400) = 400 000 9254.40 q + 192000 = 400 000 oq = 22.48 MPa answer Problem 241 As shown in Fig. P-241, three steel wires, each 0.05 in.2 in area, are used to lift a load W = 1500 lb. Their unstressed lengths are 74.98 ft, 74.99 ft, and 75.00 ft. (a) What stress exists in the longest wire? (b) Determine the stress in the shortest wire if W = 500 lb. Solution 241 y Click here to expand or collapse this section Let L, = 74.98 ft; Lo = 74.99 ft; and Lz = 75.00 ft Part (a) Bring Ly and Lz into L3 = 75 ft length: (For steel: E = 29 x 10° psi) PL 6=— AE For Ly: P, (74.98 x 12) 5 — 7A, = te eves) 0.05(29 x 10°) P, = 386.77 1b For Ly: Po (74.99 x 12) (75 — 74.99) (12) = —————_—_ 0.05(29 x 10°) P2 = 193.36 1b Let P = P3 (Load carried by L3) P + P (Total load carried by Ly) P + Py (Total load carried by L4) DFy =0 (P+P,)+(P+P,)+P=Ww 3P + 386.77 + 193.36 = 1500 P = 306.62lb = P; Ps 306.62 A 0.05 03 = o3 = 6132.47 psi answer Ww Figure P-241 Part (b) From the above solution: P, + Pz = 580.13 Ib > 500 Ib (L3 carries no load) Bring Ly into Ly = 74.99 ft PL - AE P, (74.98 x 12) 74.99 — 74.98)(12) = ( x 0.05(29 « 10°) P, = 193.38 lb Let P = Ps (Load carried by Lz) P+ P, (Total load carried by L4) DFy =0 (P + Pi) + P = 500 2P + 193.38 = 500 P = 183.31 1b P+ P; = 153.31 + 193.38 P+ P; = 346.69 1b P+P, 346.69 ge elie A 0.05 o = 6933.8 psi answer 5 = (9.975 x 10 4 )ow + (14.28 x 10?) og al = 350 — 0.69825 o 5 qi = 350 — 0.69825(2.4 41) 2.67580. = 350 oq = 1380.8 MPa answer Problem 243 A homogeneous rod of constant cross section is attached to unyielding supports, It carries an axial load P applied as shown in Fig. P-243. Prove that the reactions are given by Ry = Ph/L and Ry = Pa/L. Figure P-243 Solution 243 ~ Click here to expand or collapse this section EFg =0 L=a+b Ri, +R) =P RR, a <—R: Ry, =P-R, 5, = 5 =5 (=) (24) AE), AE), Ria Ryd AE AE Ria=R2b Ria=(P—R1)b Ria@=Pb— Ry b Ri (a+b) = Pb Ri L=Pb Ry = Pb/L (okay!) Ry = P— Pb/L P(L—b) Ry 2 b Ry =Pa/L (okay!) Problem 245 The composite bar in Fig. P-245 is firmly attached to unyielding supports. Compute the stress in each material caused by the application of the axial load P = 50 kips. Solution 245 Steel Aluminum: A= 2.0 in? A=1L25in? £=29 x 10° psi E= 10x 10° psi 15in ¥ Click here to expand or collapse this section BFy =0 R, +R, = 50000 R, = 50000 — Ry Sa = be = 6 (=) (=) Ap) 7) \ABY) R, (15) Rp (10) 10in Figure P-245 and P-246 Ry Rz Ee SS 15in 10in Ry <—GGs a 8. = ae Re Rl 1.25(10 x 10°) -2.0(29 x 108) Ry = 6.96R1 Ry = 6.96(50.000 — Rz) 7.96R) = 348 000 Ry = 43 718.59 Ib eat = As 2.0 Ost = 21859.30 psi R, 43718.59 answer R; = 50000 — 43 718.59 R, = 6281.41 lb Ri _ 628141 Ag 1.25 Tal = oq = 5025.12 psi answer Problem 246 Referring to the composite bar in Problem 245, what maximum axial load P can be applied if the allowable stresses are 10 ksi for aluminum and 18 ksi for steel. Solution 246 * Click here to expand or collapse this section ba =Se=8 R, == Re (2) 4 a Bla Bs gi 15 in 10in Ry o x (10) qq (15) — > F, 29x10° 10x 10° 8 ; Re <— og = 4.3504 When 6, = 10 ksi a = 4.35(10) a4 =43.5 ksi > 18 ksi (not okay!) When 6, = 18 ksi 18 = 4.350 al oy =414 ksil0 ksi (okay) Use 6g) = 4.14 ksi and 64; = 18 ksi EF, =0 P=R, +R, Pog Aq +%% Ast P = 4,14(1.25) + 18(2.0) P=41.17 kips answer Py =R, =77.60kN P,, = 150 — 77.60 = 72.40kN P,,, = 240 — 77.60 = 162.40kN a=P/A ai = 77.60(1000) /900 oq = 86.22 MPa answer as: = 72.40(1000) 2000 os = 36.20 MPa answer om = 162.40(1000)/1200 op = 135.33 MPa answer Problem 248 Solve Problem 247 if the right wall yields 0.80 mm. Solution 248 Click here to expand or collapse this section Aluminum Steel Bronze SOKN 1 L 150 kN 7 500mm 250mm 350 mm 08mm Ri aR -———hin Bale 150,000-R,<—| | te 150,000 - i Sa 240,000- 8; <—=f ms zn aol 3u 0.8 Sar = bet + (Spr +08) PL PL PL (Ze), ~ (Ge), *Ge), °°" AE) 4 AE), \AE/) y, R, (500) (150000 — R,)(250) (240.000 — R1)(350) 900(70000) —_2000(200 000) 1200(83 000) R, 150000—R, — 7(240000 — Ry) = ah + 0.8 126 000 1600000 1992000 ARi = Z(150000 — Ri) + (240000 — z,) + 1600 (a + se + aE )Ri = ge (150000) + F (240000) + 1600 R, = 143854 N = 143.854 kN Pa = R, = 143.854kN Py = 150 — Ry = 150 — 143.854 = 6.146kN Pry, = Ro = 240 — Ry = 240 — 143.854 = 96.146 kN o=P/A oa = 143.854(1000)/900 oq = 159.84 MPa — answer og = 6.146(1000) /2000 og = 3.073 MPa answer om = 96.148(1000)/1200 \sigma_{bi .122 \, \text{ MPa}} answer